JEE Main 2026 — Chemistry Physical Chemistry

2026mcqhard

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

Choose the correct answer from the options given below :

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
D and E only
B
A and D only
C
B and C only
D
C and E only

Did you get this right?

Solution

From the graph of G vs fraction of $N_2O_4$ dissociated:

At $x = 0$ (pure $N_2O_4$ ): $G = 0$

At equilibrium ( $x \approx 0.4$ ): $G = -0.84$ kJ/mol (minimum)

At $x = 1$ (complete dissociation): $G = 5.40$ kJ/mol

Statement A: $\Delta G° = 5.40$ kJ/mol (positive), not $-5.40$ kJ/mol. False.

Statement B: Even with positive $\Delta G°$ , equilibrium is established with partial dissociation. False.

Statement C: Reverse reaction reaches equilibrium, not completion. False.

Statement D: $\Delta G = G_{eq} - G_{N_2O_4} = -0.84 - 0 = -0.84$ kJ/mol. True.

Statement E: $\Delta G = G_{eq} - G_{NO_2} = -0.84 - 5.40 = -6.24$ kJ/mol. True.

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

From the graph of G vs fraction of $N_2O_4$ dissociated:

At $x = 0$ (pure $N_2O_4$ ): $G = 0$

At equilibrium ( $x \approx 0.4$ ): $G = -0.84$ kJ/mol (minimum)

At $x = 1$ (complete dissociation): $G = 5.40$ kJ/mol

Statement A: $\Delta G° = 5.40$ kJ/mol (positive), not $-5.40$ kJ/mol. False.

Statement B: Even with positive $\Delta G°$ , equilibrium is established with partial dissociation. False.

Statement C: Reverse reaction reaches equilibrium, not completion. False.

Statement D: $\Delta G = G_{eq} - G_{N_2O_4} = -0.84 - 0 = -0.84$ kJ/mol. True.

Statement E: $\Delta G = G_{eq} - G_{NO_2} = -0.84 - 5.40 = -6.24$ kJ/mol. True.

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