Question

m 4.65 Water-Gas Shift Problem At low to moderate pressures, the equilibrium state of the water-gas shift reaction CO H20 CO2 H2 is approximately described by the relation co2 yH Keq T 0.0247 exp 4020/T K where Tis the reactor temperature, Keq s the reaction equilibrium constant, and yi is the mole fraction of species i in the reactor contents at equilibrium The feed to a batch shift reactor contains 20.0 mol% CO, 10.0 mol% CO2, 30.0 mol% H20, and the balance an inert gas. The reactor is maintained at T 1623 K.

Answer 1

First, calculate K:

K = 0.0247*exp(4020/1623)

K = 0.2940

now..

initially:

y-CO2 = 0.10

y-H2 = 0

y-H2O = 0.3

y-CO = 0.20

after reaction:

y-CO2 = 0.10 +x

y-H2 = 0+x

y-H2O = 0.3-x

y-CO = 0.20-x

so

(y-CO2 * y-H2 ) /(Y-CO * Y-H2O) = 0.2940

(0.10 +x)(x) / ((0.3-x)(0.20-x)) = 0.2940

0.1x + x^2 = 0.2940*(0.3*0.2 + 0.5x + x^2)

0.01764 + 0.147x + 0.2940x^2 = 0.1x + x^2

(1-0.2940)*x^2 + (0.1-0.147)x - 0.01764 = 0

0.706*x^2 - 0.047x -0.01764 = 0

x = 0.1948

y-CO2 = 0.10 +x = 0.1+0.1948 = 0.2948

y-H2 = 0+x = 0.1948

y-H2O = 0.3-x = 0.3-0.1948 = 0.1052

y-CO = 0.20-x = 0.2-0.1948 = 0.0052

inerts = 1- (0.2948+0.1948+0.1052+0.0052) = 0.4