Skip to main content

For the reaction, N_2 O_4 ⇌ 2 NO_2, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is…

JEE Main 2026 — Chemistry Physical Chemistry

2026mcqhard

For the reaction, N2O4⇌2NO2\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}N2​O4​⇌2NO2​, graph is plotted as shown below. Identify correct statements.

A. Standard free energy change for the reaction is −5.40 kJ mol−1-5.40 \mathrm{~kJ} \mathrm{~mol}^{-1}−5.40 kJ mol−1.

B. As ΔG⊖\Delta \mathrm{G}^{\ominus}ΔG⊖ in graph is positive, N2O4\mathrm{N}_{2} \mathrm{O}_{4}N2​O4​ will not dissociate into NO2\mathrm{NO}_{2}NO2​ at all.

C. Reverse reaction will go to completion.

D. When 1 mole of N2O4\mathrm{N}_{2} \mathrm{O}_{4}N2​O4​ changes into equilibrium mixture, value of ΔG⊖=−0.84 kJ mol−1\Delta \mathrm{G}^{\ominus}=-0.84 \mathrm{~kJ} \mathrm{~mol}^{-1}ΔG⊖=−0.84 kJ mol−1

E. When 2 mole of NO2\mathrm{NO}_{2}NO2​ changes into equilibrium mixture, ΔG⊖\Delta \mathrm{G}^{\ominus}ΔG⊖ for equilibrium mixture is −6.24 kJ mol−1-6.24 \mathrm{~kJ} \mathrm{~mol}^{-1}−6.24 kJ 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

  1. A

    D and E only

  2. B

    A and D only

  3. C

    B and C only

  4. D

    C and E only

Did you get this right?

Sign in to track your attempts and accuracy.

Your note

Sign in to keep a private note on this question. Nothing you write is ever public.

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

2026medium

![JEE Main 2026 Chemistry, Structure of Atom — question figure 1](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Structure_of_Atom/6983c1ae8bfdbb24356397c7/question_1__q_6983c1ae8bfdbb24356397c7__cdn-question-pool.getmarks.app__28S1_q_64_1__80fd8624a7_final_ppt_sync.png) ![JEE Main 2026 Chemistry, Structure of Atom — question figure 2](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Structure_of_Atom/6983c1ae8bfdbb24356397c7/question_2__q_6983c1ae8bfdbb24356397c7__cdn-question-pool.getmarks.app__28S1_q_64_2__04a1587871_final_ppt_sync.png) Which of the following point in Figure 2 most accurately represents the nodal surface as shown in Figure 1?

2026medium

At 298 K, the mole percentage of $\mathrm{N}_{2}(\mathrm{~g})$ in air is $80 \%$. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $\mathrm{N}_{2}(\mathrm{~g})$ in water at 298 K ? ($\mathrm{K}_{\mathrm{H}}$ for $\mathrm{N}_{2}$ is $6.5 \times 10^{7} \mathrm{~mm} \mathrm{Hg}$)

2026medium

Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is $\_\_\_\_$ mL. (Nearest integer) [Given: $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \mathrm{E}_{\text {red }}^{\mathrm{o}}=+0.34 \mathrm{~V}$ $\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\text {red }}^{\mathrm{o}}=+1.23 \mathrm{~V}$ Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{O}_{2}=32 \mathrm{~g} \mathrm{~mol}^{-1}$ Faraday Constant $=96500 \mathrm{C} \mathrm{mol}^{-1}$ Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$]

2026hard

Arrange the following resultant mixtures in increasing order of their pH values A. $10$ mL $0.2$ M Ca(OH)$_2$ + $25$ mL $0.1$ M HCl B. $10$ mL $0.01$ M H$_2$SO$_4$ + $10$ mL $0.01$ M Ca(OH)$_2$ C. $10$ mL $0.1$ M H$_2$SO$_4$ + $10$ mL $0.1$ M KOH Choose the correct answer from the options given below:

2026medium

Other Chemistry topics for JEE Main

Keep practicing Physical Chemistry

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

Browse all Chemistry topics