For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is \underline\hspace2cm . (Integer answer)

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2021 · 31 Aug · Shift 1 · Q17 with Solution

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Correct answer: 2

For a first-order reaction,

kt = \ln\frac{[A]_0}{[A]_t}

where $[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time $t$ .

At $50\%$ completion, half of the reactant remains:

[A]_t = \frac{[A]_0}{2}

So,

kt_{50} = \ln\frac{[A]_0}{[A]_0/2} = \ln 2

Hence,

t_{50} = \frac{\ln 2}{k}

At $75\%$ completion, only $25\%$ of the reactant remains:

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

So,

kt_{75} = \ln\frac{[A]_0}{[A]_0/4} = \ln 4 = 2\ln 2

Hence,

t_{75} = \frac{\ln 4}{k} = \frac{2\ln 2}{k}

\frac{t_{75}}{t_{50}} = \frac{(2\ln 2)/k}{(\ln 2)/k} = 2

Therefore, the required integer answer is:

\boxed{2}