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(12)Kff

i0

iii

The expression of K in terms of fugacity coefficient is:

The standard state for a gas is the ideal-gas state of the

pure gas at the standard-state pressure P0of 1 bar.

Since the fugacity of an ideal gas is equal to its pressure,fi

0= P0for each species i .

Thus for gas-phase reactions , and Eq. (12)

becomes:

0

i

0

ii Pfff

KP

f i

0

i

i

(26)

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The equilibrium constant K is a function of temperature

only.

However, Eq. (26) relates K to fugacities of the reacting

species as they exist in the real equilibrium mixture.

These fugacities reflect the nonidealities of the equili-brium mixture and are functions of temperature,

pressure, and composition.

This means that for a fixed temperature the

composition at equilibrium must change with pressure

in such a way that remains constanti0

ii

Pf

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The fugacity is related to the fugacity coefficient by

Substitution of this equation into Eq. (26) provides an

equilibrium expression displaying the pressure and the

composition:

Py

f

iii

KP

Py

0iii

i

Where = iiand P0i s the standard-state pressure of 1bar, expressed in the same units used for P.

(27)

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If the assumption that the equilibrium mixture is an ideal

solution is justified, then each becomes i, the fugacitycoefficient of pure species i at T and P.In this case, Eq. (27) becomes:

i

KPP

y 0iiii

(27)For pressures sufficiently low or temperatures sufficiently

high, the equilibrium mixture behaves essentially as an

ideal gas. In this event, each i= 1, and Eq. (27) reduces to:K

P

Py

0ii

i

(28)

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Although Eq. (28) holds only for an ideal-gas reaction,

we can base some conclusions on it that are true in

general:According to Eq. (20), the effect of temperature on

the equilibrium constant K is determined by the sign

of H0:

o H0> 0 (the reaction is endothermic) T>> K >>.Eq. (28) shows that K >> at constant P >>>>

o H0< 0 (the reaction is exothermic) T>> K

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If the total stoichiometric number ( ii) isnegative, Eq. (28) shows that am increase in P atconstant T causes an increase in , implying a

shift of the reaction to the right.

If the total stoichiometric number ( ii) ispositive, Eq. (28) shows that am increase in P at

constant T causes a decrease in , implying a

shift of the reaction to the left, and a decrease in e.

i

ii

y

ii

i

y

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For a reaction occurring in the liquid phase, we return to

(12)Kff i0ii

i

For the usual standard state for liquids f0iis the fugacity

of pure liquid i at the temperature of the system and at 1

bar.

The activity coefficient is related to fugacity according to:

iiii fxf (29)

The fugacity ratio can now be expressed

0

i

i

ii0

i

iii

0

i

i

f

fx

f

fx

f

f(30)

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(31)

Gibbs free energy for pure species i in its standard state at

the same temperature:

0

ii

0

i flnRTTG (7)

Gibbs free energy for pure species i at P and the same

temperature:

iii flnRTTG (7.a)

The difference between these two equations is:

0

i

i0

iif

flnRTGG

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Fundamental equation for Gibbs energy:

dTSdPVdGiii

(32)For a constant-temperature process:

dPVdGii

(32)For a pure substance undergone a constant-temperature

process from P0to P, the Gibbs free energy change is:

PP

i

G

G 0

i

0

i

dPVdG (33)

(34)0

i

0

ii PPVGG

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Combining eqs. (31) and (34) yields:

(35)

RTPPV

ffln

0

i0

i

i

RT

PPVexpx

f

f 0

i

ii0

i

i

(36)

RTPPV

expf

f 0

i

0i

ior


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KRTPPVexpx0

iii

i


KRT

PPV

expx

u

i

0

i

iiii

uiRT

PPVexpKx

0

i

iii

i

i

ii

0

iii

V

RT

PPexpKx i (37)

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For low and moderate pressure, the exponential term is

close to unity and may be omitted. Then,

Kx i

iii (38)

and the only problem is determination of the activity

coefficients.

An equation such as the Wilson equation or the UNIFAC

method can in principle be applied, and the compositions

can be found from eq. (38) by a complex iterative

computer program.However, the relative ease of experimental investigation

for liquid mixtures has worked against the application of

Eq. (38).

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If the equilibrium mixture is an ideal solution, then iisunity, and Eq. (38) becomes:

Kx i

ii

(39)This relation is known as THE LAW OF MASS ACTION.

Since liquids often form non-ideal solutions, Eq. (39) can

be expected in many instances to yield poor results.

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Suppose a single reaction occurs in a homogeneous

system, and suppose the equilibrium constant is known.

In this event, the calculation of the phase composition

at equilibrium is straightforward if the phase is

assumed an ideal gas [Eq. (28)] or an ideal solution [Eq.

(27) or (39)].

When an assumption of ideality is not reasonable, theproblem is still tractable for gas-phase reactions

through application of an equation of state and solution

by computer.

For heterogeneous systems, where more than one

phase is present, the problem is more complicated and

requires the superposition of the criterion for phase

equilibrium developed in Sec. 11.6

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Single-Phase Reactions

The water-gas shift reaction,

CO (g) + H2O (g) CO2(g) + H2(g)Is carried out under the different set of conditions below.Calculate the fraction of steam reacted in each case.

Assume the mixture behaves as an ideal gas.

The reactants consist of 1 mol of H2

O vapor and 1 mol of

CO. The temperature is 1100 K and the pressure is 1 bar.

Example

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Solution

CO H2O CO2 H2 1 1 + 1 + 1A 3,376 3,470 5,547 3,249

B 103 0,557 1,450 1,045 0,422C 106 4,392 0 0 0D 10-5 0,031 0,121 1,157 0,083

H0

f,298

110.525

241.818

393.509 0G0f,298 137.169 228.572 394.359 0

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950,1A3

10540,0B

610392,4C

5

10164,1D 10

298 molJ166.41H

10

298 molJ618.28G

260.10315,298314,8

618.28exp

RT

GexpK

0

0

0

0

20

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TT

1RT

HexpK 0

0

0

0

1

610527,5110015,298115,298314,8 166.41exp

6894,315,298

1100

T

T

0

189,2K

2 189,28354,310527,5261.103KKKK 6

210

21

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01111i

i

Since the reaction mixture is an ideal gas:

Ky iii

189,2Kyy

yy

OHCO

HCO

2

22

eiii 0nn

e0

nn

The number of each species at equilibrium is:

While total number of all species at equilibrium is:

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eCO 1n

eOH 1n

2

eCO2n eH2

n

2n

2

1y e

CO

2

1y e

OH2

2y e

CO2

2

y eH2

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189,2Kyy

yy

OHCO

HCO

2

22

189,21 22

222 21189,21189,2 0189,2378,4189,1

2 5436,0

e

Therefore the fraction of the steam that reacts is 0.5

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Estimate the maximum conversion of ethylene to ethanol

by vapor-phase hydration at 523.15 K and 1.5 bars for aninitial steam-to-ethylene ratio of 5.

Example

Solution

Reaction:

C2H4(g) + H2O (g) C2H5OH (g)

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C2H4 H2O C2H5OH 1 1 + 1A 1,424 3,470 3,518

B 14,394 10-3 1,450 10-3 20,001 10-3C 4,392 10-6 0 6,002 10-6D 0 0,121 105 0

H0f,298 52.510 241.818 235.100

G0f,298 68.460 228.572 168.490

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376,1A 3

10157,4B

610610,1C

5

10121,0D 10

298 molJ792.45H

10

298 molJ378.8G

366,29

15,298314,8

378.8exp

RT

GexpK

0

0

0

0

27

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TT

1RT

HexpK 0

0

0

0

1

4105,315,523 15,298115,298314,8 792.45exp

7547,115,298

15,523

T

T

0

9778,0K2

34210

1005.109778,0105,3366,29KKKK

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1111i

i

KP

Py

0ii

i

KPPyy y 0OHHC

OHHC

242

52

KPPy 0ii i

For hign temperature and sufficiently low pressure:

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eiii 0nn

ee0

6nn

The number of each species at equilibrium is:

While total number of all species at equilibrium is:

eHC 5n

42

eOH 1n 2 eOHHC 52

n

e

e

HC6

5y

42

e

eOH

61y

2

e

e

OHHC

6

y52

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KPPyy y 0OHHC

OHHC

242

52

33ee

ee 1007.151005.105.115

6

ee

3

ee 151007.156

00754.009045.601507.1e

2

e

0135.0e

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The gas-phase oxidation of SO2to SO3is carried out at a

pressure of 1 bar with 20% excess air in an adiabaticreactor. Assuming that the reactants enter at 298.15 K and

that equilibrium is attained at the exit, determine the

composition and temperature of the product stream from

the reactor.

Example

Solution

Reaction:

SO2(g) + O2(g) SO3(g)

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Basis: 1 mole of SO2entering the reactor:

moles of O2entering = (0.5) (1.2) = 0.6

moles of N2entering = (0.6) (79/21) = 2.257

The amount of each species in the product stream is:

eiii 0nn

eSO 1n

2

eO 5.06.0n 2

eSO3n

257.2n3N

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e

5.0857.3n Total amount of all species:

Mole fraction of each species:

e

e

SO5.0857.3

1y

2

e

e

O5.0857.3

5.06.0y

2

e

e

SO5.0857.3

y3

e

N5.0857.3

257.2y

2

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Energy balance:

Reactant

T = 298.15 K

SO2= 1

O2= 0.6

N2= 2.257

Product

T = 298.15 KSO2= 1eO2= 0.60.5 eN2= 2.257

Reactione

Product

T

SO2= 1eO2= 0.60.5 eN2= 2.257

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0HHH 0

Pe

0

298

i

T

T

P

i

0

P

0

i dTRCRnH

i

T

T

2

iiii

0

dTTDTBARn

T

T

2

iii

iii

iii

0

P

0

dTTDnTBnAnRH

0i

ii20

2i

ii

0i

iiT1

T1DnTT

2

Bn

TTAnR

(a)

(b)

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SO2

O2 SO

3

1 1 + 1A 5.699 3.639 8.060

B 0.801 10-3 0.506 10-3 1.056 10-3C 0 0 0

D 1.015 105 0.227 105 2.028 105H0f,298 296 830 0 395 720

G0f,298 300 194 0 371 060

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278.1A 3

10251,0B

0C 5

10786.0D 10

298 molJ98890H

10

298 molJ70866G

12

0

0

0

0 106054,2

15,298314,8

70866exp

RT

GexpK

38

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TT

1RT

HexpK 0

0

0

0

1

T15,298

115,298314,8

98890exp

TfK2

210 KKKK

T15,298

1894.39exp (c)

(d)

(e)

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5.0

e

e

e

e05

OSO

SO

5.06.05.0857.3

1yyyK

22

3

KKPPy

0ii

i

For hign temperature and pressure of 1 bar:

(f)

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Algorithm:

1. Assume a starting value of T

2. Evaluate K1[eq. (c)], K2 [eq. (d)], and K [eq. (e)]

3. Solve for e[eq. (f)]4. Evaluate T [eq. (a)]

5. Find a new value of T as the arithmatic mean value

just calculated and the initial value; return to step

2.

The scheme converges on the value e= 0.77 and T =855.7 K

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The composition of the product is:

0662.0472.3

23.0

77.05.0857.3

77.01

5.0857.3

1

ye

e

SO2

0619.0472.3

215.0

77.05.0857.3

77.05.06.0

y 2O

2218.0472.3

77.0y

3SO

6501.0472.3

257.2y

2N

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