Start free

Chemistry · Chemical Kinetics

JEE Main 2025 — 8 April, Evening Shift — Question 11

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are t1t_{1} and t2(s)t_{2}(s), respectively. The ratio t1/t2t_{1} / t_{2} will be:

  1. Option A:

    43\frac{4}{3}

  2. Option B:

    32\frac{3}{2}

  3. Option C:

    34\frac{3}{4}

  4. Option D:

    23\frac{2}{3}

    Correct

Answer: D

Step-by-step solution

t1=t14=1kln⁡A0 A04=1kln⁡4\mathrm{t}_{1}=\mathrm{t}_{\frac{1}{4}}=\frac{1}{\mathrm{k}} \ln \frac{\mathrm{A}_{0}}{\frac{\mathrm{~A}_{0}}{4}}=\frac{1}{\mathrm{k}} \ln 4

t2=t18=1kln⁡A0 A08=1kln⁡8\mathrm{t}_{2}=\mathrm{t}_{\frac{1}{8}}=\frac{1}{\mathrm{k}} \ln \frac{\mathrm{A}_{0}}{\frac{\mathrm{~A}_{0}}{8}}=\frac{1}{\mathrm{k}} \ln 8

t1t2=ln⁡4ln⁡8=2ln⁡23ln⁡2=23\frac{t_{1}}{t_{2}}=\frac{\ln 4}{\ln 8}=\frac{2 \ln 2}{3 \ln 2}=\frac{2}{3}

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 2025
Paper
8 April, Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Integrated Rate Laws