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 K_p(298)=10^-5. The value of x at 298 K(to a precision of 0.00001)

Answer 1

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