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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2026 · 21 January · Morning Shift · Chemistry · Chemical Equilibrium

For the reaction, N_ 2 O_ 4 leftharpoons 2 NO_ 2 , graph is plotted...

For the reaction, N2O4⇌2NO2\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 kJ mol−1-5.40 \mathrm{~kJ} \mathrm{~mol}^{-1}.
B. As ΔG⊖\Delta \mathrm{G}^{\ominus} in graph is positive, N2O4\mathrm{N}_{2} \mathrm{O}_{4} will not dissociate into NO2\mathrm{NO}_{2} at all.
C. Reverse reaction will go to completion.
D. When 1 mole of N2O4\mathrm{N}_{2} \mathrm{O}_{4} changes into equilibrium mixture, value of ΔG⊖=−0.84 kJ mol−1\Delta \mathrm{G}^{\ominus}=-0.84 \mathrm{~kJ} \mathrm{~mol}^{-1}
E. When 2 mole of NO2\mathrm{NO}_{2} changes into equilibrium mixture, ΔG⊖\Delta \mathrm{G}^{\ominus} for equilibrium mixture is −6.24 kJ mol−1-6.24 \mathrm{~kJ} \mathrm{~mol}^{-1}.
For the reaction, N_ 2 O_ 4 leftharpoons 2 NO_ 2 , graph is plotted as shown below. Identify correct statements. A.
Choose the correct answer from the options given below :

Options

  1. AD and E only
  2. BA and D only
  3. CB and C only
  4. DC and E only

Correct answer

A. D and E only

Step-by-step solution

From the graph of G vs fraction of N2O4N_2O_4 dissociated:
At x=0x = 0 (pure N2O4N_2O_4): G=0G = 0
At equilibrium (x≈0.4x \approx 0.4): G=−0.84G = -0.84 kJ/mol (minimum)
At x=1x = 1 (complete dissociation): G=5.40G = 5.40 kJ/mol
Statement A: ΔG°=5.40\Delta G° = 5.40 kJ/mol (positive), not −5.40-5.40 kJ/mol. False.
Statement B: Even with positive ΔG°\Delta G°, equilibrium is established with partial dissociation. False.
Statement C: Reverse reaction reaches equilibrium, not completion. False.
Statement D: ΔG=Geq−GN2O4=−0.84−0=−0.84\Delta G = G_{eq} - G_{N_2O_4} = -0.84 - 0 = -0.84 kJ/mol. True.
Statement E: ΔG=Geq−GNO2=−0.84−5.40=−6.24\Delta G = G_{eq} - G_{NO_2} = -0.84 - 5.40 = -6.24 kJ/mol. True.
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