1. Chemical Reaction Equilibria
Part V

2. Multireaction equilibria
The phase rule is modified including the number of chemical reactions:
F=2-p+N-r
There is one equilibrium constant Kj for each reaction; one extent of reaction ej for each reaction.

3. Example 1
A feedstock of pure n-butane is cracked at 750 K and 1.2 bar to produce olefins. The rxns are:
C4H10C2H4 + C2H (I)
C4H10C3H6 + CH (II)
The equilibrium constants are KI = and KII = If the rxns reach equilibrium, what is the product composition?

4. solution
nI = 1; nII = 1; n0 = 1 yC4H10 = (1-eI-eII)/(1+eI+eII) yC2H4=yC2H6= eI/(1+eI+eII) yC3H6=yCH4= eII/(1+eI+eII) Assuming ideal gas mixture (low pressure):
eII =(KII/KI)1/2 eI

5. Using the expressions of yi in terms of eI and eII, we get
Solving: eII = ; eI =
And the product compositions are:
yC4H10 = 0.001; yC2H4 = yC2H6 = ;
yC3H6=yCH4=0.4461

6. 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 DG of formation for each compound are available at each temperature

7. DGo data (J/mol) The feed contains 1 mol of steam and 2.38 mol of air:
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

8. 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/fio = 1
We can write the equilibrium constants for the three rxns:

9. Initially, nH2O=1, nO2 =0.5, nN2 = 1.88 nI = -1/2, nII = 1/2, nIII = 0
Then we substitute these expressions in the KI, KII, KIII equations

10. Since we know the DGs 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

11. Therefore we can reformulate the rxns removing O2
C + CO2  2CO (a)
H2O + C  H2 + CO (b)
Then re-write
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

12. Write mole fractions as functions of ea and eb
C + CO2  2CO (a)
H2O + C  H2 + CO (b)

13. Now write the equilibrium constants in terms of ea and eb
From the data given calculate DGo at 1000 K = -4,690 J/mol
Calculate Ka =1.758 and Kb = 2.561
for P/Po =20 use the two above expressions to get ea and eb

15. 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?

16. solution At 298 K from table C for the 1st rxn
DH298K = J/mol; DG 298K= J/mol, and calculate DG1300 K = x 105 J/mol,
K1 = 13845
For the 2nd rxn, see Problem 13.32, DH298K = J/mol; DG 298K= J/mol, and calculate DG1300 K = x 103 J/mol,
K2 =
Since K1 >> K2, 1st rxn is primary rxn.
Since n = 2 (>0), rxn shifts to left when P increases

17. 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???

18. 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 = 5
yCO=yH2O =(1-e)/5, yCO2 =e/5, yH2=(3+e)/5
Solve for e, get yH2/yCO

19. 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.

20. 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+CO2
Calculate DG at 1300 K = 5.674x104 J/mol
K = 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.