ater-Gas At low to moderate pressures, the equilibrium state of the water-gas sh

Question

ater-Gas At low to moderate pressures, the equilibrium state of the water-gas shift reaction CO+H20Co2H2 is approximately described by the relation Keg (T) = 0.0247 expl 4020/T(K = yco Vit2o where T is the reactor temperature, Keq is the reaction equilibrium constant, and y is the mole fraction of species i in the reactor contents at equilibrium The feed to a batch shift reactor contains 20.0 m01% co, 15.0 mol% CO2, 50.0 mol% H20, and the balance an inert gas. The reactor s maintained at T: 423 K Degrees of Freedom Stoichiometry Equilibrium Constant Equilibrium Mole Fractions Calculate the mole fractions in the reactor at equilibrium. H2: H20: CO: Co2: Inerts:

Explanation / Answer

# Taking basis as 100 moles of feed initially

Moles of CO = 20 moles

Moles of CO2= 15 moles

Moles of H2O = 50 moles

Moles of Inert = 100-(20+15+50) = 15 moles

# at T = 1423 K, equilibrium constant Keq=0.0247exp[4020/1423] = 0.4165

# at equilibrium- let moles of CO and H2O consumed be x, then

moles of CO = 20 - x

moles of H2O = 50 - x

moles of CO2 = 15 + x

moles of H2 = x

putting these value in above equation we get

[x][15+x]/[20-x][50-x] = 0.4165

# solving for x, we get x = 8.482

# hence, at equilibrium

moles of CO = 20 - x = 11.518

moles of H2O = 50 - x = 41.518

moles of CO2 = 15 + x = 23.482

moles of H2 = x = 8.482

moles of inert = 15

# Total moles = 11.518 + 41.518 + 23.482 + 8.482 + 15 = 100

Mole fraction of CO = yco = 0.11518

Mole fraction of H2O = yH2O = 0.41518

Mole fraction of CO2 = 0.23482

Mole fraction of H2 = 0.08482

Mole fraction of inert = 0.15000

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