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) $[\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}]$

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Chemical Kinetics

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

ln \frac{k _{2}}{k _{1}}

will be (nearest integer)

[R = 8.3 J K^{- 1} mol^{- 1}]

Answer: 8

Solution

With a common pre-exponential factor,

ln k = ln A - \frac{E _{a}}{R T}

for each reaction, so

ln \frac{k _{2}}{k _{1}} = \frac{E _{1} - E _{2}}{R T}

. The first activation energy exceeds the second by

20 kJ mo l^{- 1}

, giving

ln \frac{k _{2}}{k _{1}} = \frac{20000}{8.3 \times 300} = \frac{20000}{2490} = 8.03

. To the nearest integer

ln \frac{k _{2}}{k _{1}} = 8

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Integer answer

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