20-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 20 Thermodynamics: Entropy, Free Energy and the

20-1

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 20

Thermodynamics:

Entropy, Free Energy and the Direction of Chemical Reactions

20-2


Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions

20.1 The Second Law of Thermodynamics: Predicting Spontaneous Change

20.2 Calculating the Change in Entropy of a Reaction

20.3 Entropy, Free Energy, and Work

20.4 Free Energy, Equilibrium, and Reaction Direction

20-3


First law of thermodynamicsFirst law of thermodynamics

Law of conservation of energy: Energy can Law of conservation of energy: Energy can be neither created nor destroyed.be neither created nor destroyed.

∆∆E= q + w (q=heat, w=work, E=internal E= q + w (q=heat, w=work, E=internal energy)energy)

E univ= E sys + E surrE univ= E sys + E surr Heat gained by system is lost by Heat gained by system is lost by

surroundings and vice-versa.surroundings and vice-versa. Total energy of Universe is constant ∆E Total energy of Universe is constant ∆E

sys + ∆E surr=0=∆E univsys + ∆E surr=0=∆E univ

20-4


Limitations of the First Law of Thermodynamics

E = q + w

Euniverse = Esystem + Esurroundings

Esystem = -Esurroundings

The total energy-mass of the universe is constant.

However, this does not tell us anything about the direction of change in the universe.

Esystem + Esurroundings = 0 = Euniverse

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Spontaneous Spontaneous

It occurs without outside intervention It occurs without outside intervention

20-6


Figure 20.1 A spontaneous endothermic chemical reaction.

water

Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l).

H0rxn = +62.3 kJ

20-7


EnthalpyEnthalpy

H=E+PV,H=E+PV, E=internal energy, E=internal energy, P=pressure, P=pressure, V=volume V=volume

20-8


The Concept of Entropy (S)

Entropy refers to the state of order.

A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process.

solid liquid gasmore order less order

crystal + liquid ions in solution

more order less order

more order less order

crystal + crystal gases + ions in solution

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1877 Ludwig Boltzman S = k ln W

where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/NA (R = universal gas constant, NA = Avogadro’s number.

•A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy.

•A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy.

Suniverse = Ssystem + Ssurroundings > 0

This is the second law of thermodynamics.

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Entropy:Entropy: S, A measure of molecular randomness or S, A measure of molecular randomness or

disorder.disorder. Thermodynamic function that describes number Thermodynamic function that describes number

of arrangements that are available to a system of arrangements that are available to a system existing in a given state.existing in a given state.

Probability of occurrence of a particular Probability of occurrence of a particular arrangement(state) depends on the number of arrangement(state) depends on the number of ways(microstates) in which it can be arranged.ways(microstates) in which it can be arranged.

S=k ln W, k= Boltzmann constant=R/Av S=k ln W, k= Boltzmann constant=R/Av number=1.38 x 10-23J/K.number=1.38 x 10-23J/K.

Un its of S are J/KUn its of S are J/K

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EntropyEntropy

More the microstates available, higher is More the microstates available, higher is the entropy. Ex: solid →liquid→gasthe entropy. Ex: solid →liquid→gas

Entropy is a state function depends only Entropy is a state function depends only on present state and not its path.on present state and not its path.

∆ ∆ Ssys= Sfinal – Sinitial , ∆ Ssys > 0, as Ssys= Sfinal – Sinitial , ∆ Ssys > 0, as entropy increases, entropy increases,

∆ ∆ Ssys < 0, as entropy decreases.Ssys < 0, as entropy decreases.

20-12


EntropyEntropy

If there are n number of molecules, If there are n number of molecules, relative probability of finding all molecules relative probability of finding all molecules in one place is: (1/2)nin one place is: (1/2)n

∆ ∆ Ssys= 5.76 J/K for 1 molSsys= 5.76 J/K for 1 mol ∆ ∆ Ssys=qrev/T , T=temp, q=heat Ssys=qrev/T , T=temp, q=heat

absorbed, rev=reversible process absorbed, rev=reversible process

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Figure 20.2

1 atm evacuated

Spontaneous expansion of a gas

stopcock closed

stopcock opened

0.5 atm 0.5 atmSpontaneous expansion of gas

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Figure 20.3

Expansion of a gas and the increase in number of microstates.

20-15


Entropy and Second Law of Entropy and Second Law of ThermodynamicsThermodynamics::

Second Law of Thermo: In any spontaneous Second Law of Thermo: In any spontaneous process there is always an increase in the process there is always an increase in the entropy of the universe. OR The entropy of the entropy of the universe. OR The entropy of the universe is increasing.universe is increasing.

∆∆Suniv= ∆Ssys + ∆SsurrSuniv= ∆Ssys + ∆Ssurr If ∆Suniv= +ve, process is spontaneous in If ∆Suniv= +ve, process is spontaneous in

direction written.direction written. ∆ ∆Suniv= -ve, process is spontaneous in the Suniv= -ve, process is spontaneous in the

opposite direction.opposite direction. ∆ ∆Suniv= 0, process has no tendency to Suniv= 0, process has no tendency to

occur, system is at equilibrium occur, system is at equilibrium

20-16


Third Law of ThermodynamicsThird Law of Thermodynamics

the entropy of a perfect crystal at 0 K is the entropy of a perfect crystal at 0 K is zero zero

20-17


Figure 20.4 Random motion in a crystal

The third law of thermodynamics.

A perfect crystal has zero entropy at a temperature of absolute zero.

Ssystem = 0 at 0 K

20-18


Predicting Relative S0 Values of a System

1. Temperature changes

2. Physical states and phase changes

3. Dissolution of a solid or liquid

5. Atomic size or molecular complexity

4. Dissolution of a gas

S0 increases as the temperature rises.

S0 increases as a more ordered phase changes to a less ordered phase.

S0 of a dissolved solid or liquid is usually greater than the S0 of the pure solute. However, the extent depends upon the nature of the solute and solvent.

A gas becomes more ordered when it dissolves in a liquid or solid.

In similar substances, increases in mass relate directly to entropy.

In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.

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Figure 20.5 The increase in entropy from solid to liquid to gas.

20-20


Figure 20.6

The entropy change accompanying the dissolution of a salt.

pure solid

pure liquid

solution

MIX

20-21


Figure 20.7

Ethanol Water Solution of ethanol

and water

The small increase in entropy when ethanol dissolves in water.

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Figure 20.8

The large decrease in entropy when a gas dissolves in a liquid.

O2 gas

O2 gas in H2O

20-23


Figure 20.9

NO

NO2

N2O4

Entropy and vibrational motion.

20-24


Answer the following:Answer the following:

1. Will solid or gaseous CO2 have higher 1. Will solid or gaseous CO2 have higher positional entropy?positional entropy?

2. Will N2 gas at 1 atm or 1.0 x 10-2 atm have 2. Will N2 gas at 1 atm or 1.0 x 10-2 atm have higher positional entropy?higher positional entropy?

3. When solid sugar is dissolved in water to form 3. When solid sugar is dissolved in water to form a solution, what will be the sign of entropy a solution, what will be the sign of entropy change?change?

4. When iodine vapors are condensed to form 4. When iodine vapors are condensed to form crystals, what will be the sign of entropy crystals, what will be the sign of entropy change?change?

20-25


Sample Problem 20.1:

SOLUTION:

Predicting Relative Entropy Values

PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]:

(a) 1mol of SO2(g) or 1mol of SO3(g)(b) 1mol of CO2(s) or 1mol of CO2(g)(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)(d) 1mol of KBr(s) or 1mol of KBr(aq)(e) Seawater in midwinter at 20C or in midsummer at 230C(f) 1mol of CF4(g) or 1mol of CCl4(g)

PLAN: In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature.

(a) 1mol of SO3(g) - more atoms

(b) 1mol of CO2(g) - gas > solid

(c) 3mol of O2(g) - larger #mols

(d) 1mol of KBr(aq) - solution > solid

(e) 230C - higher temperature

(f) CCl4 - larger mass

20-26


Entropy Changes in the System

S0rxn - the entropy change that occurs when all reactants

and products are in their standard states.

S0rxn = S0

products - S0reactants

The change in entropy of the surroundings is directly related to an opposite change in the heat of the system and inversely related to the temperature at which the heat is transferred.

Ssurroundings = - Hsystem

T

20-27


Sample Problem 20.2: Calculating the Standard Entropy of Reaction, S0rxn

PROBLEM: Calculate S0rxn for the combustion of 1mol of propane at 250C.

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

PLAN: Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6 mols of gas to 3 mols of gas.

SOLUTION: Find standard entropy values in the Appendix or other table.

S = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)] - [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)]

S = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) + (5 mol)(205.0J/mol*K)]

S = - 374 J/K

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Entropy changes in the Entropy changes in the surrounding: surrounding:

Exothermic change: Heat is lost by the system Exothermic change: Heat is lost by the system and gained by the surroundings.and gained by the surroundings.

q sys< 0 , q surr >0, ∆ Ssurr > 0q sys< 0 , q surr >0, ∆ Ssurr > 0 Endothermic Change: Heat gained by the Endothermic Change: Heat gained by the

system is lost by the surroundings.system is lost by the surroundings. q sys> 0 , q surr <0, ∆ Ssurr < 0q sys> 0 , q surr <0, ∆ Ssurr < 0 The change is entropy of the surroundings is α The change is entropy of the surroundings is α

opposite charge in the heat of a system 1/α opposite charge in the heat of a system 1/α temperature at which heat is transferred.temperature at which heat is transferred.

20-29


Entropy changes in the surrounding Entropy changes in the surrounding

∆∆Ssurr = -q sys/TSsurr = -q sys/T ∆∆Ssurr = -∆H sys/TSsurr = -∆H sys/T Sign of ∆Ssurr depends on the direction of heat Sign of ∆Ssurr depends on the direction of heat

flow. When an exothermic process occurs flow. When an exothermic process occurs ∆Ssurr is + ve. Opposite is true for endothermic ∆Ssurr is + ve. Opposite is true for endothermic process. Nature tends to seek the lowest process. Nature tends to seek the lowest possible energy.possible energy.

Magnitude of ∆Ssurr depends on the Magnitude of ∆Ssurr depends on the temperature. Transfer of heat at low temperature temperature. Transfer of heat at low temperature produces a greater percent change in the produces a greater percent change in the randomness of the surroundings.randomness of the surroundings.

20-30



SOLUTION:

Determining Reaction Spontaneity

PROBLEM: At 298K, the formation of ammonia has a negative S0sys;

Calculate S0rxn, and state whether the reaction occurs

spontaneously at this temperature.

N2(g) + 3H2(g) 2NH3(g) S0sys = -197 J/K

PLAN: S0universe must be > 0 in order for this reaction to be spontaneous, so

S0surroundings must be > 197 J/K. To find S0

surr, first find Hsys; Hsys = Hrxn which can be calculated using H0

f values from tables. S0universe =

S0surr + S0

sys.

H0rx = [(2 mol)(H0

fNH3)] - [(1 mol)(H0fN2) + (3 mol)(H0

fH2)]

H0rx = -91.8 kJ

S0surr = -H0

sys/T = -(-91.8x103J/298K) = 308 J/K

S0universe = S0

surr + S0sys = 308 J/K + (-197 J/K) = 111 J/K

S0universe > 0 so the reaction is spontaneous.

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ProblemProblem

P.3. Sb2 S3(s) + 3Fe(s)______ 2Sb (s) + P.3. Sb2 S3(s) + 3Fe(s)______ 2Sb (s) + 3FeS (s) ∆H=-125kJ3FeS (s) ∆H=-125kJ

Sb4 O6(s) + 6C(s)______ 4Sb (s) + 6 Sb4 O6(s) + 6C(s)______ 4Sb (s) + 6 CO (s) ∆H= 778 kJCO (s) ∆H= 778 kJ

Calculate ∆Ssurr for both of these Calculate ∆Ssurr for both of these reactions at 25 C and 1 atm reactions at 25 C and 1 atm

20-32


Figure B20.1

Lavoisier studying human respiration as a form of combustion.

Figure B20.2

A whole-body calorimeter.

20-33


Figure 20.10 Components of S0universe for spontaneous reactions

exothermic

system becomes more disordered

exothermic

system becomes more ordered

endothermic

system becomes more disordered

∆Ssurr must be larger

∆Ssurr must be smaller

20-34


G0system = H0

system - TS0system

G0rxn = mG0

products - nG0reactants

Gibbs Free Energy (G)

G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it.

G < 0 for a spontaneous processG > 0 for a nonspontaneous processG = 0 for a process at equilibrium

20-35


Free EnergyFree Energy

G= H-TS, H= enthalpy, T=temp in K, S= entropy.G= H-TS, H= enthalpy, T=temp in K, S= entropy. Also called as Gibbs Free Energy.Also called as Gibbs Free Energy. Free energy change (∆G) is a measure of the Free energy change (∆G) is a measure of the

spontaneity of a process and of the useful spontaneity of a process and of the useful energy available from it.energy available from it.

∆ ∆ G= ∆H-T∆S(constant temp)G= ∆H-T∆S(constant temp) ∆∆Suniv = -∆G/TSuniv = -∆G/T Process is spontaneous in the direction in which Process is spontaneous in the direction in which

the free energy decreases. (constant T and P)the free energy decreases. (constant T and P)

20-36


Free EnergyFree Energy

∆∆S univ =0, at equilibriumS univ =0, at equilibrium ∆∆S univ > 0, for a spontaneous process.S univ > 0, for a spontaneous process. ∆∆S univ < 0, for a nonspontaneous S univ < 0, for a nonspontaneous

process.process. ∆∆G=0, at equilibriumG=0, at equilibrium ∆∆G > 0, for a nonspontaneous process.G > 0, for a nonspontaneous process. ∆∆G< 0, for a spontaneous process.G< 0, for a spontaneous process.

20-37


Standard free energy of Standard free energy of formationformation

∆∆Gf0 is defined as a change in free energy that Gf0 is defined as a change in free energy that accompanies the formation of 1 mole of that accompanies the formation of 1 mole of that substance from its constituent elements with all substance from its constituent elements with all reactants and products in their standard states.reactants and products in their standard states.

Standard free energy of formation of an element Standard free energy of formation of an element in its standard state is zero.in its standard state is zero.

An equation coefficient x ∆Gf0 by the number.An equation coefficient x ∆Gf0 by the number. Reversing a reaction changes ∆Gf0 Reversing a reaction changes ∆Gf0

20-38


ProblemProblem

P.4. For the equation 2CH3OH (g) + 3 P.4. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO 2(g) + 4 H2O (g) O2(g) _____ 2 CO 2(g) + 4 H2O (g)

Calculate ∆G0 -163 Calculate ∆G0 -163 0 -394 -229 ∆Gf0 0 -394 -229 ∆Gf0 kJ/molkJ/mol

20-39



PROBLEM:

PLAN:

SOLUTION:

Determining the Effect of Temperature on G0

An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g):

2SO2(g) + O2(g) 2SO3(g)

At 298K, G0 = -141.6kJ; H0 = -198.4kJ; and S0 = -187.9J/K

(a) Use the data to decide if this reaction is spontaneous at 250C, and predict how G0 will change with increasing T.

(b) Assuming H0 and S0 are constant with increasing T, is the reaction spontaneous at 900.0C?

The sign of G0 tells us whether the reaction is spontaneous and the signs of H0 and S0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b).

(a) The reaction is spontaneous at 250C because G0 is (-). Since H0 is (-) but S0 is also (-), G0 will become less spontaneous as the temperature increases.

20-40


Sample Problem 20.6: Determining the Effect of Temperature on G0

continued

(b) G0rxn

= H0rxn - T S0

rxn

G0rxn

= -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/103J)

G0rxn

= 22.0 kJ; the reaction will be nonspontaneous at 900.0C

20-41


ProblemsProblems

P.6. Use ∆Gf0 values to calculate free energy change at 25 P.6. Use ∆Gf0 values to calculate free energy change at 25 C for the following:C for the following:

2NO (g) + O2(g) → 2NO2 (g)2NO (g) + O2(g) → 2NO2 (g)

P.7. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO P.7. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO 2(g) + 4 H2O (g) 2(g) + 4 H2O (g)

Calculate ∆G0 -163 0 -Calculate ∆G0 -163 0 -394 -229 ∆Gf0 kJ/mol394 -229 ∆Gf0 kJ/mol

P.8. Is the reaction C2H4 (g) + H2O (l) _______ C2H5OH P.8. Is the reaction C2H4 (g) + H2O (l) _______ C2H5OH (l) spontaneous under standard conditions?(l) spontaneous under standard conditions?

20-42


G and the Work a System Can Do

For a spontaneous process, G is the maximum work obtainable from the system as the process takes place: G = - workmax

For a nonspontaneous process, G is the maximum work that must be done to the system as the process takes place: G = workmax

An example

20-43


∆∆G and work a system can do:G and work a system can do:

∆∆G < 0, for a spontaneous process., ∆G=-wmax, G < 0, for a spontaneous process., ∆G=-wmax, maximum useful work obtainable from the system.maximum useful work obtainable from the system.

∆∆G > 0, for a nonspontaneous process., minimum work G > 0, for a nonspontaneous process., minimum work that must be done to the system to make the process that must be done to the system to make the process take place.take place.

In any real process work is performed irreversibly and In any real process work is performed irreversibly and some of the free energy is lost to the surroundings as some of the free energy is lost to the surroundings as heat.heat.

You can never obtain maximum possible work.You can never obtain maximum possible work. You need more than minimum energy to start the You need more than minimum energy to start the

process for a nonspontaneous type.process for a nonspontaneous type. At equilibrium a reaction can no longer do any work.At equilibrium a reaction can no longer do any work.

20-44


Effect of Temperature on Effect of Temperature on Reaction spontaneityReaction spontaneity

1.Reaction is spontaneous at all temperatures: 1.Reaction is spontaneous at all temperatures: ∆∆H < 0, ∆S >0, -T ∆S= -ve, ∆G =-veH < 0, ∆S >0, -T ∆S= -ve, ∆G =-ve 2. Reaction is nonspontaneous at all temperatures:2. Reaction is nonspontaneous at all temperatures: ∆ ∆H > 0, ∆S <0, -T ∆S= +ve,∆G =+veH > 0, ∆S <0, -T ∆S= +ve,∆G =+ve 3. Reaction is spontaneous at higher temperatures: 3. Reaction is spontaneous at higher temperatures: ∆∆H > 0, ∆S >0, -T ∆S favors spontaneity at high temp , ∆G H > 0, ∆S >0, -T ∆S favors spontaneity at high temp , ∆G

=-ve and reaction is spontaneous.=-ve and reaction is spontaneous. 4. Reaction is spontaneous at lower temperatures: 4. Reaction is spontaneous at lower temperatures: ∆∆H < 0, ∆S <0, ∆H favors spontaneity at low temp , H < 0, ∆S <0, ∆H favors spontaneity at low temp ,

Reaction occurs if -T ∆S< ∆H Reaction occurs if -T ∆S< ∆H

20-45


Table 20.1 Reaction Spontaneity and the Signs of H0, S0, and G0

H0 S0 -TS0 G0 Description

- + - -

+ - + +

+ + - + or -

- - + + or -

Spontaneous at all T

Nonspontaneous at all T

Spontaneous at higher T;nonspontaneous at lower T

Spontaneous at lower T;nonspontaneous at higher T

20-46


Figure 20.11 The effect of temperature on reaction spontaneity.

Reaction between Cu2O and C.Relatively const ∆H and steadily increasing T∆S

20-47


Free Energy, Equilibrium and Reaction Direction

•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (G < 0)

•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (G > 0)

•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (G = 0)

G = RT ln Q/K = RT lnQ - RT lnK

Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so

G0 = - RT lnK

20-48


FO

RW

AR

D R

EA

CT

ION

RE

VE

RS

E R

EA

CT

ION

Table 20.2 The Relationship Between G0 and K at 250C

G0(kJ) K Significance

200

100

50

10

1

0

-1

-10

-50

-100

-200

9x10-36

3x10-18

2x10-9

2x10-2

7x10-1

1

1.5

5x101

6x108

3x1017

1x1035

Essentially no forward reaction; reverse reaction goes to completion

Forward and reverse reactions proceed to same extent

Forward reaction goes to completion; essentially no reverse reaction

20-49


Free Energy and Equilibrium:Free Energy and Equilibrium:

GG = = GG + + RTRT ln( ln(QQ)) QQ = reaction quotient from the law of mass = reaction quotient from the law of mass

action.action. Equilibrium point occurs at the lowest Equilibrium point occurs at the lowest

value of free energy available to the value of free energy available to the reaction system.reaction system.

GG = = RTRT ln( ln(KK) )

20-50



PROBLEM:

Calculating G at Nonstandard Conditions

The oxidation of SO2, which we considered in Sample Problem 20.6

2SO2(g) + O2(g) 2SO3(g)

is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature.

(a) Calculate K at 298K and at 973K. (G0298 = -141.6kJ/mol of reaction as

written using H0 and S0 values at 973K. G0973 = -12.12kJ/mol of reaction

as written.)(b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO2, 0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature?(c) Calculate G for the system in part (b) at each temperature.

PLAN: Use the equations and conditions found on slide .

20-51


SOLUTION:

Sample Problem 20.7: Calculating G at Nonstandard Conditions

continued (2 of 3)

(a) Calculating K at the two temperatures:

G0 = -RTlnK so

K e (G 0 /RT)

At 298, the exponent is -G0/RT = -(-141.6kJ/mol)(103J/kJ)

(8.314J/mol*K)(298K)= 57.2

K e (G 0 /RT) = e57.2 = 7x1024

At 973, the exponent is -G0/RT(-12.12kJ/mol)(103J/kJ)

(8.314J/mol*K)(973K)= 1.50

K e (G 0 /RT) = e1.50 = 4.5

20-52


Sample Problem 20.7: Calculating G at Nonstandard Conditions

continued (3 of 3)

(b) The value of Q =pSO3

2

(pSO2)2(pO2)=

(0.100)2

(0.500)2(0.0100)= 4.00

Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight.

(c) The nonstandard G is calculated using G = G0 + RTlnQ

G298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(ln4.00)

G298 = -138.2kJ/mol

G973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(ln4.00)

G298 = -0.9kJ/mol

20-53


Figure 20.12

The relation between free energy and the extent of reaction.

G0 < 0 K >1

G0 > 0 K <1

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20-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 20 Thermodynamics: Entropy, Free Energy and the