1 November 2011

Objective You will be able to describe enthalpy and differentiate

between endothermic and exothermic reactions and calculate the enthalpy of a reaction

Homework p 263 21 22 23 24 25 tomorrow

Thermochemistry

Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy

can be converted from one form to another but can not be created or destroyed

We can not accurately measure total energy of a system

Instead we measure changes in energy ΔE

Exothermic process is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Specifically wersquore interested in heat flow that occurs under constant pressure

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)

Hproducts lt HreactantsDH lt 0

Hproducts gt HreactantsDH gt 0 64

Enthalpies of Reaction

Expressed in kJ or kJmol endothermic ∆H is always positive

endothermic changes absorb heat exothermic ∆H is always negative

exothermic changes liberate heat Calculating ∆H is the same for a wide variety

of processes (see table)

Thermochemical Equations

H2O (s) H2O (l) DHfus = 601 kJ

Is DH negative or positive

System absorbs heat

Endothermic

DH gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Thermochemistry

Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy

can be converted from one form to another but can not be created or destroyed

We can not accurately measure total energy of a system

Instead we measure changes in energy ΔE

Exothermic process is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Specifically wersquore interested in heat flow that occurs under constant pressure

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)

Hproducts lt HreactantsDH lt 0

Hproducts gt HreactantsDH gt 0 64

Enthalpies of Reaction

Expressed in kJ or kJmol endothermic ∆H is always positive

endothermic changes absorb heat exothermic ∆H is always negative

exothermic changes liberate heat Calculating ∆H is the same for a wide variety

of processes (see table)

Thermochemical Equations

H2O (s) H2O (l) DHfus = 601 kJ

Is DH negative or positive

System absorbs heat

Endothermic

DH gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy

can be converted from one form to another but can not be created or destroyed

We can not accurately measure total energy of a system

Instead we measure changes in energy ΔE

Exothermic process is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Specifically wersquore interested in heat flow that occurs under constant pressure

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)

Hproducts lt HreactantsDH lt 0

Hproducts gt HreactantsDH gt 0 64

Enthalpies of Reaction

Expressed in kJ or kJmol endothermic ∆H is always positive

endothermic changes absorb heat exothermic ∆H is always negative

exothermic changes liberate heat Calculating ∆H is the same for a wide variety

of processes (see table)

Thermochemical Equations

H2O (s) H2O (l) DHfus = 601 kJ

Is DH negative or positive

System absorbs heat

Endothermic

DH gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Exothermic process is any process that gives off heat ndash transfers thermal energy from the system to the surroundings

Endothermic process is any process in which heat has to be supplied to the system from the surroundings

2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)

Specifically wersquore interested in heat flow that occurs under constant pressure

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)

Hproducts lt HreactantsDH lt 0

Hproducts gt HreactantsDH gt 0 64

Enthalpies of Reaction

Expressed in kJ or kJmol endothermic ∆H is always positive

endothermic changes absorb heat exothermic ∆H is always negative

exothermic changes liberate heat Calculating ∆H is the same for a wide variety

of processes (see table)

Thermochemical Equations

H2O (s) H2O (l) DHfus = 601 kJ

Is DH negative or positive

System absorbs heat

Endothermic

DH gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Specifically wersquore interested in heat flow that occurs under constant pressure

Enthalpy of Chemical Reactions

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)

Hproducts lt HreactantsDH lt 0

Hproducts gt HreactantsDH gt 0 64

Enthalpies of Reaction

Expressed in kJ or kJmol endothermic ∆H is always positive

endothermic changes absorb heat exothermic ∆H is always negative

exothermic changes liberate heat Calculating ∆H is the same for a wide variety

of processes (see table)

Thermochemical Equations

H2O (s) H2O (l) DHfus = 601 kJ

Is DH negative or positive

System absorbs heat

Endothermic

DH gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)

Hproducts lt HreactantsDH lt 0

Hproducts gt HreactantsDH gt 0 64

Enthalpies of Reaction

Expressed in kJ or kJmol endothermic ∆H is always positive

endothermic changes absorb heat exothermic ∆H is always negative

exothermic changes liberate heat Calculating ∆H is the same for a wide variety

of processes (see table)

Thermochemical Equations

H2O (s) H2O (l) DHfus = 601 kJ

Is DH negative or positive

System absorbs heat

Endothermic

DH gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Enthalpies of Reaction

Expressed in kJ or kJmol endothermic ∆H is always positive

endothermic changes absorb heat exothermic ∆H is always negative

exothermic changes liberate heat Calculating ∆H is the same for a wide variety

of processes (see table)

Thermochemical Equations

H2O (s) H2O (l) DHfus = 601 kJ

Is DH negative or positive

System absorbs heat

Endothermic

DH gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Thermochemical Equations

H2O (s) H2O (l) DHfus = 601 kJ

Is DH negative or positive

System absorbs heat

Endothermic

DH gt 0

601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ

Is DH negative or positive

System gives off heat

Exothermic

DH lt 0

8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ

bull The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

bull If you reverse a reaction the sign of DH changesH2O (l) H2O

(s)DH = -601 kJ

bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)

DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be

specified in thermochemical equations

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Note on ldquoimplied signsrdquo

In a sentence the sign of ∆H is often implied

ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ

ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Phase transitions

solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ

gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Example 1

Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol

calculate the heat evolved when 879 g of SO2 is converted to SO3

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Example 2

Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air

P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Example 3

How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Problem

Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Why are phase changes from solid to liquid and from liquid to gas always endothermic

Is the process of sublimation endothermic or exothermic Explain

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

2 November 2011

Objective You will be able to describe calorimetry and calculate

heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off

when 126x104 g of NO2 are produced according to the equation

2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions

lab notebook set upp

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Calorimetry

Calorimetry measurement of heat flow Calorimeter an apparatus that

measures heat flow Heat capacity C heat required to raise

the temperature of an object by 1 K (units are JK)

Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Calorimetry

q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance

the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)

q=msΔtEx 1 A 466 g sample of water is heated from

850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Ex 2

1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Ex 3

A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Ex 4

What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K

heat capacity of the calorimeter is 758 kJK

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Pre-Lab

Molar heat of crystallization aka Latent heat of fusion

Set up your lab notebook and answer the pre-lab questions (refer to your textbook)

Make a procedure summary and set up a data table

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Homework

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

3 November 2011

Objective You will be able to calculate the molar heat of crystallization

for a chemical handwarmer Homework Quiz A 100 gram piece of

copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter

specific heat of water = 4184 J g oC

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Lab

1 With your partner carefully follow the directions

Hand warmer pouch and metal disk 342 grams

2 Collect all data results and answers to questions in your lab notebook

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

7 November 2011

Objective You will be able to calculate the standard enthalpy of

formation and reaction for compounds

Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from

94oC to 5oC Calculate the heat released (in kJ) by the metal

specific heat of iron=0444 JgoC

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

I Homework QuizII Return TestsIII Standard Enthalpy of

FormationReaction notesproblems Homework p 264 46 49 51 53 56

57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

8 November 2011

Objective You will be able to review thermochemistry

Homework Quiz (Week of Nov 7)Calculate the heat of combustion for

the following reaction from the standard enthalpies of formation in Appendix 3

2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

I Homework quizII Collect lab notebooksIII Homework answers and more

enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

9 November 2022

Objective You will be able to practice thermochemistry for a

test Homework QuizCalculate the standard enthalpy of

formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

Homework Quiz Problem Set work timeHomework Problem set due

tomorrowThermochemistry test tomorrow

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Test Corrections

Required 2nd quarter quiz grade Show work including for multiple

choice questions On a separate sheet of paper or

using a different color pen Due Mon

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Standard Enthalpy of Formation and Reaction

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Formation Reaction

a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state

frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC

f

The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f

DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f

DH0 (C diamond) = 190 kJmolf

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Example

Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Hessrsquos Law

Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Example

Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)

from the enthalpies of these related reactions

S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ

2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f=S DH0 (reactants)fS-

Standard Heat of Reaction

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Example 1

Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol

(Hint Be really careful with your math and the + and minus signs)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Example 2

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ

-6535 kJ2 mol = - 3267 kJmol C6H6

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Practice Problem 1

Calculate the heat of combustion for the following reaction

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of

ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene

combusted

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Problem 2

The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron

This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al

∆Hof for Fe(l) is 1240 kJmol

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

What if the reaction is not a single step Practice Problem 3

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1 Write the enthalpy of formation reaction for CS2

2 Add the given rxns so that the result is the desired rxn

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Calculate the standard enthalpy of formation of CS2 (l) given that

C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Practice Problem 4

Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements

2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935

kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

DH0= -25988 kJmol

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Homework

p 264 46 49 51 53 56 57 59 62 64 Tuesday

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry

for a test on Monday Do now Calculate the standard enthalpy of

formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)

Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Chapter 6 Problem Set

Due Mon p 263 16 20 33 37 38 52 54b

57 64 70 73 80 Please show all your work

BONUS p 269 119 Tues (Up to +5 on your quarter grade)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Read p 258-261 Heats of solution and dilution on your own

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

8 Nov 2010

Hand in Ch 6 Problem Set Objective SWBAT show what you know

about thermochemistry on a test Do Now Review the sign conventions

for q and w A reaction absorbs heat from its

surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due

tomorrow p 269 119Read p 801-814

p 8291-5

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

9 Nov 2010

Take out homework Bonus problem Objective SWBAT predict and

calculate entropy changes to systems

Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

I Do nowII Notes on spontaneous processes

and entropyIII Practice ProblemsHomework p 829 10-14 (SR)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Thermodynamics Part 2

1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of

heat given off or absorbed by a system

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

2nd Law of Thermodynamics explains why chemical processes tend to favor one direction

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Spontaneous Processes

Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Does a decrease in enthalpy mean a reaction proceeds spontaneously

H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol

H2O (s) H2O (l) ∆H0 = 601 kJmol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ∆H0 = 25 kJmolH2O

Spontaneous reactions

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Does a decrease in enthalpy mean a reaction proceeds spontaneously

No Being exothermic (ldquoexothermicityrdquo) favors

spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are

spontaneous So how do we predict if a reaction is

spontaneous under given conditions if ∆H doesnt help us

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Entropy (S) is a measure of the randomness or disorder of a system

order SdisorderS

If the change from initial to final results in an increase in randomness∆S gt 0

For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas

H2O (s) H2O (l)∆S gt 0

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

74

Processes that lead to an increase in entropy (∆S gt 0)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

∆S gt 0

Example Br2(l) Br2(g)

∆S gt 0

Example I2(s) I2(g)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Entropy

State functions are properties that are determined by the state of the system regardless of how that condition was achieved

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths

energy enthalpy pressure volume temperature entropy

Review

Examples

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

77

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

How does the entropy of a system change for each of the following processes

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

How does the entropy of a system change for each of the following processes

(a) Condensing water vaporRandomness decreases

Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated

solutionRandomness decreases

Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C

Randomness increases

Entropy increases (∆S gt 0)(d) Subliming dry ice

Randomness increases

Entropy increases (∆S gt 0)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Practice

Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid

bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to

20oC

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

∆S0rxn nS0(products)= Σ mS0(reactants)Σ-

The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C

rxn

What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol

S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol

DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]

DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Practice

From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)

bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0

bull If the total number of gas molecules diminishes ∆S0 lt 0

bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number

What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is

negative

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Practice

Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Homework

p 829 10-14 (SR)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

10 Nov 2010

Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction

Do now Give one example of a change that increases entropy

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

I Do nowII Homework SRIII Gibbs Free Energy and Predicting

SpontaneityHW p 829 15 16 17 18 19 20 (TTL)

(Tues)Read lab and answer pre-lab questions in

lab notebook (Mon)AP Test Thermo questions (Tues)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Entropy Changes in the Surroundings (∆Ssurr)

Exothermic Process∆Ssurr gt 0

Endothermic Process∆Ssurr lt 0

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

91

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature

S = k ln W

W = 1

S = 0

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process

∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process

Gibbs Free Energy

For a constant-temperature process

∆G = ∆Hsys -T ∆SsysGibbs free energy (G)

∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction

∆G = 0 The reaction is at equilibrium

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

aA + bB cC + dD

∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]

∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-

The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions

rxn

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states

f

∆G0 of any element in its stable form is zerof

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0

rxn nDG0 (products)f= S mDG0 (reactants)fS-

What is the standard free-energy change for the following reaction at 25 0C

DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]

DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol

Is the reaction spontaneous at 25 0C

DG0 = -6405 kJmollt 0

spontaneous

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Practice

Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +

2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

∆G = ∆H - T ∆S

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Recap Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Homework

p 829 15 16 17 18 19 20 (TTL) (Tues)

Read lab and answer pre-lab questions in lab notebook (Mon)

AP Test Thermo questions (Tues)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

16 Nov 2010

Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems

Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Agenda

I Do nowII Homework solutions (TTL)III AP Test question solutions -

collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter

paper + M2CO3

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

CaCO3(s) CaO(s) + CO2(g)

CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control

It is made by decomposing limestone (CaCO3) in a kiln at high temperature

But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to

shift equilibrium to favor the formation of CaO

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

At what temperature does this reaction favor the formation of CaO

Calculate ∆Ho and ∆So for the reaction at 25oC

Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go

What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which

equilibrium occurs At what temperature can we expect this

reaction to proceed spontaneously Choose a temperature plug in and check

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Keep in mind that this does not mean that CaO is only formed above 835oC

Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Practice Problems

1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous

a ∆H = -126 kJmol∆S = 84 JKmiddotmol

b ∆H = -117 kJmol∆S = -105 JKmiddotmol

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Solutions to AP test problems

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

17 Nov 2010

Objective SWBAT review enthalpy and entropy for a quiz

Do nowThe reaction between nitrogen and hydrogen to

form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change

∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in

part (a) is negative at low temperatures but positive at high temperatures Explain

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Do now solutions

(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2

(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Thermochemistry review

p 831 42 52 54 56 57 60 78 Quiz Thursday

Entropy (∆S) Free energy (∆G)

Study homework assignments from chapter 18 AP test questions these review problems

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

stop

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)

How do you calculate free energy (and thus the spontaneity of the reaction)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ

R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium

∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

What is Q

Reaction quotient

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

What is K

equilibrium constant

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

∆G0 = - RT lnK

186

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius

C = ms

Heat (q) absorbed or released

q = msDt

q = CDt

Dt = tfinal - tinitial

65

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm

118

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

118

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)

Carbon Dioxide

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Mel

ting

118Fr

eezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Subl

imat

ion

118

Depo

sitio

n

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid

DHsub = DHfus + DHvap

( Hessrsquos Law)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance

118

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

118

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C

Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ

Step 2 Convert the solid to liquid ΔH fusion

Q = 20 mol x 601 kJmol = 12 kJ

Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)

Sample Problem How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization

Q = 20 mol x 4401 kJmol = 88 kJ

Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ

Now add all the steps together

059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ

• 1 November 2011
• Slide 2
• Thermochemistry
• Slide 4
• Slide 5
• Enthalpy of Chemical Reactions
• Slide 7
• Enthalpies of Reaction
• Slide 9
• Slide 10
• Slide 11
• Note on ldquoimplied signsrdquo
• Phase transitions
• Example 1
• Example 2
• Example 3
• Problem
• Slide 18
• Slide 19
• 2 November 2011
• Agenda
• Calorimetry
• Slide 23
• Calorimetry (2)
• Ex 2
• Ex 3
• Ex 4
• Pre-Lab
• Homework
• 3 November 2011
• Lab
• 7 November 2011
• Agenda (2)
• 8 November 2011
• Agenda (3)
• 9 November 2022
• Agenda (4)
• Test Corrections
• Standard Enthalpy of Formation and Reaction
• Formation Reaction
• Slide 41
• Slide 42
• Example
• Slide 44
• Hessrsquos Law
• Example (2)
• Standard Heat of Reaction
• Example 1 (2)
• Example 2 (2)
• Slide 50
• Practice Problem 1
• Problem 2
• What if the reaction is not a single step Practice Problem 3
• Slide 54
• Slide 55
• Practice Problem 4
• Homework (2)
• 4 Nov 2010
• Agenda (5)
• Chapter 6 Problem Set
• Slide 61
• 8 Nov 2010
• Agenda (6)
• 9 Nov 2010
• Agenda (7)
• Thermodynamics Part 2
• Slide 67
• Spontaneous Processes
• Slide 69
• Slide 70
• Slide 71
• Slide 72
• Slide 73
• Slide 74
• Slide 75
• Slide 76
• Slide 77
• Slide 78
• Slide 79
• Practice
• Slide 81
• Slide 82
• Slide 83
• Practice (2)
• Slide 85
• Practice (3)
• Homework (3)
• 10 Nov 2010
• Agenda (8)
• Slide 90
• Slide 91
• Slide 92
• Slide 93
• Slide 94
• Slide 95
• Slide 96
• Practice (4)
• Slide 98
• Recap Signs of Thermodynamic Values
• Homework
• 16 Nov 2010
• Agenda (9)
• CaCO3(s) CaO(s) + CO2(g)
• Slide 104
• Slide 105
• Practice Problems
• Slide 107
• 17 Nov 2010
• Do now solutions
• Thermochemistry review
• Slide 111
• Slide 112
• Slide 113
• What is Q
• What is K
• Slide 116
• Slide 117
• Slide 118
• Slide 119
• Wherersquos Waldo
• Slide 121
• Slide 122
• Slide 123
• Slide 124
• Sample Problem
• Sample Problem (2)