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  4. Consider the following case of completing 1 st order reactions <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/chemistry/4e347b42e053bfa16c7196b7fe7ff6bf-.png /> After the start of the reaction at t=0 with only A, the [B] is equal to the [C] at all times. The time in which all three concentrations will be equal is given by

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

Chemical Kinetics

Solution:

$k_1=k_2=\frac{2}{3}$ rd of $A$ has reacted for $[A]=[B]=$ [C]
$\therefore k_1+k_2=\frac{1}{t} \ln \frac{[A]_0}{\frac{1}{3}[A]_0}$
Or $t=\frac{1}{k_1+k_2} \ln 3=\frac{1}{2 k_1} \ln 3=\frac{1}{2 k_2} \ln 3$