Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mol -1 . If k 1 and k 2 are the rate constants of first and second reaction respectively at 300 K , then In frac k 2 k 1 will be (nearest integer) [ R =8.3 J K -1 mol -1 ]

Chemistry · Chemical Kinetics

JEE Main 2026 — 21 January, Morning Shift — Question 63

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by $20 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . If $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$ are the rate constants of first and second reaction respectively at 300 K , then In $\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}$ will be (nearest integer) $\left[\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{A} \xrightarrow{\mathrm{RX}^{\mathrm{n}}(1)}$ product $\mathrm{E}_{1}$ $\mathrm{B} \xrightarrow{\mathrm{RX}^{\mathrm{n}}(2)}$ product $\mathrm{E}_{2}$ Assuming 'A' same for both reaction. $\ln \mathrm{k}_{1}=\ln \mathrm{A}-\frac{\mathrm{E}_{1}}{300 \mathrm{R}}$ $\ln \mathrm{k}_{1}=\ln \mathrm{A}-\frac{\mathrm{E}_{2}}{300 \mathrm{R}}$ $\ln \left(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\right)=\frac{\mathrm{E}_{1}-\mathrm{E}_{2}}{300 \mathrm{R}}=\frac{20 \times 1000}{300 \mathrm{R}}$ $=8.032$

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 2026
Paper: 21 January, Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Arrhenius Equation