JEE Main 2026 · ChemistryNumerical answerNumerical valueHardMulti-step

JEE Main 4 April 2026, Shift 2, Chemistry Q72

Question 72 of 75 in this shift, Chemistry question 22 of 25, Section B.

For the following reaction at 50

^\circ

C and at 2 atm pressure,

\mathrm{2N_2O_5(g)\rightleftharpoons2N_2O_4(g)+O_2(g)}

\mathrm{N_2O_5}

is 50% dissociated. The magnitude of standard free energy change at this temperature is

x

x=

________

\mathrm{J\ mol^{-1}}

[Nearest integer]. Given :

R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}

\log2=0.30

\log3=0.48

\ln10=2.303

^\circ\mathrm{C}+273=\mathrm{K}

Official answer

2474

NTA final key.

Chapter: Thermodynamics (Chemistry)
Topic: Gibbs Free Energy
Idea tested: Gibbs Energy and Equil

Same idea in other shifts

Asked 5× in all

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31 Jan 2024, Shift 1 · Q84Consider the following reaction at 298 K. $\frac{3}{2}\mathrm{O_{2(g)}} \rightleftharpoons \mathrm{O_{3(g)}}$ .…EasyNumerical value
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24 Jun 2022, Shift 1 · Q82 $\mathrm{2O_3(g)\rightleftharpoons 3O_2(g)}$ At 300 K, ozone is fifty percent dissociated. The standard free energy change at this…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.