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:

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2025 · 8 Apr · Shift 2 · Q4 with Solution

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Correct answer: D

For a first-order reaction,

[A]_t = [A]_0 e^{-kt}

Taking logarithm,

t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right)

Given,

[A]_t = \frac{[A]_0}{4}

So,

t_1 = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_0/4}\right) = \frac{1}{k} \ln 4

Given,

[A]_t = \frac{[A]_0}{8}

So,

t_2 = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_0/8}\right) = \frac{1}{k} \ln 8

\frac{t_1}{t_2} = \frac{\ln 4}{\ln 8}

Now,

\ln 4 = \ln(2^2) = 2\ln 2

and

\ln 8 = \ln(2^3) = 3\ln 2

Therefore,

\frac{t_1}{t_2} = \frac{2\ln 2}{3\ln 2} = \frac{2}{3}

\frac{t_1}{t_2} = \frac{2}{3}

So the correct option is D.