The water-gas shift reaction can be written as CO + H2O CO2 + H2. In a continuou

Question

The water-gas shift reaction can be written as CO + H2O CO2 + H2. In a continuous, steady-state reactor 1 mol CO and 2 mol H2O are fed to the system. The equilibrium constant, K at 100oC for the reaction is equal to 1. What is the composition of the product stream?

a. Following the general procedure from class (PFD, convert all units to same basis, write all species balances, DOF analysis, etc.), determine if it possible to solve this problem. If so, derive the algebraic equations that would allow you to solve the problem, but do not solve them. Instead, describe how you would use them to solve for the composition of the product stream—this can be a bulleted list.

Explanation / Answer

Basis 1 L of reaction mixture

Initial moles of CO = 1 mol

Initial moles of H2O = 2 mol

Let the moles of CO converted = x mol

Moles of H2O converted = 2x mol

Moles of CO2 formed = x mol

Moles of H2 formed = x mol

At equilibrium composition of product stream

[H2] = x mol

[CO2] = x mol

[CO] = (1 - x) mol

[H2O] = ( 2 - 2*x) mol

The reaction with ICE TABLE

CO + H2O = CO2 + H2

I 1 2

C -x -2x +x +x

E (1-x) (2-2x) x x

Equilibrium constant expression of the reaction

K = [H2] [CO2] / [CO] [H2O]

1 = [H2] [CO2] / [CO] [H2O]

[CO] [H2O] = [H2] [CO2]

(1-x)(2-2x) = (x) (x)

2 - 2x - 2x + 2x2 - x2 = 0

2 - 4x + x2 = 0

x = 0.586

At equilibrium composition of product stream

[H2] = x = 0.586 mol

[CO2] = x = 0.586 mol

[CO] = 1 - x = 1 - 0.586 = 0.414 mol

[H2O] = 2 - 2*x = 2 - 2*0.586 = 0.828 mol

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