equilibrium constant?

Question

1 mole of CO(g) reacts with 1 mole of steam to form CO2(g) and hydrogen(g) in a 10L closed container. when equlibrium was reached, 40% of initial concentration of CO changed into products. Calculate the value of equlibrium constant.(show the method)

Answer 1

CO(g) + H2O(g) <===> CO2(g) + H2(g)

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

If 40% of the 1 mole CO reacted, then 0.4 mole reacted, and there is 0.6 mole CO left. The balanced equation shows that the figure is identical for H2O. The 0.4 moles each of CO and H2O gave rise to 0.40 mole each of CO2 and H2.

(0.4)(0.4)/(0.6)(0.6) = 0.16/0.36 = 4/9 = 0.44