Q25 / 75 · JEE Main 2026 (21 January Shift 1) · Chemistry
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 $\ln \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}$ will be $\_\_\_\_$. (nearest integer) $\left[\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
Correct answer: 8
Solution
Given: $A_1 = A_2$, $E_{a1} = E_{a2} + 20$ kJ/mol, T = 300 K, R = 8.3 J K⁻¹ mol⁻¹
Using Arrhenius equation: $k = Ae^{-E_a/RT}$
$\frac{k_2}{k_1} = \frac{e^{-E_{a2}/RT}}{e^{-E_{a1}/RT}} = e^{(E_{a1} - E_{a2})/RT}$
$E_{a1} - E_{a2} = 20$ kJ/mol = 20000 J/mol
$\frac{k_2}{k_1} = e^{\frac{20000}{8.3 \times 300}} = e^{\frac{20000}{2490}}$
$\ln\frac{k_2}{k_1} = \frac{20000}{2490} = 8.03 \approx 8$
Using Arrhenius equation: $k = Ae^{-E_a/RT}$
$\frac{k_2}{k_1} = \frac{e^{-E_{a2}/RT}}{e^{-E_{a1}/RT}} = e^{(E_{a1} - E_{a2})/RT}$
$E_{a1} - E_{a2} = 20$ kJ/mol = 20000 J/mol
$\frac{k_2}{k_1} = e^{\frac{20000}{8.3 \times 300}} = e^{\frac{20000}{2490}}$
$\ln\frac{k_2}{k_1} = \frac{20000}{2490} = 8.03 \approx 8$
Hence, the correct answer is 8.