Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics

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Chemical Equilibrium

In this part of the chapter, the equilibrium criterion dG]

T, p

5 0 introduced in Sec. 14.1

is used to study the equilibrium of reacting mixtures. The objective is to establish the

composition present at equilibrium for a specified temperature and pressure. An

important parameter for determining the equilibrium composition is the equilibrium

constant. The equilibrium constant is introduced and its use illustrated by several

solved examples. The discussion is concerned only with equilibrium states of reacting

systems, and no information can be deduced about the rates of reaction. Whether an

equilibrium mixture would form quickly or slowly can be determined only by con-

sidering the chemical kinetics, a topic that is not treated in this text.

14.2 Equation of Reaction Equilibrium

In Chap. 13 the conservation of mass and conservation of energy principles are applied

to reacting systems by assuming that the reactions can occur as written. However, the

extent to which a chemical reaction proceeds is limited by many factors. In general, the

composition of the products actually formed from a given set of reactants, and the relative

amounts of the products, can be determined only from experiment. Knowledge of the

composition that would be present were a reaction to proceed to equilibrium is frequently

useful, however. The equation of reaction equilibrium introduced in the present section

provides the basis for determining the equilibrium composition of a reacting mixture.

14.2.1

Introductory Case

Consider a closed system consisting initially of a gaseous mixture of hydrogen and

oxygen. A number of reactions might take place, including

O (14.18)

2H (14.19)

2O (14.20)

Let us consider for illustration purposes only the first of the reactions given above,

in which hydrogen and oxygen combine to form water. At equilibrium, the system

will consist in general of three components: H

, O

, and H

O, for not all of the hydro-

gen and oxygen initially present need be reacted. Changes in the amounts of these

components during each differential step of the reaction leading to the formation of

an equilibrium mixture are governed by Eq. 14.18. That is

52dn





(14.21a)

where dn denotes a differential change in the respective component. The minus signs

signal that the amounts of hydrogen and oxygen present decrease when the reaction

proceeds toward the right. Equations 14.21a can be expressed alternatively as

2dn

(14.21b)

which emphasizes that increases and decreases in the components are proportional

to the stoichiometric coefficients of Eq. 14.18.

Equilibrium is a condition of balance. Accordingly, as suggested by the direction of

the arrows in Eq. 14.18, when the system is at equilibrium, the tendency of the hydrogen

and oxygen to form water is just balanced by the tendency of water to dissociate into

oxygen and hydrogen. The equilibrium criterion dG]

T, p

5 0 can be used to determine

the composition at an equilibrium state where the temperature is T and the pressure is p.

This requires evaluation of the differential dG]

T, p

in terms of system properties.

14.2 Equation of Reaction Equilibrium 853

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854 Chapter 14

Chemical and Phase Equilibrium

For the present case, Eq. 14.10 giving the difference in the Gibbs function of the

mixture between two states having the same temperature and pressure, but composi-

tions that differ infinitesimally, takes the following form:

dG4

T, p

5 m

1 m

(14.22)

The changes in the mole numbers are related by Eqs. 14.21. Hence

dG4

T, p

5 121m

1 1m

2 dn

At equilibrium, dG]

T, p

5 0, so the term in parentheses must be zero. That is

21m

1 1m

5 0

When expressed in a form that resembles Eq. 14.18, this becomes

5 1m

(14.23)

Equation 14.23 is the equation of reaction equilibrium for the case under consideration.

The chemical potentials are functions of temperature, pressure, and composition. Thus,

the composition that would be present at equilibrium for a given temperature and

pressure can be determined, in principle, by solving this equation. The solution proce-

dure is described in Sec. 14.3.

14.2.2

General Case

The foregoing development can be repeated for reactions involving any number of

components. Consider a closed system containing five components, A, B, C, D, and E,

at a given temperature and pressure, subject to a single chemical reaction of the form

A 1 n

C 1 n

D (14.24)

where the n’s are stoichiometric coefficients. Component E is assumed to be inert

and thus does not appear in the reaction equation. As we will see, component E does

influence the equilibrium composition even though it is not involved in the chemical

reaction. The form of Eq. 14.24 suggests that at equilibrium the tendency of A and

B to form C and D is just balanced by the tendency of C and D to form A and B.

The stoichiometric coefficients n

, n

, and n

do not correspond to the respec-

tive number of moles of the components present. The amounts of the components

present are designated n

, n

, and n

. However, changes in the amounts of the

components present do bear a definite relationship to the values of the stoichiomet-

ric coefficients. That is

2dn

(14.25a)

where the minus signs indicate that A and B would be consumed when C and D are

produced. Since E is inert, the amount of this component remains constant, so dn

5 0.

Introducing a proportionality factor d´, Eqs. 14.25a take the form

2dn

5 de

from which the following expressions are obtained:

52n

de, dn

52n

5 n

de, dn

5 n

(14.25b)

The parameter ´ is sometimes referred to as the extent of reaction.

extent of reaction

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For the system under present consideration, Eq. 14.10 takes the form

dG4

T, p

5 m

1 m

Introducing Eqs. 14.25b and noting that dn

5 0, this becomes

dG4

T, p

5 12n

2 n

1 n

2 de

At equilibrium, dG]

T, p

5 0, so the term in parentheses must be zero. That is

2 n

1 n

5 0

or when written in a form resembling Eq. 14.24

1 n

5 n

1 n

(14.26)

For the present case, Eq. 14.26 is the equation of reaction equilibrium. In principle,

the composition that would be present at equilibrium for a given temperature and

pressure can be determined by solving this equation. The solution procedure is simpli-

fied through the equilibrium constant concept introduced in the next section.

equation of reaction

equilibrium

14.3 Calculating Equilibrium Compositions

The objective of the present section is to show how the equilibrium composition of a

system at a specified temperature and pressure can be determined by solving the equation

of reaction equilibrium. An important part is played in this by the equilibrium constant.

14.3.1

Equilibrium Constant for Ideal Gas Mixtures

The first step in solving the equation of reaction equilibrium, Eq. 14.26, for the equi-

librium composition is to introduce expressions for the chemical potentials in terms

of temperature, pressure, and composition. For an ideal gas mixture, Eq. 14.17 can be

used for this purpose. When this expression is introduced for each of the components

A, B, C, and D, Eq. 14.26 becomes

ag 8

1 RT ln

ref

b1 n

ag 8

1 RT ln

ref

5 n

ag 8

1 RT ln

ref

b1 n

ag 8

1 RT ln

ref

(14.27)

where g8

is the Gibbs function of component i evaluated at temperature T and the

pressure p

ref

5 1 atm. Equation 14.27 is the basic working relation for chemical equi-

librium in a mixture of ideal gases. However, subsequent calculations are facilitated

if it is written in an alternative form, as follows.

Collect like terms and rearrange Eq. 14.27 as

1 n

2 n

RT an

ref

1 n

ref

2 n

ref

2 n

ref

(14.28)

The term on the left side of Eq. 14.28 can be expressed concisely as DG8. That is

¢G85n

1 n

2 n

(14.29a)

which is the change in the Gibbs function for the reaction given by Eq. 14.24 if

each reactant and product were separate at temperature T and a pressure of 1 atm.

14.3 Calculating Equilibrium Compositions 855

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856 Chapter 14 Chemical and Phase Equilibrium

This expression can be written alternatively in terms of specific enthalpies and

entropies as

¢G8 5 n

2 T s8

21 n

2 Ts8

22 n

2 Ts8

22 n

2 T s 8

5 1n

1 n

2 n

22 T1n

1 n

s 8

2 n

s 8

2 n

s 8

2 (14.29b)

Since the enthalpy of an ideal gas depends on temperature only, the h’s of Eq. 14.29b

are evaluated at temperature T. As indicated by the superscript 8, each of the entro-

pies is evaluated at temperature T and a pressure of 1 atm.

Introducing Eq. 14.29a into Eq. 14.28 and combining the terms involving loga-

rithms into a single expression gives

¢G8

5 ln c

ref

(14.30)

Equation 14.30 is simply the form taken by the equation of reaction equilibrium,

Eq. 14.26, for an ideal gas mixture subject to the reaction Eq. 14.24. As illustrated

by subsequent examples, similar expressions can be written for other reactions.

Equation 14.30 can be expressed concisely as

¢G8

5 ln K1T2

(14.31)

where K is the equilibrium constant defined by

K1T25

ref

(14.32)

Given the values of the stoichiometric coefficients, n

, n

, and n

and the tem-

perature T, the left side of Eq. 14.31 can be evaluated using either of Eqs. 14.29

together with the appropriate property data. The equation can then be solved for the

value of the equilibrium constant K. Accordingly, for selected reactions K can be

evaluated and tabulated against temperature. It is common to tabulate log

K or ln

K versus temperature, however. A tabulation of log

K values over a range of tem-

peratures for several reactions is provided in Table A-27, which is extracted from a

more extensive compilation.

The terms in the numerator and denominator of Eq. 14.32 correspond, respectively,

to the products and reactants of the reaction given by Eq. 14.24 as it proceeds from

left to right as written. For the inverse reaction n

C 1 n

A 1 n

B, the equi-

librium constant takes the form

ref

(14.33)

Comparing Eqs. 14.32 and 14.33, it follows that the value of K

is just the reciprocal

of K: K

5 1/K. Accordingly,

log

52log

K (14.34)

Hence, Table A-27 can be used both to evaluate K for the reactions listed proceeding

in the direction left to right and to evaluate K

for the inverse reactions proceeding

in the direction right to left.

Example 14.1 illustrates how the log

K values of Table A-27 are determined. Sub-

sequent examples show how the log

K values can be used to evaluate equilibrium

compositions.

equilibrium constant

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Evaluating the Equilibrium Constant at a Specified Temperature

c c c c EXAMPLE 14.1 c

Evaluate the equilibrium constant, expressed as log

K, for the reaction CO 1

at (a) 298 K and

(b) 2000 K. Compare with the value obtained from Table A-27.

SOLUTION

Known:

The reaction is CO 1

Find: Determine the equilibrium constant for T 5 298 K 1258C2 and T 5 2000 K.

Engineering Model: The ideal gas model applies.

Analysis: The equilibrium constant requires the evaluation of ¢G8 for the reaction. Invoking Eq. 14.29b for this

purpose, we have

¢G851h

2 h

22 T1s8

2 s8

s 8

where the enthalpies are evaluated at temperature T and the absolute entropies are evaluated at temperature T

and a pressure of 1 atm. Using Eq. 13.9, the enthalpies are evaluated in terms of the respective enthalpies of

formation, giving

¢G8531h8

2 1h8

1h8

41 31¢h2

2 1¢h2

1¢h2

42 T1s8

2 s8

s 8

where the ¢h terms account for the change in specific enthalpy from T

ref

5 298 K to the specified temperature

T. The enthalpy of formation of oxygen is zero by definition.

(a) When T 5 298 K, the ¢

terms of the above expression for DG8 vanish. The required enthalpy of formation

and absolute entropy values can be read from Table A-25, giving

¢G85312393,52022 12110,53022

10242 2983213.69 2 197.54 2

1205.0324

52257,253 kJ

kmol

With this value for DG8, Eq. 14.31 gives

ln K 52

2257,253 kJ

kmol

8.314 kJ

kmol ? K

298 K

5 103.83

which corresponds to log

K 5 45.093.

Table A-27 gives the logarithm to the base 10 of the equilibrium constant for the inverse reaction:

CO 1

. That is, log

5 245.066. Thus, with Eq. 14.34, log

K 5 45.066, which agrees closely with

the calculated value.

(b) When T 5 2000 K, the ¢h and s8 terms for O

, CO, and CO

required by the above expression for DG8 are

evaluated from Tables A-23. The enthalpy of formation values are the same as in part (a). Thus

¢G85312393,52022 12110,53022

10241 31100,804 2 936422 165408 2 86692

167,881 2 8682242 20003309.210 2 258.600 2

1268.65524

52282,990 1 5102 1 167,435 52110,453 kJ

kmol

With this value, Eq. 14.31 gives

ln K 52

12110,453

18.3142120002

5 6.643

which corresponds to log

K 5 2.885.

14.3 Calculating Equilibrium Compositions 857

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858 Chapter 14 Chemical and Phase Equilibrium

If ln K 5 23.535 for the given reaction, use Table A-27 to deter-

mine T, in K. Ans. 1000 K.

At 2000 K, Table A-27 gives log

5 22.884. With Eq. 14.34, log

K 5 2.884,

which is in agreement with the calculated value.

Using the procedures described above, it is straightforward to determine log

versus temperature for each of several specified reactions and tabulate the results

as in Table A-27.

Ability to…

❑

evaluate log

K based on

Eq. 14.31 and data from

Tables A-23 and A-25.

❑

use the relation of Eq.

14.34 for inverse reactions.

✓Skills Developed

14.3.2

Illustrations of the Calculation of Equilibrium

Compositions for Reacting Ideal Gas Mixtures

It is often convenient to express Eq. 14.32 explicitly in terms of the number of moles

that would be present at equilibrium. Each mole fraction appearing in the equation has

the form y

5 n

/n, where n

is the amount of component i in the equilibrium mixture

and n is the total number of moles of mixture. Hence, Eq. 14.32 can be rewritten as

K 5

ref

(14.35)

The value of n must include not only the reacting components A, B, C, and D but

also all inert components present. Since inert component E has been assumed pres-

ent, we would write n 5 n

1 n

Equation 14.35 provides a relationship among the temperature, pressure, and compo-

sition of an ideal gas mixture at equilibrium. Accordingly, if any two of temperature,

pressure, and composition are known, the third can be found by solving this equation.

suppose that the temperature T and pressure p are known and

the object is the equilibrium composition. With temperature known, the value of K

can be obtained from Table A-27. The n’s of the reacting components A, B, C, and

D can be expressed in terms of a single unknown variable through application of

the conservation of mass principle to the various chemical species present. Then,

since the pressure is known, Eq. 14.35 constitutes a single equation in a single

unknown, which can be solved using an equation solver or iteratively with a hand

calculator. b b b b b

In Example 14.2, we apply Eq. 14.35 to study the effect of pressure on the equi-

librium composition of a mixture of CO

, CO, and O

Determining Equilibrium Composition Given Temperature and Pressure

c c c c EXAMPLE 14.2 c

One kilomole of carbon monoxide, CO, reacts with

kmol of oxygen, O

, to form an equilibrium mixture of CO

, CO,

and O

at 2500 K and (a) 1 atm, (b) 10 atm. Determine the equilibrium composition in terms of mole fractions.

SOLUTION

Known:

A system initially consisting of 1 kmol of CO and

kmol of O

reacts to form an equilibrium mixture

of CO

, CO, and O

. The temperature of the mixture is 2500 K and the pressure is (a) 1 atm, (b) 10 atm.

Find: Determine the equilibrium composition in terms of mole fractions.

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Engineering Model: The equilibrium mixture is modeled as an ideal gas mixture.

Analysis: Equation 14.35 relates temperature, pressure, and composition for an ideal gas mixture at equilibrium.

If any two are known, the third can be determined using this equation. In the present case, T and p are known,

and the composition is unknown.

Applying conservation of mass, the overall balanced chemical reaction equation is

1CO 1

S zCO 1

1 11 2 z2CO

where z is the amount of CO, in kmol, present in the equilibrium mixture. Note that 0 # z # 1.

The total number of moles n in the equilibrium mixture is

n 5 z 1

1 11 2 z25

2 1 z

Accordingly, the molar analysis of the equilibrium mixture is

2 1 z



2 1 z



211 2 z

2 1 z

At equilibrium, the tendency of CO and O

to form CO

is just balanced by the tendency of CO

to form

CO and O

, so we have CO

CO 1

. Accordingly, Eq. 14.35 takes the form

K 5

z1z

1 2 z

ref

2 1 z

111

221

1 2 z

2 1 z

ref

At 2500 K, Table A-27 gives log

K 5 21.44. Thus, K 5 0.0363. Inserting this value into the last expression

0.0363 5

1 2 z

2 1 z

ref

(a)

(a) When p 5 1 atm, Eq. (a) becomes

0.0363 5

1 2 z

2 1 z

Using an equation solver or iteration with a hand calculator, z 5 0.129. The equilibrium composition in terms

of mole fractions is then

210.129

2.129

5 0.121,



0.129

2.129

5 0.061,



211 2 0.129

2.129

5 0.818

(b) When p 5 10 atm, Eq. (a) becomes

0.0363 5

1 2 z

2 1 z

1102

➊

Solving, z 5 0.062. The corresponding equilibrium composition in terms of

mole fractions is y

5 0.06, y

5 0.03, y

5 0.91.

➊

Comparing the results of parts (a) and (b) we conclude that the extent to

which the reaction proceeds toward completion (the extent to which CO

is formed) is increased by increasing the pressure.

If z 5 0.0478 (corresponding to p 5 22.4 atm, T 5 2500 K),

what would be the mole fraction of each constituent of the equilibrium

mixture? Ans. y

5 0.0467, y

5 0.0233, y

5 0.9300.

Ability to…

❑

apply Eq. 14.35 to deter-

mine equilibrium composition

given temperature and

pressure.

❑

retrieve and use data from

Table A-27.

✓

Skills Developed

14.3 Calculating Equilibrium Compositions 859

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860 Chapter 14 Chemical and Phase Equilibrium

In Example 14.3, we determine the temperature of an equilibrium mixture when

the pressure and composition are known.

In Example 14.4, we consider the effect of an inert component on the equilibrium

composition.

Considering the Effect on Equilibrium of an Inert Component

c c c c EXAMPLE 14.4 c

One kilomole of carbon monoxide reacts with the theoretical amount of air to form an equilibrium mixture of

, CO, O

, and N

at 2500 K and 1 atm. Determine the equilibrium composition in terms of mole fractions,

and compare with the result of Example 14.2.

Measurements show that at a temperature T and a pressure of 1 atm, the equilibrium mixture for the system of

Example 14.2 has the composition y

5 0.298, y

5 0.149, y

5 0.553. Determine the temperature T of the

mixture, in K.

SOLUTION

Known:

The pressure and composition of an equilibrium mixture of CO, O

, and CO

are specified.

Find: Determine the temperature of the mixture, in K.

Engineering Model: The mixture can be modeled as an ideal gas mixture.

Analysis: Equation 14.35 relates temperature, pressure, and composition for an ideal gas mixture at equilibrium.

If any two are known, the third can be found using this equation. In the present case, composition and pressure

are known, and the temperature is the unknown.

Equation 14.35 takes the same form here as in Example 14.2. Thus, when p 5 1 atm, we have

K1T25

1 2 z

2 1 z

where z is the amount of CO, in kmol, present in the equilibrium mixture and T is the temperature of the

mixture.

The solution to Example 14.2 gives the following expression for the mole

fraction of the CO in the mixture: y

5 2z/(2 1 z). Since y

5 0.298,

z 5 0.35.

Inserting this value for z into the expression for the equilibrium constant

gives K 5 0.2078. Thus, log

K 5 20.6824. Interpolation in Table A-27 then gives

T 5 2881 K.

➊

Comparing this example with part (a) of Example 14.2, we conclude that the

extent to which the reaction proceeds to completion (the extent to which

is formed) is decreased by increasing the temperature.

Determining Equilibrium Temperature Given Pressure and Composition

c c c c EXAMPLE 14.3 c

Determine the temperature, in K, for a pressure of 2 atm if the

equilibrium composition were unchanged. Ans. <2970 K.

Ability to…

❑

apply Eq. 14.35 to deter-

mine temperature given

pressure and equilibrium

composition.

❑

retrieve and use data from

Table A-27.

✓

Skills Developed

➊

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Determine the amounts, in kmol, of each component of the equi-

librium mixture. Ans. n

5 0.175, n

5 0.0875, n

5 0.8250, n

5 1.88.

SOLUTION

Known:

A system initially consisting of 1 kmol of CO and the theoretical amount of air reacts to form an equi-

librium mixture of CO

, CO, O

, and N

. The temperature and pressure of the mixture are 2500 K and 1 atm.

Find: Determine the equilibrium composition in terms of mole fractions, and compare with the result of Example

14.2.

Engineering Model: The equilibrium mixture can be modeled as an ideal gas mixture wherein N

is inert.

Analysis: For a complete reaction of CO with the theoretical amount of air

CO 1

1 1.88N

Accordingly, the reaction of CO with the theoretical amount of air to form CO

, CO, O

, and N

CO 1

1 1.88N

zCO 1

1 11 2 z2CO

1 1.88N

where z is the amount of CO, in kmol, present in the equilibrium mixture.

The total number of moles n in the equilibrium mixture is

n 5 z 1

1 11 2 z21 1.88 5

5.76 1 z

The composition of the equilibrium mixture in terms of mole fractions is

5.76 1



5.76 1



1 2 z

5.76 1



3.76

5.76 1

At equilibrium we have CO

CO 1

. So Eq. 14.35 takes the form

K 5

z1z

1 2 z

ref

5.76 1 z

The value of K is the same as in the solution to Example 14.2, K 5 0.0363.

Thus, since p 5 1 atm, we have

0.0363 5

1 2 z

5.76 1 z

Solving, z 5 0.175. The corresponding equilibrium composition is y

5 0.059,

5 0.278, y

5 0.029, y

5 0.634.

Comparing this example with Example 14.2, we conclude that the presence

of the inert component nitrogen reduces the extent to which the reaction pro-

ceeds toward completion at the specified temperature and pressure (reduces

the extent to which CO

is formed).

Ability to…

❑

apply Eq. 14.35 to deter-

mine equilibrium composition

for given temperature and

pressure in the presence of

an inert component.

❑

retrieve and use data from

Table A-27.

✓

Skills Developed

In the next example, the equilibrium concepts of this chapter are applied together

with the energy balance for reacting systems developed in Chap. 13.

c c c c EXAMPLE 14.5 c

Using Equilibrium Concepts with the Energy Balance

Carbon dioxide at 258C, 1 atm enters a reactor operating at steady state and dissociates, giving an equilibrium

mixture of CO

, CO, and O

that exits at 3200 K, 1 atm. Determine the heat transfer to the reactor, in kJ per

kmol of CO

entering. The effects of kinetic and potential energy can be ignored and W

5 0.

14.3 Calculating Equilibrium Compositions 861

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862 Chapter 14

Chemical and Phase Equilibrium

SOLUTION

Known:

Carbon dioxide at 258C, 1 atm enters a reactor at steady state. An equilibrium mixture of CO

, CO, and

exits at 3200 K, 1 atm.

Find: Determine the heat transfer to the reactor, in kJ per kmol of CO

entering.

Schematic and Given Data:

Analysis:

The required heat transfer can be determined from an energy rate balance for the control volume, but

first the composition of the exiting equilibrium mixture must be determined.

Applying the conservation of mass principle, the overall dissociation reaction is described by

zCO

1 11 2 z2CO 1

1 2 z

where z is the amount of CO

, in kmol, present in the mixture exiting the control volume, per kmol of CO

entering. The total number of moles n in the mixture is then

n 5 z 1 11 2 z21

1 2 z

3 2 z

The exiting mixture is assumed to be an equilibrium mixture (assumption 3). Thus, for the mixture we have

CO 1

. Equation 14.35 takes the form

K 5

11 2 z2311 2 z2

ref

3 2 z

111

221

Rearranging and noting that p 5 1 atm

K 5

1 2 z

3 2 z

At 3200 K, Table A-27 gives log

K 5 20.189. Thus, K 5 0.647, and the equilibrium constant expression becomes

0.647 5

1 2 z

3 2 z

Solving, z 5 0.422. The composition of the exiting equilibrium mixture, in kmol per kmol of CO

entering, is

then 0.422CO

, 0.578CO, 0.289O

When expressed per kmol of CO

entering the control volume, the energy rate balance reduces by assumption 1 to

0 5

1 h

2 10.422h

1 0.578h

1 0.289h

Solving for the heat transfer per kmol of CO

entering, and evaluating each enthalpy in terms of the respective

enthalpy of formation

5 0.4221h

81¢h2

1 0.5781h

81¢h2

1 0.2891h

1 ¢h2

2 1h

81¢h2

Engineering Model:

The control volume shown on the accompanying sketch by

a dashed line operates at steady state with W

5 0. Kinetic

energy and potential energy effects can be ignored.

2. The entering CO

is modeled as an ideal gas.

3. The exiting mixture of CO

, CO, and O

is an equilibrium

ideal gas mixture.

Fig. E14.5

25°C, 1 atm

(CO

, CO, O

)

3200 K, 1 atm

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