A reaction has a half life of 1 min. The time required for 99.9% completion of the reaction is \underline\hspace2cm min. (Round off to the Nearest Integer). [Use : ln 2 = 0.69; ln 10 = 2.3]

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2021 · 18 Mar · Shift 2 · Q20 with Solution

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Correct answer: 10

For a first-order reaction, the half-life is constant and given by

$t_{1/2} = \frac{0.693}{k}$

Given:

$t_{1/2} = 1\ \text{min}$

So,

$k = \frac{0.693}{1} \approx 0.69\ \text{min}^{-1}$

For a first-order reaction,

$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$

For 99.9% completion, only $0.1\%$ reactant remains. Thus,

$\frac{[A]}{[A]_0} = 0.001 = 10^{-3}$

Hence,

$\frac{[A]_0}{[A]} = 10^3$

So,

$t = \frac{2.303}{0.69} \log(10^3)$

Since

$\log(10^3)=3$

we get

$t = \frac{2.303}{0.69} \times 3$

Using the given approximation $2.303 \approx 2.3$ ,

$t = \frac{2.3}{0.69} \times 3 = \frac{6.9}{0.69} = 10\ \text{min}$

$\boxed{10\ \text{min}}$

Rounded to the nearest integer, the answer is 10.