Page 252 - Biomass Gasification, Pyrolysis And Torrefaction Practical Design and Theory

228 Biomass Gasification, Pyrolysis and Torrefaction

Example 7.3
For shift reaction CO 1 H 2 O-CO 2 1 H 2 , assume that the reaction begins with
1 mol of CO, 1 mol of H 2 O, and 1 mol of nitrogen. Find:
The equilibrium constant at 1100 K and 1 atm.
The equilibrium mole fraction of carbon dioxide.
Whether the reaction is endothermic or exothermic.
If pressure is increased to 100 atm, the impact of the equilibrium constant at
1100 K.
Solution
Part (a): For the shift reaction, the Gibbs free energy at a certain temperature
can be calculated from Eq. (7.57):
0
ΔG 5232:197 1 0:031T 2 ð1774:7=TÞ
0
at 1100 K, ΔG 5 0.2896 kJ/mol.
The equilibrium constant can be calculated from Eq. (7.56):
0
k for 2 ΔG
K equilibrium 5 5 exp
k back RT

20:2896
K equilibrium 5 exp
0:008314 3 1100
K equilibrium 5 0:9688
Part (b): At equilibrium, the rate of the forward reaction will be equal to the
rate of the backward reaction. So, using the definition of the equilibrium
constant, we have
p
K equilibrium 5 5 0:9688
p CO 2 H 2
p CO p H 2 O
where p denotes the partial pressure of the various species. In this reaction,
nitrogen stays inert and does not react. Thus, 1 mol of nitrogen comes out
from it. If x moles of CO and H 2 O react to form x moles of CO 2 and H 2 ,
then at equilibrium, (1 2 x) moles of CO and H 2 O remain unreacted. We
can list the component mole fraction as:
Species Mole Mole Fraction
CO (1 x) (1 x)/3
H 2 O (1 x) (1 x)/3
x x/3
CO 2
x x/3
H 2
1 1/3
N 2
The mole fraction y is related to the partial pressure, p, by the relation
yP 5 p, where P stands for total pressure.
Substituting the values for the partial pressures of the various species, we get:

ððx=3ÞPÞððx=3ÞPÞ
5 0:9688
ðð1 2 x=3ÞPÞðð1 2 x=3ÞPÞ