OnlineCBT
Chemistry · EQUILIBRIUM
हिन्दीIn the reaction \( \text{CO}(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons \text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \), if the volume of the equilibrium mixture is suddenly halved at constant temperature, how does the reaction quotient \( Q_c \) change immediately, and in which direction does the reaction shift?
Correct answer · Option 1
\( Q_c \) becomes \( \frac{1}{4}K_c \), reaction shifts in the forward direction
Explanation
Halving volume doubles all concentrations: \( Q_c = \frac{(2[\text{CH}_4])(2[\text{H}_2\text{O}])}{(2[\text{CO}])(2[\text{H}_2])^3} = \frac{4}{16} K_c = \frac{1}{4} K_c \). Since \( Q_c < K_c \), the reaction shifts forward.
