JEE Main 2026 Memory Based Question |21 Jan Shift 1|Chemistry

JEE MAIN 2026

21-01-2025 SHIFT-1

Question

For two chemical reactions $A$ and $B$, if the difference between their activation energy is 20 kJ at $300 \mathrm{~K}\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$. Determine $\ln \frac{\mathrm{k}_2}{\mathrm{k}_1}$.

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Solution

\begin{aligned} & \ln \frac{K_2}{K_1}=\frac{\Delta E_{\mathrm{a}}}{R T}=\frac{20 \times 1000}{8.3 \times 300} \\ & =\frac{200}{8.3 \times 3}=8.03 \end{aligned}

Question Tags

JEE Main

Chemistry

Easy

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