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

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14.3.3

Equilibrium Constant for Mixtures and Solutions

The procedures that led to the equilibrium constant for reacting ideal gas mixtures

can be followed for the general case of reacting mixtures by using the fugacity and

activity concepts introduced in Sec. 11.9. In principle, equilibrium compositions of

such mixtures can be determined with an approach paralleling the one for ideal gas

mixtures.

Equation 11.141 can be used to evaluate the chemical potentials appearing in the

equation of reaction equilibrium (Eq. 14.26). The result is

1g8

1 RT ln a

21 v

1g8

1 RT ln a

25 v

1g 8

1 RT ln a

21 v

1g 8

1 RT ln a

(14.36)

where g8

is the Gibbs function of pure component i at temperature T and the pres-

sure p

ref

5 1 atm, and a

is the activity of that component.

Collecting terms and employing Eq. 14.29a, Eq. 14.36 becomes

¢G8

5 ln

(14.37)

This equation can be expressed in the same form as Eq. 14.31 by defining the equi-

librium constant as

K 5

(14.38)

The enthalpy of formation of O

is zero by definition;

¢h

for the CO

at the inlet vanishes because CO

enters

at 258C.

With enthalpy of formation values from Tables A-25 and ¢

values for O

, CO, and CO

, from Table A-23

5 0.42232393,520 1 1174,695 2 9364241 0.57832110,530 1 1109,667 2 866924

1 0.289

114,809 2 8682

2393,520

➊

5 322,385 kJ

kmol

➊

For comparison, let us determine the heat transfer if we assume no dissociation—

namely, when CO

alone exits the reactor. With data from Table A-23, the heat

transfer is

5 h

13200 K22 h

1298 K2

5 174,695 2 9364 5 165,331 kJ

kmol

The value is much less than the value obtained in the solution above because

the dissociation of CO

requires more energy input (an endothermic reac-

tion).

Determine the heat transfer rate, in kW, and the molar flow

rate of mixture exiting, in kmol/s, for a flow rate of 3.1 3 10

kmol/s of

entering. Ans. 10 kW, 4 3 10

kmol/s.

Ability to…

❑

apply Eq. 14.35 together

with the energy balance for

reacting systems to determine

heat transfer for a reactor.

❑

retrieve and use data from

Tables A-23, A-25, and A-27.

✓

Skills Developed

TAKE NOTE...

Study of Sec. 14.3.3

requires content from

Sec. 11.9

14.3 Calculating Equilibrium Compositions 863

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864 Chapter 14 Chemical and Phase Equilibrium

Since Table A-27 and similar compilations are constructed simply by evaluating

2¢

RT for specified reactions at several temperatures, such tables can be employed

to evaluate the more general equilibrium constant given by Eq. 14.38. However,

before Eq. 14.38 can be used to determine the equilibrium composition for a known

value of K, it is necessary to evaluate the activity of the various mixture compo-

nents. Let us illustrate this for the case of mixtures that can be modeled as ideal

solutions.

IDEAL SOLUTIONS. For an ideal solution, the activity of component i is given by

(11.142)

where f

is the fugacity of pure i at the temperature T and pressure p of the mixture,

and f8

is the fugacity of pure i at temperature T and the pressure p

ref

. Using this

expression to evaluate a

, a

, and a

, Eq. 14.38 becomes

K 5

f 8

(14.39a)

which can be expressed alternatively as

K 5 c

1f 8

ref

1f 8

ref

f 8

ref

f 8

ref

(14.39b)

The ratios of fugacity to pressure in this equation can be evaluated, in principle,

from Eq. 11.124 or the generalized fugacity chart, Fig. A-6, developed from it. In

the special case when each component behaves as an ideal gas at both T, p and T,

ref

, these ratios equal unity and Eq. 14.39b reduces to the underlined term, which

is just Eq. 14.32.

While carbon dioxide is often mentioned by the media

because of its effect on global climate change, and

rightly so, other gases released to the atmosphere also

contribute to climate change but get less publicity. In particular,

methane, CH

, which receives little notice as a greenhouse gas,

has a Global Warming Potential of 25, compared to carbon diox-

ide’s GWP of 1 (see Table 10.1).

Sources of methane related to human activity include fossil-fuel

(coal, natural gas, and petroleum) production, distribution, com-

bustion, and other uses. Wastewater treatment, landfills, and agri-

culture, including ruminant animals raised for food, are also human-

related sources of methane. Natural sources of methane include

wetlands and methane hydrate deposits in seafloor sediments.

For decades, the concentration of methane in the atmosphere

has increased significantly. But some observers report that the

increase has slowed recently and may be ceasing. While this

could be only a temporary pause, reasons have been advanced

to explain the development. Some say governmental actions

aimed at reducing release of methane have begun to show

results. Changes in agricultural practices, such as the way rice

is produced, also may be a factor in the reported reduction of

methane in the atmosphere.

Another view is that the plateau in atmospheric methane may

at least in part be due to chemical equilibrium: Methane released

to the atmosphere is balanced by its consumption in the atmo-

sphere. Methane is consumed in the atmosphere principally by its

reaction with the hydroxyl radical (OH), which is produced through

decomposition of atmospheric ozone by action of solar radiation.

For instance, OH reacts with methane to yield water and CH

, a

methyl radical, according to

H S H

. Other reac-

tions follow this, leading eventually to water-soluble products that

are washed out of the atmosphere by rain and snow.

Understanding the reasons for the apparent slowing rate of

growth of methane in the atmosphere will take effort, including

quantifying changes in the various sources of methane and pin-

pointing natural mechanisms by which it is removed from the

atmosphere. Better understanding will enable us to craft mea-

sures aimed at curbing release of methane, allowing the atmo-

sphere’s natural ability to cleanse itself to assist in maintaining

a healthier balance.

Methane, Another Greenhouse Gas

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14.4 Further Examples of the Use

of the Equilibrium Constant

Some additional aspects of the use of the equilibrium constant are introduced in this

section: the equilibrium flame temperature, the van’t Hoff equation, and chemical

equilibrium for ionization reactions and simultaneous reactions. To keep the presen-

tation at an introductory level, only the case of ideal gas mixtures is considered.

14.4.1

Determining Equilibrium Flame Temperature

In this section, the effect of incomplete combustion on the adiabatic flame tempera-

ture, introduced in Sec. 13.3, is considered using concepts developed in the present

chapter. We begin with a review of some ideas related to the adiabatic flame tem-

perature by considering a reactor operating at steady state for which no significant

heat transfer with the surroundings takes place.

Let carbon monoxide gas entering at one location react completely with the theo-

retical amount of air entering at another location as follows:

CO 1

1 1.88N

As discussed in Sec. 13.3, the products would exit the reactor at a temperature we

have designated the maximum adiabatic flame temperature. This temperature can be

determined by solving a single equation, the energy equation. At such an elevated

temperature, however, there would be a tendency for CO

to dissociate

CO 1

Since dissociation requires energy (an endothermic reaction), the temperature of the

products would be less than the maximum adiabatic temperature found under the

assumption of complete combustion.

When dissociation takes place, the gaseous products exiting the reactor would not

be CO

and N

, but a mixture of CO

, CO, O

, and N

. The balanced chemical reac-

tion equation would read

CO 1

1 1.88N

S z CO 1 11 2 z2CO

1 1.88N

(14.40)

where z is the amount of CO, in kmol, present in the exiting mixture for each kmol

of CO entering the reactor.

Accordingly, there are two unknowns: z and the temperature of the exiting stream.

To solve a problem with two unknowns requires two equations. One is provided by

an energy equation. If the exiting gas mixture is in equilibrium, the other equation

is provided by the equilibrium constant, Eq. 14.35. The temperature of the products

may then be called the equilibrium flame temperature. The equilibrium constant used

to evaluate the equilibrium flame temperature would be determined with respect to

CO 1

Although only the dissociation of CO

has been discussed, other products of com-

bustion may dissociate, for example

OH 1

When there are many dissociation reactions, the study of chemical equilibrium is

facilitated by the use of computers to solve the simultaneous equations that result.

equilibrium flame

temperature

14.4 Further Examples of the Use of the Equilibrium Constant 865

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866 Chapter 14 Chemical and Phase Equilibrium

Simultaneous reactions are considered in Sec. 14.4.4. The following example illustrates

how the equilibrium flame temperature is determined when one dissociation reaction

occurs.

Determining the Equilibrium Flame Temperature

c c c c EXAMPLE 14.6 c

Carbon monoxide at 258C, 1 atm enters a well-insulated reactor and reacts with the theoretical amount of air

entering at the same temperature and pressure. An equilibrium mixture of CO

, CO, O

, and N

exits the reactor

at a pressure of 1 atm. For steady-state operation and negligible effects of kinetic and potential energy, determine

the composition and temperature of the exiting mixture in K.

SOLUTION

Known:

Carbon monoxide at 258C, 1 atm reacts with the theoretical amount of air at 258C, 1 atm to form an

equilibrium mixture of CO

, CO, O

, and N

at temperature T and a pressure of 1 atm.

Find: Determine the composition and temperature of the exiting mixture.

Schematic and Given Data:

Analysis:

The overall reaction is the same as in the solution to Example 14.4

CO 1

1 1.88N

S z CO 1

1 11 2 z2CO

1 1.88N

By assumption 3, the exiting mixture is an equilibrium mixture. The equilibrium constant expression developed

in the solution to Example 14.4 is

K1T25

z1z

11 2 z2

ref

15.76 1 z2

(a)

Since p 5 1 atm, Eq. (a) reduces to

K1T25

11 2 z2

5.76 1 z

(b)

This equation involves two unknowns: z and the temperature T of the exiting equilibrium mixture.

Another equation involving the two unknowns is obtained from an energy rate balance of the form Eq. 13.12b,

which reduces with assumption 1 to

5 h

(c)

Engineering Model:

The control volume shown on the accompany-

ing sketch by a dashed line operates at steady

state with Q

5 0, W

5 0, and negligible

effects of kinetic and potential energy.

2. The entering gases are modeled as ideal gases.

3. The exiting mixture is an ideal gas mixture at

equilibrium wherein N

is inert.

Fig. E14.6

Air

25°C, 1 atm

Insulation

(CO

, CO, O

, N

)

T, 1 atm

25°

C, 1 atm

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As illustrated by Example 14.7, the equation solver and property retrieval features

of Interactive Thermodynamics: IT allow the equilibrium flame temperature and com-

position to be determined without the iteration required when using table data.

c c c c EXAMPLE 14.7 c

Solve Example 14.6 using Interactive Thermodynamics: IT and plot equilibrium flame temperature and z, the

amount of CO present in the exiting mixture, each versus pressure ranging from 1 to 10 atm.

SOLUTION

Known:

See Example 14.6.

Find: Using IT, plot the equilibrium flame temperature and the amount of CO present in the exiting mixture of

Example 14.6, each versus pressure ranging from 1 to 10 atm.

Engineering Model: See Example 14.6.

Analysis: Equation (a) of Example 14.6 provides the point of departure for the IT solution

K1T 25

z1z

11 2 z2

ref

15.76 1 z2

(a)

For a given pressure, this expression involves two unknowns: z and T.

Also, from Example 14.6, we use the energy balance, Eq. (c)

5 h

(c)

where

5 1h

1 1.881h

Determining the Equilibrium Flame Temperature Using Software

If the CO and air each entered at 5008C, would the equilibrium

flame temperature increase, decrease, or stay constant? Ans. Increase.

where

5 1h8

1 ¢h

1h8

1 ¢h

1 1.881h8

1 ¢h

and

5 z1h8

1 ¢h2

1h8

1 ¢h2

1 11 2 z21h8

1 ¢h2

1 1.881h8

1 ¢h2

The enthalpy of formation terms set to zero are those for oxygen and nitrogen.

Since the reactants enter at 258C, the corresponding

¢h

terms also vanish. Col-

lecting and rearranging, we get

z1¢h2

1¢h2

1 11 2 z21¢h2

1 1.881¢h2

1 11 2 z231h8

2 1h8

45 0 (d)

Equations (b) and (d) are simultaneous equations involving the unknowns

z and T. When solved iteratively using tabular data, the results are z 5 0.125

and T 5 2399 K, as can be verified. The composition of the equilibrium mixture,

in kmol per kmol of CO entering the reactor, is then 0.125CO, 0.0625O

, 0.875CO

1.88N

Ability to…

❑

apply Eq. 14.35 together

with the energy balance

for reacting systems to

determine equilibrium flame

temperature.

❑

retrieve and use data from

Tables A-23, A-25, and

A-27.

✓

Skills Developed

14.4 Further Examples of the Use of the Equilibrium Constant 867

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868 Chapter 14 Chemical and Phase Equilibrium

and

5 z1h

1 1z

221h

1 11 2 z21h

1 1.881h

where the subscripts R and P denote reactants and products, respectively, and z denotes the amount of CO in

the products, in kmol per kmol of CO entering.

With pressure known, Eqs. (a) and (c) can be solved for T and z using the following IT code. Choosing SI

from the Units menu and amount of substance in moles, and letting hCO_R denote the specific enthalpy of CO

in the reactants, and so on, we have

// Given data

TR 5 25 1 273.15 // K

p 5 1 // atm

pref 5 1 // atm

// Evaluating the equilibrium constant using Eq. (a)

K 5 ((z * (z/2)^0.5) / (1 – z)) * ((p / pref) / ((5.76 1 z) / 2))^0.5

// Energy balance: Eq. (c)

hR 5 hP

hR 5 hCO_R 1 (1/2) * hO2_R 1 1.88 * hN2_R

hP 5 z * hCO_P 1 (z /2) * hO2_P 1 (1 2 z) * hCO2_P 1 1.88 * hN2_P

hCO_R 5 h_T(“CO”,TR)

hO2_R 5 h_T(“O2”,TR)

hN2_R 5 h_T(“N2”,TR)

hCO_P 5 h_T(“CO”,T)

hO2_P 5 h_T(“O2”,T)

hCO2_P 5 h_T (“CO2”,T)

hN2_P 5 h_T (“N2”,T)

/* To obtain data for the equilibrium constant use the Look-up Table

option under the Edit menu. Load the file “eqco2.lut”. Data for

➊ CO2

CO 1 1/2 O2 from Table A-27 are stored in the look-up table

as T in column 1 and log10(K) in column 2. To retrieve the data use */

log(K) 5 lookupvall(eqco2, 1, T,2)

Obtain a solution for p 5 1 using the Solve button. To ensure rapid convergence, restrict T and K to positive

values, and set a lower limit of 0.001 and an upper limit of 0.999 for z. The results are T 5 2399 K and z 5

0.1249, which agree with the values obtained in Example 14.6.

Now, use the Explore button and sweep p from 1 to 10 atm in steps of 0.01. Using the Graph button, construct

the following plots:

Fig. E14.7

T (K)

2500

2450

2400

2350

2300

24681013579

p (atm)

0.2

0.15

0.1

0.05

24681013579

p (atm)

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14.4.2

Van’t Hoff Equation

The dependence of the equilibrium constant on temperature exhibited by the values

of Table A-27 follows from Eq. 14.31. An alternative way to express this dependence

is given by the van’t Hoff equation, Eq. 14.43b.

The development of this equation begins by introducing Eq. 14.29b into Eq. 14.31

to obtain on rearrangement

RT ln K 5231n

1 n

2 n

22 T 1n

s 8

1 n

s 8

2 n

s 8

2 n

(14.41)

Each of the specific enthalpies and entropies in this equation depends on temperature

alone. Differentiating with respect to temperature

d ln K

1 R ln K 52cn

2 T

s 8

b1 n

d h

2 T

d s 8

2 n

d h

2 T

d s 8

b2 n

d h

2 T

d s 8

1 1n

s 8

1 n

s 8

2 n

s 8

2 n

s 8

From the definition of s 81T2 (Eq. 6.19), we have d s 8

dT 5 c

T. Moreover, dh

dT 5 c

Accordingly, each of the underlined terms in the above equation vanishes identically,

leaving

d ln

R ln K 5 1n

s 8

1 n

s 8

2 n

s 8

2 n

s 8

(14.42)

Using Eq. 14.41 to evaluate the second term on the left and simplifying the resulting

expression, Eq. 14.42 becomes

d ln K

1 n

2 n

(14.43a)

or, expressed more concisely

d ln K

¢H

(14.43b)

which is the van’t Hoff equation.

In Eq. 14.43b, ¢H is the enthalpy of reaction at temperature T. The van’t Hoff equa-

tion shows that when ¢H is negative (exothermic reaction), K decreases with tempera-

ture, whereas for ¢H positive (endothermic reaction), K increases with temperature.

From Fig. E14.7, we see that as pressure increases more CO is oxidized to CO

(z decreases) and temperature increases.

➊ Similar files are included in IT for each of the reactions in Table A-27.

If the CO and air each entered at 5008C, determine the equi-

librium flame temperature in K using Interactive Thermodynamics: IT.

Ans. 2575.

Ability to…

❑

apply Eq. 14.35 together

with the energy balance

for reacting systems to

determine equilibrium flame

temperature.

❑

perform equilibrium calcula-

tions using Interactive

Thermodynamics: IT.

✓Skills Developed

van’t Hoff equation

14.4 Further Examples of the Use of the Equilibrium Constant 869

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870 Chapter 14 Chemical and Phase Equilibrium

The enthalpy of reaction ¢H is often very nearly constant over a rather wide

interval of temperature. In such cases, Eq. 14.43b can be integrated to yield

¢H

(14.44)

where K

and K

denote the equilibrium constants at temperatures T

and T

, respec-

tively. This equation shows that ln K is linear in 1/T. Accordingly, plots of ln K versus

1/T can be used to determine ¢H from experimental equilibrium composition data.

Alternatively, the equilibrium constant can be determined using enthalpy data.

14.4.3

Ionization

The methods developed for determining the equilibrium composition of a reactive

ideal gas mixture can be applied to systems involving ionized gases, also known as

plasmas. In previous sections we considered the chemical equilibrium of systems where

dissociation is a factor. For example, the dissociation reaction of diatomic nitrogen

can occur at elevated temperatures. At still higher temperatures, ionization may take

place according to

1 e

(14.45)

That is, a nitrogen atom loses an electron, yielding a singly ionized nitrogen atom N

and a free electron e

. Further heating can result in the loss of additional electrons

until all electrons have been removed from the atom.

For some cases of practical interest, it is reasonable to think of the neutral atoms,

positive ions, and electrons as forming an ideal gas mixture. With this idealization,

ionization equilibrium can be treated in the same manner as the chemical equilibrium

of reacting ideal gas mixtures. The change in the Gibbs function for the equilibrium

ionization reaction required to evaluate the ionization-equilibrium constant can be

calculated as a function of temperature by using the procedures of statistical thermo-

dynamics. In general, the extent of ionization increases as the temperature is raised

and the pressure is lowered.

Example 14.8 illustrates the analysis of ionization equilibrium.

Considering Ionization Equilibrium

c c c c EXAMPLE 14.8 c

Consider an equilibrium mixture at 36008R consisting of Cs, Cs

, and e

, where Cs denotes neutral cesium atoms,

singly ionized cesium ions, and e

free electrons. The ionization-equilibrium constant at this temperature for

1 e

is K 5 15.63. Determine the pressure, in atmospheres, if the ionization of Cs is 95% complete, and plot percent

completion of ionization versus pressure ranging from 0 to 10 atm.

SOLUTION

Known:

An equilibrium mixture of Cs, Cs

, e

is at 36008R. The value of the equilibrium constant at this tem-

perature is known.

Find: Determine the pressure of the mixture if the ionization of Cs is 95% complete. Plot percent completion

versus pressure.

Engineering Model: Equilibrium can be treated in this case using ideal gas mixture equilibrium considerations.

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14.4.4

Simultaneous Reactions

Let us return to the discussion of Sec. 14.2 and consider the possibility of more than

one reaction among the substances present within a system. For the present applica-

tion, the closed system is assumed to contain a mixture of eight components A, B, C,

D, E, L, M, and N, subject to two independent reactions

(1) n

A 1 n

C 1 n

D (14.24)

(2) n

A 1 n

M 1 n

N (14.46)

As in Sec. 14.2, component E is inert. Also, note that component A has been taken

as common to both reactions but with a possibly different stoichiometric coefficient

is not necessarily equal to n

14.4 Further Examples of the Use of the Equilibrium Constant 871

Solving Eq. (a) for z, determine the percent of ionization of

Cs at T 5 28808R (K 5 0.78) and p 5 1 atm. Ans. 66.2%.

Analysis: The ionization of cesium to form a mixture of Cs, Cs

, and e

is described by

Cs S 11 2 z2Cs 1 z Cs

1 ze

where z denotes the extent of ionization, ranging from 0 to 1. The total number of moles of mixture n is

n 5 11 2 z21 z 1 z 5 1 1 z

At equilibrium, we have Cs

1 e

, so Eq. 14.35 takes the form

K 5

1z21z

11 2 z2

ref

11 1 z2

11121

5 a

1 2 z

b a

ref

(a)

Solving for the ratio p

ref

and introducing the known value of K

ref

5 115.632 a

1 2 z

For p

ref

5 1 atm and z 5 0.95 (95%), p 5 1.69 atm. Using an equation-solver and plotting package, the following

plot can be constructed:

Ability to…

❑

apply Eq. 14.35 to deter-

mine the extent of ioniza-

tion of cesium given tem-

perature and pressure.

✓

Skills Developed

Figure E14.8 shows that ionization tends to occur to a lesser extent as pressure is raised. Ionization also tends

to occur to a greater extent as temperature is raised at fixed pressure.

Fig. E14.8

z (%)

100

0246810

p (atm)

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872 Chapter 14 Chemical and Phase Equilibrium

The stoichiometric coefficients of the above equations do not correspond to the

numbers of moles of the respective components present within the system, but changes

in the amounts of the components are related to the stoichiometric coefficients by

2dn

(14.25a)

following from Eq. 14.24, and

2dn

A¿

2dn

(14.47a)

following from Eq. 14.46. Introducing a proportionality factor de

, Eqs. 14.25a may

be represented by

52n





52n

5 n

, dn

5 n

(14.25b)

Similarly, with the proportionality factor de

, Eqs. 14.47a may be represented by

52n

A¿



52n

5 n

, dn

5 n

(14.47b)

Component A is involved in both reactions, so the total change in A is given by

52n

2 n

A¿

(14.48)

Also, we have dn

5 0 because component E is inert.

For the system under present consideration, Eq. 14.10 is

dG4

T, p

5 m

1 m

(14.49)

Introducing the above expressions giving the changes in the n’s, this becomes

dG4

T, p

5 12n

2 n

1 n

2 de

1 12n

A¿

2 n

1 n

2 de

(14.50)

Since the two reactions are independent, de

and de

can be independently varied.

Accordingly, when dG4

T, p

5 0, the terms in parentheses must be zero and two equations

of reaction equilibrium result, one corresponding to each of the foregoing reactions:

1 n

5 n

1 n

(14.26b)

A¿

1 n

5 n

1 n

(14.51)

The first of these equations is exactly the same as that obtained in Sec. 14.2. For

the case of reacting ideal gas mixtures, this equation can be expressed as

¢G8

5 ln c

ref

(14.52)

Similarly, Eq. 14.51 can be expressed as

¢G 8

5 ln c

A¿

ref

2 n

A¿

(14.53)

In each of these equations, the DG8 term is evaluated as the change in Gibbs function

for the respective reaction, regarding each reactant and product as separate at tem-

perature T and a pressure of 1 atm.

From Eq. 14.52 follows the equilibrium constant

ref

(14.54)

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