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In a reversible reaction, study of its mechanism says that both the forward and reverse reaction follows first order kinetics. If the half life of forward and backward reactions are 2 min and 4 min respectively, the equilibrium constant of the reaction is ____

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$\frac{1}{2}$

answer is C.

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Detailed Solution

$t_{1 / 2} = \frac{0 .693}{K}$ ( for first order kinetics) $K_{f} = \frac{0 .693}{2} s^{- 1}; K_{b} = \frac{0 .693}{4} s^{- 1} K = \frac{K_{f}}{K_{b}} = \frac{4}{2} = 2$

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