If $k_{f}$ and $k_{b}$ are the rate constants of forward and backward reactions in an equilibrium reaction, the equilibrium constant of the reaction is given by

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$K_{eq} = k_{b} / k_{f}$

$K_{eq} = 1 / k_{f} k_{b}$

$K_{eq} = k_{f} / k_{b}$

$K_{eq} = k_{f} k_{b}$

answer is A.

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Detailed Solution

$\begin{aligned} K_{f} = \frac{[C] [D]}{[A] [B]} \\ K_{b} = \frac{[A] [B]}{[C] [D]} \end{aligned}$
Therefore, $K_{f} = \frac{1}{K_{b}}$
Additionally, $K_{f} \times K_{b} = 1$ .

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