The reaction, CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kcfor the reaction at the given temperature is 3.90.
Given:
The reaction given is CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
No. of moles of CO = 0.30 mol
No. of moles of H2 = 0.10 mol
No. of moles of H2O = 0.02 mol
The equilibrium constant of the given reaction (Kc) = 3.90
As we know that molar concentration = No. of moles / Volume of flask. Hence,
Molar concentration of [CO] = 
= 0.30 M
Molar concentration of [H2] = 
= 0.10 M
Molar concentration of [H2O] = 
= 0.02 M
To find out the concentration of CH4, we will apply the formula
Kc = 
Where X and Y are the products and A and B are the reactants.
Hence, for the given reaction,
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
Equilibrium constant (Kc) = 
As the Kc = 3.90 (given)
[CO] = 0.30 M
[H2] = 0.10 M
[H2O] = 0.02 M
∴ 3.90 = 
⇒CH4 = 0.0585 M = 5.85 × 10-2 M
Thus, the concentration of CH4 in the mixture is 5.85 × 10-2 M