At $27^{\circ} \mathrm{C}$ in presence of a catalyst, activation energy of a reaction is lowered by $10 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The logarithm ratio of $\frac{\mathrm{k} \text{ (catalysed) }}{\mathrm{k} \text{ (uncatalysed) }}$ is .... (Consider that the frequency factor for both the reactions is same)

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Chemical Kinetics

2 7^{\circ} C

in presence of a catalyst, activation energy of a reaction is lowered by

10 kJ mol^{- 1}

. The logarithm ratio of

\frac{k (catalysed)}{k (uncatalysed)}

is ....
(Consider that the frequency factor for both the reactions is same)

Answer: (B)

With the same frequency factor, the Arrhenius equation gives

\frac{k _{c a t}}{k _{u n c a t}} = e^{Δ E_{a} / R T}

, so

lo g \frac{k _{c a t}}{k _{u n c a t}} = \frac{Δ E _{a}}{2.303 R T}

. Substituting

Δ E_{a} = 10000 J mo l^{- 1}

R = 8.314 J mo l^{- 1} K^{- 1}

and

T = 300 K

gives

\frac{10000}{2.303 \times 8.314 \times 300}

. The logarithm of the rate-constant ratio is

1.741

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option