JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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 of ratio of $\frac{\mathrm{k} \text { (catalysed) }}{\mathrm{k} \text { (uncatalysed) }}$ is....

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

Official previous-year question

Held on 24 Jan 2026 · Verified 6 Jul 2026.

Options

A
1.741
B
17.41
C
3.482
D
0.1741

Did you get this right?

Solution

Using the Arrhenius equation: $k = A e^{-E_a/RT}$

The ratio of rate constants is: $\frac{k_{catalysed}}{k_{uncatalysed}} = \frac{A e^{-E_a^c/RT}}{A e^{-E_a^u/RT}} = e^{(E_a^u - E_a^c)/RT} = e^{\Delta E_a/RT}$

Since the catalyst lowers activation energy by 10 kJ/mol: $\Delta E_a = 10$ kJ/mol = 10000 J/mol

At T = 27°C = 300 K, R = 8.314 J/(mol·K):

$\frac{k_c}{k_u} = e^{10000/(8.314 \times 300)} = e^{10000/2494.2} = e^{4.009}$

Taking logarithm base 10: $\log_{10}\frac{k_c}{k_u} = 4.009 \times \log_{10}(e) = 4.009 \times 0.4343 = 1.741$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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

Official previous-year question

Held on 24 Jan 2026 · Verified 6 Jul 2026.

Options

A
1.741
B
17.41
C
3.482
D
0.1741

Did you get this right?

Solution

Using the Arrhenius equation: $k = A e^{-E_a/RT}$

The ratio of rate constants is: $\frac{k_{catalysed}}{k_{uncatalysed}} = \frac{A e^{-E_a^c/RT}}{A e^{-E_a^u/RT}} = e^{(E_a^u - E_a^c)/RT} = e^{\Delta E_a/RT}$

Since the catalyst lowers activation energy by 10 kJ/mol: $\Delta E_a = 10$ kJ/mol = 10000 J/mol

At T = 27°C = 300 K, R = 8.314 J/(mol·K):

$\frac{k_c}{k_u} = e^{10000/(8.314 \times 300)} = e^{10000/2494.2} = e^{4.009}$

Taking logarithm base 10: $\log_{10}\frac{k_c}{k_u} = 4.009 \times \log_{10}(e) = 4.009 \times 0.4343 = 1.741$

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry