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Question

At $500 \, \text{K}$, for a reversible reaction $A_{2}(g) + B_{2}(g) \rightleftharpoons 2 \, AB(g)$ in a closed container, $K_{C} = 2 \times 10^{-5}$. In the presence of a catalyst, the equilibrium is attained 10 times faster. The equilibrium constant $K_{C}$ in the presence of a catalyst at the same temperature is

$2 \times 10^{-10}$

$2 \times 10^{-5}$

$2 \times 10^{-4}$

$2 \times 10^{-6}$

Solution

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