For the reversible reaction : $A \underset{K_{2}}{\overset{K_{1}}{⇌}} B, k_{1} = 0.02 \min^{- 1}$ and $k_{2} = 0.03 \min^{- 1}$ The reaction is started with A only. If the half life of such reaction is defined as the time in which half of the equilibrium amount of B is formed, then, what is the half life of reaction?

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Detailed Solution

The half life of the reaction,

$t_{1 / 2} = \frac{\ln 2}{K_{1} + K_{2}} t_{1 / 2} = \frac{0.693}{0.02 + 0.03} t_{1 / 2} = \frac{0.693}{0.05} t_{1 / 2} = 13.86$

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