At 500 K, for a reversible reaction $A_{2_{(g)}}+B_{2_{(g)}}⇌2AB_{(g)}$ in a closed container, K C = 2 × 10 -5 . In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant K C in the presence of catalyst at the same temperature is

Question

At 500 K, for a reversible reaction $A_{2_{(g)}}+B_{2_{(g)}}⇌2AB_{(g)}$  in a closed container, K C = 2 × 10 -5 . In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant K C in the presence of catalyst at the same temperature is

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Equilibrium Constant with Catalyst  Given: At 500 K, $ K_c = 2 \times 10^{-5} $ Equilibrium is attained 10 times faster in the presence of a catalyst   Objective: Determine the equilibrium constant $ K_c $ in the presence of a catalyst at the same temperature. The equilibrium constant $ K_c $ for the reaction: $$ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) $$ is given by: $$ K_c = \frac{[AB]^2}{[A_2][B_2]} $$ The presence of a catalyst affects the rate of reaction but does not change the equilibrium constant. Therefore, the equilibrium constant $ K_c $ remains: $$ \boxed{2 \times 10^{-5}} $$

Answer

Answer

Answer

2 × 10^-10

Answer

At 500 K, for a reversible reaction \(A_{2_{(g)}}+B_{2_{(g)}}⇌2AB_{(g)}\) in a closed container, K_C = 2 × 10^-5. In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant K_C in the presence of catalyst at the same temperature is

The Correct Option is D

Solution and Explanation

Equilibrium Constant with Catalyst

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At 500 K, for a reversible reaction \(A_{2_{(g)}}+B_{2_{(g)}}⇌2AB_{(g)}\) in a closed container, KC = 2 × 10-5. In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant KC in the presence of catalyst at the same temperature is