Chemistry · Chemical Kinetics

JEE Main 2024 — 6 April, Shift 2 — Question 84

Consider the two different first order reactions given below

$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$ (Reaction 1 $)$

$\mathrm{P} \rightarrow \mathrm{Q}$ (Reaction 2)

The ratio of the half life of Reaction 1: Reaction 2 is $5: 2$ . If $t_{1}$ and $t_{2}$ represent the time taken to complete $2 / 3^{\text {rd }}$ and $4 / 5$ of Reaction 1 and Reaction 2, respectively, then the value of the ratio $\mathrm{t}_{1}: \mathrm{t}_{2}$ is $\qquad$ $\times 10^{-1}$ (nearest integer). [Given: $\log _{10}(3)=0.477$ and $\log _{10}(5)=0.699$ ]

Answer: 17

Numerical answer — enter this value.

Step-by-step solution

$\frac{\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{I}}}{\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{II}}}=\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{5}{2}$ $\therefore \mathrm{K}_{1} \mathrm{t}_{1}=\ln \frac{1}{1-\frac{2}{3}}=\ln 3$ $\mathrm{K}_{2} \mathrm{t}_{2}=\ln \frac{1}{1-\frac{4}{5}}=\ln 5$

$\Rightarrow \frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}} \times \frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}=\frac{0.477}{0.699}$ $\Rightarrow \frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}=\frac{0.477}{0.699} \times \frac{5}{2}=1.7=17 \times 10^{-1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws