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.