energy and chemical reactions l. scheffler 1. heat and temperature heat is energy that is...

Energy and Chemical Energy and Chemical ReactionsReactions

L. SchefflerL. Scheffler

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Heat and TemperatureHeat and Temperature

HeatHeat is energy that is transferred from one is energy that is transferred from one object to another due to a difference in object to another due to a difference in temperaturetemperature

TemperatureTemperature is a measure of the average is a measure of the average kinetic energy of a bodykinetic energy of a body

Heat is always transferred from objects at a Heat is always transferred from objects at a higher temperature to those at a lower higher temperature to those at a lower temperaturetemperature

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Factors Affecting Heat Factors Affecting Heat QuantitiesQuantities

The amount of heat contained by an object The amount of heat contained by an object depends primarily on three factors:depends primarily on three factors:– The mass of materialThe mass of material– The temperatureThe temperature– The kind of material and its ability to The kind of material and its ability to

absorb or retain heat. absorb or retain heat.

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Heat QuantitiesHeat Quantities

The heat required to raise the temperature of 1.00 g The heat required to raise the temperature of 1.00 g of water 1 of water 1 ooC is known as a C is known as a caloriecalorie

The SI unit for heat is the The SI unit for heat is the joulejoule. It is based on the . It is based on the mechanical energy requirements. mechanical energy requirements.

1.00 calorie = 4.184 Joules1.00 calorie = 4.184 Joules

The energy required to raise 1 pound of water of 1 The energy required to raise 1 pound of water of 1 ooF F is called a is called a British Thermal UnitBritish Thermal Unit or or BTUBTU

The BTU is widely used in the USA to compute The BTU is widely used in the USA to compute energy capacities of heating and air conditioning energy capacities of heating and air conditioning equipmentequipment

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CalorimetryCalorimetry

Calorimetry involves the measurement of Calorimetry involves the measurement of heat changes that occur in chemical heat changes that occur in chemical processes or reactions. processes or reactions.

The heat change that occurs when a The heat change that occurs when a substance absorbs or releases energy is substance absorbs or releases energy is really a function of three quantities:really a function of three quantities:– The massThe mass– The temperature changeThe temperature change– The heat capacity of the materialThe heat capacity of the material

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Heat Capacity and Specific Heat Capacity and Specific HeatHeat

The ability of a substance to absorb or retain heat The ability of a substance to absorb or retain heat varies widely.varies widely.

The heat capacity depends on the nature of the The heat capacity depends on the nature of the material.material.

The The specific heatspecific heat of a material is the amount of heat of a material is the amount of heat required to raise the temperature of 1 gram of a required to raise the temperature of 1 gram of a substance 1 substance 1 ooC (or Kelvin)C (or Kelvin)

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SubstanceSubstance CCJ gJ g-1-1 K K-1-1 C C J molJ mol-1-1KK-1-1

Water (liquid)Water (liquid) 4.184 4.184 75.327 75.327

Water (steam)Water (steam) 2.080 2.080 37.47 37.47

Water (ice) Water (ice) 2.050 2.050 38.09 38.09

CopperCopper 0.385 0.385 24.47 24.47

AluminumAluminum 0.897 0.897 24.2 24.2

Ethanol Ethanol 2.44 2.44 112 112

Lead Lead 0.127 0.127 26.4 26.4

Specific Heat values for Specific Heat values for Some Common Some Common

SubstancesSubstances

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Heat ExchangeHeat Exchange

When two systems When two systems are put in contact are put in contact with each other, with each other, there will be a net there will be a net exchange of energy exchange of energy between them between them unless they are at unless they are at thermal equilibrium, thermal equilibrium, i.e. at the same i.e. at the same temperature.temperature.

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Heat will flow from the substance at the higher temperature Heat will flow from the substance at the higher temperature to that at a lower temperatureto that at a lower temperature

Heat ChangesHeat Changes

The heat equation may be stated asThe heat equation may be stated as

Q = m C Q = m C TTwhere:where:

Q = Change in heatQ = Change in heat

m = mass in gramsm = mass in grams

C = specific heat in J gC = specific heat in J g-1-1 ooCC-1-1

T = Temperature changeT = Temperature change

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Temperature Changes Temperature Changes

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Measuring the temperature Measuring the temperature change in a calorimetry change in a calorimetry experiment can be difficult experiment can be difficult since the system is losing since the system is losing heat to the surroundings heat to the surroundings even as it is generating even as it is generating heat.heat.

By plotting a graph of time By plotting a graph of time v temperature it is possible v temperature it is possible to extrapolate back to what to extrapolate back to what the maximum temperature the maximum temperature would have been had the would have been had the system not been losing system not been losing heat to the surroundings.heat to the surroundings.

A time v temperature graph

Heat Transfer Problem 1Heat Transfer Problem 1

Calculate the heat that would be required an aluminum Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20cooking pan whose mass is 400 grams, from 20ooC to C to 200200ooC. The specific heat of aluminum is 0.902 J gC. The specific heat of aluminum is 0.902 J g-1-1 ooCC-1-1..

Solution

Q = mCT

= (400 g) (0.902 J g0.902 J g-1-1 ooCC-1-1)(200)(200ooC – 20C – 20ooCC))

= 64,944 J= 64,944 J

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Heat Transfer Problem 2Heat Transfer Problem 2 What is the final temperature when 50 grams of water at What is the final temperature when 50 grams of water at

2020ooC is added to 80 grams water at 60C is added to 80 grams water at 60ooC? Assume that C? Assume that the loss of heat to the surroundings is negligible. The the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 specific heat of water is 4.184 J g-1 ooCC-1-1

Solution:Solution: Q (Cold) = Q (hot) mCT= mCT

Let T = final temperatureLet T = final temperature

(50 g) (4.184 J g(50 g) (4.184 J g-1-1 ooCC-1-1)(T- 20)(T- 20ooC)C) = (80 g) (4.184 J g= (80 g) (4.184 J g-1-1 ooCC-1-1)(60)(60ooC- T) C- T)

(50 g)(T- 20(50 g)(T- 20ooC) = (80 g)(60C) = (80 g)(60ooC- T) C- T) 50T -1000 = 4800 – 80T50T -1000 = 4800 – 80T 130T =5800130T =5800 T = 44.6 T = 44.6 ooCC

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Phase Changes & HeatEnergy is required to change the phase of a Energy is required to change the phase of a substancesubstanceThe amount of heat necessary to melt a The amount of heat necessary to melt a substance is called the substance is called the Heat of fusionHeat of fusion ((HHfusfus).The heat of fusion is expressed in ).The heat of fusion is expressed in terms of 1 mole or 1 gramterms of 1 mole or 1 gram

It takes 6.00 kJ of energy to melt 1 mole It takes 6.00 kJ of energy to melt 1 mole (18 grams) of ice into liquid water. This (18 grams) of ice into liquid water. This is equivalent to about 335 J per gramis equivalent to about 335 J per gram

The amount of heat necessary to boil a substance is The amount of heat necessary to boil a substance is called the called the Heat of vaporizationHeat of vaporization ( (HHvapvap))

It may be expressed in terms of 1 mole or 1 gramIt may be expressed in terms of 1 mole or 1 gram

It It takes 40.6 kJ of energy to boil away 1 mole (18 takes 40.6 kJ of energy to boil away 1 mole (18 grams)grams) of water. This is equivalent to about 2240 of water. This is equivalent to about 2240 J per gram.J per gram. 13

SubstanceSubstance QQfusfus QQvapvap

Mercury, HgMercury, Hg 2.29kJ/mol2.29kJ/mol 59.1kJ/mol59.1kJ/mol

Ethanol, CEthanol, C22HH55OHOH 5.02kJ/mol5.02kJ/mol 38.6kJ/mol38.6kJ/mol

Water, HWater, H22OO 6.00kJ/mol6.00kJ/mol 40.6kJ/mol40.6kJ/mol

Ammonia, NHAmmonia, NH33 5.65kJ/mol5.65kJ/mol 23.4kJ/mol23.4kJ/mol

Helium, HeHelium, He 0.02kJ/mol0.02kJ/mol 0.08kJ/mol0.08kJ/mol

AcetoneAcetone 5.72kJ/mol5.72kJ/mol 29.1kJ/mol29.1kJ/mol

Methanol, CHMethanol, CH33OHOH 3.16kJ/mol3.16kJ/mol 35.3kJ/mol35.3kJ/mol

Molar Heat Data for Molar Heat Data for Some Common Some Common

SubstancesSubstances

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mCmCliquid waxliquid waxTTQQtotatota==

(50g)(2.31J g(50g)(2.31J g-1-1˚C˚C-1-1)(62˚C-85˚C) )(62˚C-85˚C)

+ (50g/352.7gmol+ (50g/352.7gmol-1-1)(-70,500J mol)(-70,500J mol-1-1))

+ (50g)(2.18J g+ (50g)(2.18J g-1-1˚C˚C-1-1)(25˚C-62˚C))(25˚C-62˚C)

How much energy must be lost for 50.0 g of liquid wax at How much energy must be lost for 50.0 g of liquid wax at 85.0˚C to cool to room temperature at 25.0˚C?85.0˚C to cool to room temperature at 25.0˚C?

(C(Csolid waxsolid wax= 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, C= 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, C liquid waxliquid wax=2.31 =2.31 J/g˚C; MM = 352.7 g/mol, J/g˚C; MM = 352.7 g/mol, HHfusionfusion=70,500 J/mol)=70,500 J/mol)


Q =Q =

QQtotal total = = QQliquid wax liquid wax + + QQsolidification solidification + + QQsolid waxsolid wax

+ n(+ n(QQfusionfusion))

+ mC+ mCsolid waxsolid waxTT

QQtotal= total= = = mCmCliquid waxliquid waxT +n(T +n(QQfusionfusion) + ) + mCmCsolid waxsolid waxTT

QQtotaltotal== (-2656.5 J) + (-9994.3 J)+ (-4033 J)(-2656.5 J) + (-9994.3 J)+ (-4033 J)

QQtotaltotal== -16,683.8 J-16,683.8 J

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Steam at 175°C that occupies a volume of 32.75 dmSteam at 175°C that occupies a volume of 32.75 dm3 3

and a pressure of 2.60 atm. How much energy would and a pressure of 2.60 atm. How much energy would it need to lose to end as liquid water at 20 it need to lose to end as liquid water at 20 ooC?C?


Solution: n = PV/RT = (2.60 atm)(32.75 dm3)

(0.0821 dm3 atm mol-1 K-1)(448 K-1) = 2.315 mol

Q = (2.315 mol) (37.47 J mol-1K-1)(175oC-100oC) +(2.315 mol)(40600 J mol-1) +(2.315 mol)(75.327 J mol-1K-1)(100oC-20oC)

Q = 6505.7J + 93989 J + 13950.6 J = 114445.3 J = 114.445 kJ

Chemical ReactionsChemical Reactions

In a chemical reactionIn a chemical reaction

Chemical bonds are brokenChemical bonds are broken

Atoms are rearrangedAtoms are rearranged

New chemical bonds are formedNew chemical bonds are formed

These processes always involve These processes always involve energy changesenergy changes

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Energy ChangesEnergy Changes

Breaking chemical bonds requires energyBreaking chemical bonds requires energy

Forming new chemical bonds releases Forming new chemical bonds releases energyenergy

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Exothermic and Endothermic Exothermic and Endothermic ProcessesProcesses

Exothermic processes release energyExothermic processes release energy

CC33HH88 (g) + 5 O (g) + 5 O22 (g) (g) 3 CO 3 CO22 (g) + 4H (g) + 4H22O (g)O (g)

+ 2043 + 2043 kJkJ

Endothermic processes absorb energy Endothermic processes absorb energy

C(s) + HC(s) + H22O (g)O (g) +113 kJ+113 kJ CO(g) + H CO(g) + H22 (g) (g)

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Energy Changes in Energy Changes in endothermic and endothermic and

exothermic processesexothermic processesIn an In an endothermic endothermic reaction there is reaction there is more energy more energy required to required to break bonds break bonds than is released than is released when bonds are when bonds are formed. formed.

The opposite is The opposite is true in an true in an exothermic exothermic reaction.reaction.

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Enthalpy and Hess’ LawEnthalpy and Hess’ Law

EnthalpyEnthalpy

Enthalpy Enthalpy is the heat absorbed or released is the heat absorbed or released during a chemical reaction where the only during a chemical reaction where the only work done is the expansion of a gas at work done is the expansion of a gas at constant pressureconstant pressure

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EnthalpyEnthalpy

Not all energy changes that occur as a result Not all energy changes that occur as a result of chemical reactions are expressed as heatof chemical reactions are expressed as heat

Energy = Heat + WorkEnergy = Heat + Work

Work is a force applied over a distance.Work is a force applied over a distance.

Most energy changes resulting from Most energy changes resulting from chemical reactions are expressed in a chemical reactions are expressed in a special term known as special term known as enthalpyenthalpy

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EnthalpyEnthalpy

It is nearly impossible to set up a It is nearly impossible to set up a chemical reaction where there is no chemical reaction where there is no work performed.work performed.

The conditions for a chemical reaction The conditions for a chemical reaction are often set up so that work in are often set up so that work in minimized.minimized.

Enthalpy and heat are nearly equal Enthalpy and heat are nearly equal under these conditions.under these conditions.

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Enthalpy ChangesEnthalpy Changes

The change in enthalpy is designated by the The change in enthalpy is designated by the symbol symbol H.H.– If If H < 0 the process is exothermic.H < 0 the process is exothermic.– If If H > 0 the process is endothermic.H > 0 the process is endothermic.

Sometimes the symbol for enthalpy (Sometimes the symbol for enthalpy (H) is used H) is used for heat (for heat (Q) In many cases where work is Q) In many cases where work is minimal heat is a close approximation for minimal heat is a close approximation for enthalpy. One must always remember that while enthalpy. One must always remember that while they are closely related, heat and enthalpy are not they are closely related, heat and enthalpy are not the same thingthe same thing

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Energy and Enthalpy Energy and Enthalpy ChangesChanges

It is impractical to measure absolute It is impractical to measure absolute amounts of energy or enthalpy.amounts of energy or enthalpy.

Enthalpy is always measured relative to Enthalpy is always measured relative to previous conditions.previous conditions.

Hence we measure changes in enthalpy Hence we measure changes in enthalpy rather than total enthalpyrather than total enthalpy

Enthalpy is measured relative to the system.Enthalpy is measured relative to the system.

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Measuring EnthalpyMeasuring Enthalpy

The amount of heat absorbed or released The amount of heat absorbed or released during a chemical reaction depends on the during a chemical reaction depends on the conditions under which the reaction is conditions under which the reaction is carried out including:carried out including:– the temperaturethe temperature– the pressurethe pressure– the physical state of the reactants and the physical state of the reactants and

productsproducts

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Standard ConditionsStandard Conditions

For most thermodynamic measurements For most thermodynamic measurements standard conditions are established asstandard conditions are established as– 25 25 ooC or 298 KC or 298 K– 1.0 atmosphere of pressure1.0 atmosphere of pressure

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Standard StateStandard State

The pure form of a substance at standard The pure form of a substance at standard conditions is said to be in the standard conditions is said to be in the standard state.state.

The most stable form of an element at The most stable form of an element at standard conditions represents the standard conditions represents the standard state for that element.standard state for that element.

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Hess’ LawHess’ Law

If a series of reactions are added If a series of reactions are added together, the enthalpy change for the together, the enthalpy change for the net reaction will be the sum of the net reaction will be the sum of the enthalpy change for the individual enthalpy change for the individual steps steps

Hess’ Law provides a way to calculate Hess’ Law provides a way to calculate enthalpy changes even when the enthalpy changes even when the reaction cannot be performed directly.reaction cannot be performed directly.

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Hess’ Law: Example 1Hess’ Law: Example 1

NN22 (g) + O (g) + O22 (g) (g) 2 NO (g) 2 NO (g) HH11 = +181 kJ = +181 kJ

2 NO2 NO (g) + O(g) + O22 (g) (g) 2 NO 2 NO22 (g) (g) HH22 = -113 kJ = -113 kJ

Find the enthalpy change for Find the enthalpy change for NN22 (g) + 2 O (g) + 2 O22 (g) (g) 2 NO 2 NO2 2 (g) (g)

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Hess’ Law: Example 1Hess’ Law: Example 1

Solution:Solution:

NN22 (g) + O (g) + O22 (g) (g) 2 NO (g) 2 NO (g) HH11 = +181 kJ = +181 kJ

2 NO2 NO (g) + O(g) + O22 (g) (g) 2 NO 2 NO22 (g) (g) HH22 = -113 kJ = -113 kJ

--------------------------------------------------------------------------------------------------------------------------

NN22 (g) +2O (g) +2O22 (g)+ 2 NO (g)+ 2 NO (g) (g) 2 NO (g) + 2 NO 2 NO (g) + 2 NO22 (g) (g)

NN22 (g) +2O (g) +2O22 (g) (g) + 2 NO + 2 NO22 (g) (g)

H =H = HH1 1 + + HH2 2 = +181 kJ +(-113)= +181 kJ +(-113) = + 68 kJ = + 68 kJ

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Hess Law: Example 2Hess Law: Example 2 From the following reactions and enthalpy changes:From the following reactions and enthalpy changes:

2 SO2 SO22 (g) + O (g) + O22 (g) (g) 2 SO 2 SO33 (g) (g) H = -196 kJ H = -196 kJ

2 S (s) +3 O2 S (s) +3 O22 (g) (g) 2 SO 2 SO33 (g) (g) H = -790 kJH = -790 kJ

Find the enthalpy change for the following reaction:Find the enthalpy change for the following reaction:

S (s) + OS (s) + O22 (g) (g) SO SO22 (g) (g)

Solution:Solution:

2 SO2 SO33 (g) (g) 2 SO 2 SO22 (g) + O (g) + O22 (g) (g) H = +196 kJ H = +196 kJ

2 S (s) +3 O2 S (s) +3 O22 (g) (g) 2 SO 2 SO33 (g) (g) H = -790 kJH = -790 kJ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Reversing the order of the first equation reverses the sign of Reversing the order of the first equation reverses the sign of HH

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Hess Law Example 2Hess Law Example 2 From the following reactions and enthalpy changes:From the following reactions and enthalpy changes:




S (s) + OS (s) + O22 (g) (g) SO SO22 (g) (g)



2 SO2 SO33(g) +2 S(s) + (g) +2 S(s) + 22 3 O 3 O22 (g) (g) 2 SO 2 SO33 (g)+2 SO (g)+2 SO22 (g) + O (g) + O22 (g) (g)

H = -594H = -594 kJ kJ

2 S(s) + 2 O2 S(s) + 2 O22 (g) (g) 2 SO 2 SO22 (g) (g) H = -594H = -594 kJ kJ

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Hess Law: Example 2Hess Law: Example 2 From the following reactions and enthalpy changes:From the following reactions and enthalpy changes:




S (s) + OS (s) + O22 (g) (g) SO SO22 (g) (g)



2 SO2 SO33((gg) +2 S() +2 S(ss) + ) + 22 3 O 3 O22 ( (gg) ) 2 SO 2 SO33 ( (gg)+2 SO)+2 SO22 (g) + O (g) + O22 (g) (g)

H = -594H = -594 kJ kJ

2 S(s) + 2 O2 S(s) + 2 O22 (g) (g) 2 SO 2 SO22 (g) (g) H = -594H = -594 kJ kJ

S(s) + OS(s) + O22 (g) (g) SO SO22 (g) (g) H = -297H = -297 kJ kJ35

Standard Enthalpy Standard Enthalpy ChangesChanges

The enthalpy change that occurs when the The enthalpy change that occurs when the reactants are converted to products, both reactants are converted to products, both being in their standard states is known as the being in their standard states is known as the standard enthalpy change.standard enthalpy change.

It is designated as It is designated as HHoo..

HHo o reactionreaction = = HHo o products -products - HHo o reactantsreactants

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Calculating Enthalpy from Calculating Enthalpy from tablestables

The enthalpy of formation for compound The enthalpy of formation for compound is equal to the enthalpy change that is equal to the enthalpy change that occurs when a compound is formed from occurs when a compound is formed from its elementsits elementsThe symbol for the bond enthalpy of The symbol for the bond enthalpy of formation is formation is HHff

Enthalpies of formation have been Enthalpies of formation have been measured and tabulated for a large measured and tabulated for a large number of compoundsnumber of compounds

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Enthalpies of FormationEnthalpies of Formation

Some enthalpies of formation for common compoundsSome enthalpies of formation for common compounds

BaCOBaCO33 -1219-1219 HH22O (g)O (g) -242-242 HCl (g)HCl (g) -93-93

Ba(OH)Ba(OH)22 -998-998 HH22O (l)O (l) -286-286 HCl (aq)HCl (aq) -167-167

BBaOaO -554-554 HH22OO22 -188-188 NHNH3 3 (g)(g) -46-46

CaCOCaCO33 -1207-1207 CC33HH88 -104-104 NONO +90+90

CaOCaO -636-636 CC44HH1010 -126-126 NONO22 +33.8+33.8

Ca(OH)Ca(OH)22 -987-987 COCO -110-110 SOSO22 -297-297

CaClCaCl22 -796-796 COCO22 -394-394 AlAl22OO33(s)(s) -1670-1670

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See your text: Brown, LeMay and Bursten, Chemistry the Central Science, 7th edition pages 984-987 for addition values

Calculating Enthalpy from Calculating Enthalpy from tablestables

Enthalpies of formation represent the Enthalpies of formation represent the enthalpy changes when compound forms enthalpy changes when compound forms from its elementsfrom its elementsThe enthalpy of formation for an uncombined The enthalpy of formation for an uncombined element is therefore = 0element is therefore = 0The enthalpy of formation for a chemical The enthalpy of formation for a chemical reaction can be expressed as the difference reaction can be expressed as the difference between the enthalpy state of the products between the enthalpy state of the products and that of the reactantsand that of the reactants

Hreaction = Hreaction = HHooproductsproducts – –HHoo

reactantsreactants

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Sample Problem 1Sample Problem 1Calcium carbonate reacts with hydrochloric acid Calcium carbonate reacts with hydrochloric acid according to the following equation:according to the following equation:

CaCOCaCO33 (s)(s) + 2HCl + 2HCl (aq)(aq) CaCl CaCl2 2 (aq)(aq) + H + H22O O (l)(l) + CO + CO2 2 (g)(g)

Calculate the enthalpy change for this reactionCalculate the enthalpy change for this reaction

HHooreactionreaction = = HHoo

productsproducts – –HHooreactantsreactants

HHooCaCOCaCO33 -1207-1207

HHo o HCl HCl (aq)(aq) -167-167HHooCaClCaCl22 -796-796HHo o HH22O O (l)(l) -286-286HHo o COCO2 2 (g)(g) -394-394

SolutionSolutionHHoo

products products =(-796)+(-286)+(-394) =(-796)+(-286)+(-394)

= -1476 kJ= -1476 kJ

HHooreactants reactants =(=(-1207-1207)+(2)(-167))+(2)(-167)

== -1541 kJ -1541 kJ HHoo

reaction reaction = -1476-(-1541) = = -1476-(-1541) = +75 kJ+75 kJ

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Sample Problem 2Sample Problem 2Calculate the enthalpy change for the burning Calculate the enthalpy change for the burning of 11 grams of propaneof 11 grams of propane

CC33HH88 (g) + 5 O (g) + 5 O22 (g) (g) 3 CO 3 CO2 2 (g) + 4 H(g) + 4 H22O (g)O (g)

HHooreactionreaction = = HHoo

productsproducts – –HHooreactantsreactants

HHo o CC33HH88 -104-104

HHo o OO22 (g) (g) 00

HHo o HH22O O (g)(g) -242-242HHo o COCO2 2 (g)(g) -394-394

SolutionSolutionHHoo

products products =(3)(-394)+(4)(-242) =(3)(-394)+(4)(-242)

= -2150 kJ= -2150 kJHHoo

reactants reactants =(=(-104-104)+(5)(0))+(5)(0)

== -104 kJ -104 kJ HHoo

reaction reaction = -2150-(-104) = = -2150-(-104) = -2046 kJmol-2046 kJmol-1-1

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Now 11 grams = 0.25 mole of propane (11 g/44 g mol-1)(0.25 mol )(-2046 kJ mol-1) = 511.5 kJ

Bond EnthalpiesBond Enthalpies

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Bond EnthalpiesBond EnthalpiesAnother approach to determining an enthalpy Another approach to determining an enthalpy change for a chemical reactionchange for a chemical reaction

The energy to required to break a covalent The energy to required to break a covalent bond in the gaseous phase is called a bond in the gaseous phase is called a bond bond enthalpy.enthalpy.

Bond enthalpy tables give the average Bond enthalpy tables give the average energy to break a chemical bond. Actually energy to break a chemical bond. Actually there are slight variations depending on the there are slight variations depending on the environment in which the chemical bond is environment in which the chemical bond is locatedlocated

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Bond Enthalpy TableBond Enthalpy TableThe average bond enthalpies for several types of The average bond enthalpies for several types of chemical bonds are shown in the table below:chemical bonds are shown in the table below:

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Bond EnthalpiesBond EnthalpiesBond enthalpies can be used to calculate the Bond enthalpies can be used to calculate the enthalpy change for a chemical reactionenthalpy change for a chemical reaction..Energy is required toEnergy is required to break chemical bondsbreak chemical bonds. . Therefore when a chemical bond is broken Therefore when a chemical bond is broken its its enthalpy changeenthalpy change carries a carries a positive signpositive sign..Energy is released when chemical bonds Energy is released when chemical bonds formform. When a chemical bond is formed its . When a chemical bond is formed its enthalpy changeenthalpy change is expressed as a is expressed as a negative negative valuevalueBy combining the enthalpy required and the By combining the enthalpy required and the enthalpy released for the breaking and enthalpy released for the breaking and forming chemical bonds, one can calculate forming chemical bonds, one can calculate the enthalpy change for a chemical reactionthe enthalpy change for a chemical reaction

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Bond Enthalpy Bond Enthalpy CalculationsCalculations

Example 1: Calculate the enthalpy change for the Example 1: Calculate the enthalpy change for the reaction Nreaction N22 + 3 H + 3 H22 2 NH 2 NH33

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Bonds broken

1 N=N: = 9453 H-H: 3(435) = 1305 Total = 2250 kJ

Bonds formed

2x3 = 6 N-H: 6 (390) = - 2340 kJ

Net enthalpy change

= + 2250 - 2340 = - 90 kJ

Born Haber CycleBorn Haber Cycle

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Born-Haber CycleBorn-Haber Cycle

Born-Haber Cycles are energy cycles for the Born-Haber Cycles are energy cycles for the formation of certain ionic compoundsformation of certain ionic compounds

Application of Hess’ LawApplication of Hess’ Law

A Born-Haber cycle can be used to calculate A Born-Haber cycle can be used to calculate quantities that are difficult to measure quantities that are difficult to measure directly such as lattice energiesdirectly such as lattice energies

The formation of an ionic compound as a The formation of an ionic compound as a sequence of steps whose energies can be sequence of steps whose energies can be determined.determined.

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Some DefinitionsSome DefinitionsThe The enthalpy of atomizationenthalpy of atomization is the enthalpy change that is the enthalpy change that occurs when one mole of gaseous atoms is formed occurs when one mole of gaseous atoms is formed from the element in the standard state under standard from the element in the standard state under standard conditionsconditions

Example: ½ ClExample: ½ Cl22 (g) (g) Cl (g) Cl (g) HHooatat = 121 kJ mol = 121 kJ mol-1-1

The The electron affinityelectron affinity is the enthalpy change that occurs is the enthalpy change that occurs when an electron is added to an isolated atom in the when an electron is added to an isolated atom in the gaseous state:gaseous state:

O (g) + e- O (g) + e- O O-- (g) (g) HHoo = -142 kJ mol = -142 kJ mol-1-1

OO-- (g) + e- (g) + e- O O2- 2- (g) (g) HHoo = +844 kJ mol = +844 kJ mol-1-1

The The lattice enthalpylattice enthalpy is the enthalpy change that occurs is the enthalpy change that occurs from the conversion of an ionic compound in the from the conversion of an ionic compound in the gaseous state into its gaseous ionsgaseous state into its gaseous ions

LiCl (g) LiCl (g) Li Li++ (g) + Cl (g) + Cl-- (g) (g) HHoo = +846 kJ mol = +846 kJ mol-1-1

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Born Haber Cycle Born Haber Cycle DiagramDiagram

The stepwise energy changes for the formation of NaClThe stepwise energy changes for the formation of NaCl50

Born Haber Cycle for Born Haber Cycle for NaClNaCl

The formation of NaCl can be considered as a five The formation of NaCl can be considered as a five step processstep processNa (s) + 1/2 ClNa (s) + 1/2 Cl22 (g) (g) NaCl (s)NaCl (s)

1.1. The vaporization of sodium metal to form the gaseous The vaporization of sodium metal to form the gaseous element. element.

2.2. The dissociation of chlorine gas to gaseous chlorine The dissociation of chlorine gas to gaseous chlorine atoms is equal to one half of the bond energy for a Cl-Cl atoms is equal to one half of the bond energy for a Cl-Cl covalent bond covalent bond

3.3. The ionization of gaseous sodium atoms to Na(g) The ionization of gaseous sodium atoms to Na(g) NaNa++

4.4. The ionization of chlorine atoms. (This quantity is the The ionization of chlorine atoms. (This quantity is the negative electron affinity for the element chlorine.)negative electron affinity for the element chlorine.)

5.5. The lattice energy on the formation of sodium chloride The lattice energy on the formation of sodium chloride from the gaseous ionsfrom the gaseous ions

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Born-Haber Cycle for Born-Haber Cycle for NaClNaCl

The stepwise energy changes for the formation of NaCl:The stepwise energy changes for the formation of NaCl:The vaporization of sodium metal to form the gaseous The vaporization of sodium metal to form the gaseous elementelement..

Na (s) Na (s) Na (g) Na (g) ∆H° ∆H°sublimationsublimation = + 109 kJ mol = + 109 kJ mol-1-1

The dissociation of chlorine gas to gaseous chlorine The dissociation of chlorine gas to gaseous chlorine atoms is equal to one half of the bond energy for a Cl-Cl atoms is equal to one half of the bond energy for a Cl-Cl covalent bond covalent bond

1/2 Cl1/2 Cl2 2 (g) (g) Cl (g) Cl (g) ∆H° ∆H°dissdiss = + 122 kJ mol = + 122 kJ mol-1-1

The ionization of gaseous sodium atoms to:The ionization of gaseous sodium atoms to: Na (g) Na (g) Na Na++ (g) + e (g) + e-- ∆H° ∆H°ionizationionization = + 496 kJ mol = + 496 kJ mol-1-1

The ionization of chlorine atoms. (This quantity is the The ionization of chlorine atoms. (This quantity is the negative electron affinity for the element chlorine.)negative electron affinity for the element chlorine.)

Cl (g)Cl (g) + e+ e-- Cl Cl-- (g) ∆H° (g) ∆H°elect.affinityelect.affinity = - 368 kJ mol = - 368 kJ mol-1-1

The lattice energy on the formation of sodium chloride The lattice energy on the formation of sodium chloride from the gaseous ionsfrom the gaseous ions

NaNa++ (g) + Cl (g) + Cl-- (g) (g) NaCl (s) ∆H° NaCl (s) ∆H° latticelattice = - 770 kJ mol = - 770 kJ mol-1-152

EntropyEntropy

EntropyEntropy

EntropyEntropy is defined is defined as a state of as a state of disorder or disorder or randomness.randomness.

In general the In general the universe tends to universe tends to move toward move toward release of energy release of energy and greater entropy.and greater entropy.

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EntropyEntropy

The statistical The statistical interpretation of interpretation of thermodynamics was thermodynamics was pioneered by James pioneered by James Clerk Maxwell (1831–Clerk Maxwell (1831–1879) and brought to 1879) and brought to fruition by the Austrian fruition by the Austrian physicist Ludwig physicist Ludwig Boltzmann (1844–1906).Boltzmann (1844–1906).

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EntropyEntropy

Spontaneous chemical Spontaneous chemical processes often result in a processes often result in a final state is more final state is more DisorderedDisordered or or Random Random than the original.than the original.

The Spontaneity of a The Spontaneity of a chemical process is related to chemical process is related to a change in randomness.a change in randomness.

Entropy Entropy is ais a thermodynamic thermodynamic property related to the property related to the degree of randomness or degree of randomness or disorder in a system.disorder in a system.

Reaction of potassium Reaction of potassium metal with water. The metal with water. The products are more products are more randomly distributed randomly distributed than the reactantsthan the reactants 56

Entropy is DisorderEntropy is Disorder Disorder Disorder in a system can take many forms. in a system can take many forms.

Each of the following represent an increase in Each of the following represent an increase in disorder and therefore in entropy:disorder and therefore in entropy:

1.1. Mixing different types of particles. i.e. Mixing different types of particles. i.e. dissolving salt in water.dissolving salt in water.

2.2. A change is state where the distance between A change is state where the distance between particles increases. Evaporation of water.particles increases. Evaporation of water.

3.3. Increased movement of particles. Increase in Increased movement of particles. Increase in temperature.temperature.

4.4. Increasing numbers of particles. Ex.Increasing numbers of particles. Ex.

2 KClO2 KClO33 2 KCl + 3O 2 KCl + 3O22

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Entropy StatesEntropy StatesThe greatest increase in The greatest increase in entropy entropy is usually is usually found when there is an increase of found when there is an increase of particles in the gaseous state.particles in the gaseous state.

The symbol for the change in disorder or The symbol for the change in disorder or entropy is given by the symbol, entropy is given by the symbol, SS..

The The more disorderedmore disordered a system becomes a system becomes the the more positivemore positive the value for the value for SS will be. will be.

Systems that become Systems that become more orderedmore ordered have have negative negative SS values. values.

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The entropy of a substance depends The entropy of a substance depends on its state: on its state:

S (gases) > S (liquids) > S (solids)S (gases) > S (liquids) > S (solids)

Entropy, SEntropy, S

SSoo (J/K (J/K-1-1molmol-1-1))

HH22O (liquid)O (liquid) 69.9569.95

HH22O (gas)O (gas) 188.8188.8

SSoo (J/K (J/K-1-1molmol-1-1))

HH22O (liquid)O (liquid) 69.9569.95

HH22O (gas)O (gas) 188.8188.8 59

Entropy and States of Entropy and States of MatterMatter

S˚(BrS˚(Br22 liquid) < S˚(Br liquid) < S˚(Br22 gas) gas) S˚(HS˚(H22O solid) < S˚(HO solid) < S˚(H22O liquid)O liquid)

60

Entropy, Phase & Entropy, Phase & TemperatureTemperature

Entropy, Phase & Entropy, Phase & TemperatureTemperature

S increases slightly with T

S increases a large amount with phase changes

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The Entropy of a substance increases with The Entropy of a substance increases with temperature.temperature.

Molecular motions Molecular motions of heptane, Cof heptane, C77HH1616

Molecular motions of Molecular motions of heptane at different temps.heptane at different temps.

Entropy and TemperatureEntropy and Temperature

62

Entropy - BoltzmanEntropy - Boltzman

Ludwig BoltzmanLudwig Boltzman

S = k Ln W S = k Ln W Entropy is proportional Entropy is proportional to the number of to the number of degrees of freedom or degrees of freedom or possible configurations possible configurations in a system.in a system.

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Standard Entropy ValuesStandard Entropy Values

The standard entropy, The standard entropy, SSoo, of a substance , of a substance is the is the entropy change per moleentropy change per mole that occurs that occurs when heating a substance from when heating a substance from 0 K0 K to the to the standard temperature of standard temperature of 298 K.298 K.Unlike enthalpy, absolute entropy changes Unlike enthalpy, absolute entropy changes can be measured.can be measured.Like enthalpy, Like enthalpy, entropy is a state functionentropy is a state function. . The change in entropy is the difference The change in entropy is the difference between the products and the reactantsbetween the products and the reactants

SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

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Standard Entropy ValuesStandard Entropy Values

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The amount of entropy in a The amount of entropy in a pure substance depends on pure substance depends on the temperature, pressure, the temperature, pressure, and the number of molecules and the number of molecules in the substance. in the substance.

Values for the entropy of Values for the entropy of many substances at have many substances at have been measured and tabulated. been measured and tabulated.

The standard entropy is also The standard entropy is also measured at 298 K. measured at 298 K.

Some standard enthalpy values

Factors That Determine Factors That Determine Entropy StatesEntropy States

The greater the disorder or randomness in a system The greater the disorder or randomness in a system the larger the entropy. the larger the entropy.

Some generalizations Some generalizations

1.1. The entropy of a substance always The entropy of a substance always increases increases as it as it changes changes from solid to liquid to gasfrom solid to liquid to gas and vice versa. and vice versa.

2.2. When a pure solid or liquid dissolves in a solvent, When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases.the entropy of the substance increases.

3.3. When gas molecules escape from a solvent, the When gas molecules escape from a solvent, the entropy increases.entropy increases.

4.4. Entropy generally Entropy generally decreases decreases with increasing with increasing molecular complexitymolecular complexity

Gibbs Free EnergyGibbs Free Energy

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SpontaneitySpontaneity

A chemical reaction is spontaneous if it A chemical reaction is spontaneous if it results in the system moving form a less results in the system moving form a less stable to a more stable state.stable to a more stable state.Decreases in enthalpy and increases in Decreases in enthalpy and increases in entropy move a system to greater stabilityentropy move a system to greater stabilityThe combination of the enthalpy factor and The combination of the enthalpy factor and the entropy factor can be expressed as the the entropy factor can be expressed as the Gibbs Free EnergyGibbs Free Energy

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Gibbs Free EnergyGibbs Free Energy

The standard free energy change is defined The standard free energy change is defined by this equationby this equation

GGoo = = HHoo – T – T SSoo

WhereWhere

HHo o = the enthalpy change = the enthalpy changeSSoo = the entropy change = the entropy change T = Kelvin temperatureT = Kelvin temperature

A chemical reaction is A chemical reaction is spontaneous if it results spontaneous if it results in a negative free energy change.in a negative free energy change.

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Gibbs Free EnergyGibbs Free EnergyPossible Combinations for free energy change:Possible Combinations for free energy change:

GGoo = = HHoo – T – T SSoo

GG HH SS H-TH-TSS

Always Always SpontaneousSpontaneous

< 0 (-)< 0 (-) < 0 (-)< 0 (-) > 0 (+)> 0 (+) Always (-)Always (-)

Never Never SpontaneousSpontaneous

> 0 (+)> 0 (+) > 0 (+)> 0 (+) < 0 (-)< 0 (-) Always (+)Always (+)

Spontaneous at Spontaneous at High TemperatureHigh Temperature

< 0 (-)< 0 (-) > 0 (+)> 0 (+)

> 0 (+)> 0 (+)

> 0 (+)> 0 (+)(-) if T large(-) if T large

(+) if T small(+) if T small

Spontaneous at Spontaneous at Low TemperatureLow Temperature

> 0 (+)> 0 (+)

< 0 (-)< 0 (-)< 0 (-)< 0 (-) < 0 (-)< 0 (-)

(+) if T large(+) if T large

(-) if T small(-) if T small

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Free Energy Problem 1Free Energy Problem 1A certain chemical reaction is exothermic with a A certain chemical reaction is exothermic with a standard enthalpy of - 400 kJ molstandard enthalpy of - 400 kJ mol-1-1. . The entropy The entropy change for this reaction is +44 J molchange for this reaction is +44 J mol-1-1 K K-1-1. . Calculate the free energy change for this reaction Calculate the free energy change for this reaction at 25 at 25 ooC. Is the reaction spontaneous?C. Is the reaction spontaneous?

SolutionSolutionConvert the entropy value to kJ. 44 J molConvert the entropy value to kJ. 44 J mol-1-1 K K-1-1 = 0.044 = 0.044

kJ molkJ mol-1-1 K K-1-1

G = - 400 G = - 400 kJ molkJ mol-1-1 – (298 K)(0.044 kJ – (298 K)(0.044 kJ molmol-1-1 K K-1-1) ) G = - 400 G = - 400 kJ molkJ mol-1 -1 – 13.1 kJ – 13.1 kJ molmol-1-1 G G = - 413.1 = - 413.1 kJ molkJ mol-1-1 . . Since Since G is negative the G is negative the reaction is spontaneous.reaction is spontaneous.

Note. Because Note. Because H <0 and H <0 and S >0, this reaction is spontaneous S >0, this reaction is spontaneous at all temperatures.at all temperatures.

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Free Energy Problem 2Free Energy Problem 2A certain chemical reaction is endothermic with a A certain chemical reaction is endothermic with a standard enthalpy of +300 kJ molstandard enthalpy of +300 kJ mol-1-1. . The entropy The entropy change for this reaction is +25 J molchange for this reaction is +25 J mol-1-1 K K-1-1. . Calculate the free energy change for this reaction Calculate the free energy change for this reaction at 25 at 25 ooC. Is the reaction spontaneous?C. Is the reaction spontaneous?

SolutionSolutionConvert the entropy value to kJ. 25 J molConvert the entropy value to kJ. 25 J mol-1-1 K K-1-1 = 0.025 = 0.025

kJ molkJ mol-1-1 K K-1-1

G = + 300 G = + 300 kJ molkJ mol-1-1 – (298 K)(0.025 kJ – (298 K)(0.025 kJ molmol-1-1 K K-1-1) ) G = + 300 G = + 300 kJ molkJ mol-1 -1 – 7.45 kJ – 7.45 kJ molmol-1-1 G G = = + 292.55 + 292.55 kJ molkJ mol-1-1 . . Since Since G is positive the G is positive the reaction is non-spontaneous.reaction is non-spontaneous.

Note. Because Note. Because H >0 and H >0 and S >0, this reaction is non-S >0, this reaction is non-spontaneous at low temperatures. It the temperature were spontaneous at low temperatures. It the temperature were substantially increased it would become spontaneous.substantially increased it would become spontaneous.

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energy and chemical reactions l. scheffler 1. heat and temperature heat is energy that is...

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