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and from Eq. 14.53 follows

A¿

ref

A¿

(14.55)

The equilibrium constants K

and K

can be determined from Table A-27 or a

similar compilation. The mole fractions appearing in these expressions must be eval-

uated by considering all the substances present within the system, including the inert

substance E. Each mole fraction has the form y

5 n

/n, where n

is the amount of

component i in the equilibrium mixture and

n 5 n

(14.56)

The n’s appearing in Eq. 14.56 can be expressed in terms of two unknown variables

through application of the conservation of mass principle to the various chemical

species present. Accordingly, for a specified temperature and pressure, Eqs. 14.54 and

14.55 give two equations in two unknowns. The composition of the system at equilib-

rium can be determined by solving these equations simultaneously. This procedure is

illustrated by Example 14.9.

The procedure discussed in this section can be extended to systems involving sev-

eral simultaneous independent reactions. The number of simultaneous equilibrium

constant expressions that results equals the number of independent reactions. As

these equations are nonlinear and require simultaneous solution, the use of a com-

puter is usually required.

c c c c EXAMPLE 14.9 c

As a result of heating, a system consisting initially of 1 kmol of CO

kmol of O

, and

kmol of N

forms an

equilibrium mixture of CO

, CO, O

, N

, and NO at 3000 K, 1 atm. Determine the composition of the equilibrium

mixture.

SOLUTION

Known:

A system consisting of specified amounts of CO

, O

, and N

is heated to 3000 K, 1 atm, forming an

equilibrium mixture of CO

, CO, O

, N

, and NO.

Find: Determine the equilibrium composition.

Engineering Model: The final mixture is an equilibrium mixture of ideal gases.

Analysis: The overall reaction has the form

➊ 1CO

aCO 1 bNO 1 c CO

1 d O

1 eN

Applying conservation of mass to carbon, oxygen, and nitrogen, the five unknown coefficients can be expressed

in terms of any two of the coefficients. Selecting a and b as the unknowns, the following balanced equation

results:

1CO

S aCO 1 bNO 1

1 2 a

1 1 a 2 b

1 2 b

The total number of moles n in the mixture formed by the products is

n 5 a 1 b 1 11 2 a21

11 1 a 2 b21

11 2 b25

1 a

At equilibrium, two independent reactions relate the components of the product mixture:

1. CO

CO 1

Considering Equilibrium with Simultaneous Reactions

14.4 Further Examples of the Use of the Equilibrium Constant 873

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874 Chapter 14 Chemical and Phase Equilibrium

Phase Equilibrium

In this part of the chapter the equilibrium condition dG

5 0 introduced in Sec. 14.1

is used to study the equilibrium of multicomponent, multiphase, nonreacting systems.

The discussion begins with the elementary case of equilibrium between two phases of

a pure substance and then turns to the general case of several components present in

several phases.

Determine the mole fractions of the components of the equi-

librium mixture. Ans. y

5 0.171, y

5 0.031, y

5 0.28

, y

5 0.299,

5 0.213.

For the first of these reactions, the form taken by the equilibrium constant when p 5 1 atm is

11 1 a 2 b24

1 2 a

4 1 a

111

221

1 2 a

1 1 a 2 b

4 1 a

Similarly, the equilibrium constant for the second of the reactions is

1 1 a 2 b

1 2 b

14 1 a2

121

221

1 1 a 2 b

1 2 b

At 3000 K, Table A-27 provides lo

520.485 and lo

520.913, giving K

5 0.3273 and K

5 0.1222.

Accordingly, the two equations that must be solved simultaneously for the two

unknowns a and b are

0.3273 5

1 2 a

1 1 a 2 b

4 1 a



0.1222 5

1 1 a 2 b

1 2 b

The solution is a 5 0.3745, b 5 0.0675, as can be verified. The composition of

the equilibrium mixture, in kmol per kmol of CO

present initially, is then

0.3745CO, 0.0675NO, 0.6255CO

, 0.6535O

, 0.4663N

➊ If high enough temperatures are attained, nitrogen can combine with oxy-

gen to form components such as nitric oxide. Even trace amounts of oxides

of nitrogen in products of combustion can be a source of air pollution.

Ability to…

❑

apply Eq. 14.35 to deter-

mine equilibrium composition

given temperature and

pressure for two simultane-

ous equilibrium reactions.

❑

retrieve and use data from

Table A-27.

✓

Skills Developed

14.5 Equilibrium between Two Phases

of a Pure Substance

Consider the case of a system consisting of two phases of a pure substance at equi-

librium. Since the system is at equilibrium, each phase is at the same temperature

and pressure. The Gibbs function for the system is

G 5 n¿g¿

T, p

1 n–g–

T, p

(14.57)

where the primes 9 and 0 denote phases 1 and 2, respectively.

Forming the differential of G at fixed T and p

dG4

5 g¿ dn¿ 1 g– dn– (14.58)

Since the total amount of the pure substance remains constant, an increase in the

amount present in one of the phases must be compensated by an equivalent decrease

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in the amount present in the other phase. Thus, we have dn0 5 2dn9, and Eq. 14.58

becomes

dG4

5 1g¿ 2 g–2 dn¿

At equilibrium, dG

5 0, so

¿ 5

– (14.59)

At equilibrium, the molar Gibbs functions of the phases are equal.

CLAPEYRON EQUATION. Equation 14.59 can be used to derive the Clapeyron

equation, obtained by other means in Sec. 11.4. For two phases at equilibrium, varia-

tions in pressure are uniquely related to variations in temperature: p 5 p

sat

; thus,

differentiation of Eq. 14.59 with respect to temperature gives

g¿

sat

g–

sat

With Eqs. 11.30 and 11.31, this becomes

2s¿ 1 y ¿

sat

52s– 1 y–

sat

Or on rearrangement

sat

s– 2 s¿

– 2

This can be expressed alternatively by noting that, with

5 h 2 T s, Eq. 14.59

becomes

h¿ 2 T s¿ 5 h– 2 T s–

s– 2 s¿ 5

h– 2 h¿

(14.60)

Combining results, the Clapeyron equation is obtained

sat

– 2

y–

y¿

(14.61)

An application of the Clapeyron equation is provided in Example 11.4.

A special form of Eq. 14.61 for a system at equilibrium consisting of a liquid or

solid phase and a vapor phase can be obtained simply. If the specific volume of the

liquid or solid, y¿, is negligible compared with the specific volume of the vapor, y–,

and the vapor can be treated as an ideal gas, y

– 5

R T

sat

, Eq. 14.61 becomes

sat

h– 2 h¿

sat

d ln p

sat

– 2

(14.62)

which is the Clausius–Clapeyron equation. The similarity in form of Eq. 14.62 and the

van’t Hoff equation, Eq. 14.43b, may be noted. The van’t Hoff equation for chemi-

cal equilibrium is the counterpart of the Clausius–Clapeyron equation for phase

equilibrium.

Clapeyron equation

Clausius–Clapeyron

equation

14.5 Equilibrium between Two Phases of a Pure Substance 875

TAKE NOTE...

Equations 11.40 and

11.42 are special cases

of Eqs. 14.61 and 14.62,

respectively.

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876 Chapter 14 Chemical and Phase Equilibrium

14.6 Equilibrium of Multicomponent,

Multiphase Systems

The equilibrium of systems that may involve several phases, each having a number

of components present, is considered in this section. The principal result is the Gibbs

phase rule, which summarizes important limitations on multicomponent, multiphase

systems at equilibrium.

14.6.1

Chemical Potential and Phase Equilibrium

Figure 14.1 shows a system consisting of two components A and B in two phases

1 and 2 that are at the same temperature and pressure. Applying Eq. 14.10 to

each of the phases

dG¿4

T,p

5 m¿

dn¿

1 m¿

dn¿

dG–4

T,p

5 m–

dn–

1 m–

dn–

(14.63)

where as before the primes identify the two phases.

When matter is transferred between the two phases in the absence of chem-

ical reaction, the total amounts of A and B must remain constant. Thus, the

increase in the amount present in one of the phases must be compensated by

an equivalent decrease in the amount present in the other phase. That is

–

52dn



–

52dn

(14.64)

With Eqs. 14.63 and 14.64, the change in the Gibbs function for the system is

5 dG¿

1 dG–

–

dn¿

–

dn¿

(14.65)

Since n9

and n9

can be varied independently, it follows that when dG4

T, p

5 0, the

terms in parentheses are zero, resulting in

5 m

–



and



5 m

–

(14.66)

At equilibrium, the chemical potential of each component is the same in each phase.

The significance of the chemical potential for phase equilibrium can be brought

out simply by reconsidering the system of Fig. 14.1 in the special case when the

chemical potential of component B is the same in both phases:

m¿

m–

. With this

constraint, Eq. 14.65 reduces to

m¿

2 m–

dn¿

Any spontaneous process of the system taking place at a fixed temperature and pres-

sure must be such that the Gibbs function decreases: dG

, 0. Thus, with the above

expression we have

1m¿

2 m–

2dn¿

, 0

Accordingly,

c when the chemical potential of A is greater in phase 1 than in phase 2

m¿

. m–

it follows that dn

, 0. That is, substance A passes from phase 1 to phase 2.

c when the chemical potential of A is greater in phase 2 than in phase 1

–

it follows that dn

. 0. That is, substance A passes from phase 2 to phase 1.

At equilibrium, the chemical potentials are equal

–

, and there is no net

transfer of A between the phases.

With this reasoning, we see that the chemical potential can be regarded as a measure

of the escaping tendency of a component. If the chemical potential of a component is not

the same in each phase, there will be a tendency for that component to pass from the

Fig. 14.1 System consisting of two

components in two phases.

Phase 2

Component A, n´´

, ´´

Component B, n´´

, ´´

Phase 1

Component A, n´

, ´

Component B, n´

, ´

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phase having the higher chemical potential for that component to the phase having the

lower chemical potential. When the chemical potential is the same in both phases, there

is no tendency for a net transfer to occur from one phase to the other.

In Example 14.10, we apply phase equilibrium principles to provide a rationale for

the model introduced in Sec. 12.5.3 for moist air in contact with liquid water.

14.6 Equilibrium of Multicomponent, Multiphase Systems 877

c c c c EXAMPLE 14.10 c

A closed system at a temperature of

8F and a pressure of 1 atm consists of a pure liquid water phase in equi-

librium with a vapor phase composed of water vapor and dry air. Determine the departure, in percent, of the

partial pressure of the water vapor from the saturation pressure of water at 70°F.

SOLUTION

Known:

A phase of liquid water only is in equilibrium with moist air at 70°F and 1 atm.

Find: Determine the percentage departure of the partial pressure of the water vapor in the moist air from the

saturation pressure of water at 70°F.

Schematic and Given Data:

Evaluating Equilibrium of Moist Air in Contact with Liquid Water

Engineering Model:

The gas phase can be modeled as an ideal

gas mixture.

2. The liquid phase is pure water only.

Analysis: For phase equilibrium, the chemical potential of the water must have the same value in both phases:

5 m

, where m

and m

denote, respectively, the chemical potentials of the pure liquid water in the liquid phase

and the water vapor in the vapor phase.

The chemical potential m

is the Gibbs function per mole of pure liquid water (Eq. 14.12)

5 g 1T, p2

Since the vapor phase is assumed to form an ideal gas mixture, the chemical potential m

equals the Gibbs func-

tion per mole evaluated at temperature T and the partial pressure p

of the water vapor (Eq. 14.16)

5 g 1T, p

For phase equilibrium, m

5 m

, or

g 1T, p

25 g 1T, p2

With g 5 h 2 T s, this can be expressed alternatively as

h 1T, p

22 T s 1T, p

25 h 1T, p22 T s 1T, p2

The water vapor is modeled as an ideal gas. Thus, the enthalpy is given closely by the saturated vapor value

at temperature T

h 1T, p

2< h

Fig. E14.10

70°F

Gas phase of

water vapor

and dry air

Liquid water

p = 1 atm

Sat. liquid

Sat. vapor

State of

the liquid

State of

the water

vapor

sat

➊

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878 Chapter 14

Chemical and Phase Equilibrium

Furthermore, the relationship between the specific entropy of the water vapor and the specific entropy at the

corresponding saturated vapor state is

s 1T, p

25 s

1T 22 R ln

sat

where p

sat

is the saturation pressure at temperature T (see Sec. 12.5.2 for discussion).

With Eq. 3.13, the enthalpy of the liquid is closely

h 1T, p2< h

1 y

1p 2 p

sat

where y

and h

are the saturated liquid specific volume and enthalpy at temperature T. Furthermore, with Eq. 6.5

s 1T, p2< s

1T 2

where s

is the saturated liquid specific entropy at temperature T.

Collecting the foregoing expressions, we have

2 T as

2 R ln

sat

b5 h

1 y

1p 2 p

sat

22 T s

RT ln

sat

1p 2 p

sat

22 31h

2 h

22 T 1s

2 s

The underlined term vanishes by Eq. 14.60, leaving

sat

1p 2 p

sat





sat

5 exp

1p 2 p

sat

With data from Table A-2E at 70°F, y

5 0.01605 ft

/lb and p

sat

5 0.3632 lbf/in.

, we have on a mass basis

p 2 p

sat

0.01605 ft

lb 114.696 2 0.36322 1lbf

in.

2 0144 in.

154

18.02

ft ? lbf

lb ? 8R

b 15308R2

5 7.29 3 10

Finally

sat

5 exp 17.29 3 10

25 1.00073

When expressed as a percentage, the departure of p

from p

sat

➋

2 p

sat

b110025 11.00073 2 12110025 0.073%

➊ For phase equilibrium, there would be a small, but finite, concentration of air

in the liquid water phase. However, this small amount of dissolved air is

ignored in the present development.

➋ The departure of p

from p

sat

is negligible at the specified conditions. This sug-

gests that at normal temperatures and pressures the equilibrium between the

liquid water phase and the water vapor is not significantly disturbed by the

presence of the dry air. Accordingly, the partial pressure of the water vapor can

be taken as equal to the saturation pressure of the water at the system tem-

perature. This model, introduced in Sec. 12.5.3, is used extensively in Chap. 12.

Ability to…

❑

apply the concept of phase

equilibrium expressed by

Eqs. 14.66 to an air–water

vapor mixture in equilibrium

with liquid water.

✓

Skills Developed

Using the methods of Sec. 12.5.2, determine the humidity ratio,

v, of the air–water vapor mixture. Ans. 0.01577 lb(vapor)/lb(dry air).

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14.6.2

Gibbs Phase Rule

The requirement for equilibrium of a system consisting of two components and two

phases, given by Eqs. 14.66, can be extended with similar reasoning to nonreacting

multicomponent, multiphase systems. At equilibrium, the chemical potential of each

component must be the same in all phases. For the case of N components that are

present in P phases we have, therefore, the following set of N(P 2 1) equations:

P 2

5 m

. . .

5 m

. . .

5 m

. . .

5 m

(14.67)

where

denotes the chemical potential of the ith component in the jth phase. This

set of equations provides the basis for the Gibbs phase rule, which allows the deter-

mination of the number of independent intensive properties that may be arbitrarily

specified in order to fix the intensive state of the system. The number of independent

intensive properties is called the degrees of freedom (or the variance).

Since the chemical potential is an intensive property, its value depends on the

relative proportions of the components present and not on the amounts of the com-

ponents. In other words, in a given phase involving N components at temperature T

and pressure p, the chemical potential is determined by the mole fractions of the

components present and not the respective n’s. However, as the mole fractions add

to unity, at most N 2 1 of the mole fractions can be independent. Thus, for a system

involving N components, there are at most N 2 1 independently variable mole frac-

tions for each phase. For P phases, therefore, there are at most P(N 2 1) indepen-

dently variable mole fractions. In addition, the temperature and pressure, which are

the same in each phase, are two further intensive properties, giving a maximum of

P(N 2 1) 1 2 independently variable intensive properties for the system. But because

of the N(P 2 1) equilibrium conditions represented by Eqs. 14.67 among these

properties, the number of intensive properties that are freely variable, the degrees

of freedom F, is

F 5 3P1N 2 121 242 N1P 2 125 2 1 N 2 P (14.68)

which is the Gibbs phase rule.

In Eq. 14.68, F is the number of intensive properties that may be arbitrarily spec-

ified and that must be specified to fix the intensive state of a nonreacting system at

equilibrium.

let us apply the Gibbs phase rule to a liquid solution consisting

of water and ammonia such as considered in the discussion of absorption refrigeration

(Sec. 10.5). This solution involves two components and a single phase: N 5 2 and P 5 1.

Equation 14.68 then gives F 5 3, so the intensive state is fixed by giving the values of

three intensive properties, such as temperature, pressure, and the ammonia (or water)

mole fraction. b b b b b

The phase rule summarizes important limitations on various types of systems. For

example, for a system involving a single component such as water, N 5 1 and Eq. 14.68

becomes

F 5 3 2 P (14.69)

c The minimum number of phases is one, corresponding to P 5 1. For this case, Eq. 14.69

gives F 5 2. That is, two intensive properties must be specified to fix the intensive

degrees of freedom

Gibbs phase rule

14.6 Equilibrium of Multicomponent, Multiphase Systems 879

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880 Chapter 14 Chemical and Phase Equilibrium

state of the system. This requirement is familiar from our use of the steam tables

and similar property tables. To obtain properties of superheated vapor, say, from

such tables requires that we give values for any two of the tabulated properties,

for example, T and p.

c When two phases are present in a system involving a single component, N 5 1 and

P 5 2. Equation 14.69 then gives F 5 1. That is, the intensive state is determined

by a single intensive property value. For example, the intensive states of the sepa-

rate phases of an equilibrium mixture of liquid water and water vapor are com-

pletely determined by specifying the temperature.

c The minimum allowable value for the degrees of freedom is zero: F 5 0. For a

single-component system, Eq. 14.69 shows that this corresponds to P 5 3, a three-

phase system. Thus, three is the maximum number of different phases of a pure

component that can coexist in equilibrium. Since there are no degrees of freedom,

both temperature and pressure are fixed at equilibrium. For example, there is only

a single temperature 0.01°C (32.02°F) and a single pressure 0.6113 kPa (0.006 atm)

for which ice, liquid water, and water vapor are in equilibrium.

The phase rule given here must be modified for application to systems in which

chemical reactions occur. Furthermore, the system of equations, Eqs. 14.67, giving the

requirements for phase equilibrium at a specified temperature and pressure can be

expressed alternatively in terms of partial molal Gibbs functions, fugacities, and

activities, all of which are introduced in Sec. 11.9. To use any such expression to

determine the equilibrium composition of the different phases present within a system

at equilibrium requires a model for each phase that allows the relevant quantities—

the chemical potentials, fugacities, and so on—to be evaluated for the components

present in terms of system properties that can be determined. For example, a gas

phase might be modeled as an ideal gas mixture or, at higher pressures, as an ideal

solution.

In this chapter, we have studied chemical equilibrium and phase

equilibrium. The chapter opens by developing criteria for equilib-

rium and introducing the chemical potential. In the second part

of the chapter, we study the chemical equilibrium of ideal gas

mixtures using the equilibrium constant concept. We also utilize

the energy balance and determine the equilibrium flame tem-

perature as an application. The final part of the chapter concerns

phase equilibrium, including multicomponent, multiphase sys-

tems and the Gibbs phase rule.

The following list provides a study guide for this chapter.

When your study of the text and end-of-chapter exercises has

been completed, you should be able to

write out the meaning of the terms listed in the margin

throughout the chapter and understand each of the related

concepts. The subset of key concepts listed below is particu-

larly important.

apply the equilibrium constant relationship, Eq. 14.35, to

determine the third quantity when any two of temperature,

pressure, and equilibrium composition of an ideal gas mixture

are known. Special cases include applications with simultane-

ous reactions and systems involving ionized gases.

use chemical equilibrium concepts with the energy balance,

including determination of the equilibrium flame temperature.

apply Eq. 14.43b, the van’t Hoff equation, to determine the

enthalpy of reaction when the equilibrium constant is known,

and conversely.

apply the Gibbs phase rule, Eq. 14.68.

c CHAPTER SUMMARY AND STUDY GUIDE

c KEY ENGINEERING CONCEPTS

Gibbs function, p. 849

equilibrium criterion, p. 849

chemical potential, p. 850

equation of reaction equilibrium, p. 855

equilibrium constant, p. 856

equilibrium flame temperature, p. 865

Gibbs phase rule, p. 879

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c EXERCISES: THINGS ENGINEERS THINK ABOUT

1. Why is using the Gibbs function advantageous when studying

chemical and phase equilibrium?

2. For Eq. 14.6 to apply at equilibrium, must a system attain

equilibrium at fixed T and p?

3. Show that (dA)

T,V

5 0 is a valid equilibrium criterion, where

A 5 U 2 TS is the Helmholtz function.

4. A mixture of 1 kmol of CO and

kmol of O

is held at

ambient temperature and pressure. After 100 hours only an

insignificant amount of CO

has formed. Why?

5. Why might oxygen contained in an iron tank be treated as

inert in a thermodynamic analysis even though iron oxidizes

in the presence of oxygen?

6. For

1 H

, how does pressure affect the

equilibrium composition?

7. For each of the reactions listed in Table A-27, the value of log

increases with increasing temperature. What does this imply?

8. For each of the reactions listed in Table A-27, the value of

the equilibrium constant K at 298 K is relatively small. What

does this imply?

9. If a system initially containing CO

and H

O were held at

fixed T, p, list chemical species that might be present at

equilibrium.

10. Using Eq. 14.12 together with phase equilibrium consi-

derations, suggest how the chemical potential of a mixture

component could be evaluated.

11. Note 1 of Example 14.10 refers to the small amount of air

that would be dissolved in the liquid phase. For equilibrium,

what must be true of the chemical potentials of air in the

liquid and gas phases?

12. Water can exist in a number of different solid phases. Can

liquid water, water vapor, and two phases of ice exist in

equilibrium?

Exercises: Things Engineers Think About 881

c KEY EQUATIONS

dG4

T, p

5 0 (14.6) p. 849 Equilibrium criterion

5 a

T, p, n

(14.8) p. 850 Chemical potential of component i in a mixture.

G 5

i51

1T, p

(14.15) p. 851

5 g

1T, p

2 (14.16) p. 851

5 g

81RT ln

ref

(14.17) p. 852

K1T25

ref

(14.32) p. 856

K 5

ref

(14.35) p. 858

d ln K

¢H

(14.43b) p. 869 van’t Hoff equation.

g¿ 5 g–

(14.59) p. 875 Phase equilibrium criterion for a pure substance

m¿

5 m–

m¿

m–

(14.66) p. 876

F 5 2 1

2 P (14.68) p. 879 Gibbs phase rule.

Equilibrium constant expressions for an equilibrium mixture

of ideal gases.

Gibbs function and chemical potential relations for ideal

gas mixtures.

Phase equilibrium criteria for two-component,

two-phase systems.

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882 Chapter 14

Chemical and Phase Equilibrium

c PROBLEMS: DEVELOPING ENGINEERING SKILLS

Working with the Equilibrium Constant

14.1 Determine the change in the Gibbs function DG8 at 258C

in kJ/kmol, for the reaction

1g21 2O

1 2H

O1g2

using

(a) Gibbs function of formation data.

(b) enthalpy of formation and absolute entropy data.

14.2 Calculate the equilibrium constant, expressed as log

for CO

CO 1

at (a) 500 K, (b) 18008R. Compare with

values from Table A-27.

14.3 Calculate the equilibrium constant, expressed as

for the water-gas shift reaction CO 1 H

1 H

(a) 298 K, (b) 1000 K. Compare with values from Table A-27.

14.4 Calculate the equilibrium constant, expressed as

for H

at (a) 298 K, (b) 36008R. Compare

with values from Table A-27.

14.5 Using data from Table A-27, determine log

K at 2500 K for

(a) H

(b) H

1 O

14.6 In Table A-27, log

K is nearly linear in 1yT: log

K 5

1 C

T, where C

and C

are constants. For selected reac-

tions listed in the table

(a) verify this by plotting log

K versus 1/T for temperature

ranging from 2100 to 2500 K.

(b) evaluate C

and C

for any pair of adjacent table entries

in the temperature range of part (a).

14.7 Determine the relationship between the ideal gas equilibrium

constants K

and K

for the following two alternative ways of

expressing the ammonia synthesis reaction:

2. N

1 3H

14.8 Consider the reactions

1. CO 1 H

1 CO

2. 2CO

2CO 1 O

3. 2H

1 O

Show that K

14.9 Consider the reactions

1. CO

1 H

CO 1 H

2. CO

CO 1

3. H

(a) Show that K

5 K

(b) Evaluate 1og

at 298 K, 1 atm using the expression from

part (a), together with 1og

K data from Table A-27.

obtained in part (b) by

applying Eq. 14.31 to reaction 1.

14.10 Evaluate the equilibrium constant at 2000 K for

1 H

1 CO. At 2000 K, log

K 5 7.469 for

C 1

CO, and log

K 523.408 for C 1 2H

14.11 For each of the following dissociation reactions, determine

the equilibrium compositions:

(a) One kmol of N

dissociates to form an equilibrium

ideal gas mixture of N

and NO

at 25°C, 2 atm. For

2NO

, ¢G855400 kJ

kmol at 25°C.

(b) One kmol of CH

dissociates to form an equilibrium

ideal gas mixture of CH

, H

, and C at 1000 K, 5 atm. For

C 1 2H

, log

K 5 1.011 at 1000 K.

14.12 Determine the extent to which dissociation occurs in the

following cases: One lbmol of H

O dissociates to form an

equilibrium mixture of H

O, H

, and O

at 47408F, 1.25 atm.

One lbmol of CO

dissociates to form an equilibrium mixture

of CO

, CO, and O

at the same temperature and pressure.

14.13 One lbmol of carbon reacts with 2 lbmol of oxygen (O

)

to form an equilibrium mixture of CO

, CO, and O

at 49408F,

1 atm. Determine the equilibrium composition.

14.14 The following exercises involve oxides of nitrogen:

(a) One kmol of N

dissociates at 258C, 1 atm to form an

equilibrium ideal gas mixture of N

and NO

in which the

amount of N

present is 0.8154 kmol. Determine the

amount of N

that would be present in an equilibrium

mixture at 258C, 0.5 atm.

(b) A gaseous mixture consisting of 1 kmol of NO, 10 kmol

of O

, and 40 kmol of N

reacts to form an equilibrium

ideal gas mixture of NO

, NO, and O

at 500 K, 0.1 atm.

Determine the composition of the equilibrium mixture. For

NO 1

, K 5 120 at 500 K.

and N

reacts to form an

equilibrium ideal gas mixture of O

, N

, and NO. Plot the mole

fraction of NO in the equilibrium mixture versus equilibrium

temperature ranging from 1200 to 2000 K.

Why are oxides of nitrogen of concern?

14.15 One kmol of CO

dissociates to form an equilibrium

ideal gas mixture of CO

, CO, and O

at temperature T and

pressure p.

(a) For T 5 3000 K, plot the amount of CO present, in kmol,

versus pressure for 1 # p # 10 atm.

(b) For p 5 1 plot the amount of CO present, in kmol, versus

temperature for 2000 # T # 3500 K.

14.16 One lbmol of H

O dissociates to form an equilibrium

ideal gas mixture of H

O, H

, and O

at temperature T and

pressure p.

(a) For T 5 54008R, plot the amount of H

present, in lbmol,

versus pressure ranging from 1 to 10 atm.

(b) For p 5 1 atm, plot the amount of H

present, in lbmol,

versus temperature ranging from 3600 to 63008R.

14.17 One lbmol of H

O together with x lbmol of N

(inert)

forms an equilibrium mixture at 54008R, 1 atm consisting of

O, H

, O

, and N

. Plot the amount of H

present in the

equilibrium mixture, in lbmol, versus x ranging from 0 to 2.

14.18 An equimolar mixture of CO and O

reacts to form an

equilibrium mixture of CO

, CO, and O

at 3000 K.

Determine the effect of pressure on the composition of the

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