Chemical Reaction Equilibria
Part V
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.
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 + C2H6
(I)
C4H10C3H6 + CH4
(II)
The equilibrium constants are KI = 3.856 and KII =
268.4. If the rxns reach equilibrium, what is
the product composition?
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):
1
yC 2 H 4 yC 2 H 6  P 
  o  KI
yC 4 H 10
P 
1
yC 3 H 6 yCH 4  P 
  o  K II
yC 4 H 10
P 
eII =(KII/KI)1/2 eI
Using the expressions of yi in terms of eI and eII, we get
e
2
I
e
2
II
1
 P 
  o  KI
(1  e I  e II )(1  e I  e II )  P 
1
 P 
  o  K II
(1  e I  e II )(1  e I  e II )  P 
Solving:
eII = 0.8914; eI = 0.1068
And the product compositions are:
yC4H10 = 0.001; yC2H4 = yC2H6 = 0.0534;
yC3H6=yCH4=0.4461
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
DGo 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
C+ ½ O2  CO
C+O2  CO2
(I)
(II)
(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:
 y H 2O  P 
 0 
K I   1/ 2
 yO 2 y H 2  P 
 yCO  P 
K II   1/ 2  0 
 yO 2  P 
1/ 2
K III
 yCO 2  P 
 0 
 
 yO 2  P 
0
1 / 2
Initially, nH2O=1, nO2 =0.5, nN2 = 1.88
nI = -1/2, nII = 1/2, nIII = 0
yH 2
eI
1 / 2(1  e I  e II )  e III

; yO 2 
3.38  (e II  e I ) / 2
3.38  (e II  e I ) / 2
e III
e II
yCO 2 
; yCO 
3.38  (e II  e I ) / 2
3.38  (e II  e I ) / 2
1 e I
1.88
y H 2O 
; yN 2 
3.38  (e II  e I ) / 2
3.38  (e II  e I ) / 2
Then we substitute these expressions in the KI, KII, KIII equations
(1  e I )( 2(3.38  (e II  e I ) / 2)1/ 2 ( P / P 0 ) 1/ 2
KI 
(1  e I  e II  2e III )1/ 2  1
2e II ( P / P 0 )1/ 2
K II 
(1  e I  e II  2e III )1/ 2  (3.38  (e II  e I ) / 2)1/ 2
K III
2e III

(1  e I  e II  2e III )
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
Therefore we can reformulate the rxns
removing O2
• C + CO2  2CO
• H2O + C  H2 + CO
• Then re-write
(a)
(b)
y  P
y H 2 yCO
Ka 
 0 ; K b 
yCO 2  P 
y H 2O
2
CO
 P
 0
P 
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 ea
and eb
C + CO2  2CO
H2O + C  H2 + CO
(a)
(b)
eb
2e a  e b
yH 2 
; yCO 
3.38  e a  e b
3.38  e a  e b
1 eb
0.5  e a
y H 2O 
; yCO 2 
3.38  e a  e b
3.38  e a  e b
yN 2
1.88

3.38  e a  e b
Now write the equilibrium constants in
terms of ea and eb
(2e a  e b ) 2
 P 
Ka 
 0
(0.5  e a )(3.38  e a  e b )  P 
e b (2e a  e b )
 P 
Kb 
 0
(1  e b )(3.38  e a  e b )  P 
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
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
DH298K = 205813 J/mol; DG 298K= 141863 J/mol, and
calculate DG1300 K = -1.031 x 105 J/mol,
K1 = 13845
For the 2nd rxn, see Problem 13.32, DH298K = -41166
J/mol; DG 298K= -28618 J/mol, and calculate
DG1300 K = 5.892 x 103 J/mol,
K2 = 0.5798
(a) Since K1 >> K2, 1st rxn is primary rxn.
Since n = 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 = 5
yCO=yH2O =(1-e)/5, yCO2 =e/5, yH2=(3+e)/5
Solve for e, 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+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.