JEE Main 2026 — Chemistry Physical Chemistry

2026integerhard

$\mathrm{A} \longrightarrow \mathrm{B}$ (first reaction)

$\mathrm{C} \longrightarrow \mathrm{D}$ (second reaction)

Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At $500 \mathrm{~K}, 50 \%$ of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is $\_\_\_\_$ $\times 10^{-1}$ hour $^{-1}$ (nearest integer).

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

For $A \xrightarrow{K_1} B$

$\ln(2) = \dfrac{E_{a_1}}{R} \left[ \dfrac{1}{300} - \dfrac{1}{500} \right]$

$E_{a_1} = \dfrac{\ln 2 \times R \times 1500}{2}$

$E_{a_2} = \dfrac{E_{a_1}}{2} = \dfrac{\ln 2 \times R \times 1500}{4}$

$(K_1)_{\text{at 500 K}} = \dfrac{\ln 2}{2}$

$(K_2)_{\text{at 500 K}} = \ln 2$

Now for $C \xrightarrow{K_2} D$

$\ln \left[ \dfrac{(K_2)_{\text{at 500K}}}{(K_2)_{\text{at 300K}}} \right] = \left( \dfrac{\ln 2 \times R \times 1500}{4} \right) \times \dfrac{1}{R} \times \left[ \dfrac{1}{300} - \dfrac{1}{500} \right]$

$(K_2)_{\text{at 300 K}} = \dfrac{\ln 2}{\sqrt{2}} = 0.49$

$(K_2)_{\text{at 300 K}} = 4.9 \times 10^{-1}$ .

So, answer is 5.

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

Solution

For $A \xrightarrow{K_1} B$

$\ln(2) = \dfrac{E_{a_1}}{R} \left[ \dfrac{1}{300} - \dfrac{1}{500} \right]$

$E_{a_1} = \dfrac{\ln 2 \times R \times 1500}{2}$

$E_{a_2} = \dfrac{E_{a_1}}{2} = \dfrac{\ln 2 \times R \times 1500}{4}$

$(K_1)_{\text{at 500 K}} = \dfrac{\ln 2}{2}$

$(K_2)_{\text{at 500 K}} = \ln 2$

Now for $C \xrightarrow{K_2} D$

$(K_2)_{\text{at 300 K}} = \dfrac{\ln 2}{\sqrt{2}} = 0.49$

$(K_2)_{\text{at 300 K}} = 4.9 \times 10^{-1}$ .

So, answer is 5.

JEE Main 2026 — Chemistry Physical Chemistry

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

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