m 4.65 Water-Gas Shift Problem At low to moderate pressures, the equilibrium sta
ID: 482047 • Letter: M
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.Explanation / Answer
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
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