Chemical Reaction Equilibria

Part V

Multireaction equilibria

• The phase rule is modified including the number of chemical reactions:

F=2-+N-r

There is one equilibrium constant Kj for each reaction; one extent of reaction j for each reaction.

Example 1A feedstock of pure n-butane is cracked at 750 K

and 1.2 bar to produce olefins. The rxns are:C4H10C2H4 + C2H6 (I)

C4H10C3H6 + CH4 (II)

The equilibrium constants are KI = 3.856 and KII = 268.4. If the rxns reach equilibrium, what is the product composition?

solutionI = 1; = 1; n0 = 1

yC4H10 = (1-I-II)/(1+I+II)

yC2H4=yC2H6= I/(1+I+II)

yC3H6=yCH4= II/(1+I+II)

Assuming ideal gas mixture (low pressure):

IIoHC

CHHC

IoHC

HCHC

KP

P

y

yy

KP

P

y

yy

1

104

463

1

104

6242

II =(KII/KI)1/2 I

Solving: II = 0.8914; I = 0.1068

And the product compositions are:yC4H10 = 0.001; yC2H4 = yC2H6 = 0.0534; yC3H6=yCH4=0.4461

Using the expressions of yi in terms of I and II, we get

IIoIIIIII

II

IoIIIIII

I

KP

P

KP

P

12

12

)1)(1(

)1)(1(

Example 2

• A bed of coal (assume pure carbon) in a coal gasifier is fed with steam and air, and produces a gas stream containing H2, CO, O2, H2O, CO2, and N2. If the feed contains 1 mol of steam and 2.38 mol of air, calculate the equilibrium composition of the gas stream at 20 bar for 1,000, 1,100, 1,200, 1,300, 1,400, and 1,500 K. The G of formation for each compound are available at each temperature

Go data (J/mol)T(K) H2O CO CO2

1,000 -192,420 -200,240 -395,790

1,100 -187,000 -209,110 -395,960

1,200 -181,380 -217,830 -396,020

1,300 -175,720 -226,530 -396,080

1,400 -170,020 -235,130 -396,130

1,500 -164,310 -243,740 -396,160

The feed contains 1 mol of steam and 2.38 mol of air:O2 = 0.21 x2.38 = 0.5 mol; N2 = 0.79 x2.38 =1.88 mol

Species present at equilibrium: C, H2, CO, O2, H2O, CO2, and N2

Formation reactions:H2 + ½ O2 H2O (I)C+ ½ O2 CO (II)C+O2 CO2 (III)

All species are present as gases except carbon which is a pure solid phase; its ratio of fugacities fi/fi

o = 1We can write the equilibrium constants for the three rxns:

0

02

2

2/1

02/12

2/1

02

2/12

2

P

P

y

yK

P

P

y

yK

P

P

yy

yK

O

COIII

O

COII

HO

OHI

Initially, nH2O=1, nO2 =0.5, nN2 = 1.88

I = -1/2, II = 1/2, III = 0

2/)(38.3

88.1;2/)(38.3

1

2/)(38.3;2/)(38.3

2/)(38.3

)1(2/1;2/)(38.3

22

2

22

IIIN

III

IOH

III

IICO

III

IIICO

III

IIIIIIO

III

IH

yy

yy

yy

Then we substitute these expressions in the KI, KII, KIII equations

)21(

2

)2/)(38.3()21(

)/(2

1)21(

)/()2/)(38.3(2)(1(

2/12/1

2/10

2/1

2/102/1

IIIIII

IIIIII

IIIIIIIII

IIII

IIIIII

IIIII

K

PPK

PPK

Since we know the Gs we can calculate the equilibrium constants

All the calculated values are huge, KI = 106, K2 = 108,KIII=1014

That means that the mole fraction of O2 is really small

Therefore we can reformulate the rxns removing O2

• C + CO2 2CO (a)

• H2O + C H2 + CO (b)

• Then re-write

02

20

2

2

;P

P

y

yyK

P

P

y

yK

OH

COHb

CO

COa

In the feed: 1 mol H2O, 0.5 mol O2 and 1.88 mol N2

Since we eliminated O2, we substitute 0.5 mol O2 by 0.5 mol CO2

Write mole fractions as functions of a and b

baN

ba

aCO

ba

bOH

ba

baCO

ba

bH

y

yy

yy

38.3

88.1

38.3

5.0;

38.3

1

38.3

2;

38.3

2

22

2

C + CO2 2CO (a)H2O + C H2 + CO (b)

Now write the equilibrium constants in terms of a and b

0

0

2

)38.3)(1(

)2(

)38.3)(5.0(

)2(

P

PK

P

PK

bab

babb

baa

baa

From the data given calculate Go at 1000 K = -4,690 J/molCalculate Ka =1.758 and Kb = 2.561 for P/Po =20 use the two above expressions to get a and b

Example 3

• Synthesis gas may be produced by the catalytic reforming of methane with steam:

CH4 (g) + H2O(g) CO(g) +3H2(g)

CO(g) + H2O(g) CO2 (g) +H2(g)

Assume equilibrium is obtained at 1 bar and 1,300 K for both reactions

a) Would it be better to carry out the rxn at pressures above 1 bar?

solution• At 298 K from table C for the 1st rxn H298K = 205813 J/mol; G 298K= 141863 J/mol, and

calculate G1300 K = -1.031 x 105 J/mol,

K1 = 13845

For the 2nd rxn, see Problem 13.32, H298K = -41166 J/mol; G 298K= -28618 J/mol, and calculate G1300 K = 5.892 x 103 J/mol,

K2 = 0.5798

(a) Since K1 >> K2, 1st rxn is primary rxn. Since = 2 (>0), rxn shifts to left when P increases

Solution (cont)(b) Would it be better to carry out the rxn at T <

1,300 K?No, because the primary rxn is endotermic, so, it

shifts left when T decreases(c) Estimate the molar ratio of hydrogen to CO in

the synthesis gas if the feed consists of an equimolar mixture of steam and methane

CH4 (g) + H2O(g) CO(g) +3H2(g)

CO(g) + H2O(g) CO2 (g) +H2(g)

If the feed is equimolar, no more water for the 2nd rxn. Ratio H2/CO???

Solution (cont.)(d) Repeat (c) for a steam to methane mole ratio

of 2 in the feed.Rxn 2 may proceed. However rxn 1 proceeds to

completion and provides feed to rxn 2.CH4 (g) + H2O(g) CO(g) +3H2(g)CO(g) + H2O(g) CO2 (g) +H2(g)

1 mole CH4 and 2 mol H2O initially,Therefore for rxn 2 there is 1 mole H2O, 1 mol CO and 3 mol H2, no = 5yCO=yH2O =(1-)/5, yCO2 =/5, yH2=(3+)/5Solve for , get yH2/yCO

Solution (cont.)

• (e) how could the feed comp. be altered to yield a lower ratio of H2 to CO in the synthesis gas than that found in (c).

CH4 (g) + H2O(g) CO(g) +3H2(g)CO(g) + H2O(g) CO2 (g) +H2(g)

Add CO2 to the feed. Some H2 reacts with CO2 and generates more CO, lowering the H2/CO ratio.

Solution (cont)• Is there any danger that C will deposit by the rxn

2CO C+CO2 under the conditions of part (c)? In part d? If so, how could the feed be altered to prevent C deposition?

2CO C+CO2Calculate G at 1300 K = 5.674x104 J/molK = 5.255x10-3

Deposition of C depends on the ratio yCO2/(yCO)2

When for the actual compositions this ratio is higher than that given by K, there is no carbon deposition. If yCO2 0, there is danger of C deposition.