20-1

CHEM 1B:

GENERAL

CHEMISTRY

Chapter 20:

Thermodynamics:

Entropy, Free

Energy, and the

Direction of

Chemical

Reactions

Instructor: Dr. Orlando E. Raola

Santa Rosa Junior College

20-2

Chapter 20

Thermodynamics: Entropy, Free Energy,

and the Direction of Chemical Reactions

20-3

Thermodynamics: Entropy, Free Energy, and the

Direction of Chemical Reactions

20.1 The Second Law of Thermodynamics: Predicting

Spontaneous Change

20.2 Calculating Entropy Change of a Reaction

20.3 Entropy, Free Energy, and Work

20.4 Free Energy, Equilibrium, and Reaction Direction

20-4

Spontaneous Change

A spontaneous change is one that occurs without a

continuous input of energy from outside the system.

All chemical processes require energy (activation energy)

to take place, but once a spontaneous process has begun,

no further input of energy is needed.

A nonspontaneous change occurs only if the

surroundings continuously supply energy to the system.

If a change is spontaneous in one direction, it will be

nonspontaneous in the reverse direction.

20-5

The First Law of Thermodynamics Does

Not Predict Spontaneous Change

Energy is conserved. It is neither created nor destroyed,

but is transferred in the form of heat and/or work.

DE = q + w

The total energy of the universe is constant:

DEsys = -DEsurr or DEsys + DEsurr = DEuniv = 0

The law of conservation of energy applies to all

changes, and does not allow us to predict the direction

of a spontaneous change.

20-6

DH Does Not Predict Spontaneous Change

A spontaneous change may be exothermic or endothermic.

Spontaneous exothermic processes include:

• freezing and condensation at low temperatures,

• combustion reactions,

• oxidation of iron and other metals.

Spontaneous endothermic processes include:

• melting and vaporization at higher temperatures,

• dissolving of most soluble salts.

The sign of DH does not by itself predict the direction

of a spontaneous change.

20-7

Figure 20.1 A spontaneous endothermic chemical reaction.

water

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

DH°rxn = +62.3 kJ

This reaction occurs spontaneously when the solids are mixed. The

reaction mixture absorbs heat from the surroundings so quickly that

the beaker freezes to a wet block.

20-8

Freedom of Particle Motion

All spontaneous endothermic processes result in an

increase in the freedom of motion of the particles in the

system. solid → liquid → gas

crystalline solid + liquid → ions in solution

A change in the freedom of motion of particles in a system

is a key factor affecting the direction of a spontaneous

process.

less freedom of particle motion more freedom of particle motion

localized energy of motion dispersed energy of motion

20-9

Microstates and Dispersal of Energy

• Just as the electronic energy levels within an atom are

quantized, a system of particles also has different

allowed energy states.

• Each quantized energy state for a system of particles is

called a microstate.

– At any instant, the total energy of the system is dispersed

throughout one microstate.

• At a given set of conditions, each microstate has the

same total energy as any other.

– Each microstate is therefore equally likely.

• The larger the number of possible microstates, the larger

the number of ways in which a system can disperse its

energy.

20-10

Entropy

The number of microstates (W) in a system is related to

the entropy (S) of the system.

A system with fewer microstates has lower entropy.

A system with more microstates has higher entropy.

All spontaneous endothermic processes exhibit an

increase in entropy.

Entropy, like enthalpy, is a state function and is therefore

independent of the path taken between the final and initial

states.

S = k lnW

20-11

Figure 20.2 Spontaneous expansion of a gas.

When the stopcock is opened, the gas spontaneously expands to

fill both flasks.

Increasing the volume increases the number of translational

energy levels the particles can occupy. The number of microstates

– and the entropy – increases.

20-12

Figure 20.3 The entropy increase due to the expansion of a gas.

Opening the stopcock increases the number of possible energy

levels, which are closer together on average. More distributions of

particles are possible.

20-13

Figure 20.4 Expansion of a gas and the increase in number of

microstates.

When the stopcock opens,

the number of microstates

is 2n, where n is the

number of particles.

20-14

DS for a Reversible Process

DS = qrev

T

A reversible process is one that occurs in such tiny

increments that the system remains at equilibrium, and

the direction of the change can be reversed by an

infinitesimal reversal of conditions.

20-15

Figure 20.5 Simulating a reversible process.

A sample of Ne gas is confined to a

volume of 1 L by the “pressure” of a

beaker of sand on the piston.

We remove one grain of sand (an

“infinitesimal” decrease in pressure),

causing the gas to expand a tiny amount.

The gas does work on its surroundings,

absorbing a tiny increment of heat, q, from

the heat reservoir.

This simulates a reversible process, since

it can be reversed by replacing the grain

of sand.

20-16

The Second Law of Thermodynamics

The sign of ΔS for a reaction does not, by itself, predict the

direction of a spontaneous reaction.

If we consider both the system and the surroundings, we

find that all real processes occur spontaneously in the

direction that increases the entropy of the universe.

DSuniv = DSsys + DSsurr > 0

For a process to be spontaneous, a decrease in the

entropy of the system must be offset by a larger

increase in the entropy of the surroundings.

20-17

Comparing Energy and Entropy

The total energy of the universe remains constant.

DEsys + DEsurr =DEuniv = 0

DH is often used to approximateDE.

For enthalpy there is no zero point; we can only measure

changes in enthalpy.

The total entropy of the universe increases.

DSuniv =DSsys +DSsurr > 0

For entropy there is a zero point, and we can determine

absolute entropy values.

20-18

The Third Law of Thermodynamics

A perfect crystal has zero entropy at absolute zero.

Ssys = 0 at 0 K

A “perfect” crystal has flawless alignment of all its particles.

At absolute zero, the particles have minimum energy, so

there is only one microstate.

S = k lnW = k ln 1 = 0

To find the entropy of a substance at a given temperature, we cool it

as close to 0 K as possible. We then heat in small increments,

measure q and T, and calculate DS for each increment. The sum of

these DS values gives the absolute entropy at the temperature of

interest.

20-19

Standard Molar Entropies

S° is the standard molar entropy of a substance,

measured for a substance in its standard state in

units of J/mol·K.

The conventions for defining a standard state include:

• 1 bar for gases

• 1 mol/L for solutions, and

• the pure substance in its most stable form for

solids and liquids.

20-20

Factors Affecting Entropy

• Entropy depends on temperature.

– For any substance, S° increases as temperature increases.

• Entropy depends on the physical state of a substance.

– S° increases as the phase changes from solid to liquid to gas.

• The formation of a solution affects entropy.

• Entropy is related to atomic size and molecular

complexity.

– Remember to compare substances in the same physical state.

20-21

Figure 20.6A Visualizing the effect of temperature on entropy.

Computer simulations show each particle in a crystal moving about its

lattice position. Adding heat increases T and the total energy, so the

particles have greater freedom of motion, and their energy is more

dispersed. S therefore increases.

20-22

Figure 20.6B Visualizing the effect of temperature on entropy.

At any T, there is a range of occupied energy levels and therefore a

certain number of microstates. Adding heat increases the total energy

(area under the curve), so the range of occupied energy levels

becomes greater, as does the number of microstates.

20-23

Figure 20.6C Visualizing the effect of temperature on entropy.

A system of 21 particles occupy energy levels (lines) in a box whose

height represents the total energy. When heat is added, the total

energy increases and becomes more dispersed, so S increases.

20-24

Figure 20.7 The increase in entropy during phase changes

from solid to liquid to gas.

20-25

Figure 20.8 The entropy change accompanying the dissolution

of a salt.

pure solid

pure liquid

solution

MIX

The entropy of a salt solution is usually greater than that of the solid

and of water, but it is affected by the organization of the water molecules

around each ion.

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

20-26

Figure 20.9 The small increase in entropy when ethanol

dissolves in water.

Ethanol (A) and water (B) each have many H bonds between their own

molecules. In solution (C) they form H bonds to each other, so their

freedom of motion does not change significantly.

20-27

Figure 20.10 The entropy of a gas dissolved in a liquid.

20-28

Entropy and Atomic Size

S° is higher for larger atoms or molecules of the same type.

Li Na K Rb Cs

Atomic radius (pm) 152 186 227 248 265

Molar mass (g/mol) 6.941 22.99 39.10 85.47 132.9

S°(s) 29.1 51.4 64.7 69.5 85.2

HF HCl HBr HI

Molar mass (g/mol) 20.01 36.46 80.91 127.9

S°(s) 173.7 186.8 198.6 206.3

20-29

Entropy and Structure

NaCl AlCl3 P4O10 NO NO2 N2O4

S°(s) 72.1 167 229

S°(g) 211 240 304

For allotropes, S° is higher in the form that allows the

atoms more freedom of motion.

S° of graphite is 5.69 J/mol·K, whereas S° of diamond is 2.44 J/mol·K.

For compounds, S° increases with chemical complexity.

These trends only hold for substances in the same

physical state.

20-30

Figure 20.11 Entropy, vibrational motion, and molecular

complexity.

20-31

Sample Problem 20.1 Predicting Relative Entropy Values

PLAN: In general, particles with more freedom of motion have more

microstates in which to disperse their kinetic energy, so they

have higher entropy. Raising the temperature or having more

particles increases entropy.

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) 1 mol of SO2(g) or 1 mol of SO3(g)

(b) 1 mol of CO2(s) or 1 mol of CO2(g)

(c) 3 mol of O2(g) or 2 mol of O3(g)

(d) 1 mol of KBr(s) or 1 mol of KBr(aq)

(e) seawater at 2°C or at 23°C

(f) 1 mol of CF4(g) or 1 mol of CCl4(g)

20-32

Sample Problem 20.1

SOLUTION:

(a) 1 mol of SO3(g). For equal numbers of moles of substances with

the same types of atoms in the same physical state, the more

atoms in the molecule the higher the entropy.

(b) 1 mol of CO2(g). For a given substance, entropy increases as the

phase changes from solid to liquid to gas.

(c) 3 mol of O2(g). The two samples contain the same number of

oxygen atoms but different numbers of molecules. Although each

O3 molecule is more complex than each O2 molecule, the greater

number of molecules dominates because there are many more

microstates possible for 3 mol of particles than for 2.

20-33

Sample Problem 20.1

(d) 1 mol of KBr(aq). The two samples have the same number of

ions, but their motion is more limited and their energy less

dispersed in the solid than in the solution. An ionic substance in

solution usually has a higher entropy than the solid.

(e) Seawater at 23°C. Entropy increases with rising temperature.

(f) 1 mol of CCl4(g). For similar compounds, entropy increases with

molar mass.

20-34

Entropy Changes in the System

The standard entropy of reaction, DS°rxn, is the entropy

change that occurs when all reactants and products are in

their standard states.

DS°rxn = SmS°products - SnS°reactants

where m and n are the amounts (mol) of products and

reactants, given by the coefficients in the balanced equation.

20-35

Predicting the Sign of DS°rxn

We can often predict the sign of DS°rxn for processes that

involve a change in the number of moles of gas.

• DS°rxn is positive if the amount of gas increases;

– e.g., H2(g) + I2(s) → 2HI(g).

• DS°rxn is negative if the amount of gas decreases;

– e.g.,

• DS° is likely to be positive if a new structure forms that

has more freedom of motion.

N2(g) + 3H2(g) 2NH3(g)

20-36

Sample Problem 20.2 Calculating the Standard Entropy of

Reaction, DS°rxn

PROBLEM: Predict the sign of DS°rxn and calculate its value for the

combustion of 1 mol of propane at 25°C.

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

PLAN: From the change in the number of moles of gas (6 mol yields

3 mol), the entropy should decrease (DS°rxn < 0). To find

DS°rxn, we apply Equation 20.4, using the S° values from

Appendix B.

SOLUTION:

ΔS°rxn = [(3 mol CO2)(S° of CO2) + (4 mol H2O)(S° of H2O)]

- [(1 mol C3H8)(S° of C3H8) + (5 mol O2)(S° of O2)

= [(3mol)(213.7J/K·mol) + (4 mol)(69.9J/K·mol)]

– [(1mol)(269.9 J/K·mol) + (5 mol)(205.0 J/K·mol) = -374 J/K

20-37

Entropy Changes in the Surroundings

A decrease in the entropy of the system is outweighed

by an increase in the entropy of the surroundings.

The surroundings function as a heat source or heat sink.

In an exothermic process, the surroundings absorbs the

heat released by the system, and Ssurr increases.

qsys < 0; qsurr > 0 and DSsurr > 0

In an endothermic process, the surroundings provides the

heat absorbed by the system, and Ssurr decreases.

qsys > 0; qsurr < 0 and DSsurr < 0

20-38

Temperature at which Heat is Transferred

Since entropy depends on temperature, DS°surr is also

affected by the temperature at which heat is transferred.

The impact on the surroundings is larger when the

surroundings are at lower temperature, because there is

a greater relative change in Ssurr.

For any reaction, qsys = -qsurr. The heat transferred is

specific for the reaction and is the same regardless of the

temperature of the surroundings.

DSsurr = - qsys

T DSsurr = -

DHsys

T for a process at constant P.

20-39

Sample Problem 20.3 Determining Reaction Spontaneity

PROBLEM: At 298 K, the formation of ammonia has a negative DS°sys;

N2(g) + 3H2(g) → 2NH3(g); DS°sys = -197 J/K

Calculate DS°univ, and state whether the reaction occurs

spontaneously at this temperature.

PLAN: For the reaction to occur spontaneously, DS°univ > 0, so DS°surr

must be greater than +197 J/K. To find DS°surr we need DH°sys,

which is the same as DH°rxn. We use the standard enthalpy

values from Appendix B to calculate this value.

DH°rxn = [(2 mol)(DH°f of NH3)] - [(1 mol)(DH°f of N2) + (3 mol)(DH°f of H2)]

SOLUTION:

= [(2 mol)(-45.9 kJ/mol)] - [(1 mol)(0 kJ/mol) + (3 mol)(0 kJ/mol)]

= 91.8 kJ

20-40

Sample Problem 20.3

DSsurr = - DHsys

T = -

-91.8 kJ x 1000 J

1 kJ

298 K = 308 J/K

DS°univ = DS°sys + DS°surr = -797 J/K + 308 J/K = 111 J/K

Since DS°univ > 0, the reaction occurs spontaneously at 298 K.

Although the entropy of the system

decreases, this is outweighed by

the increase in the entropy of the

surroundings.

20-41

Figure 20.12 A whole-body calorimeter.

In this room-sized apparatus, a person exercises while respiratory gases,

energy input and output, and other physiological variables are monitored.

Biological systems also obey the laws of thermodynamics. For all

processes, however complex, the entropy of the university increases.

20-42

DS and the Equilibrium State

For any process approaching equilibrium, DS°univ > 0.

At equilibrium, there is no further net change, and DS°sys is

balanced by DS°surr.

DS°univ = DS°sys + DS°surr = 0, so DS°sys = - DS°surr

When a system reaches equilibrium, neither the forward

nor the reverse reaction is spontaneous, so there is no

net reaction in either direction.

20-43

Figure 20.13AB Components of DS°univ for spontaneous reactions.

exothermic

For an exothermic reaction in

which DSsys > 0, the size of DSsurr

is not important since DSsurr > 0

for all exothermic reactions. The

reaction will always be

spontaneous.

exothermic

For an exothermic reaction in

which DSsys < 0, DSsurr must be

larger than DSsys for the reaction

to be spontaneous.

20-44

Figure 20.13C Components of DS°univ for spontaneous reactions.

endothermic

For an endothermic reaction in which

DSsys > 0, DSsurr must be smaller

than DSsys for the reaction to be

spontaneous. DSsurr is always < 0 for

endothermic reactions.

20-45

Gibbs Free Energy

The Gibbs free energy (G) combines the enthalpy and

entropy of a system; G = H – TS.

The free energy change (DG) is a measure of the

spontaneity of a process and of the useful energy available

from it.

DGsys = DHsys - T DSsys

DG < 0 for a spontaneous process

DG > 0 for a nonspontaneous process

DG = 0 for a process at equilibrium

20-46

Calculating DG°

DG°sys = DH°sys - T DS°sys

DG°rxn can be calculated using the Gibbs equation:

DG°rxn can also be calculated using values for the

standard free energy of formation of the components.

DG°rxn = SmDG°products - SnDG°reactants

20-47

Sample Problem 20.4 Calculating DG°rxn from Enthalpy and

Entropy Values

PROBLEM: Potassium chlorate, a common oxidizing agent in

fireworks and matchheads, undergoes a solid-state

disproportionation reaction when heated.

4KClO3(s) 3KClO4(s) + KCl(s)

+7 -1 +5 Δ

PLAN: To solve for DG°, we need values from Appendix B. We use

DHf° to calculate DH°rxn, use S° values to calculate DS°rxn and

then apply the Gibbs equation.

SOLUTION: DH°rxn = SmDHf°(products) - SnDHf°(reactants)

= [(3 mol KClO4)(DHf° of KClO4) + (1 mol KCl)(DHf° of KCl)]

‒ [(4 mol KClO3)(DHf° of KClO3)]

= [(3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol)

– [(4 mol)(-397.7 kJ/mol)] = 144 kJ

20-48

Sample Problem 20.4

DS°rxn = SmS°(products) - SS°(reactants)

= [(3 mol of KClO4)(S° of KClO4) + (1 mol KCl)(S° of KCl)]

‒ [(4 mol KClO3)(S° of KClO3)]

= [(3 mol)(151.0 J/mol·K) + (1 mol)(82.6 J/mol·K)

– [(4 mol)(-397.7 kJ/mol)] = -36.8 J/K

DG°sys = DH°sys - T DS°sys = 144 kJ – (298 K)(-36.8 J/K) 1000 J

1 kJ

= -133 kJ

20-49

Sample Problem 20.5 Calculating DG°rxn from DG°f Values

PLAN: We apply Equation 20.8 and use the values from Appendix

B to calculate DG°rxn.

SOLUTION:

PROBLEM: Use DG°f values to calculate DG°rxn for the reaction in

Sample Problem 20.4:

4KClO3(s) 3KClO4(s) + KCl(s) Δ

DG°rxn = SmDG°products - SnDG°reactants

= [(3 mol KClO4)(DGf° of KClO4) + (1 mol KCl)(DGf° of KCl)]

‒ [(4 mol KClO3)(DGf° of KClO3)]

= [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)]

‒ [(4 mol)(-296.3 kJ/mol)]

= -134 kJ

20-50

DG° and Useful Work

DG is the maximum useful work done by a system

during a spontaneous process at constant T and P.

In practice, the maximum work is never done. Free energy not used

for work is lost to the surroundings as heat.

DG is the minimum work that must be done to a system

to make a nonspontaneous process occur at constant T

and P.

A reaction at equilibrium (DGsys = 0) can no longer do any

work.

20-51

Figure 20.14 An expanding gas lifting a weight.

20-52

Effect of Temperature on Reaction Spontaneity

Reaction is spontaneous at

all temperatures

If DH < 0 and DS > 0

DG < 0 for all T

Reaction is nonspontaneous

at all temperatures

If DH > 0 and DS < 0

DG > 0 for all T

DGsys = DHsys - T DSsys

20-53

Effect of Temperature on Reaction Spontaneity

Reaction becomes spontaneous

as T increases

If DH > 0 and DS > 0

DG becomes more negative as T

increases.

Reaction becomes spontaneous

as T decreases

If DH < 0 and DS < 0

DG becomes more negative as T

decreases.

20-54

Table 20.1 Reaction Spontaneity and the Signs of DH, DS, and DG

DH DS -TDS DG Description

– + – – Spontaneous at all T

+ – + + Nonspontaneous at all T

+ + – + or – Spontaneous at higher T;

nonspontaneous at lower T

– – + + or – Spontaneous at lower T;

nonspontaneous at higher T

20-55

Sample Problem 20.6 Using Molecular Scenes to Determine the

Signs of DH, DS, and DG

PROBLEM: The following scenes represent a familiar phase change

for water (blue spheres).

(a) What are the signs of DH and DS for this process? Explain.

(b) Is the process spontaneous at all T, no T, low T, or high T? Explain.

PLAN: (a) From the scenes, we determine any change in the amount

of gas, which indicates the sign of DS, and any change in

the freedom of motion of the particles, which indicates

whether heat is absorbed or released.

(b) To determine reaction spontaneity we need to consider the

sign of DG at different temperatures.

20-56

Sample Problem 20.6

SOLUTION:

(a) The scene represents the condensation of water vapor, so the

amount of gas decreases dramatically, and the separated molecules

give up energy as they come closer together.

DS < 0 and DH > 0

(b) Since DS is negative and DH is positive, the –TDS term is positive.

In order for DG to be < 0, the temperature must be low.

The process is spontaneous at low temperatures.

20-57

Sample Problem 20.7 Determining the Effect of Temperature on ΔG

PROBLEM: A key step in the production of sulfuric acid is the oxidation

of SO2(g) to SO3(g):

2SO2(g) + O2(g) → 2SO3(g)

At 298 K, DG = -141.6 kJ; DH = -198.4 kJ; and DS = -187.9 J/K

(a) Use the data to decide if this reaction is spontaneous at 25°C, and

predict how DG will change with increasing T.

(b) Assuming DH and DS are constant with increasing T, is the

reaction spontaneous at 900.° C?

PLAN: We note the sign of DG to see if the reaction is spontaneous

and the signs of DH and DS to see the effect of T. We can then

calculate whether or not the reaction is spontaneous at the

higher temperature.

20-58

Sample Problem 20.7

SOLUTION:

(a) DG < 0 at 209 K (= 25°C), so the reaction is spontaneous.

With DS < 0, the term -TDS > 0 and this term will become more

positive at higher T.

DG will become less negative, and the reaction less spontaneous,

with increasing T.

(b) DG = DH - TDS

DG = -198.4 kJ – [1173 K)(-0.187.9 kJ/K) = 22.0 kJ

Convert T to K: 900. + 273.15 = 1173 K

Convert S to kJ/K: -187.9 J/K = 0.1879 kJ/K

DG > 0, so the reaction is nonspontaneous at 900.°C.

20-59

Figure 20.15 The effect of temperature on reaction spontaneity.

DH

ΔS T =

The sign of DG

switches at

20-60

Sample Problem 20.8 Finding the Temperature at Which a

Reaction Becomes Spontaneous

PROBLEM: At 25°C (298 K), the reduction of copper(I) oxide is

nonspontaneous DG = 8.9 kJ). Calculate the temperature

at which the reaction becomes spontaneous.

PLAN: We need to calculate the temperature at which DG crosses

over from a positive to a negative value. We set DG equal to

zero, and solve for T, using the values for DH and DS from

the text.

SOLUTION:

DH

DS T = =

58.1 kJ

0.165 kJ/K = 352 K

At any temperature above 352 K (= 79°C), the reaction becomes

spontaneous.

20-61

Figure B20.1 The coupling of a nonspontaneous reaction to

the hydrolysis of ATP.

Chemical Connections

A spontaneous reaction can be coupled to a nonspontaneous reaction

so that the spontaneous process provides the free energy required to

drive the nonspontaneous process. The coupled processes must be

physically connected.

20-62

Figure B20.2 The cycling of metabolic free energy.

Chemical Connections

20-63

Figure B20.3 ATP is a high-energy molecule.

Chemical Connections

When ATP is hydrolyzed to ADP, the decrease in charge repulsion (A)

and the increase in resonance stabilization (B) causes a large

amount of energy to be released.

20-64

DG, Equilibrium, and Reaction Direction

A reaction proceeds spontaneously to the right if

Q < K; < 1 so ln < 0 and DG < 0 Q

K

Q

K

A reaction is at equilibrium if

Q = K; = 1 so ln = 0 and DG = 0 Q

K

Q

K

A reaction proceeds spontaneously to the left if

Q > K; > 1 so ln > 0 and DG > 0 Q

K

Q

K

20-65

DG, Q, and K

Q

K DG = RT ln = RT lnQ – RT lnK

If Q and K are very different, DG has a very large value

(positive or negative). The reaction releases or absorbs a

large amount of free energy.

If Q and K are nearly the same, DG has a very small value

(positive or negative). The reaction releases or absorbs very

little free energy.

20-66

ΔG and the Equlibrium Constant

For standard state conditions, Q = 1 and

As DG° becomes more positive, K becomes smaller.

As DG° becomes more negative, K becomes larger.

A small change in DG° causes a large change in K, due to

their logarithmic relationship.

DG = DG° + RT lnQ

To calculate DG for any conditions:

DG° = -RT lnK

20-67

Table 20.2 The Relationship Between DG° and K at 298 K

DG° (kJ) K Significance

200 9x10-36 Essentially no forward reaction;

reverse reaction goes to completion 100 3x10-18

50 2x10-9

10 2x10-2

1 7x10-1

0 1 Forward and reverse reactions proceed

to same extent -1 1.5

-10 5x101

-50 6x108

-100 3x1017 Forward reaction goes to completion;

essentially no reverse reaction -200 1x1035

FO

RW

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{

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20-68

Sample Problem 20.9 Using Molecular Scenes to Find DG for a

Reaction at Nonstandard Conditions PROBLEM: These molecular scenes represent three mixtures in which

A2 (black) and B2 (green) are forming AB. Each molecule

represents 0.10 atm. The equation is

A2(g) + B2(g) 2AB(g) DGo = -3.4 kJ/mol

(a) If mixture 1 is at equilibrium, calculate K.

(b) Which mixture has the most negative DG, and which has the most

positive?

(c) Is the reaction spontaneous at the standard state, that is P = P

= PAB = 1.0 atm? A2 B2

20-69

Sample Problem 20.9

PLAN:

(a) Mixture 1 is at equilibrium, so we first write the expression for Q

and then find the partial pressure of each substance from the

number of molecules and calculate K.

(b) To find DG, we apply equation 20.13. We can calculate the value

of T using K from part (a), and substitute the partial pressure of

each substance to get Q.

(c) Since 1.0 atm is the standard state for gases, DG = DG°.

A2(g) + B2(g) 2AB(g)

SOLUTION:

(a) Q = PAB

2

P x P A2 B2

K = (0.40)2

(0.20)(0.20) = 4.0

20-70

Sample Problem 20.9

(b) DG° = -RT lnK

= - -3.4 kJ

mol

T ln 4.0 8.314 J

mol·K

T =

-3.4 kJ

mol

1000 J

1 kJ

8.314 J

mol·K ln 4.0 -

= 295 K

Mixture 1 is at equilibrium, so ΔG = 0;

DG = DG° + RT lnQ

= -3400 J/mol + (8.314 J/mol·K)(295 K)(ln 4.0)

= -3400 J + 3400 J = 0.0 J

Mixture 2:

(0.20)2

(0.30)(0.30) Q = = 0.44

DG = DG° + RT lnQ

= -3400 J/mol + (8.314 J/mol·K)(295 K)(ln 0.44)

= -3400 J – 2010 J = -5.4x10-3 J

20-71

Sample Problem 20.9

Mixture 3:

(0.60)2

(0.10)(0.10) Q = = 36

DG = DG° + RT lnQ

= -3400 J/mol + (8.314 J/mol·K)(295 K)(ln 36)

= -3400 J + 8800 J = +5.4x10-3 J

Mixture 2 has the most negative DG, and mixture 3 has the

most positive DG.

(c) Under standard conditions, DG = DG° = -3.4 kJ/mol.

Yes, the reaction is spontaneous when the components are in

their standard states.

20-72

Sample Problem 20.10 Calculating DG at Nonstandard Conditions

PROBLEM: The oxidation of SO2(g) is too slow at 298 K to be useful

in the manufacture of sulfuric acid, so the reaction is run

at high T.

2SO2(g) + O2(g) → 2SO3(g)

(a) Calculate K at 298 K and at 973 K, (DG°298 = -141.6 kJ/mol of

reaction as written; using DH° and DS°values at 973 K,

DG°973 = -12.12 kJ/mol for reaction as written.)

(b) Two containers are filled with 0.500 atm of SO2, 0.0100 atm of

O2, and 0.100 atm of SO3; one is kept at 25°C and the other at

700.°C. In which direction, if any, will the reaction proceed to

reach equilibrium at each temperature?

(c) Calculate DG for the system in part (b) at each temperature.

20-73

Sample Problem 20.10

PLAN:

(a) We know DG°, T, and R, so we can calculate the values of K at

each temperature.

(b) To determine if a net reaction will occur, we find Q from the given

partial pressures and compare Q to K from part (a). of each

substance to get Q.

(c) These are not standard-state pressures, so we find DG at each

temperature using the equation DG = DG° + RT lnQ.

SOLUTION:

(a) DG° = -RT lnK so K = e -(DG°/RT)

At 298 K, -(DG°)/RT = -141.6x103 J/mol

8.314 J/mol·K x 298 K = 57.2

K at 298 K = e57.2 = 7x1024

20-74

Sample Problem 20.10

At 973 K, -(DG°)/RT = -12.12x103 J/mol

8.314 J/mol·K x 973 K = 1.50

K at 973 K = e1.50 = 4.5

(b) Calculating the value of Q

P2

P2 x P

SO3

SO2 O2

Q = = (0.100)2

(0.500)2(0.0100) = 4.00

Since Q < K at both temperatures, the reaction will proceed

toward the products (to the right) in both cases until Q equals K.

At 298 K, the system is far from equilibrium and will proceed far to

the right. At 973 K the system is closer to equilibrium and will

proceed only slightly toward the right.

20-75

(c)

Sample Problem 20.10

DG298 = DG° + RT ln Q

= -141.6 kJ/mol + (8.314x10-3 kJ/mol·K x 298 K x ln 4.00)

= -138.2 kJ/mol

DG973 = DG° + RT ln Q

= -12.12 kJ/mol + (8.314x10-3 kJ/mol·K x 973 K x ln 4.00)

= -0.9 kJ/mol

20-76

Figure 20.16 Free energy and the extent of reaction.

Each reaction proceeds spontaneously (green curved arrows) from

reactants or products to the equilibrium mixture, at which point DG = 0.

After that, the reaction is nonspontaneous in either direction (red curved

arrows). Free energy reaches a minimum at equilibrium.