In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and…

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_1$ and $t_2(s)$, respectively. The ratio $t_1 / t_2$ will :
  1. $\frac{4}{3}$
  2. $\frac{3}{2}$
  3. $\frac{3}{4}$
  4. $\frac{2}{3}$

Solution

For $\mathrm{I}^{\text {st }}$ order reaction
When $\mathrm{C}_{\mathrm{t}}=\mathrm{Co} / 4$
$\begin{aligned}
& \mathrm{t}_1=2 \mathrm{t}_{50 \%}. \\
& \text { when } \mathrm{C}_{\mathrm{t}}=\mathrm{Co} / 8 \\
& \mathrm{t}_2=3 \mathrm{t}_{50 \%} \\
& \text { so } \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{2}{3}
\end{aligned}$

Asked in: JEE Main 2025 (08 Apr Shift 2)

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