Verified Solution • 2025–26 Session moderate

QJEEM24Ch5b41: JEE Main 2024 Chemistry - Class 12 JEEM24Ch5b41 Solution: The rate of First order reaction is at 10 minutes and at 20…

The rate of First order reaction is 0.04 mol L−1 s−10.04 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} at 10 minutes and 0.03 mol L−1 s−10.03 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} at 20 minutes after initiation. Half life of the reaction is _______ minutes.

(Given log⁡2=0.3010,log⁡3=0.4771\log 2=0.3010, \log 3=0.4771 )

Step-by-Step Verified Solution

amp;0.04=k[A]0e−k×10×60..... (1)amp;0.03=k[A]0e−k×20×60..... (2)\begin{aligned}& 0.04=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 10 \times 60} \quad \text{..... (1)}\\& 0.03=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 20 \times 60} \quad \text{..... (2)}\end{aligned}

43amp;=e600k(2−1)/(2)43amp;=e600kln⁡43amp;=600kln⁡43amp;=600×ln⁡2t1/2t1/2amp;=600ln⁡2ln⁡43sec⁡t1/2amp;=600×log⁡2log⁡4−log⁡3sec.=10×0.30100.6020−0.477 mint1/2amp;=24.08 min\begin{aligned}\frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}(2-1)} /(2) \\\frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}} \\\ln \frac{4}{3} & =600 \mathrm{k} \\\ln \frac{4}{3} & =600 \times \frac{\ln 2}{\mathrm{t}_{1 / 2}} \\\mathrm{t}_{1 / 2} & =600 \frac{\ln 2}{\ln \frac{4}{3}} \sec \\\mathrm{t}_{1 / 2} & =600 \times \frac{\log 2}{\log 4-\log 3} \mathrm{sec} .=10 \times \frac{0.3010}{0.6020-0.477} \mathrm{~min} \\\mathrm{t}_{1 / 2} & =24.08 \mathrm{~min}\end{aligned}

Ans. 24

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring