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} ]
Let’s calculate (K):
The equilibrium expression for the reaction involving carbon monoxide (CO), water (H₂O), carbon dioxide (CO₂), and hydrogen (H₂) is:
\[ K = \dfrac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} \]
Plugging in the given concentrations:
\[ K = \dfrac{(0.30)(0.30)}{(0.180)(0.12)} \]
\[ K = \dfrac{0.09}{0.0216} \]
\[ K = 4.17 \]
Therefore, the equilibrium constant (K) for the reaction is 4.17.