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Hydrogen and carbon dioxide can be used to produce carbon monoxide according to

ID: 552224 • Letter: H

Question

Hydrogen and carbon dioxide can be used to produce carbon monoxide according to the water-gas shift reaction given by: H2 + CO2 = CO + H2O.

If a reaction is set-up initially with n moles each of H2 and CO2 and none of either of the products, CO and H2O, and the variable x represents the fraction of H2 dissociated at equilibrium. Complete the following table in terms of n and x.

H2

CO2

CO

H2O

Total moles

Moles at reaction start

n

n

0

0

2n

Moles at equilibrium

N(1-x)

2n

At T=298 K, the equilibrium constant for this reaction is Kp(298)=10-5. The value of x at 298 K(to a precision of 0.00001)

H2

CO2

CO

H2O

Total moles

Moles at reaction start

n

n

0

0

2n

Moles at equilibrium

N(1-x)

2n

Explanation / Answer

The reaction is H2(g)+ CO2(g)----àCO(g)+ H2O(g),

The reaction is carried out at a total pressure of 1 atm

Let x= fraction of H2 converted

H2

CO2

H2O

CO

Total

Moles at the start of reaction

n

n

0

0

2n

change

-nx

nx

nx

nx

Equilibrium

n-nx

n-nx

nx

nx

2n

Mole fractions= moles of component/total moles

n*(1-x)/2n =(1-x)/2

(1-x)/2

x/2

x/2

1

Partial pressure =mole fraction* total pressure

P*(1-x)/2

P*(1-x)/2

Px/2

Px/2

P

Kp= [PCO] [PH2O]/ [PH2] [PCO2] = 10-5,

(Px/2)*(Px/2)/{(P*(1-x)/2}{P*(1-x)/2= 10-5

Hence x2/(1-x)2= 10-5

taking square root, x/(1-x)= 0.0032

x= 0.0032*(1-x)

x*(1.0032)=0.0032

x= 0.0032/1.0032= 0.00319

H2

CO2

H2O

CO

Total

Moles at the start of reaction

n

n

0

0

2n

change

-nx

nx

nx

nx

Equilibrium

n-nx

n-nx

nx

nx

2n

Mole fractions= moles of component/total moles

n*(1-x)/2n =(1-x)/2

(1-x)/2

x/2

x/2

1

Partial pressure =mole fraction* total pressure

P*(1-x)/2

P*(1-x)/2

Px/2

Px/2

P