Consider the following case of completing 1 $^{s t}$ order reactions.
After the start of the reaction at t $=$ 0 with only A, the [C] is equal to the [D] at all times. The time in which all three concentrations will be equal is given by:

t = \frac{1}{3 k_{1}} l n 3

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t = \frac{1}{2 k_{2}} l n 3

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t = \frac{1}{3 k_{1}} l n 2

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t = \frac{1}{3 k_{2}} l n 2

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Solution

The correct option is B $t = \frac{1}{2 k_{2}} l n 3$
$k_{1} = k_{2}$
$\Rightarrow \frac{2}{3}$ rd of A has reacted for [A] = [C] = [D]
$∴ k_{1} + k_{2} = \frac{1}{t} l n \frac{[A]_{0}}{\frac{1}{2} [A]_{0}}$
$\Rightarrow t = \frac{1}{k_{1} + k_{2}} l n 3 = \frac{1}{2 k_{1}} l n 3 = \frac{1}{2 k_{2}} l n 3$

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