For the reaction CO(g) + H_2O(g) leftharpoons CO_2(g) + H_2(g) ,...

The reaction quotient $Q_c$ is calculated using the given concentrations of the species at the given instant:

$Q_c = \dfrac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}$

Substituting the given values:

$Q_c = \dfrac{0.4 \times 0.4}{0.1 \times 0.1} = \dfrac{0.16}{0.01} = 16$

The equilibrium constant is given as $K_c = 4.0$.

Comparing $Q_c$ and $K_c$, we find that $Q_c > K_c$. Since the reaction quotient is greater than the equilibrium constant, the system is not at equilibrium. To reach equilibrium, the reaction will proceed in the reverse direction, consuming the products $\text{CO}_2$ and $\text{H}_2$ to form more reactants $\text{CO}$ and $\text{H}_2\text{O}$.

Answer: The reaction will proceed in the reverse direction, consuming $\text{CO}_2$ and $\text{H}_2$.