Report Issue

JEE MAIN_2026
_210126_S1
Question
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 $k_2$ are the rate constants of first and second reaction respectively at 300 K , then $\ln \frac{k_2}{k_1}$ will be $\_\_\_\_$ (nearest integer) $\left[\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right]$
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Solution Image
Question Tags
JEE Main
Chemistry
Easy
Start Preparing for JEE with Competishun
Video Solution
BY competishun
Video Solution
Watch Solution