KESEIMBANGANREAKSI KIMIA

(2)

RELATION OFEQUILIBRIUM CONSTANTS

TO COMPOSITION

(12) Kff i0ii

i

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 P0 of 1 bar.

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

0 = P0 for each species i .

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

0i

0ii Pfff

KPf i

0i

i

(26)

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 constant i0

ii

Pf

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:

Pyˆf iii

KPPˆy 0ii

i

i

Where = ii and P0 i s the standard-state pressure of 1 bar, expressed in the same units used for P.

(27)

If the assumption that the equilibrium mixture is an ideal solution is justified, then each becomes i, the fugacity coefficient of pure species i at T and P.

In this case, Eq. (27) becomes:

i

KPPy 0ii

i

i

(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:

KPPy 0i

i

i

(28)

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 <<. Eq. (28) shows that K << at constant P << e <<

i

ii

y

i

ii

y

If the total stoichiometric number ( ii) is negative, Eq. (28) shows that am increase in P at constant T causes an increase in , implying a shift of the reaction to the right.

If the total stoichiometric number ( ii) is positive, 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

i

ii

y

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

LIQUID PHASE REACTION

(12) Kff i0ii

i

For the usual standard state for liquids f0i is 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

iii0

i

iii0

i

i

ffx

ffx

ff

(30)

(31)

Gibbs free energy for pure species i in its standard state at the same temperature:

0ii

0i 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:

0i

i0ii f

flnRTGG

Fundamental equation for Gibbs energy:

dTSdPVdG iii (32)

For a constant-temperature process:

dPVdG ii (32)

For a pure substance undergone a constant-temperature process from P0 to P, the Gibbs free energy change is:

P

Pi

G

G 0

i

0i

dPVdG (33)

(34) 0i

0ii PPVGG

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

(35)

RT

PPVffln

0i

0i

i

RT

PPVexpxff 0

iii0

i

i

(36)

RT

PPVexpff 0

i0

i

ior

Combining eqs. (31) and (35) yields:

K

RTPPVexpx

0i

iii

Combining eqs. (36) and (12) yields:

K

RTPPVexpx

u

i

0i

iiii

u

i

RTPPVexpKx

0i

iiii

iii

0

iiiV

RTPPexpKx i (37)

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).

If the equilibrium mixture is an ideal solution, then i is unity, 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.

EQUILIBRIUM CONVERSIONSFOR SINGLE REACTIONS

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, the problem 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

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 H2O vapor and 1 mol of CO. The temperature is 1100 K and the pressure is 1 bar.

Example

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 0,083H0

f,298 – 110.525 – 241.818 – 393.509 0

G0f,298 – 137.169 – 228.572 – 394.359 0

950,1A

310540,0B

610392,4C

510164,1D

10298 molJ166.41H

10298 molJ618.28G

260.103

15,298314,8618.28exp

RTGexpK0

00

0

20

TT1

RTHexpK 0

0

00

1

610527,5

110015,2981

15,298314,8166.41exp

6894,315,298

1100TT

0

189,2K2

189,28354,310527,5261.103KKKK 6210

21

01111i

i

Since the reaction mixture is an ideal gas:

Ky i

ii

189,2Kyyyy

OHCO

HCO

2

22

eiii 0nn

e0nn

The number of each species at equilibrium is:

While total number of all species at equilibrium is:

eCO 1n

eOH 1n2

eCO2n

eH2n

2n

21y e

CO

21y e

OH2

2y e

CO2

2y e

H2

189,2Kyyyy

OHCO

HCO

2

22

189,2

1 2

2

222 21189,21189,2

0189,2378,4189,1 2

5436,0e

Therefore the fraction of the steam that reacts is 0.5

Estimate the maximum conversion of ethylene to ethanol by vapor-phase hydration at 523.15 K and 1.5 bars for an initial steam-to-ethylene ratio of 5.

Example

Solution

Reaction:C2H4 (g) + H2O (g) C2H5OH (g)

C2H4 H2O C2H5OH – 1 – 1 + 1A 1,424 3,470 3,518B 14,394 10-3 1,450 10-3 20,001 10-3

C – 4,392 10-6 0 – 6,002 10-

6

D 0 0,121 105 0H0

f,298 52.510 – 241.818 – 235.100G0

f,298 68.460 – 228.572 – 168.490

376,1A

310157,4B

610610,1C

510121,0D

10298 molJ792.45H

10298 molJ378.8G

366,29

15,298314,8378.8exp

RTGexpK0

00

0

27

TT1

RTHexpK 0

0

00

1

4105,3

15,52315,2981

15,298314,8792.45exp

7547,115,29815,523

TT

0

9778,0K2

34210 1005.109778,0105,3366,29KKKK

1111i

i

KPPy 0i

i

i

KPP

yyy

0OHHC

OHHC

242

52

KPPy 0i

i

i

For hign temperature and sufficiently low pressure:

eiii 0nn

ee0 6nn

The number of each species at equilibrium is:

While total number of all species at equilibrium is:

eHC 5n42

eOH 1n2

eOHHC 52n

e

eHC 6

5y42

e

eOH 6

1y2

e

eOHHC 6

y52

KPP

yyy

0OHHC

OHHC

242

52

33

ee

ee 1007.151005.105.115

6

ee3

ee 151007.156

00754.009045.601507.1 e2e

0135.0e

The gas-phase oxidation of SO2 to SO3 is carried out at a pressure of 1 bar with 20% excess air in an adiabatic reactor. 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) + ½ O2 (g) SO3 (g)

Basis: 1 mole of SO2 entering the reactor:moles of O2 entering = (0.5) (1.2) = 0.6moles of N2 entering = (0.6) (79/21) = 2.257

The amount of each species in the product stream is:eiii 0

nn

eSO 1n2

eO 5.06.0n2

eSO3n

257.2n3N

e5.0857.3n

Total amount of all species:

Mole fraction of each species:

e

eSO 5.0857.3

1y2

e

eO 5.0857.3

5.06.0y2

e

eSO 5.0857.3

y3

eN 5.0857.3

257.2y2

Energy balance:

ReactantT = 298.15 KSO2 = 1O2 = 0.6N2 = 2.257

ProductT = 298.15 KSO2 = 1 – e O2 = 0.6 – 0.5 e

N2 = 2.257

Reactione

ProductTSO2 = 1 – e O2 = 0.6 – 0.5 e

N2 = 2.257

0HHH 0Pe

0298

i

T

T

Pi

0P

0

i dTRC

RnH

i

T

T

2iiii

0

dTTDTBARn

T

T

2

iii

iii

iii

0P

0

dTTDnTBnAnRH

0iii

20

2iii

0i

ii T1

T1DnTT

2

BnTTAnR

(a)

(b)

SO2 O2 SO3

– 1 – 1 + 1A 5.699 3.639 8.060B 0.801 10-3 0.506 10-3 1.056 10-3

C 0 0 0D – 1.015 105 – 0.227 105 – 2.028 105

H0f,298 – 296 830 0 – 395 720

G0f,298 – 300 194 0 – 371 060

278.1A

310251,0B

0C

510786.0D

10298 molJ98890H

10298 molJ70866G

12

0

00

0 106054,215,298314,8

70866expRT

GexpK

38

TT1

RTHexpK 0

0

00

1

T15,2981

15,298314,898890exp

TfK2

210 KKKK

T15,2981894.39exp (c)

(d)

(e)

5.0

e

e

e

e05

OSO

SO

5.06.05.0857.3

1yyy

K22

3

KKPPy 0i

i

i

For hign temperature and pressure of 1 bar:

(f)

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

The composition of the product is:

0662.0

472.323.0

77.05.0857.377.01

5.0857.31y

e

eSO2

0619.0472.3215.0

77.05.0857.377.05.06.0y

2O

2218.0472.377.0y

3SO

6501.0472.3257.2y

2N