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Chemistry · Chemical Kinetics

JEE Main 2026 — 5 April, Morning Shift — Question 68

If the half life of a first order reaction is 6.93 minutes then the time required for completion of 99%99 \% of the reaction will be ____\_\_\_\_ minutes. (Given : log⁡2=0.3010\log 2=0.3010 ).

Answer: 46

Numerical answer — enter this value.

Step-by-step solution

t1/2=ℓn2 K⇒ K1=0.6936.93⇒0.1 min−1\quad \mathrm{t}_{1 / 2}=\frac{\ell \mathrm{n} 2}{\mathrm{~K}} \Rightarrow \mathrm{~K}_{1}=\frac{0.693}{6.93} \Rightarrow 0.1 \mathrm{~min}^{-1}

t99%=1 Kln⁡[ A0 At]=10.1ln⁡[1001]⇒2ln⁡100.1=2×2.3030.1⇒46.06min⁡≈46\begin{aligned} & \mathrm{t}_{99 \%}=\frac{1}{\mathrm{~K}} \ln \left[\frac{\mathrm{~A}_{0}}{\mathrm{~A}_{\mathrm{t}}}\right] & =\frac{1}{0.1} \ln \left[\frac{100}{1}\right] \Rightarrow \frac{2 \ln 10}{0.1} & =\frac{2 \times 2.303}{0.1} & \Rightarrow 46.06 \min \approx 46 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Integrated Rate Laws

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