The reaction, CO(g) + 3H_{2}(g) ↔ CH_{4}(g) + H_{2}O(g) is at equilibrium at 1300 K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of H_{2} and 0.02 mol of H_{2}O and an unknown amount of CH_{4} in the flask. Determine the concentration of CH_{4} in the mixture. The equilibrium constant, *K*_{c} for the reaction at the given temperature is 3.90.

#### Solution

Let the concentration of methane at equilibrium be *x*.

`CO_(g) + 3H_(2(g)) ↔ CH_(4(g)) + H_2O_(g)`

At equilibrium 0.3/1 = 0.3 M 0.1/1 = 0.1 M x 0.02/1 = 0.02 M

It is given that* K*_{c}_{ }= 3.90.

Therefore

`([CH_(4(g))][H_2O_(g)])/([CO_((g))][H_(2(g))]^3) = K_c`

`=> (x xx 0.02)/0.3 xx (0.1)^3 = 3.90`

`=> x = (3.90 xx 0.3 xx (0.1)^3)/0.02`

`= 0.00117/0.02`

= 0.0585 M

`= 5.85 xx 10^(-2) M`

Hence, the concentration of CH_{4} at equilibrium is 5.85 × 10^{–2} M.