Chapter 16 Spontaneity, entropy and free energy. Spontaneous A reaction that will occur without outside intervention. A reaction that will occur without.

Chapter 16Chapter 16

Spontaneity, entropy and free Spontaneity, entropy and free energyenergy

SpontaneousSpontaneous

A reaction that will occur without outside A reaction that will occur without outside intervention.intervention.

We can’t determine how fast.We can’t determine how fast.We need both thermodynamics and We need both thermodynamics and

kinetics to describe a reaction completely.kinetics to describe a reaction completely.Thermodynamics compares initial and final Thermodynamics compares initial and final

states.states.Kinetics describes pathway between.Kinetics describes pathway between.

ThermodynamicsThermodynamics

1st Law- the energy of the universe is 1st Law- the energy of the universe is constant.constant.

Keeps track of thermodynamics doesn’t Keeps track of thermodynamics doesn’t correctly predict spontaneity.correctly predict spontaneity.

EntropyEntropy (S) is disorder or randomness (S) is disorder or randomness2nd Law the entropy of the universe 2nd Law the entropy of the universe

increases.increases.

EntropyEntropy

Defined in terms of probability.Defined in terms of probability.Substances take the arrangement that is Substances take the arrangement that is

most likely.most likely.The most likely is the most random.The most likely is the most random.Calculate the number of arrangements for Calculate the number of arrangements for

a system.a system.

2 possible 2 possible arrangementsarrangements

50 % chance of 50 % chance of finding the left finding the left emptyempty

4 possible arrangements

25% chance of finding the left empty

50 % chance of them being evenly dispersed

4 Molecules create 12 possible arrangements

8% chance of finding the left empty

50 % chance of them being evenly dispersed

GasesGases

Gases completely fill their chamber Gases completely fill their chamber because there are many more ways to do because there are many more ways to do that than to leave half empty.that than to leave half empty.

SSsolid solid <S<Sliquid liquid <<S<<Sgasgas there are many more ways for the there are many more ways for the

molecules to be arranged as a liquid than molecules to be arranged as a liquid than a solid.a solid.

Gases have a huge number of positions Gases have a huge number of positions possible.possible.

Positional EntropyPositional Entropy

Probability Equation:Probability Equation:P =( P =( 11 ))# of Particles# of Particles

# of Positions# of PositionsExample Problem. 5 Molecules of Carbon Example Problem. 5 Molecules of Carbon

Dioxide are place in a 4 bulb container. Dioxide are place in a 4 bulb container. What is the probability that all of the What is the probability that all of the Molecules will be in one bulb? Molecules will be in one bulb?

SolutionSolution

P = (P = (11 ) ) # of Particles# of Particles

# of Positions# of PositionsP = (P = (11 ))55

44P=1/1024P=1/1024

Positional EntropyPositional Entropy

Which would have the higher positional Which would have the higher positional entropy, one Mole of Solid Carbon Dioxide entropy, one Mole of Solid Carbon Dioxide and gaseous Carbon Dioxide?and gaseous Carbon Dioxide?

A mole of gaseous Carbon Dioxide has a A mole of gaseous Carbon Dioxide has a greater volume then a mole of solid greater volume then a mole of solid Carbon Dioxide, so by logic gaseous Carbon Dioxide, so by logic gaseous Carbon Dioxide would have a higher Carbon Dioxide would have a higher positional energypositional energy

Predicting Entropy ChangesPredicting Entropy Changes

Solid sugar is added to water to form a Solid sugar is added to water to form a solution, would the entropy be positive or solution, would the entropy be positive or negative? (greater or less)negative? (greater or less)

Sugar Molecules become randomly Sugar Molecules become randomly dispersed in the water, so they have a dispersed in the water, so they have a larger number of positions, so entropy larger number of positions, so entropy would be greater (Positive)would be greater (Positive)

HomeworkHomework

Book Exercises 16-21 to 16.24Book Exercises 16-21 to 16.24

EntropyEntropySolutions form because there are many Solutions form because there are many

more possible arrangements of more possible arrangements of dissolved pieces than if the solute and dissolved pieces than if the solute and solvent stay separate.solvent stay separate.

2nd Law2nd LawSSunivuniv = = SSsyssys + + SSsurrsurr IfIfSSuniv univ is positive the process is is positive the process is

spontaneous.spontaneous. IfIfSSuniv univ is negative the process is is negative the process is

spontaneous in the opposite direction.spontaneous in the opposite direction.

For exothermic processes For exothermic processes SSsurr surr is is

positive. positive. For endothermic processes For endothermic processes SSsurr surr is is

negative.negative. Consider this processConsider this process

HH22O(l)O(l)HH22O(g)O(g)

SSsyssys is positive is positive

SSsurr surr is negativeis negative

SSuniv univ depends on temperature.depends on temperature.

Example of the Second LawExample of the Second Law

Is the organized state of life consistent with Is the organized state of life consistent with the second law?the second law?

SSuniv univ must be positive for this to be true.must be positive for this to be true.

SSsyssys is negative in this case, but…. is negative in this case, but….

SSsurr surr is positive and much great then is positive and much great then SSsyssys

so….so….SSuniv univ is greater then zero, life is consistent is greater then zero, life is consistent

with the second law.with the second law.

HomeworkHomework

Book Problems 16.11 and 16.12Book Problems 16.11 and 16.12

Temperature and Temperature and SpontaneitySpontaneity

Entropy changes in the surroundings are Entropy changes in the surroundings are determined by the heat flow.determined by the heat flow.

An exothermic process is favored because by An exothermic process is favored because by giving up heat the entropy of the surroundings giving up heat the entropy of the surroundings increases.increases.

The size of The size of SSsurr surr depends on temperaturedepends on temperature

SSsurr surr = -= -H/TH/T

SSsurr surr = - (Change in enthalpy)/Temp in K= - (Change in enthalpy)/Temp in K

SsysSsurr Suniv

Spontaneous?

+ + +

- --

+ - ?

+- ?

Yes

No, Reverse

At Low temp.

At High temp.

Example ProblemExample Problem

In the metallurgy of antimony, the pure In the metallurgy of antimony, the pure metal is recovered via different reactions, metal is recovered via different reactions, depending on the composition of the ore. depending on the composition of the ore. For example, iron is used to reduce For example, iron is used to reduce antimony in sulfide ores:antimony in sulfide ores:

SbSb22SS33(s) + 3 Fe(s) (s) + 3 Fe(s) 2Sb(s) + 3FeS(s) 2Sb(s) + 3FeS(s)H = -125 kJH = -125 kJ

Example ContinuedExample Continued

Carbon is used as the reducing agent for Carbon is used as the reducing agent for oxide ores:oxide ores:

SbSb44OO66(s) + 6 C(s) (s) + 6 C(s) 4 Sb(s) + 6 CO(g) 4 Sb(s) + 6 CO(g)H = 778 kJH = 778 kJCalculate Calculate SSsurr surr for each of these reactions for each of these reactions

at 25 C and 1 atm.at 25 C and 1 atm.

SolutionSolution

SSsurr surr = -= -H/TH/T T = 25 + 273 = 298 KT = 25 + 273 = 298 K Sulfide Ore ReactionSulfide Ore Reaction

SSsurr surr = -(-125 kJ/298 K) = 0.419 kJ/K = 419 J/K= -(-125 kJ/298 K) = 0.419 kJ/K = 419 J/K Notice it is positive, so it is exotermic, heat is released Notice it is positive, so it is exotermic, heat is released

into the surroundings, increasing randomness of the into the surroundings, increasing randomness of the surroundings.surroundings.

Oxide Ore ReactionOxide Ore Reaction SSsurr surr = -(778 kJ/298 K) = -2.61 kJ/K = -2610 J/K= -(778 kJ/298 K) = -2.61 kJ/K = -2610 J/K Notice it is negative so heat flows from the Notice it is negative so heat flows from the

surroundings to the system.surroundings to the system.

HomeworkHomework

Book Questions 16.25 and 16.26Book Questions 16.25 and 16.26

Gibb's Free EnergyGibb's Free Energy G=H-TSG=H-TS Never used this way.Never used this way. G=G=H-TH-TS at constant temperatureS at constant temperature Divide by –TDivide by –T G/-T = G/-T = H/-T-TH/-T-TS/-TS/-T --G/T = -G/T = -H/T+H/T+SS --G/T = G/T = SSsurrsurr + + S S --G/T = G/T = SSunivuniv If If G is negative at constant T and P, the G is negative at constant T and P, the

Process is spontaneous.Process is spontaneous.

Let’s CheckLet’s Check

For the reaction HFor the reaction H22O(s) O(s) H H22O(l)O(l) Sº = 22.1 J/K mol Sº = 22.1 J/K mol Hº =6030 J/molHº =6030 J/mol Calculate Calculate G at 10ºC and -10ºCG at 10ºC and -10ºC Look at the equation Look at the equation G=G=H-TH-TSS G=G=J/mol –(283K*22.1 J/K*mol)J/mol –(283K*22.1 J/K*mol) G= 6030 J/mol – 6254 J/mol = -224G= 6030 J/mol – 6254 J/mol = -224 G=G=J/mol –(263K*22.1 J/K*mol)J/mol –(263K*22.1 J/K*mol) G= 6030 J/mol – 5812 J/mol = 218G= 6030 J/mol – 5812 J/mol = 218 Spontaneity can be predicted from the sign of Spontaneity can be predicted from the sign of

H and H and S.S.

G=G=H-TH-TSS

HS Spontaneous?

+ - At all Temperatures

+ + At high temperatures,

“entropy driven”

- - At low temperatures,

“enthalpy driven”

+- Not at any temperature,

Reverse is spontaneous

Sample ProblemSample Problem

At what temperature is the following At what temperature is the following process spontaneous at 1 atm?process spontaneous at 1 atm?

BrBr22 (l) (l) Br Br22 (g) (g)

HHoo = 31.0 kJ/mol = 31.0 kJ/mol SSoo = 93.0 J/K mol = 93.0 J/K molWhat is the normal boiling point?What is the normal boiling point?

SolutionSolution

If If G is negative, vaporization is G is negative, vaporization is spontaneous.spontaneous.

SSoo is positive, so it favors vaporization. is positive, so it favors vaporization.HHoo is positive, so it favors condensation. is positive, so it favors condensation.Opposite tendencies exactly balance at Opposite tendencies exactly balance at

boiling point.boiling point.We can determine the boiling point by We can determine the boiling point by

setting setting GGoo = 0. = 0.

EquationEquation

GGoo = =HHoo -T -TSSoo

==HHoo –T –TSSoo

HHoo = T = TSSoo

==HHoo / / SS (3.10 x 10(3.10 x 1044 J/mol) / 93.0 J/K mol = 333 K J/mol) / 93.0 J/K mol = 333 K> 333K vaporization is dominant> 333K vaporization is dominant< 333K condensation is dominant< 333K condensation is dominant= 333K both coexist, neither is dominant= 333K both coexist, neither is dominant

HomeworkHomework

16.29 and 16.3116.29 and 16.31

Third Law of ThermoThird Law of Thermo

The entropy of a pure crystal at 0 K is 0.The entropy of a pure crystal at 0 K is 0.Gives us a starting point.Gives us a starting point.All others must be>0.All others must be>0.Standard Entropies Sº ( at 298 K and 1 Standard Entropies Sº ( at 298 K and 1

atm) of substances are listed.atm) of substances are listed.Products - reactants to find Products - reactants to find Sº (a state Sº (a state

function).function).More complex molecules higher Sº.More complex molecules higher Sº.

ExampleExample

Predict the sign of Predict the sign of Sº for the following Sº for the following reaction:reaction:

CaCOCaCO33(s) (s) CaO(s) + CO CaO(s) + CO2 2 (g)(g)Since in this reaction a gas is produced Since in this reaction a gas is produced

from a solid reactant, the positional from a solid reactant, the positional entropy increases, and entropy increases, and Sº is positive.Sº is positive.

Another ExampleAnother Example

In the oxidation of Sulfur Dioxide in air, is In the oxidation of Sulfur Dioxide in air, is the the Sº positive or negative?Sº positive or negative?

2 SO2 SO22 (g) + O (g) + O22 (g) (g) 2 SO 2 SO33 (g) (g) In this case, three molecules of reactants In this case, three molecules of reactants

produce 2 molecules of Products. Since produce 2 molecules of Products. Since the number of gas molecules decreases, the number of gas molecules decreases, positional entropy decreases, and positional entropy decreases, and Sº is Sº is negativenegative

More HomeworkMore Homework

16.33 and 16.3416.33 and 16.34

Free Energy in ReactionsFree Energy in Reactions

Gº = standard free energy change.Gº = standard free energy change.Free energy change that will occur if Free energy change that will occur if

reactants in their standard state turn to reactants in their standard state turn to products in their standard state.products in their standard state.

Can’t be measured directly, can be Can’t be measured directly, can be calculated from other measurements.calculated from other measurements.

Gº=Gº=Hº-THº-TSºSºUse Hess’s Law with known reactions.Use Hess’s Law with known reactions.

Free Energy in ReactionsFree Energy in Reactions

There are tables of There are tables of GºGºff . .

Products-reactants because it is a state Products-reactants because it is a state function.function.

The standard free energy of formation The standard free energy of formation for any element in its standard state is for any element in its standard state is 0.0.

Remember- Spontaneity tells us Remember- Spontaneity tells us nothing about rate.nothing about rate.

G=G=H-TH-TSS

HS Spontaneous?

+ - At all Temperatures

+ + At high temperatures,

“entropy driven”

- - At low temperatures,

“enthalpy driven”

+- Not at any temperature,

Reverse is spontaneous

Third Law of ThermoThird Law of Thermo

The entropy of a pure crystal at 0 K is 0.The entropy of a pure crystal at 0 K is 0.Gives us a starting point.Gives us a starting point.All others must be>0.All others must be>0.Standard Entropies Sº ( at 298 K and 1 Standard Entropies Sº ( at 298 K and 1

atm) of substances are listed.atm) of substances are listed.Products - reactants to find Products - reactants to find Sº (a state Sº (a state

function)function)More complex molecules higher Sº.More complex molecules higher Sº.

Yet Another EquationYet Another Equation

Entropy is a state function of the systemEntropy is a state function of the system It is not pathway-dependentIt is not pathway-dependent

Entropy for a chemical reaction can be Entropy for a chemical reaction can be calculated by taking the difference between the calculated by taking the difference between the values of the products and the reactants.values of the products and the reactants.

SSooreactionreaction = = nnppSSoo

productsproducts – – nnrrSSooreactantsreactants

nnpp = moles of products = moles of products nnrr = moles of reactants = moles of reactants Entropy is an extensive propertyEntropy is an extensive property

It depends on how much of the substance is present.It depends on how much of the substance is present.

ExampleExample

Calculate Calculate SSoo at 25 C for the reaction: at 25 C for the reaction:

2 NiS (s) + 3 O2 NiS (s) + 3 O22 (g) (g) 2 SO 2 SO22 (g) + 2 NiO (s) (g) + 2 NiO (s) Given the following standard entropy values:Given the following standard entropy values:

SubstanceSubstance SSoo (J/K mol) (J/K mol)SOSO22 (g) (g) 248248NiO (s)NiO (s) 3838OO22 (g) (g) 205205NiS (s)NiS (s) 5353

SolutionSolution

SSooreactionreaction = = nnppSSoo

productsproducts – – nnrrSSooreactantsreactants

SSooreactionreaction = (2S = (2Soo

SO2 (g) SO2 (g) + 2S+ 2SooNiS (s)NiS (s)) – (2S) – (2Soo

NiS (s)NiS (s) + +

3S3SooO2(g)O2(g)))

SSooreactionreaction = [2 mol(248 J/K mol)+2 mol(38 J/K mol)] – = [2 mol(248 J/K mol)+2 mol(38 J/K mol)] –

[2mol(53 J/K mol) + 3 mol(205 J/K mol)][2mol(53 J/K mol) + 3 mol(205 J/K mol)] SSoo

reactionreaction = (496 J/K + 76 J/K) – (106 J/K + 615 J/K) = (496 J/K + 76 J/K) – (106 J/K + 615 J/K)

SSooreactionreaction = - 149 J/K = - 149 J/K

We would expect a negative We would expect a negative SSoo since the since the number of gaseous molecules decreases.number of gaseous molecules decreases.

Another exampleAnother example

Calculate Calculate SSoo for the reduction of aluminum oxide by for the reduction of aluminum oxide by hydrogen gas:hydrogen gas:

AlAl22OO33(s) + 3H(s) + 3H22(g) (g) 2Al(s) + 3H 2Al(s) + 3H22O(g)O(g) Given the following standard entropy values:Given the following standard entropy values:

SubstanceSubstance SSoo (J/K mol) (J/K mol)AlAl22OO33(s)(s) 5151

HH22(g)(g) 131131Al(s)Al(s) 2828

HH22O(g)O(g) 189189

SolutionSolution

SSooreactionreaction = = nnppSSoo

productsproducts – – nnrrSSooreactantsreactants

SSooreactionreaction = (2S = (2Soo

(Al(s) (Al(s) + 3S+ 3SooH2O(g)H2O(g)) – (3S) – (3Soo

H2(g)H2(g) + +

SSooAl2O3(s)Al2O3(s)))

SSooreactionreaction = [2 mol(28 J/K mol)+3 mol(189 J/K mol)] – = [2 mol(28 J/K mol)+3 mol(189 J/K mol)] –

[3mol(131 J/K mol) + 1 mol(51 J/K mol)][3mol(131 J/K mol) + 1 mol(51 J/K mol)] SSoo

reactionreaction = [56 J/K + 567 J/K] – [393 J/K + 51 J/K] = [56 J/K + 567 J/K] – [393 J/K + 51 J/K]

SSooreactionreaction = 179 J/K = 179 J/K

We would expect a positive number since the moles of We would expect a positive number since the moles of solid went up.solid went up.

HomeworkHomework

Problems 16.37, 16.39 and 16.40Problems 16.37, 16.39 and 16.40

Free Energy in ReactionsFree Energy in Reactions

Gº = standard free energy change.Gº = standard free energy change.Free energy change that will occur if Free energy change that will occur if

reactants in their standard state turn to reactants in their standard state turn to products in their standard state.products in their standard state.

Can’t be measured directly, can be Can’t be measured directly, can be calculated from other measurements.calculated from other measurements.

Gº=Gº=Hº-THº-TSºSºUse Hess’s Law with known reactions.Use Hess’s Law with known reactions.

Free Energy in ReactionsFree Energy in Reactions

There are tables of There are tables of GºGºff

Products-reactants because it is a state Products-reactants because it is a state function.function.

The standard free energy of formation The standard free energy of formation for any element in its standard state is for any element in its standard state is 0.0.

Remember- Spontaneity tells us Remember- Spontaneity tells us nothing about rate.nothing about rate.

Example ProblemExample Problem

Consider the reaction:Consider the reaction:

2SO2SO22(g) + O(g) + O22(g) (g) 2SO 2SO33(g)(g) Carried out at 25Carried out at 25oo C and 1 atm. Calculate C and 1 atm. Calculate HHoo, ,

SSoo, and , and GGoo using the following data: using the following data:

SubstanceSubstance HHff (kJ/mol) (kJ/mol) SSff (J/K*mol) (J/K*mol)

SOSO22(g)(g) -297-297 248248

OO22(g)(g) -396-396 257257

SOSO33(g)(g) 00 205205

SolutionSolution

HHooreactionreaction = = nnppHHff

ooproductsproducts – – nnrrHHff

ooreactantsreactants

HHooreactionreaction = (2 = (2HHff

oo(SO3(g)(SO3(g) ) – (2 ) – (2HHff

oo(SO2(g)(SO2(g)+ + HHff

oo(O2(g)(O2(g)))

HHooreactionreaction = [2 mol (-396 kJ/mol)] – [2 mol(-297 kJ/mol) + = [2 mol (-396 kJ/mol)] – [2 mol(-297 kJ/mol) +

(1 mol * 0 kJ/mol)](1 mol * 0 kJ/mol)]

HHooreactionreaction = [-792 kJ] – [-594 kJ + 0 kJ] = [-792 kJ] – [-594 kJ + 0 kJ]

HHooreactionreaction = -792 kJ + 594 kJ = -792 kJ + 594 kJ

HHooreactionreaction = -198 kJ = -198 kJ

S SolutionS Solution

SSooreactionreaction = = nnppSSoo

productsproducts – – nnrrSSooreactantsreactants

SSooreactionreaction = = SSoo

SO3 (g)SO3 (g) – ( – (SSooSO2 (g)SO2 (g) + S + Soo

O2 (g)O2 (g))) SSoo

reactionreaction = [ = [mol(257 J/K mol)] – [2 mol(248 J/K mol) + mol(257 J/K mol)] – [2 mol(248 J/K mol) +

1 mol(205 J/K mol)]1 mol(205 J/K mol)]

SSooreactionreaction = [ = [ J/K] – [496 J/K + 205 J/K] J/K] – [496 J/K + 205 J/K]

SSooreactionreaction = [ = [ J/K] – [701 J/K] J/K] – [701 J/K]

SSooreactionreaction = -187 = -187J/KJ/K

G SolutionG Solution

GGoo = = HHoo – T – TSSoo

SSo = o = (-187 J/K)(1kJ/1000 J)(-187 J/K)(1kJ/1000 J)GGoo = -198 kJ – [(298 K) (-.187 kJ)] = -198 kJ – [(298 K) (-.187 kJ)]GGoo = -198 kJ + 55.7 kJ = -142 kJ = -198 kJ + 55.7 kJ = -142 kJ

HomeworkHomework

Exercises 16.45 and 16.46Exercises 16.45 and 16.46

A Second MethodA Second Method

Another way to calculate Gibbs Free Another way to calculate Gibbs Free Energy (Energy (G) is to take advantage of the G) is to take advantage of the fact that it is fact that it is a state functiona state function..

We can use the procedure we used with We can use the procedure we used with Hess’s Law Hess’s Law ((HHoo

reactionreaction = = nnppHHffoo

productsproducts – – nnrrHHffoo

reactantsreactants))

GGooreactionreaction = = nnppGGff

ooproductsproducts – – nnrrGGff

ooreactantsreactants

ExampleExample

Find the free energy change for the following Find the free energy change for the following reaction:reaction:

2CO(g) + O2CO(g) + O22(g) (g) 2CO 2CO22(g)(g) We must use the following data:We must use the following data:

2CH2CH44(g) + 3O(g) + 3O22(g) (g) 2CO(g) + 4H 2CO(g) + 4H22O(g)O(g)GGoo = 1088 kJ = 1088 kJ

CHCH44(g) + 2O(g) + 2O22(g) (g) 2CO 2CO22(g) + 2H(g) + 2H22O(g)O(g)GGoo = -801 kJ = -801 kJ

Example continuedExample continued

Compare the first and second equation.Compare the first and second equation. They both consume oxygen and produce water They both consume oxygen and produce water

vapor.vapor. Reverse the first equation:Reverse the first equation:

2CO(g) + 4H2CO(g) + 4H22O(g) O(g) 2CH 2CH44(g) + 3O(g) + 3O22(g)(g)

GGoo = -1088 kJ = -1088 kJ Compare it to the second:Compare it to the second:

CHCH44(g) + 2O(g) + 2O22(g) (g) 2CO 2CO22(g) + 2H(g) + 2H22O(g)O(g) Now Oxygen and water vapor cancel out.Now Oxygen and water vapor cancel out.

Further with the ExampleFurther with the Example

This leaves us with the original equation:This leaves us with the original equation:2CO(g) + O2CO(g) + O22(g) (g) 2CO 2CO22(g)(g)

We now must adjust the energy, since in the first We now must adjust the energy, since in the first equation only 1 mole was produced, we must equation only 1 mole was produced, we must double the double the G = 2(-801 kJ) = -1602 kJG = 2(-801 kJ) = -1602 kJ

We now have We now have G for the reactants and products.G for the reactants and products. GGoo

reactionreaction = -1602 kJ – (-1088 kJ) = -1602 kJ – (-1088 kJ)

GGooreactionreaction = - 514 kJ = - 514 kJ

Free energy and PressureFree energy and Pressure

G = G = Gº +RTln(Q) where Q is the reaction Gº +RTln(Q) where Q is the reaction quotients (P of the products /P of the quotients (P of the products /P of the reactants).reactants).

CO(g) + 2HCO(g) + 2H22(g) (g) CH CH33OH(l)OH(l)

Would the reaction be spontaneous at 25ºC Would the reaction be spontaneous at 25ºC with the Hwith the H22 pressure of 5.0 atm and the CO pressure of 5.0 atm and the CO

pressure of 3.0 atm?pressure of 3.0 atm?GºGºff CH CH33OH(l) = -166 kJ (Appendix 4)OH(l) = -166 kJ (Appendix 4)

GºGºff CO(g) = -137 kJ CO(g) = -137 kJ GºGºff H H22(g) = 0 kJ (g) = 0 kJ

More on ExampleMore on Example

Gº = -166 kJ – (-137 kJ + 0 kJ) = -29 kJGº = -166 kJ – (-137 kJ + 0 kJ) = -29 kJSince this is the value for 1 mole of CO, Since this is the value for 1 mole of CO,

then this unit becomes kJ/mol rxnthen this unit becomes kJ/mol rxn -29 kJ/mol = -2.9 x 10-29 kJ/mol = -2.9 x 1044 J/mol rxn J/mol rxnG = G = Gº +RTln(Q)Gº +RTln(Q)

Gº = -2.9 x 10Gº = -2.9 x 1044 J/mol rxn J/mol rxnR = 8.3145 J/K mol (Since we are working in J R = 8.3145 J/K mol (Since we are working in J

we must use the R for kPawe must use the R for kPaT = 298 KT = 298 K

Calculating Q for this ReactionCalculating Q for this Reaction

Remember Q is calculated in the same Remember Q is calculated in the same fashion as Keqfashion as Keq

Q = Q = [Products][Products]

[P[PCOCO][P][PH2H222]]

Q = Q = [1][1] = 0.022= 0.022

[5][3[5][322]]Note Pure Ethanol is a liquid therefore not Note Pure Ethanol is a liquid therefore not

included in calculationsincluded in calculations

SolutionSolution

G = G = Gº + -RTln(Q)Gº + -RTln(Q)G = (-2.9 x 10G = (-2.9 x 1044 J/mol rxn) + {-(8.3145 J/K J/mol rxn) + {-(8.3145 J/K

mol rxn)(298K) ln(0.022)mol rxn)(298K) ln(0.022) G = (-2.9 x 10G = (-2.9 x 1044 J/mol rxn) – (9400 J/mol rxn) J/mol rxn) – (9400 J/mol rxn) G = -38000 J/mol rxn = -38 KJ/mol rxnG = -38000 J/mol rxn = -38 KJ/mol rxn

HomeworkHomework

16.55, 16.57, 16.5916.55, 16.57, 16.59

How far?How far?

G tells us spontaneity at current G tells us spontaneity at current conditions. When will it stop?conditions. When will it stop?

It will go to the lowest possible free energy It will go to the lowest possible free energy which may be an equilibrium.which may be an equilibrium.

At equilibrium At equilibrium G = 0, Q = KG = 0, Q = KGº = -RTlnKGº = -RTlnK

Gº KGº K

=0=0 =1=1

<0<0 >0>0

>0>0 <0<0

Temperature dependence Temperature dependence of Kof K

Gº= -RTlnK = Gº= -RTlnK = Hº - THº - TSºSº ln(K) = ln(K) = Hº/R(1/T)+ Hº/R(1/T)+ Sº/RSº/RA straight line of lnK vs 1/TA straight line of lnK vs 1/T

Free energy And WorkFree energy And Work

Free energy is that energy free to do work.Free energy is that energy free to do work.The maximum amount of work possible at The maximum amount of work possible at

a given temperature and pressure.a given temperature and pressure.Never really achieved because some of Never really achieved because some of

the free energy is changed to heat during the free energy is changed to heat during a change, so it can’t be used to do work.a change, so it can’t be used to do work.