Time required for completion of 99.9 \% of a First order reaction is times of half life ( t 1 / 2 ) of the reaction.

Chemistry · Chemical Kinetics

JEE Main 2024 — 27 January, Shift 2 — Question 76

Time required for completion of $99.9 \%$ of a First order reaction is $\qquad$ times of half life $\left(\mathrm{t}_{1 / 2}\right)$ of the reaction.

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

$\frac{\mathrm{t}_{99.9 \%}}{\mathrm{t}_{1 / 2}}=\frac{\frac{2.303}{\mathrm{k}}\left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)}{\frac{2.303}{\mathrm{k}} \log 2}=\frac{\log \left(\frac{100}{100-99.9}\right)}{\log 2}=\frac{\log 10^{3}}{\log 2}=\frac{3}{0.3}=10$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws