ThermochemistryThermochemistry

Ch. 5 in textbook Ch. 5 in textbook (omit Section 5.8)(omit Section 5.8)

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I.I. Heat vs. WorkHeat vs. WorkA. Energy (E)A. Energy (E)

EnergyEnergy = ability to transfer = ability to transfer heatheat or do or do workwork

Kinetic EnergyKinetic Energy = energy of an object = energy of an object in motion; depends on the mass and in motion; depends on the mass and velocity of the objectvelocity of the object

Potential EnergyPotential Energy= energy of an object = energy of an object due to its composition or position due to its composition or position relative to other objects; thought of as relative to other objects; thought of as stored energy because it is converted stored energy because it is converted to kinetic energy when changes occurto kinetic energy when changes occur

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B. Work B. Work A force (push/pull) acting A force (push/pull) acting

over a distanceover a distance Formula:Formula: w = F x dw = F x d Energy can never be in the Energy can never be in the

form of work ONLY, since form of work ONLY, since some heat will be lost due to some heat will be lost due to frictionfriction

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C. Heat (q)C. Heat (q)

The transfer of energy from The transfer of energy from high to low temperaturehigh to low temperature

Heat always flows from high Heat always flows from high to lowto low

Frictional heat is generated Frictional heat is generated when particles move against when particles move against one another; this heat is “lost” one another; this heat is “lost” in the sense that it can no in the sense that it can no longer be utilizedlonger be utilized

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D. Relating Energy, D. Relating Energy, Work, and HeatWork, and Heat

The overall energy change is The overall energy change is represented by the following represented by the following equation:equation:

ΔΔE = q + wE = q + w Units of energy: Joules (J), Units of energy: Joules (J),

calories (cal) or Calories (Cal)calories (cal) or Calories (Cal) Joule = Newtons x meters Joule = Newtons x meters 1 N = 1 kg x m/s1 N = 1 kg x m/s22

Therefore 1 J = 1 kg x mTherefore 1 J = 1 kg x m22/s/s22

1 cal = amount of heat needed to 1 cal = amount of heat needed to raise 1 gram of water 1 ºCraise 1 gram of water 1 ºC

1000 cal = 1 kcal = 1 Cal (dietary 1000 cal = 1 kcal = 1 Cal (dietary calorie)calorie)

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II.II. The 1The 1stst Law of Law of ThermodynamicsThermodynamics

A. DefinitionsA. Definitions SystemSystem = the specific part of = the specific part of

the universe on which we are the universe on which we are focusingfocusing

SurroundingsSurroundings = everything else = everything else in the universe other than the in the universe other than the systemsystem

Internal EnergyInternal Energy = total amount = total amount of energy (kinetic and of energy (kinetic and potential) in the potential) in the systemsystem

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B. The LawB. The Law

Although the energy of the Although the energy of the system and surroundings system and surroundings may change, the overall may change, the overall energy of the energy of the universeuniverse does does not (nor does the energy of not (nor does the energy of the the schuniverseschuniverse))

This is also known as the This is also known as the Law of Conservation of Law of Conservation of Energy: energy is neither Energy: energy is neither created nor destroyed, only created nor destroyed, only transferredtransferred

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C.C. Changes in Internal Changes in Internal EnergyEnergy

We don’t necessarily know the We don’t necessarily know the internal energy of the system internal energy of the system before and after a change. before and after a change. However, because the 1However, because the 1stst law law deals with net changes to the deals with net changes to the system, we only care about the system, we only care about the actual value of actual value of ΔΔEE

If If ΔΔE is (+), then the internal E is (+), then the internal energy of the system increasesenergy of the system increases

If If ΔΔE is (-), then the internal E is (-), then the internal energy of the system decreasesenergy of the system decreases Ctaikido.com

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D. D. ΔΔE, q, & w (again)E, q, & w (again)

If If ThenThen

q>0 q>0 Heat is transferred from surroundings Heat is transferred from surroundings to the system (endothermic)to the system (endothermic)

q<0q<0 Heat is transferred from system to Heat is transferred from system to surroundings (exothermic)surroundings (exothermic)

w>0w>0 Work is done by the surroundings on Work is done by the surroundings on the systemthe system

w<0w<0 Work is done by the system on the Work is done by the system on the surroundingssurroundings

If If ThenThen

q>0 and q>0 and w>0w>0

ΔΔE>0E>0

q>0 and q>0 and w<0w<0

The sign of The sign of ΔΔE depends on the E depends on the magnitude of q and wmagnitude of q and w

q<0 and q<0 and w>0w>0

The sign of The sign of ΔΔE depends on the E depends on the magnitude of q and wmagnitude of q and w

q<0 and q<0 and w<0w<0

ΔΔE<0E<0HW: 5.14, 15.16 (a),

15.18

E. State FunctionsE. State Functions

A state function is a property that A state function is a property that depends on the current state of a depends on the current state of a system, not the pathway or history of system, not the pathway or history of the systemthe system

q and w are NOT state functions, but q and w are NOT state functions, but their sum q + w = their sum q + w = ΔΔE is since E is since ΔΔE only E only depends on the final and initial states of depends on the final and initial states of the systemthe system

ΔΔE can be gained or lost as all heat or E can be gained or lost as all heat or heat and work (can never be just work, heat and work (can never be just work, due to friction); either way, the due to friction); either way, the ΔΔE will E will be the same at the end, regardless of be the same at the end, regardless of the combination of heat and workthe combination of heat and work

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Analogy from the textbook: Analogy from the textbook: Chicago is 596 ft above sea Chicago is 596 ft above sea level while Denver is 5280 level while Denver is 5280 ft above sea level; you can ft above sea level; you can take any route from take any route from Chicago to Denver Chicago to Denver (distance is not a state (distance is not a state function), but the altitude function), but the altitude change will always be 4684 change will always be 4684 ft (altitude is a state ft (altitude is a state function)function)

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III. Enthalpies of III. Enthalpies of ReactionReaction

When the heat lost or When the heat lost or gained by a system is at a gained by a system is at a constant pressure (qconstant pressure (qpp), it is ), it is

equal to the enthalpy equal to the enthalpy change (change (ΔΔH) of the systemH) of the system

Also, since very little work Also, since very little work is done in a chemical is done in a chemical reaction, we assume that reaction, we assume that ΔΔE = E = ΔΔH for a chemical H for a chemical systemsystem

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HW: 5.28

IV. CalorimetryIV. Calorimetry

Used to experimentally Used to experimentally determine enthalpy determine enthalpy valuesvalues

We have already used the We have already used the mathematical formula: mathematical formula:

q = mCq = mCΔΔT T Read this section for Read this section for

context, if neededcontext, if needed

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HW: 5.40

V. Hess’s LawV. Hess’s Law If a reaction is carried out in a series If a reaction is carried out in a series

of steps, the heat of reaction (of steps, the heat of reaction (ΔΔH) H) can be calculated by adding up the can be calculated by adding up the enthalpy changes of the individual enthalpy changes of the individual stepssteps

This is especially useful when This is especially useful when determining the heat of reaction determining the heat of reaction directly is difficultdirectly is difficult

Sometimes we have to rearrange Sometimes we have to rearrange and modify the given steps so that and modify the given steps so that they “add up” and “cancel out” to the they “add up” and “cancel out” to the correct final reactioncorrect final reaction

If we reverse a step, we change the If we reverse a step, we change the sign of sign of ΔΔHH

If we multiply the coefficients, we do If we multiply the coefficients, we do the same to the same to ΔΔHH

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Ex) Calculate the Ex) Calculate the ΔΔH for the following H for the following reaction:reaction:2C(s) + H2C(s) + H22(g) → C(g) → C22HH22(g)(g)

Given the following reactions:Given the following reactions:

CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) → 2CO(g) → 2CO22(g) + H(g) + H22O(l)O(l)

ΔΔH = -1299.6 kJH = -1299.6 kJ

C(s) + OC(s) + O22(g) → CO(g) → CO22(g)(g)

ΔΔH = -393.5 kJH = -393.5 kJ

HH22(g) + 1/2 O(g) + 1/2 O22(g) → H(g) → H22O(l)O(l)

ΔΔH = -285.8 kJH = -285.8 kJ

Hess is watching you!

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HW: 5.54

VI. Enthalpies of VI. Enthalpies of FormationFormation

Also known as heats of formation (Also known as heats of formation (ΔΔHHff); heat ); heat change associated with forming a compound change associated with forming a compound from its constituent elementsfrom its constituent elements

To standardize these values, we use standard To standardize these values, we use standard enthalpies (all reactants and products are in enthalpies (all reactants and products are in their most stable, standard states at their most stable, standard states at atmospheric pressure and room temp (25 ºC))atmospheric pressure and room temp (25 ºC))

The standard enthalpy of formation (The standard enthalpy of formation (ΔΔHHffºº) is the ) is the

enthalpy change that forms 1 mol of a enthalpy change that forms 1 mol of a compound from its elements, with all compound from its elements, with all substances in their standard statessubstances in their standard states

FYI, the standard state of oxygen is OFYI, the standard state of oxygen is O22 not O not O3 3 and the standard state of carbon is graphite not and the standard state of carbon is graphite not diamonddiamond

By definition, the By definition, the ΔΔHHffº º of the most stable form of of the most stable form of

an element is 0an element is 0

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We can now apply Hess’s Law to calculate We can now apply Hess’s Law to calculate the heat of reaction using the heats of the heat of reaction using the heats of formation of all reactants and products!formation of all reactants and products!

Ex) CEx) C33HH88(g) + 5O(g) + 5O22(g) → 3CO(g) → 3CO22(g) + 4H(g) + 4H22O(l)O(l)

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Hess is STILL watching you!

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HW: 5.60

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Chemical Chemical ThermodynamicsThermodynamics

Part 1Part 1

Ch. 19 In TextCh. 19 In Text

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I) SpontaneityI) SpontaneityA) DefinitionsA) Definitions

In any physical or chemical In any physical or chemical process, there is always a process, there is always a preferred direction preferred direction said to be said to be spontaneousspontaneous

We tend to think of this as a We tend to think of this as a process that occurs without process that occurs without any outside interventionany outside intervention

The direction that is The direction that is NOT NOT preferredpreferred is called is called nonspontaneousnonspontaneous

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Spontaneity is dependent on Spontaneity is dependent on temperature, not just the temperature, not just the processprocess

Ex) WaterEx) Water

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B) ReversibilityB) Reversibility

A A reversiblereversible process is one process is one in which the original state can in which the original state can be restored by be restored by exactly exactly reversing the changereversing the change

There is There is nono net change in the net change in the system or the surroundings system or the surroundings after the change is reversedafter the change is reversed

This is ONLY true for a This is ONLY true for a system at system at equilibriumequilibrium

Ex) Ice at its melting pointEx) Ice at its melting point Freshnessmag.com

An An irreversibleirreversible process is one in process is one in which a different path (different which a different path (different values of q and w) must be taken values of q and w) must be taken to return to the original stateto return to the original state

Although the Although the systemsystem may be may be restored, the restored, the surroundingssurroundings are are changedchanged

Ex) Expansion of a gasEx) Expansion of a gas Ex) Ice above or below its melting Ex) Ice above or below its melting

pointpoint

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HW: 19.6, 19.8

If a process is If a process is spontaneousspontaneous, then it must be , then it must be irreversible!irreversible!

Spontaneous processes are not necessarily Spontaneous processes are not necessarily fast, just the preferred direction…fast, just the preferred direction…

In general, In general, exothermicexothermic reactions are reactions are spontaneous; however some spontaneous; however some endothermic endothermic reactions are spontaneous, so what’s driving reactions are spontaneous, so what’s driving them to be spontaneous?them to be spontaneous?

II) Entropy II) Entropy

In an In an isothermal isothermal expansionexpansion of a gas, no of a gas, no heat or work is done to heat or work is done to expand a gas but the expand a gas but the increaseincrease in entropy in entropy causes the diffusion to be causes the diffusion to be spontaneousspontaneous

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Entropy is a state function (just like internal Entropy is a state function (just like internal energy, energy, ΔΔE, and enthalpy, E, and enthalpy, ΔΔH)H)

ΔS = SΔS = Sff – S – Sii

ΔS = qΔS = qrevrev/T (at const. T)/T (at const. T)

ΔS has the units J/KΔS has the units J/K If ΔS = +, then the products have more If ΔS = +, then the products have more

entropy than the reactants (look for gases!)entropy than the reactants (look for gases!) If ΔS = -, then the products have less entropy If ΔS = -, then the products have less entropy

than the reactants (look for solids!)than the reactants (look for solids!)

III) The 2III) The 2ndnd Law of Law of ThermodynamicsThermodynamics

The entropy of the The entropy of the universe can NEVER universe can NEVER decrease, instead it decrease, instead it must remain 0 or must remain 0 or increaseincrease

Unlike the 1Unlike the 1stst law in law in which energy is which energy is conserved, entropy is conserved, entropy is NOT conservedNOT conserved Zazzle.com

ΔSΔSunivuniv = ΔS = ΔSsys sys + ΔS+ ΔSsurr surr

Reversible process: Reversible process: ΔSΔSunivuniv = ΔS = ΔSsys sys + ΔS+ ΔSsurrsurr = 0 = 0

Irreversible process: Irreversible process: ΔSΔSunivuniv = ΔS = ΔSsys sys + ΔS+ ΔSsurr surr >0>0

ex) rustingex) rustingShare.ehs.uen.org

In an isolated system (the In an isolated system (the system cannot exchange heat, system cannot exchange heat, work, or matter with the work, or matter with the surroundings):surroundings):

ΔSΔSsys sys = 0= 0

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HW: 19.22, 19.28

IV) The 3IV) The 3rdrd Law of Law of ThermodynamicsThermodynamics

Each type of molecular motion Each type of molecular motion (vibrational, translational, and rotational) (vibrational, translational, and rotational) is related a is related a degree of freedom degree of freedom which which translates into an increase in entropytranslates into an increase in entropy

The fewer the bonds, the lower the IMFs, The fewer the bonds, the lower the IMFs, and the higher the temperature, the and the higher the temperature, the greater the number of degrees of freedomgreater the number of degrees of freedom

The greater the bonds, the higher the The greater the bonds, the higher the IMFs, and the lower the temperature, the IMFs, and the lower the temperature, the lower the number of degrees of freedomlower the number of degrees of freedom

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The Law states that the entropy The Law states that the entropy of a pure crystalline solid at of a pure crystalline solid at absolute zero is absolute zero is 0 0 (S @ 0 K = 0)(S @ 0 K = 0)

This is the theoretical definition This is the theoretical definition for “perfect” order where there for “perfect” order where there are no degrees of freedomare no degrees of freedom

NOTE: although entropy is NOTE: although entropy is temp. temp. dependentdependent, it is NOT related to , it is NOT related to enthalpyenthalpy

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V) Entropy CalcsV) Entropy Calcs

Standard Molar EntropyStandard Molar Entropy: : SSºº = J/mol•K = J/mol•K

SSº º for elements are NOT zerofor elements are NOT zero SSº º for gases are greater than for gases are greater than

those of solids and liquidsthose of solids and liquids SSº º increases with increasing increases with increasing

molar massmolar mass SSºº

increases with increasing increases with increasing number of atoms in the number of atoms in the formulaformula

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ΔSΔSºº = ΣS = ΣSººproductsproducts – ΣS – ΣSºº

reactantsreactants

Ex) Calculate the standard Ex) Calculate the standard entropy change for the entropy change for the synthesis of ammonia from synthesis of ammonia from its elements.its elements.

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HW: 19.34 (a) & (d)

Chemical Chemical ThermodynamicsThermodynamics

Part 2Part 2

(Ch. 19 in Text)(Ch. 19 in Text)

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VI) Gibbs Free EnergyVI) Gibbs Free Energy

Spontaneous changes are favored Spontaneous changes are favored by by an increase in entropy an increase in entropy and and a a decrease in energydecrease in energy (exothermic)(exothermic)

Spontaneity and its relation to Spontaneity and its relation to these 2 factors was quantified by these 2 factors was quantified by J.W. Gibbs in the late 1800sJ.W. Gibbs in the late 1800s

The state function known as free The state function known as free energy predicts the energy predicts the degreedegree of of spontaneity as follows:spontaneity as follows:

ΔG = ΔH - TΔSΔG = ΔH - TΔS Chemistry.about.com

If If ΔG = -, the forward rxn ΔG = -, the forward rxn is spon.is spon.

If ΔG = +, the forward If ΔG = +, the forward rxn is nonspon. and work rxn is nonspon. and work must be done by the must be done by the surroundings on the surroundings on the system for it to occursystem for it to occur

If ΔG = 0, the rxn is at If ΔG = 0, the rxn is at equilibriumequilibrium Cafepress.com

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Standard Free Energies of Standard Free Energies of FormationFormation

Tells us the free energy change Tells us the free energy change when a compound is formed from when a compound is formed from its elements in their standard states its elements in their standard states

As with ΔHAs with ΔHººff, ΔG, ΔGºº

ff for an element is for an element is 00

ΔGΔGººff = ΣG = ΣGºº

productsproducts – ΣG – ΣGººreactantsreactants

Can also be used to calculate the Can also be used to calculate the standard free energy change of any standard free energy change of any reaction (remember how Hess’s reaction (remember how Hess’s Law was applied to standard Law was applied to standard enthalpies of formation in the last enthalpies of formation in the last chapter)chapter)

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Ex) Using Appendix C, Ex) Using Appendix C, calculate the standard calculate the standard free energy for the free energy for the combustion of methane.combustion of methane.

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VII) Free Energy and VII) Free Energy and TemperatureTemperature

Assuming standard states Assuming standard states and conditions, the Gibbs and conditions, the Gibbs equation becomes:equation becomes:

ΔGº = ΔHº - TΔSºΔGº = ΔHº - TΔSº Notice that the spontaneity Notice that the spontaneity

(or the degree of (or the degree of spontaneity) may be spontaneity) may be dependent on the dependent on the temperature of the systemtemperature of the system Alohapoolandspaservice.

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ΔGº = ΔH º - TΔSºΔGº = ΔH º - TΔSº

Under what conditions is a Under what conditions is a reaction ALWAYS reaction ALWAYS spontaneous?spontaneous?

Under what conditions is a Under what conditions is a reaction NEVER reaction NEVER spontaneous?spontaneous?

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ΔGº = ΔH º - TΔSºΔGº = ΔH º - TΔSº

Ex) HEx) H22O(s) O(s) → H→ H22O(l)O(l)

Ex) Ex) HH22O(l) O(l) → H→ H22O(s)O(s)

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Ex) Calculate the standard Ex) Calculate the standard free energy change for the free energy change for the formation of ammonia at 500. formation of ammonia at 500. ºC.ºC.

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HW: 19.38, 19.50, 19.52

VIII) Free Energy Under VIII) Free Energy Under Nonstandard ConditionsNonstandard Conditions

ΔG = ΔGº + RT ln QΔG = ΔGº + RT ln Q where ΔG = the free where ΔG = the free

energy change under energy change under nonstandard conditionsnonstandard conditions

R = 8.31 J/molR = 8.31 J/mol T = the absolute temp.T = the absolute temp. Q = the reaction quotientQ = the reaction quotient

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Ex) Calculate Ex) Calculate ΔG at ΔG at 500. 500. ºC for a reaction mixture ºC for a reaction mixture that consists of 1.0 atm Nthat consists of 1.0 atm N22, ,

3.0 atm H3.0 atm H22, and 0.50 atm , and 0.50 atm

NHNH33..

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At equilibrium, At equilibrium, ΔG = 0 (no ΔG = 0 (no preferred direction) and Q = Kpreferred direction) and Q = K

Thus,Thus, ΔG = ΔGº + RT ln Q ΔG = ΔGº + RT ln Q 0 = ΔGº + RT ln K 0 = ΔGº + RT ln K

ΔGº = -RT ln KΔGº = -RT ln K

If ΔGº = -, then K>1 If ΔGº = -, then K>1

If ΔGº = 0, then K=1If ΔGº = 0, then K=1

If ΔGº = +, then K<1If ΔGº = +, then K<1

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HW: 19.58, 19.60

Solving for K:Solving for K: K = eK = e--ΔGº/RTΔGº/RT

Ex) Find KEx) Find Kpp at 500. ºC at 500. ºC

for the Haber process.for the Haber process.

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HW: 19.66

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