Question

Consider the reaction: H2(g) + CO2(g) ⇄ H2O(g) + CO(g)
0.80 mole of H2 and 0.80 mole of CO2 are placed in a 5-dm3 container at 1650 oC. At equilibrium, 0.25 mole of CO was found in the equilibrium mixture. Calculate the equilibrium constant Kc.

Answer 1

First, you must convert your values to concentrations as the equilibrium point is related to the relative concentrations of the compounds rather than their absolute quantity as given by moles. Also note that 1 dm^3 = 1 L. The units of concentration I'll use are moles/litre (M).

Therefore, you start with 0.8/5 = 0.16 M of H2 and 0.8/5 = 0.16 M of CO2, and you finish with 0.25/5 = 0.05 M of CO at equilibrium. Because you started with no products, 0.05 M represents the total amount of CO that you produce from the reaction. Because the stoichiometry of the reaction is 1:1:1:1, this means you also produce 0.05 M of H2O, and lose 0.05 M of both H2 and CO.

Your equilibrium concentrations are therefore 0.75 M H2, 0.75 M CO2, 0.05 M H2O, and 0.05 M CO. At 1650 oC, Kc is:

Kc = (0.05*0.05)/(0.75*0.75) = 4.4 x 10^-3

If anything is unclear I'm happy to explain things further. Hope that helps :)