Given the concentrations at equilibrium: [ [\text{CO}] = 0.180 , \text{M} ] [ [\text{H}_2\text{O}] = 0.12 , \text{M} ] [ [\text{CO}_2] = 0.30 , \text{M} ] [ [\text{H}_2] = 0.30 , \text{M} ]

Question

Let’s calculate (K):

Answer 1

The equilibrium constant (K) for the reaction
$$\text{CO}+{\text{H}}_2\text{O}\rightleftharpoons {\text{CO}}_2+{\text{H}}_2$$
is given by:
$$K=\frac{[{\text{CO}}_2][{\text{H}}_2]}{[\text{CO}][{\text{H}}_2\text{O}]}$$

Plugging in the concentrations at equilibrium:
$$K=\frac{(0.30)(0.30)}{(0.180)(0.12)}$$
$$K=\frac{0.09}{0.0216}$$
$$K=4.17$$

Therefore, the equilibrium constant (K) for the reaction is 4.17.