JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

At $2 7^{\circ} C$ 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)

A
17.41
B
1.741
C
3.482
D
0.1741

Solution

From the Arrhenius equation with identical pre-exponential factors:

$\frac{k _{c a t}}{k _{u n c a t}} = exp (\frac{Δ E _{a}}{R T})$

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

$= \frac{10 \times 1 0 ^{3}}{2.303 \times 8.314 \times 300} = \frac{10000}{5744.14} \approx 1.741$ .

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →