16 Burdge Lecture Notes.pdf

(1)

Chemistry: Atoms First

Julia Burdge & Jason Overby

Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 16

Entropy, Free Energy,

and Equilibrium

Kent L. McCorkle

Cosumnes River College

Sacramento, CA

Entropy, Free Energy, and Equilibrium

18

18.1 Spontaneous Processes 18.2 Entropy

A Qualitative Description of Entropy A Quantitative Description of Entropy 18.3 Entropy Changes in a System

Calculating ΔSsys Standard S°

Qualitatively Predicting ΔS°_sys 18.4 Entropy Changes in the Universe

Calculating ΔSsurr

The Second Law of Thermodynamics The Third Law of Thermodynamics 18.5 Predicting Spontaneity

Gibbs Free-Energy Change, ΔG Standard Free-Energy Changes, ΔG° Using ΔG and ΔG° to Solve Problems 18.6 Free Energy and Chemical Equilibrium

Relationship Between ΔG and ΔG° Relationship Between ΔG° and K 18.7 Thermodynamics in Living Systems

Spontaneous Processes

A process that

does

occur under a specific set of conditions is

called a

spontaneous process

.

A process that does

not

occur under a specific set of conditions is

called

nonspontaneous

.

18.1

Spontaneous Processes

A process that results in a decrease in the energy of a system often

is spontaneous:

The sign of Δ

H

alone

is insufficient to predict spontaneity in every

circumstance:

CH

4

(

g

) + 2O

2

(

g

) CO

2

(

g

) + 2H

2

O(

l

) Δ

H

°

=

-

890.4 kJ/mol

H

2

O(

l

) H

2

O(

s

)

T

> 0

°

C; Δ

H

°

= -6.01 kJ/mol

Entropy

18.2

To predict spontaneity, both the enthalpy and entropy must be

known.

Entropy

(S)

of a system is a measure of how

spread out

or how

dispersed

the system’s energy is.

Entropy

Spontaneity is favored by an increase in entropy.

k

is the Boltzmann constant (1.38 x 10

–23

J/K)

W

is the number of different arrangements

S

=

k

ln

W

The number of arrangements possible is given by:

X

is the number of cells in a volume

(2)

Entropy

There are three possible states for this system:

1) One molecule on each side (eight possible arrangements)

2) Both molecules on the left (four possible arrangements)

3) Both molecules on the right (four possible arrangements)

The most

probable

state has the

largest

number of arrangements.

Entropy Changes in a System

The change in entropy for a system is the difference in entropy of the

final state and the entropy of the initial state.

Alternatively:

18.3

ΔS

sys

=

S

final

– S

initial

final

initial

ln

sys

V

S

nR

V





Worked Example 18.1

Strategy This is the isothermal expansion of an ideal gas. Because the molecules spread out to occupy a greater volume, we expect there to be an

increase in the entropy of the system. Use ΔSsys = Sfinal – Sinitial to solve for ΔSsys.

R = 8.314 J/K∙mol, n = 1.0 mole, Vfinal = 5.0 L, and Vinitial = 2.5 L.

Determine the change in entropy for 1.0 mole of an ideal gas originally confined to one-half of a 5.0-L container when the gas is allowed to expand to fill the entire container at constant temperature.

Solution

ΔS_sys = nR lnVfinal

Vinitial = 1.0 mol ×

8.314 J

K ∙ mol ×ln5.0 L 2.5 L = 5.8 J/K

Think About It Remember that for a process to be spontaneous, something

must favor spontaneity. If the process is spontaneous but not exothermic (in this

case, there is no enthalpy change), then we should expect ΔSsys to be positive.

Worked Example 18.1 (cont.)

Solution These equilibrium concentrations are then substituted into the

equilibrium expression to give

Because we expect x to be very small (even smaller than 1.34×10-3M–see

above), because the ionization of CH3COOH is suppressed by the presence of

CH3COO-, we assume

(0.10 – x) M ≈ 0.10 M and (0.050 + x) M ≈ 0.050 M

Therefore, the equilibrium expression simplifies to

and x = 3.6×10-5_M_{. According to the equilibrium table, [H}+_{] =}_x_{, so}

pH = –log(3.6×10-5_{) = 4.44.}

1.8×10-5₌(x)(0.050 + x)

0.010 – x

1.8×10-5₌(x)(0.050)

0.010

Entropy Changes in a System

The

standard entropy

is the absolute entropy of a substance at 1 atm.

Temperature is

not

part of the standard state definition and must be

(3)

There are several important trends in entropy:



S

°

_liquid

>

S

°

_solid



S

°

_gas

>

S

°

_liquid



S

°

increases with molar mass



S

°

increases with molecular complexity



S

°

increases with the mobility of a phase (for an element with

two or more allotropes)

In addition to translational motion, molecules exhibit

vibrations

and

rotations.

For a chemical reaction

a

A +

b

B →

c

C +

d

D

Alternatively,

ΔS°rxn = [cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]

ΔS°_rxn = ΣnS°(products) – ΣmS°(reactants)

Worked Example 18.2

Strategy Look up standard enthalpy values and calculate ΔS°_rxn. Just as we did when we calculated standard enthalpies of reaction, we consider stoichiometric

coefficients to be dimensionless–giving ΔS°_rxn units of J/K∙mol.

From Appendix 2, S°[CaCO3(s)] = 92.9 J/K∙mol, S°[CaO(s)] = 39.8 J/K∙mol,

S°[CO2(g)] = 213.6 J/K∙mol, S°[N2(g)] = 191.5 J/K∙mol, S°[H2(g)] = 131.0

J/K∙mol, S°[NH3(g)] = 193.0 J/K∙mol, S°[Cl2(g)] = 223.0 J/K∙mol, and

S°[HCl(g)] = 187.0 J/K∙mol.

From the standard enthalpy values in Appendix 2, calculate the standard entropy

changes for the following reactions at 25°C.

(a) CaCO3(s) → CaO(s) + CO2(g) (b) N2(g) + 3H2(g) → 2NH3(g)

(c) H2(g) + Cl2(g) → 2HCl(g)

Worked Example 18.2 (cont.)

Solution

(a) S°_rxn = [S°(CaO) + S°(CO2)] – [S°(CaCO3)]

= [(39.8 J/K∙mol) + (213.6 J/K∙mol)] – (92.9 J/K∙mol) = 160.5 J/K∙mol

(b) S°_rxn = [2S°(NH3)] – [S°(N2) + 3S°(H2)]

=(2)(193.0 J/K∙mol) – [(191.5 J/K∙mol) + (3)(131.0 J/K∙mol)] = –198.5 J/K∙mol

(c) S°rxn = [2S°(HCl)] – [S°(H2) + S°(Cl2)]

= (2)(187.0 J/K∙mol) – [(131.0 J/K∙mol) + (223.0 J/K∙mol)] = 20.0 J/K∙mol

Think About It Remember to multiply each standard entropy value by the correct stoichiometric coefficient. Like Equation 10.18, Equation 18.5 can only be

used with a balanced chemical equation.

Entropy Changes in a System

Several processes that lead to an increase in entropy are:

 Melting

 Vaporization or sublimation

 Temperature increase

(4)

The process of

dissolving

a substance can lead to either an increase or

a decrease in entropy, depending on the nature of the solute.

Molecular solutes (i.e. sugar): entropy increases

Ionic compounds: entropy could decrease or increase

Worked Example 18.3

Strategy Consider the change in energy/mobility of atoms and the resulting change in number of possible positions that each particle can occupy in each case. An increase in the number of arrangements corresponds to an increase in entropy

and therefore a positive ΔS.

For each process, determine the sign of ΔS for the system: (a) decomposition of

CaCO3(s) to give CaO(s) and CO2(g), (b) heating bromine vapor from 45°C to

80°C, (c) condensation of water vapor on a cold surface, (d) reaction of NH3(g)

and HCl(g) to give NH4Cl(s), and (e) dissolution of sugar in water.

Solution Increases in entropy generally accompany solid-to-liquid, liquid-to-gas, and solid-to-gas transitions; the dissolving of one substance in another; a temperature increase; and reactions that increase the net number of moles of gas.

ΔS is (a) positive (b) positive (c) negative

(d) negative (e) positive

Think About It For reactions involving only liquids and solids,

predicting the sign of ΔS° can be more difficult, but in many such

cases an increase in the total number of molecules and/or ions is accompanied by an increase of entropy.

Entropy Changes in the Universe

Correctly predicting the spontaneity of a process requires us to

consider entropy changes in both the system and the surroundings.

An ice cube spontaneously melts in a room at 25

°

C.

A cup of hot water spontaneously cools to room temperature.

The entropy of both the system AND surroundings are important!

18.4

Perspective Components ΔS

System ice positive

Surroundings everything else negative

Perspective Components ΔS

System hot water negative

Surroundings everything else positive

Entropy Changes in the Universe

The change in entropy of the surroundings is directly proportional to

the enthalpy of the system.

The

second law of thermodynamics

states that for a process to be

spontaneous, Δ

S

universe

must be positive.

sys surr

H

S

T







Δ

S

universe

=

Δ

S

sys

+ Δ

S

surr

Entropy Changes in the Universe

The

second law of thermodynamics

states that for a process to be

spontaneous, Δ

S

universe must be positive.

Δ

S

universe > 0 for a

spontaneous

process

Δ

S

universe < 0 for a

nonspontaneous

process

Δ

S

universe = 0 for an

equilibrium process

Δ

S

universe

=

Δ

S

sys

+ Δ

S

surr

Worked Example 18.4

Strategy For each process, use ΔS°rxn = ΣnS°(products) – ΣmS°(reactants)

to determine ΔS°_sys; ΔSsurr = (–ΔHsys/T)and ΔH°_sys = ΣnΔH _f°(products) –

ΣmΔH f°(reactants) to determine ΔH°sys and ΔS°surr. At the specified

temperature, the process is spontaneous if ΔS°_sys and ΔS°_surr sum to a positive

number, nonspontaneous is they sum to a negative number, and an equilibrium

process if they sum to zero. Note that because the reaction is the system, ΔSrxn and

ΔSsys are used interchangeably.

Determine if each of the following is a spontaneous process, a nonspontaneous

process, or an equilibrium process at the specified temperature: (a) H2(g) + I2(g)

→ 2HI(g) at 0°C, (b) CaCO3(s) → CaO(s) + CO2(g) at 200°C, (c) CaCO3(s) →

CaO(s) + CO2(g) at 1000°C, (d) Na(s) → Na(l) at 98°C. (Assume that the

(5)

Worked Example 18.4 (cont.)

Solution (a) S°[H2(g)] = 131.0 J/K∙mol, S°[I2(g)] = 260.57 J/K∙mol,

S°[HI(g)] = 206.3 J/K∙mol; ΔHf°[H2(g)] = 0 kJ/mol, ΔHf°[I2(g)] = 62.25

kJ/mol, ΔHf°[HI(g)] = 25.9 kJ/mol.

ΔS°rxn = [2S°(HI)] – [S°(H2) + S°(I2)]

= (2)(206.3 J/K∙mol) – [131.0 J/K∙mol + 160.57 J/K∙mol] = 21.03 J/K∙mol

ΔH°_rxn = [2 ΔH°_f (HI)] – [ΔH°_f (H2) + ΔH°_f (I2)]

= (2)(25.9 kJ/mol) – [0 kJ/mol + 62.26 kJ/mol] = −10.5 kJ/mol

ΔSsurr = = = 0.0385 kJ/K∙mol = 38.5 J/K∙mol

ΔSuniverse = ΔSsys + ΔSsurr = 21.03 J/K∙mol + 38.5 J/K∙mol = 59.5 J/K∙mol

ΔSuniverse is positive; therefore the reaction is spontaneous at 0°C.

−(−10.5 kJ/mol) 273 K

−ΔHrxn

T

Worked Example 18.4 (cont.)

Solution (b) In Worked Example 18.2(a), we determined that for this reaction,

ΔS°rxn = 160.5 J/K∙mol; ΔH°f [CaCO3(s)] = −1206.9 kJ/mol, ΔH°f [CaO(s)] =

−635.6 kJ/mol, ΔH°_f [CO2(g)] = −393.5 kJ/mol.

(b), (c)

ΔS°_rxn = 160.5 J/K∙mol

ΔH°_rxn = [ΔH°_f (CaO) + ΔH°_f (CO2)] – [ΔH°_f (CaCO3)]

= [-635.6 kJ/mol + (–393.5 kJ/mol)] – (–1206.9 kJ/mol) = 177.8 kJ/mol

(b) T = 200°C and

ΔSsurr = = = −0.376 kJ/K∙mol = −376 J/K∙mol

ΔSuniverse = ΔSsys + ΔSsurr = 160.5 J/K∙mol + (−376 J/K∙mol) = − 216 J/K∙mol

Δsuniverse is negative, therefore the reaction is nonspontaneous at 200°C.

−(177.8 kJ/mol) 473 K

−ΔHsys

T

Worked Example 18.4 (cont.)

Solution (c) T = 1000°C and

ΔSsurr = = = −0.1397 kJ/K∙mol = −139.7 J/K∙mol

ΔSuniverse = ΔSsys + ΔSsurr = 160.5 J/K∙mol + (−139.7 J/K∙mol) = 20.8 J/K∙mol

In this case, ΔSuniverse is positive; therefore, the reaction is spontaneous at

1000°C.

−(177.8 kJ/mol) 473 K

−ΔHsys

T

Worked Example 18.4 (cont.)

Solution (d) S°[Na(s)] = 51.05 J/K∙mol, S°[Na(l)] = 57.56 J/K∙mol;

ΔHf°[Na(s)] = 0 kJ/mol, ΔHf°[Na(l)] = 2.41 kJ/mol.

ΔS°_rxn = S°[Na(l)] – S°[Na(s)]

= 57.56 J/K∙mol – 51.05 J/K∙mol = 6.51 J/K∙mol

ΔH°_rxn = ΔHf°[Na(l)] – ΔHf°[Na(s)]

= 2.41 kJ/mol – 0 kJ/mol = 2.41 kJ/mol

ΔSsurr = = = −0.0650 kJ/K∙mol = −6.50 J/K∙mol

ΔSuniverse = ΔSsys + ΔSsurr = 6.51 J/K∙mol − 6.50 J/K∙mol = 0.01 J/K∙mol ≈ 0

ΔSuniverse is zero; therefore, the reaction is an equilibrium process at 98°C. In

fact, this is the melting point of sodium. −(2.41 kJ/mol)

371 K

−ΔHrxn

T

Think About It Remember that standard enthalpies of formation have units of kJ/mol, whereas standard absolute entropies have units of J/K∙mol. Make sure that you convert kilojoules to joules, or vice versa, before combining the terms.

Entropy Changes in the Universe

The

third law of thermodynamics

states that the entropy of a perfect

crystalline substance is

zero

at absolute zero.

Entropy increases in a

substance as temperature

increases from absolute

zero.

Predicting Spontaneity

Measurements on the surroundings are seldom made, limiting the use

of the second law of thermodynamics.

Gibbs free energy (G)

or simply

free energy

can be used to express

spontaneity more directly.

The change in free energy for a system is:

18.5

G

=

H – TS

(6)

Predicting Spontaneity

Using the Gibbs free energy, it is possible to make predictions on

spontaneity.

Δ

G

< 0 The reaction is spontaneous in the forward direction.

Δ

G

> 0 The reaction is nonspontaneous in the forward direction.

Δ

G

= 0 The system is at equilibrium

Δ

G

=

Δ

H

– TΔ

S

Worked Example 18.5

Strategy The temperature that divides high from low is the temperature at

which ΔH = TΔS (ΔG = 0). Therefore, we use ΔG =ΔH – TΔS, substituting 0 for

ΔG and solving for T to determine temperature in kelvins; we then convert to

degrees Celsius.

According to Table 18.4, a reaction will be spontaneous only at high temperatures

if both ΔH and ΔS are positive. For a reaction in which ΔH = 199.5 kJ/mol and ΔS

= 476 J/K∙mol, determine the temperature (in °C) above which the reaction is

spontaneous.

Solution

ΔS = _K∙mol476 J _{1000 J}1 kJ = 0.476 kJ/K∙mol

T = ΔH

ΔS = 419 _K

199.5 kJ/mol 0.476 kJ/K∙mol =

= (419 – 273) = 146°C

Think About It Spontaneity is favored by a release of energy (ΔH

being negative) and by an increase in entropy (ΔS being positive).

When both quantities are positive, as in this case, only the entropy change favor spontaneity. For an endothermic process such as this, which requires the input of heat, it should make sense that adding more heat by increasing the temperature will shift the equilibrium to the right, thus making it “more spontaneous.”

Predicting Spontaneity

The

standard free energy of reaction (ΔG

°

_rxn

)

is free-energy

change for a reaction when it occurs under standard-state conditions.

The following conditions define the standard states of pure

substances and solutions are:

 Gases 1 atm pressure

 Liquids pure liquid

 Solids pure solid

 Elements the most stable allotropic form at 1 atm and 25

°

C

 Solutions 1 molar concentration

For a chemical reaction

a

A +

b

B →

c

C +

d

D

Alternatively,

Δ

G

°

f for any element in its most stable allotropic form at 1 atm is

defined as zero.

ΔG°rxn = [cΔG°f (C) + dΔG°f (D)] – [aΔG°f (A) +

bΔG°_f (B)]

ΔG°_rxn = ΣnΔG°_f (products) – ΣmΔG°_f (reactants)

Worked Example 18.6

Strategy Look up the ΔG°_f values for the reactants and products in each

equation, and use ΔG°_rxn_{= Σ}nΔG°_f_{(products) – Σ}mΔG°_f_{(reactants) to solve}

for ΔG°_rxn.

Calculate the standard free-energy changes for the following reactions at 25°C:

(a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

(b) 2MgO(s) → 2Mg(s) + O2(g)

Solution From Appendix 2, we have the following values: ΔG°f [CH4(g)] =

−50.8 kJ/mol, ΔG°_f [CO2(g)] = −394.4 kJ/mol, ΔG°_f [H2O(l)] = −237.2 kJ/mol,

and ΔG°f [MgO(s)] = −569.6 kJ/mol. All the other substances are elements in

their standard states and have, by definition, ΔG°_f = 0.

(a) ΔG°_rxn = (ΔG°_f [CO2(g)] + 2ΔG°_f [H2O(l)]) – (ΔG°_f [CH4(g)] + ΔG°_f

[O2(g)])

= [(−394.4 kJ/mol) + (2)(−237.2 kJ/mol)] − [(−50.8 kJ/mol) + (2)(0 kJ/mol)]

Worked Example 18.6 (cont.)

Solution From Appendix 2, we have the following values: ΔG°f [CH4(g)] =

−50.8 kJ/mol, ΔG°_f [CO2(g)] = −394.4 kJ/mol, ΔG°_f [H2O(l)] = −237.2 kJ/mol,

and ΔG°f [MgO(s)] = −596.6 kJ/mol. All the other substances are elements in

their standard states and have, by definition, ΔG°_f = 0.

(b) ΔG°rxn = (2ΔG°f [Mg(s)] + ΔG°f [O2(g)]) – (2ΔG°f [MgO(s)])

= [(2)(0 kJ/mol) + (0 kJ/mol)] − [(2)(−569.6 kJ/mol)] = 1139 kJ/mol

Think About It Note that, like standard enthalpies of formation (ΔH°f ),

standard free energies of formation (ΔG°_f ) depend on the state of matter. Using

water as an example, ΔG°f [H2O(l)] = −237.2 kJ/mol and ΔG°f [H2O(g)] =

(7)

Worked Example 18.7

Strategy The solid-liquid transition at the melting point and the liquid-vapor

transition at the boiling points are equilibrium processes. Therefore, because ΔG

is zero at equilibrium, in each case we can use ΔG = ΔH – TΔS, substituting 0 for

ΔG and solving for ΔS, to determine the entropy change associated with the

process.

The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid-to-liquid and

liquid-to-vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and

boils at 80.1°C.

Solution

= 0.0391 kJ/K∙mol or 39.1 J/K∙mol

ΔSfus = _TΔHfus 10.9 kJ/mol _{278.7 K}

melting =

= 0.0877 kJ/K∙mol or 87.7 J/K∙mol

ΔSvap = _TΔHvap 31.0 kJ/mol _{353.3 K}

boiling =

Think About It For the same substance, ΔSvap is always

significantly larger than ΔSfus. The change in number of

arrangements is always bigger in a liquid-to-gas transition than in a solid-to-liquid transition.

Free Energy and Chemical Equilibrium

It is the sign of Δ

G

(not Δ

G

°

) that determines spontaneity.

The relationship between Δ

G

and Δ

G

°

is:

R

is the gas constant (8.314 J/K·mol).

T

is the kelvin temperature.

Q

is the reaction quotient.

18.6

Δ

G

=

Δ

G

°

+

RT

ln

Q

Consider the following equilibrium:

H2(

g

) + I2(

g

)

⇌

2HI(

g

)

Δ

G

°

at 25

°

C = 2.60 kJ/mol

Δ

G

depends on the partial pressures of each chemical species.

If

P

_H2

= 2.0 atm;

P

_I2

= 2.0 atm; and

P

HI = 3.0 atm:

Then:

Free Energy and Chemical Equilibrium

 

  

H2HI I2

  

 

2 2

3.0 9.0 2.25 2.0 2.0 4.0 P P Q P P    







3 2.60 kJ 8.314 10 kJ

298 K ln2.25 4.60 kJ/mol

mol K mol



  

    

 

G

The spontaneity can be manipulated by changing the partial

pressures of the reaction components:

H2(

g

) + I2(

g

)

⇌

2HI(

g

)

Δ

G

°

at 25

°

C = 2.60 kJ/mol

If

P

_H2

= 2.0 atm;

P

_I2

= 2.0 atm; and

P

HI = 1.0 atm:

Then:

 

  

H2HI I2

  

 

2 2

1.0 1.0 0.25 2.0 2.0 4.0 P P Q P P    







·             3 2.60 kJ 8.314 10 kJ

298 K ln0.25 0.8 kJ/mol

mol K mol

G

Worked Example 18.8

Strategy Use the partial pressures of N2O4 and NO2 to calculate the reaction

quotient QP, and then use ΔG =ΔG° + RT lnQ to calculate ΔG.

The reaction quotient expression is

The equilibrium constant, KP, for the reaction

N2O4(g) ⇌ 2NO2(g)

is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.4

kJ/mol. In a certain experiment, the initial pressures are PN2O4 = 0.453 atm and

PNO2 = 0.122 atm. Calculate ΔGfor the reaction at these pressures, and predict

the direction in which the reaction will proceed spontaneously to establish equilibrium.

Q_P =(PNO2)2 (0.122)_0.4532

PN2O4 = = 0.0329

Worked Example 18.8 (cont.)

Solution

Because ΔG is negative, the reaction proceeds spontaneously from left to right to

reach equilibrium.

(298 K)(ln 0.0329)

= 5.4 kJ _mol + 8.314_K∙mol×10-3 kJ

= 5.4 kJ/mol – 8.46 kJ/mol

ΔG =ΔG° + RT lnQ

= –3.1 kJ/mol

Think About It Remember, a reaction with a positive ΔG° value can be spontaneous if the starting concentrations of reactants and products are such that

(8)

At equilibrium, Δ

G

= 0 and

Q

= K:

0 = Δ

G

°

+

RT

ln

K

Δ

G

°

= –RT

ln

K

At equilibrium, Δ

G

= 0 and

Q

= K:

0 = Δ

G

°

+

RT

ln

K

Δ

G

°

= –RT

ln

K

At equilibrium, Δ

G

= 0 and

Q

= K:

0 = Δ

G

°

+

RT

ln

K

Δ

G

°

= –RT

ln

K

Worked Example 18.9

Strategy Use data from Appendix 2 and ΔG°rxn = ΣnΔG°f (products) – ΣmΔG°f

(reactants) to calculate ΔG° for the reaction. Then use ΔG° = −RT lnK to solve for

KP.

Using data from Appendix 2, calculate the equilibrium constant, KP, for the

following reaction at 25°C:

2H2O(l) ⇌ 2H2(g) + O2(g)

Solution

ΔG° = (2ΔG°f [H2(g)] + ΔG°f [O2(g)]) – (2ΔG°f [H2O(l)])

= [2(0 kJ/mol) + (0 kJ/mol)] − [(2)(−237.2 kJ/mol)] = 474.4 kJ/mol

474.4 kJ

mol = 8.314×10 (298 K) ln KP

-3_kJ

K∙mol

−191.5 = ln KP

KP = e−191.5 = 7×10-84

ΔG° = −RT ln KP

Think About It This is an extremely small equilibrium constant,

which is consistent with the large, positive value of ΔG°. We know

from everyday experience that water does not decompose

spontaneously into its constituent elements at 25°C.

Worked Example 18.10

Strategy Use ΔG° = −RT lnK to calculate ΔG°.

The equilibrium constant, Ksp, for the dissolution of silver chloride in water at

25°C:

AgCl(s) ⇌ Ag+₍_aq_{) + Cl}-₍_aq₎

is 1.6×10-10_{. Calculate Δ}_G° for the process.

Solution

R = 8.314×10-3_{kJ/K∙mol and}T = (25 + 273) = 298 K.

(298 K) ln (1.6×10-10₎

= 8.314_K∙mol×10-3 kJ

= 55.9 kJ/mol

ΔG° = −RT ln Ksp

Think About It The relatively large, positive ΔG°, like the very small K value,

corresponds to a process that lies very far to the left. Note that the K in

ΔG° = −RT lnK can be any type of Kc (Ka, Kb, Ksp, etc.) or KP.

Thermodynamics of Living Systems

Many biological reactions have positive Δ

G

°

value, making the

reaction nonspontaneous.

Sone spontaneous reactions can be coupled with spontaneous

reactions in order to drive a process forward:

alanine + glycine → alanylglycine Δ

G

°

= 29 kJ/mol

ATP + H2O → ADP + H3PO4 Δ

G

°

= –31 kJ/mol

ATP + H2O + alanine + glycine → ADP + H3PO4 + alanylglycine

Δ

G

°

= 29 kJ/mol + –31 kJ/mol = –2 kJ/mol

(9)

Thermodynamics of Living Systems

Many biological reactions have positive Δ

G

°

value, making the

reaction nonspontaneous.

Key Concepts

18

A Qualitative Description of Entropy

A Quantitative Description of Entropy

Calculating Δ

s

sys

Standard

S

°

Qualitatively Predicting Δ

S

°

sys

Calculating Δ

s

surr

The Second Law of Thermodynamics

The Third Law of Thermodynamics

Gibbs Free-Energy Change, Δ

G

Standard Free-Energy Changes, Δ

G

°

Using Δ

G

and Δ

G

°

to Solve Problems

Relationship Between Δ

G

and Δ

G

°