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 be:

<p>For a first-order reaction, the integrated rate law is given by:</p> <p>$\ln \frac{[A]}{[A]_0} = -kt$</p> <p>where:</p> <p><p>$[A]_0$ is the initial concentration,</p></p> <p><p>$[A]$ is the concentration at time $t$,</p></p> <p><p>$k$ is the rate constant.</p></p> <p>Let's find the times $t_1$ and $t_2$ corresponding to when the concentration becomes $\frac{1}{4}[A]_0$ and $\frac{1}{8}[A]_0$ respectively.</p> <p>For concentration $\frac{1}{4}[A]_0$:</p> <p><p>Substitute into the integrated rate law:</p> <p>$\ln\left(\frac{1}{4}\right) = -kt_1$</p></p> <p><p>Recognizing that $\ln\left(\frac{1}{4}\right) = -\ln 4$, we have:</p> <p>$-\ln 4 = -kt_1 \quad \Longrightarrow \quad t_1 = \frac{\ln 4}{k}$</p></p> <p>For concentration $\frac{1}{8}[A]_0$:</p> <p><p>Substitute into the integrated rate law:</p> <p>$\ln\left(\frac{1}{8}\right) = -kt_2$</p></p> <p><p>Recognizing that $\ln\left(\frac{1}{8}\right) = -\ln 8$, we have:</p> <p>$-\ln 8 = -kt_2 \quad \Longrightarrow \quad t_2 = \frac{\ln 8}{k}$</p></p> <p>Now, to find the ratio $\frac{t_1}{t_2}$:</p> <p>$\frac{t_1}{t_2} = \frac{\frac{\ln 4}{k}}{\frac{\ln 8}{k}} = \frac{\ln 4}{\ln 8}$</p> <p>We can simplify further by expressing the logarithms in terms of $\ln 2$:</p> <p><p>$\ln 4 = \ln (2^2) = 2\ln 2$</p></p> <p><p>$\ln 8 = \ln (2^3) = 3\ln 2$</p></p> <p>Thus,</p> <p>$\frac{t_1}{t_2} = \frac{2\ln 2}{3\ln 2} = \frac{2}{3}$</p> <p>So, the ratio $\frac{t_1}{t_2}$ is $\frac{2}{3}$.</p> <p>This corresponds to Option D.</p>