An organic compound undergoes first order decomposition. The time taken for decomposition to (1/8 ) th and (1/10 ) th of its initial concentration are t 1 / 8 and t 1 / 10 respectively. What is the value of frac t 1 / 8 t 1 / 10 × 10 ? (log 2=0.3)

Chemistry · Chemical Kinetics

JEE Main 2026 — 28 January, Morning Shift — Question 49

An organic compound undergoes first order decomposition. The time taken for decomposition to $\left(\frac{1}{8}\right)^{\text {th }}$ and $\left(\frac{1}{10}\right)^{\text {th }}$ of its initial concentration are $t_{1 / 8}$ and $t_{1 / 10}$ respectively.

What is the value of $\frac{\mathrm{t}_{1 / 8}}{\mathrm{t}_{1 / 10}} \times 10$ ? $(\log 2=0.3)$

Option A:
9
Correct
Option B:
0.9
Option C:
3
Option D:
30

Answer: A

Step-by-step solution

$t=\frac{1}{k} \ln \frac{A_{0}}{A_{t}}$

\begin{aligned} & t_{1 / 8}=\frac{1}{k} \ln \frac{A_{0}}{A_{0} / 8}=\frac{1}{k} \ln 8 \\& t_{1 / 10}=\frac{1}{k} \ln \frac{A_{0}}{A_{0} / 10}=\frac{1}{k} \ln 10 \\& \frac{t_{1 / 8}}{t_{1 / 10}}=\frac{\ln 8}{\ln 10}=\frac{\log 8}{\log 10} \\& \frac{t_{1 / 8}}{t_{1 / 10}}=\log 8=3 \log 2=0.9 \\& \frac{t_{1 / 8}}{t_{1 / 10}} \times 10=9 \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws