JEE Main 2023 — Chemistry Physical Chemistry

2023integereasy

$A\rightarrow B$

The rate constants of the above reaction at $200K$ and $300K$ are $0.03{\mathrm{min}}^{-1}$ and $0.05{\mathrm{min}}^{-1}$ respectively. The activation energy for the reaction is $J$ (Nearest integer)

(Given : In $10=2.3$

$R=8.3J{K}^{-1}{\mathrm{mol}}^{-1}$

$\mathrm{log}5=0.70$

$\mathrm{log}3=0.48$

$\mathrm{log}2=0.30$

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

If $K$ is known at two different temperatures the activation energy can be calculated as:

At temperature 1: $\mathrm{ln}{K}_{1}=-\frac{{E}_{a}}{{\mathrm{RT}}_{1}}+\mathrm{ln}A$

At temperature 2: $\mathrm{ln}{K}_{2}=-\frac{{E}_{a}}{{\mathrm{RT}}_{2}}+\mathrm{ln}A$

We can subtract one of these equations from the other:

$\mathrm{ln}{K}_{1}-\mathrm{ln}{K}_{2}=(-\frac{{E}_{a}}{{\mathrm{RT}}_{1}}+\mathrm{ln}A)-(-\frac{{E}_{a}}{{\mathrm{RT}}_{2}}+\mathrm{ln}A)$

This equation can further simplified to:

$\mathrm{ln}\frac{{K}_{1}}{{K}_{2}}=\frac{{E}_{a}}{R}(\frac{1}{{T}_{2}}-\frac{1}{{T}_{1}})$

$\mathrm{ln}\frac{0.05}{0.03}=\frac{\mathrm{Ea}}{2.305\times 8.314}\times [\frac{1}{200}-\frac{1}{300}]$

$\Rightarrow 0.22=\frac{\mathrm{Ea}}{2.305\times 8.314}\times [\frac{1}{600}]$

$\Rightarrow {E}_{a}=2.3\times 8.3\times 600\times 0.22$

$\Rightarrow {E}_{a}=2519.88J\Rightarrow {E}_{a}=2520J$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 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

More JEE Main Physical Chemistry questions

The wavelength of photon ' A ' is 400 nm. The frequency of photon ' B ' is $10^{16} \mathrm{~s}^{-1}$. The wave number of photon ' $C^{\prime}$ is $10^{4} \mathrm{~cm}^{-1}$. The correct order of energy of these photons is :

2026medium

Given below are two statements: Statement I: The Henry's law constant $\mathrm{K}_{\mathrm{H}}$ is constant with respect to variations in solution's concentration over the range for which the solution is ideally dilute. Statement II: $\mathrm{K}_{\mathrm{H}}$ does not differ for the same solute in different solvents. In the light of the above statements, choose the correct answer from the options given below

2026medium

For the reaction, $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}$, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is $-5.40 \mathrm{~kJ} \mathrm{~mol}^{-1}$. B. As $\Delta \mathrm{G}^{\ominus}$ in graph is positive, $\mathrm{N}_{2} \mathrm{O}_{4}$ will not dissociate into $\mathrm{NO}_{2}$ at all. C. Reverse reaction will go to completion. D. When 1 mole of $\mathrm{N}_{2} \mathrm{O}_{4}$ changes into equilibrium mixture, value of $\Delta \mathrm{G}^{\ominus}=-0.84 \mathrm{~kJ} \mathrm{~mol}^{-1}$ E. When 2 mole of $\mathrm{NO}_{2}$ changes into equilibrium mixture, $\Delta \mathrm{G}^{\ominus}$ for equilibrium mixture is $-6.24 \mathrm{~kJ} \mathrm{~mol}^{-1}$. ![JEE Main 2026 Chemistry, Chemical Equilibrium — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Chemical_Equilibrium/6983c17e8bfdbb2435639572/question_1__q_6983c17e8bfdbb2435639572__cdn-question-pool.getmarks.app__21S1_q_65_1__9992d5614b_final_ppt_sync.png) Choose the correct answer from the options given below :

2026hard

The half-life of ${ }^{65} \mathrm{Zn}$ is 245 days. After $x$ days, $75 \%$ of original activity remained. The value of $x$ in days is $\_\_\_\_$. (Nearest integer) (Given: $\log 3=0.4771$ and $\log 2=0.3010$)

2026medium

One mole each of He and $A(g)$ are taken in a $10$ L closed flask and heated to $400$ K to establish the following equilibrium. $A(g) \rightleftharpoons B(g)$. $K_c$ for this reaction at $400$ K is $4.0$. The partial pressures (in atm) of He and $B(g)$ are respectively (at equilibrium) (Assume He, $A(g)$ and $B(g)$ behave as ideal gases) (Given: $R = 0.082$ L atm K$^{-1}$ mol$^{-1}$)

2026medium

View all Physical Chemistry questions

Keep practicing Physical Chemistry

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

Browse all Chemistry topics

Solution

If $K$ is known at two different temperatures the activation energy can be calculated as:

At temperature 1: $\mathrm{ln}{K}_{1}=-\frac{{E}_{a}}{{\mathrm{RT}}_{1}}+\mathrm{ln}A$

At temperature 2: $\mathrm{ln}{K}_{2}=-\frac{{E}_{a}}{{\mathrm{RT}}_{2}}+\mathrm{ln}A$

We can subtract one of these equations from the other:

$\mathrm{ln}{K}_{1}-\mathrm{ln}{K}_{2}=(-\frac{{E}_{a}}{{\mathrm{RT}}_{1}}+\mathrm{ln}A)-(-\frac{{E}_{a}}{{\mathrm{RT}}_{2}}+\mathrm{ln}A)$

This equation can further simplified to:

$\mathrm{ln}\frac{{K}_{1}}{{K}_{2}}=\frac{{E}_{a}}{R}(\frac{1}{{T}_{2}}-\frac{1}{{T}_{1}})$

$\mathrm{ln}\frac{0.05}{0.03}=\frac{\mathrm{Ea}}{2.305\times 8.314}\times [\frac{1}{200}-\frac{1}{300}]$

$\Rightarrow 0.22=\frac{\mathrm{Ea}}{2.305\times 8.314}\times [\frac{1}{600}]$

$\Rightarrow {E}_{a}=2.3\times 8.3\times 600\times 0.22$

$\Rightarrow {E}_{a}=2519.88J\Rightarrow {E}_{a}=2520J$

JEE Main 2023 — 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 2023 — 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