The rate of first order reaction is 0.04 mol L -1 s -1 at 10 minutes and 0.03 mol L -1 s -1 at 20 minutes after initiation. Half life of the reaction is minutes. (Given log 2=0.3010, log 3=0.4771 )

Chemistry · Chemical Kinetics

JEE Main 2024 — 30 January, Shift 1 — Question 68

The rate of first order reaction is $0.04 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ at 10 minutes and $0.03 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ at 20 minutes after

initiation. Half life of the reaction is $\qquad$ minutes. (Given $\log 2=0.3010, \log 3=0.4771$ )

Answer: 24

Numerical answer — enter this value.

Step-by-step solution

$\quad 0.04=\mathrm{k}[\mathrm{A}]_{0} \mathrm{e}^{-\mathrm{k} \times 10 \times 60}$

$\quad 0.03=\mathrm{k}[\mathrm{A}]_{0} \mathrm{e}^{-\mathrm{k} \times 20 \times 60}$

$\frac{4}{3}=\mathrm{e}^{600 \mathrm{k}(2-1)}$

$\frac{4}{3}=e^{600 k}$

$\ln \frac{4}{3}=600 \mathrm{k}$

$\ln \frac{4}{3}=600 \times \frac{\ln 2}{t_{1 / 2}}$

$\mathrm{t}_{1 / 2}=600 \frac{\ln 2}{\ln \frac{4}{3}} \mathrm{sec}$

$\mathrm{t}_{1 / 2}=600 \times \frac{\log 2}{\log 4-\log 3}$ sec.

$=10 \times \frac{0.3010}{0.6020-0.477} \mathrm{~min}$

$\mathrm{t}_{1 / 2}=24.08 \mathrm{~min}$

Answer key and solution verified before publishing.

Practise Chemical Kinetics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Rate Laws and Rate Constant