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JEE Main 2026 Chemistry question

ChemistryThermodynamicsJEE Main 2026

For the following reaction at 50°C and at 2 atm pressure, 2N₂O₅(g) ⇌ 2N₂O₄(g) + O₂(g), N₂O₅ is 50% dissociated. The magnitude of standard free energy change at this temperature is x. x = J mol⁻¹ [Nearest integer]. Given: R=8.314 J mol⁻¹K⁻¹, log2=0.30, log3=0.48, ln10=2.303, °C+273=K

Answer: 2474

Solution

Starting with 2 mol N₂O₅, 50% dissociation gives 1 mol N₂O₅, 1 mol N₂O₄, 0.5 mol O₂ (total 2.5 mol). At 2 atm total:

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Question ID VK-015406