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
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