Question

Consider the following reaction:
CO(g)+H2O(g)⇌CO2(g)+H2(g)
Kc=102 at 500 K
A reaction mixture initially contains 0.120 M COand 0.120 M H2O.

A)What will be the equilibrium concentration of [CO]?

B)What will be the equilibrium concentration of [H2O]?

C)What will be the equilibrium concentration of [CO2]?

D)What will be the equilibrium concentration of [H2]?

Answer 1

Let's prepare the ICE table

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

initial 0.12 0.12 0 0

change -1x -1x +1x +1x

equilibrium 0.12-1x 0.12-1x +1x +1x

Equilibrium constant expression is

Kc = [CO2]*[H2]/[CO]*[H2O]

102.0 = (1*x)^2/(0.12-1*x)^2

sqrt(102.0) = (1*x)/(0.12-1*x)

10.0995 = (1*x)/(0.12-1*x)

1.21194-10.0995*x = 1*x

1.21194-11.0995*x = 0

x = 0.109

At equilibrium:

[CO] = 0.12-1x = 0.12-1* 0.109 = 0.0108 M

[H2O] = 0.12-1x = 0.12-1* 0.109 = 0.0108 M

[CO2] = +1x = +1* 0.109 = 0.109 M

[H2] = +1x = +1* 0.109 = 0.109 M

A)

[CO] = 0.0108 M

B)

[H2O] = 0.0108 M

C)

[CO2] = 0.109 M

D)

[H2] = 0.109 M