16 Spontaneity.pdf

CHAPTER 16:

Changes in Energy Associated with Physical

Changes and Chemical Reactions

The

system

is a part of the universe that is of specific interest.

The

surroundings

constitute the rest of the universe outside the system.

The

system

is usually defined as the

substances involved

in chemical and physical changes

.

System

Surroundings

Thermochemistry

is the study of

heat

(the transfer

of thermal energy) in chemical reactions.

Heat

(aka

thermal energy

) is energy that flows

An

exothermic process occurs when

heat

is

transferred

from

the

system

to

the

surroundings

.

“Feels hot!”

Surroundings

Universe = System + Surroundings

heat

System

An

endothermic

process occurs when

heat

is

transferred

from

the

surroundings

to

the

system

.

“Feels cold”

energy + 2HgO(s) 2Hg(l) + O2(g)

Surroundings

Universe = System + Surroundings

heat

Units for Measuring Energy,

calories and joules

A

calorie

(

cal

) is the amount of energy needed to raise

the temperature of 1 g of water by 1°C.

The SI unit of energy is

joule

(

J

).

Both joules and calories can be reported in the larger

units kilojoules (kJ) and kilocalories (kcal)

1 cal = 4.184 J

1,000

J = 1 kJ

1 Cal = 1,000 cal = 1 kcal

Three Types of Systems

Open system:

Can

exchange mass and energy

with the surroundings.

Closed system:

Allows the

transfer of energy

but not mass.

Isolated system:

State functions

are properties that are

determined by the state of the system,

regardless of how that condition was achieved.

(State functions are indicated with a capital letter.)

The magnitude of change depends only on the

initial

and

final

states

of the system.



Energy,

D

E



Pressure,

D

P



Volume,

D

V



Temperature,

D

T

1

st

Law of Thermodynamics

Energy is conserved

1

st

Law of Thermodynamics

How much energy is involved in the change?

Does energy flow into or out of the system?

Spontaneity

•

In chemistry, a

spontaneous process

is one that, once

started, proceeds on its own without continuous

external influence.

•

A

nonspontaneous process

takes place only in the

Which processes are spontaneous?

Which are nonspontaneous?

a)

Diffusion of perfume molecules from one side of a

room to the other.

b)

Heat flow from a cold object to a hot object.

c)

Decomposition of rust (Fe

O

•H

O) to iron metal,

Spontaneity

•

Spontaneous doesn’t mean the same thing as fast.

A spontaneous reaction can be either fast or slow.

•

The gradual rusting of iron metal is a slow spontaneous reaction.

•

Thermodynamics tells us where the reaction is headed, but it

Processes that result in a decrease in the energy of a

system are often spontaneous.

The

Entropy (

S

)

of a system is a measure of

molecular randomness or disorder of a system.

S

=

k

ln

W

k

= Boltzmann constant (1.38 x 10

J/k)

W

= number of microstates (energetically

Microstates (Possible Arrangements)

1 microstate

2 microstates

Microstates (Possible Arrangements)

1 microstate

The

entropy (

S

) of a system is a measure of

molecular randomness or disorder of a system.

S

=

k

ln

W

k

= Boltzmann constant (1.38 x 10

J/k)

W

=

number of microstates (energetically equivalent

different ways the molecules in a system can be

arranged

W

=

X

N

X

= volume consisting of X cells

Entropy: The most probable state is the one with the

largest number of microstates.

Entropy Change

D

S

=

S

-

S

D

S

>

0

Entropy Changes for Ionic Compounds

Dissolved in Water

D

S

= +

43

J

K

×

mol

Predict the sign of

D

S

(Qualitative Predictions) in the

system for each of the following processes.

a)

b)

c)

When the temperature of a system is increased, the

energy of the system’s molecules increases and the

3

rd

Law of Thermodynamics

2

nd

Law of Thermodynamics

In any

spontaneous

process there is

always an increase in the entropy of the

universe.

D

S

= D

S

+D

S

Spontaneity

D

S

>

0

D

S

<

0

D

S

= D

S

+D

S

Spontaneous Process

An ice cube spontaneously melts in a room

where the temperature is 25°C.

There is no temperature change

during the phase change.

D

S

>

0

D

S

<

0

Spontaneous Process

A cup of hot water spontaneously cools to room temperature.

D

S

<

0

D

S

>

0

Spontaneous Process

Hydrogen peroxide produces water and oxygen gas.

D

S

>

0

D

S

>

0

D

S

= D

S

+D

S

>

0

NONspontaneous Process

Any process in which

D

S

and

D

S

are both

negative is a

nonspontaneous

process.

D

S

<

0

D

S

<

0

Calculating

D

S

sys

, J/K•mol

There are tables of standard molar entropies

(Table 18.2, p.767, and Appendix 2 in your textbook).

Because

S

° values are quoted on a per-mole basis, the

S

°

value for each substance must be multiplied by the

stoichiometric coefficient of that substance

in the balanced equation.

Calculate the standard entropy change,

D

S°

Compound S° [J/(Kmol)]

N₂O₄ 304.3 NO₂ 240.0

D

S

° =

2

´

S

°

(

NO

)

-

S

°

(

N

O

)

=

(

2

mol

)

240.0

J

K

×

mol

æ

è

ç

ö

ø

÷-

(

1

mol

)

304.3

J

K

×

mol

æ

è

ç

ö

ø

÷

Calculating

D

S

surr

D

S

µ

-D

H

D

S

µ

1

T

Predicting Spontaneity:

Gibbs Free-Energy Change,

D

G

D

S

=

-D

H

T

For a spontaneous process,

D

S

= D

S

+D

S

>

0

D

S

= D

S

+

-D

H

T

æ

è

ç

ö

ø

÷ >

0

Gibbs Free-Energy Change,

D

G

T

D

S

=

T

D

S

- D

H

>

0

This equation can be rearranged to give

-

T

D

S

= D

H

-

T

D

S

<

0

This equation can predict whether or not a process is

spontaneous in terms of only the

system

.

Gibbs Free-Energy Change,

D

G

-

T

D

S

= D

H

-

T

D

S

<

0

D

G

is called the Gibbs free energy.

-

T

D

S

= D

G

Spontaneity

D

G

<

0

the reaction is spontaneous

D

G

>

0

the reaction is nonspontaneous

D

G

=

0

the reaction mixture is at equilibrium

Signs of Enthapy, Entropy, and Free-Energy

D

H

D

S

D

G

Reaction Spontaneity

–

+

–

Spontaneous at all temperatures

–

–

– or +

Spontaneous at low temps where

D

H

outweighs

T

D

S

Nonspontaneous at high temps

where

T

D

S

outweighs

D

H

+

–

+

Nonspontaneous at all

temperatures

+

+

– or +

Spontaneous at high temps where

T

D

S

outweighs

D

H

Versions of Free Energy

The

standard free energy change

of a reaction is

defined as the change in free energy when all the

reactants and products are in their standard states

(1 atm pressure, 298 K, 1

M

concentration).

The

standard free energy of formation

is the free

energy change for the reaction that forms 1 mol of the

compound from its component elements, with all

substances in their standard states.

D

G

Three Ways to Calculate

D

G

°

Using a Hess’s law approach

D

G

° = D

H

°

-

T

D

S

°

D

G

° = D

G

products

(

)

-D

G

reactants

Example Problem #1

Iron metal is produced commercially by reducing iron(III) oxide in

iron ore with carbon monoxide:

Compound H°f (kJ/mol) S° [J/(Kmol)]

Fe₂O₃(s) –824.2 87.4 CO(g) –110.5 197.6

Fe(s) 0 27.3

CO₂(g) –393.5 213.6

Calculate the standard free energy change for this reaction at

25

°

C.

Example Problem #1

D

H

° = D

H

products

(

)

-D

H

reactants

(

)

=

éë

( )

2

( )

0

+

( )

3

(

-

393.5

)

ùû-

éë

( )

1

(

-

824.2

)

+

( )

3

(

-

110.5

)

ùû

D

H

° = -

24.8

kJ

Compound H°f (kJ/mol) S° [J/(Kmol)]

Fe₂O₃(s) –824.2 87.4 CO(g) –110.5 197.6

Fe(s) 0 27.3

Example Problem #1

D

S

° =

S

°

(

)

-

S

°

(

)

=

éë

( )

2

(

27.3

)

+

( )

3

(

213.6

)

ùû-

éë

( )

1

(

87.4

)

+

( )

3

(

197.6

)

ùû

D

S

° =

15.2

J

K

=

0.0152

kJ

K

Compound H°f (kJ/mol) S° [J/(Kmol)]

Fe₂O₃(s) –824.2 87.4

CO(g) –110.5 197.6

Fe(s) 0 27.3

Example Problem #1

= -

(

24.8

)

-

( )

298

(

0.0152

)

Example Problem #2

Iron metal is produced commercially by reducing iron(III) oxide in

iron ore with carbon monoxide:

Calculate the standard free energy change for this reaction at

25

°

C.

Compound G°f (kJ/mol)

Fe₂O₃(s) –742.2

CO(g) –137.2

Fe(s) 0 CO₂(g) –394.4

D

G

° = D

G

products

(

)

-D

G

reactants

Example Problem #2

Compound G°f (kJ/mol)

Fe₂O₃(s) –742.2

CO(g) –137.2

Fe(s) 0 CO₂(g) –394.4

=

éë

( )

2

( )

0

+

( )

3

(

-

394.4

)

ùû-

éë

( )

1

(

-

742.2

)

+

( )

3

(

-

137.2

)

ùû

D

G

° = -

29.4

kJ

D

G

° = D

G

products

(

)

-D

G

reactants

Example Problem #3

Iron metal is produced commercially by reducing iron(III) oxide in

iron ore with carbon monoxide:

Example Problem #3

Multiply the first equation (and its

D

G

°) by ½.

Example Problem #3

D

G

° =

( )

0.5

(

1484.4

)

+

( )

1.5

(

-

514.4

)

Is a reaction spontaneous

under standard state conditions at 25°C?

Iron metal is produced commercially by reducing iron(III) oxide in

iron ore with carbon monoxide:

Because

D

G

° is negative,

Does the reverse reaction become spontaneous

at higher temperatures?

Iron metal is produced commercially by reducing iron(III) oxide in

iron ore with carbon monoxide:

Because

D

H

° is negative

and

D

S

° is positive

,

D

G

° will be negative at all temperatures

.

The forward reaction is spontaneous at all temperatures

and the reverse reaction does not become spontaneous

D

G

°

Caveat

The sign of D

G

° tells us the direction

of spontaneous reaction when both reactants and

products are present at

standard state

conditions.

In actual reactions, however, the composition of the

reaction mixture seldom corresponds to

standard state pressures and concentrations.

It is the Sign of

D

G

(not

D

G°)

that Determines Spontaneity

The free energy change,

D

G

, for a reaction

when the reactants and products are present at nonstandard

state pressures and concentrations is given by

D

G

= D

G

°+

RT

ln

Q

D

G

°

is the free energy change under standard state

conditions

R is the gas constant (8.314 J/mol



K)

T is the Kelvin temperature

Consider the following equilibrium:

H

(

g

) + I

(

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

= 2.0 atm;

P

= 2.0 atm; and

P

= 3.0 atm:

Then:

Free Energy and Chemical Equilibrium

 

  

  

 

2 2

2 2

3.0 _9.0

2.25 2.0 2.0 4.0

P P Q P P    







2.60 kJ

8.314 10 kJ

298 K ln2.25

4.60 kJ/mol

mol

K mol









D 













The spontaneity can be manipulated by changing the partial

pressures of the reaction components:

H

(

g

) + I

(

g

)

⇌

2HI(

g

)

Δ

G

°

at 25

°

C = 2.60 kJ/mol

If

P

= 2.0 atm;

P

= 2.0 atm; and

P

= 1.0 atm:

Then:

Free Energy and Chemical Equilibrium

 

  

  

 

2 2

2 2

1.0 _1.0

0.25 2.0 2.0 4.0

P P Q P P    







·







D 







 





3

2.60 kJ

8.314 10 kJ

298 K ln0.25

0.8 kJ/mol

mol

K mol

Free Energy and Equilibrium

D

G

= D

G

°+

RT

ln

Q

For reactions that involve both gases and solutes in solution,

the reaction quotient contains partial pressures of gases in

atmospheres and molar concentrations of solutes.

D

G

= D

G

°+

RT

ln

( )

P

2

P

2

( )

P

2

( )

Calculate the free energy change for ammonia synthesis at 25°C given

the following sets of partial pressures:

a)

1.0 atm N

, 3.0 atm H

, 0.020 atm NH

D

G

= D

G

°+

RT

ln

Q

Calculate the free energy change for ammonia synthesis at 25°C given

the following sets of partial pressures:

a)

1.0 atm N

, 3.0 atm H

, 0.020 atm NH

Compound G°f (kJ/mol)

N₂(g) 0 H₂(g) 0 NH₃(g) –16.5

=

éë

( )

2

(

-

16.5

)

ùû-

éë

( )

1

( )

0

+

( )

3

( )

0

ùû=-

33.0

kJ

D

G

° = -

3.30

´

10

J

D

G

° = D

G

products

(

)

-D

G

reactants

Calculate the free energy change for ammonia synthesis at 25°C given

the following sets of partial pressures:

a)

1.0 atm N

, 3.0 atm H

, 0.020 atm NH

D

G

= D

G

°+

RT

ln

( )

P

2

P

2

( )

P

2

( )

Calculate the free energy change for ammonia synthesis at 25°C given

the following sets of partial pressures:

a)

1.0 atm N

, 3.0 atm H

, 0.020 atm NH

D

G

= -

3.30

´

10

J

+

(

8.314

J

mol

×

K

)

(

298

K

)

ln

(

0.020

)

2

1.0

( )

( )

3.0

æ

è

ç

ç

ö

ø

÷

÷

Calculate the free energy change for ammonia synthesis at 25°C given

the following sets of partial pressures:

b)

0.010 atm N

, 0.030 atm H

, 2.0 atm NH

D

G

= -

3.30

´

10

J

+

(

8.314

J

mol

×

K

)

(

298

K

)

ln

( )

2.0

2

0.010

(

)

(

0.030

)

æ

è

ç

ç

ö

ø

÷

÷

Free Energy and Equilibrium

D

G

<

0

Q

<<

1

RT

ln

Q

<<

₀

When a reaction mixture is mostly

reactants

,

D

G

>

0

Q

>>

1

RT

ln

Q

>>

₀

Free Energy and Equilibrium

Free Energy and Equilibrium

At equilibrium, D

G

= 0

D

G

= D

G

°+

RT

ln

Q

0

= D

G

°+

RT

ln

K

D

G

° = -

RT

ln

K

Free Energy and Equilibrium

D

G

° = -

RT

ln

K

D

G°

ln

K

K

Comment

< 0

> 0

> 1

The equilibrium mixture is

mainly products.

> 0

< 0

< 1

The equilibrium mixture is

mainly reactants.

= 0

= 0

= 1

The equilibrium mixture

Example Problem

Methanol, an important alcohol used in the manufacture of

adhesives, fibers, and plastics, is synthesized by the reaction

Calculate the equilibrium constant for this reaction at 25

°

C.

D

G

° = -

RT

ln

K

Start by calculating

D

G

°

Example Problem

D

G

° = -

RT

ln

K

D

G

° = -

25.1

kJ

ln

K

= -D

G

°

RT

=

- -

(

2.51

´

10

)

8.314

(

)

( )

298

=

10.1

K

=

K

=

e

=

Temperature Dependence of

K

D

G

° = -

RT

ln

K

= D

H

°-

T

D

S

°

ln

K

= - D

H

°

RT

+ D

S

°

R

ln

K

= - D

H

°

R

1

T

æ

è

ç

ö

ø

÷+

D

S

°

R

Temperature Dependence of

K

Example Problem

The equilibrium constant K of a reaction was plotted versus 1/T

and the following was obtained.

What is

D

H

°

and

D

S

°

of the reaction?

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00

0.0005 0.0006 0.0007 0.0008 0.0009 0.001 0.0011

ln

K

Temperature Dependence of

K

Example Problem

The equilibrium constant K of a reaction was plotted versus 1/T

and the following was obtained.

What is

D

H

°

and

D

S

°

of the reaction?

ln

K

= - D

H

°

R

1

T

æ

è

ç

ö

ø

÷+

D

S

°

R

slope

= - D

H

°

R

intercept

= D

S

°

R

Temperature Dependence of

K

Example Problem

Excel gives: slope = 1.07 x 10

intercept = –5.83

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00

0.0005 0.0006 0.0007 0.0008 0.0009 0.001 0.0011

ln

K

1/T

D

H

° = -

(

1.07

´

10

)

(

8.314

)

= -

8.90

´

10

J

= -

89.0

kJ

mol

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.

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

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

Worked Example 18.4

Strategy For each process, use ΔS°_rxn = ΣnS°(products) – ΣmS°(reactants) to determine ΔS°_sys; ΔS_surr = (–ΔH_sys/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, ΔS_rxn and

ΔS_sys 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

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; ΔH_f°[H₂(g)] = 0 kJ/mol, ΔH_f°[I₂(g)] = 62.25 kJ/mol, ΔH_f°[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

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

ΔS_universe = ΔS_sys + ΔS_surr = 21.03 J/K∙mol + 38.5 J/K∙mol = 59.5 J/K∙mol

ΔS_universe is positive; therefore the reaction is spontaneous at 0°C. −(−10.5 kJ/mol)

273 K −ΔH_rxn

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

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

ΔS_universe = ΔS_sys + ΔS_surr = 160.5 J/K∙mol + (−376 J/K∙mol) = − 216 J/K∙mol

Δs_universe is negative, therefore the reaction is nonspontaneous at 200°C. −(177.8 kJ/mol)

473 K −ΔH_sys

Worked Example 18.4 (cont.)

Solution (c) T = 1000°C and

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

ΔS_universe = ΔS_sys + ΔS_surr = 160.5 J/K∙mol + (−139.7 J/K∙mol) = 20.8 J/K∙mol

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

1000°C.

−(177.8 kJ/mol) 473 K

Worked Example 18.4 (cont.)

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

ΔH_f°[Na(s)] = 0 kJ/mol, ΔH_f°[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 = ΔH_f°[Na(l)] – ΔH_f°[Na(s)]

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

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

ΔS_universe = ΔS_sys + ΔS_surr = 6.51 J/K∙mol − 6.50 J/K∙mol = 0.01 J/K∙mol ≈ 0

ΔS_universe 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

−ΔH_rxn T

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 = 476 J

K∙mol = 0.476 kJ/K∙mol

1 kJ 1000 J

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

Worked Example 18.6

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

equation, and use ΔG°_rxn = ΣnΔG°_f (products) – ΣmΔG°_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)])

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 ),

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

ΔS_fus = 10.9 kJ/mol 278.7 K ΔH_fus

T_melting =

= 0.0877 kJ/K∙mol or 87.7 J/K∙mol

ΔS_vap = 31.0 kJ/mol 353.3 K ΔH_vap

T_boiling =

Think About It For the same substance, ΔS_vap is always significantly larger than ΔS_fus. The change in number of

Worked Example 18.8

Strategy Use the partial pressures of N₂O₄ and NO₂ to calculate the reaction quotient Q_P, and then use ΔG = ΔG° + RT lnQ to calculate ΔG.

The reaction quotient expression is

The equilibrium constant, K_P, 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 P_N2O4 = 0.453 atm and

P_NO2 = 0.122 atm. Calculate ΔG for the reaction at these pressures, and predict

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

Q_P = (0.122)

2

0.453 (P_NO2)2

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)

= 8.314×10

-3 _kJ

K∙mol 5.4 kJ

mol +

= 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

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

K_P.

Using data from Appendix 2, calculate the equilibrium constant, K_P, 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 = (298 K) ln KP

8.314×10-3 _kJ

K∙mol

−191.5 = ln K_P

K_P = e−191.5 = 7×10-84

ΔG° = −RT ln K_P

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

Worked Example 18.10

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

The equilibrium constant, K_sp, for the dissolution of silver chloride in water at 25°C:

AgCl(s) ⇌ Ag+_{(aq) + Cl}-_(aq)

is 1.6×10-10_{. Calculate ΔG}° _{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₎

= 8.314×10

-3 _kJ

K∙mol

= 55.9 kJ/mol

ΔG° = −RT ln K_sp

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