AktiraAktira JEE
Subscribe
ChemistryJEE Main 4 Apr 2026, Shift 2·Q22·Numerical·Moderate

Thermodynamics · Gibbs energy and equilibrium constant

For the following reaction at 50∘^\circC and 2 atm pressure,

2 NX2OX5(g)⇌2 NX2OX4(g)+OX2(g)\ce{2N2O5(g) <=> 2N2O4(g) + O2(g)}

NX2OX5\ce{N2O5} is 50% dissociated

The magnitude of standard free energy change at this temperature is x.

x = J mol−1^{-1} [Nearest integer]

Given : R=8.314R = 8.314 J mol−1^{-1} K−1^{-1}, log⁡2=0.30\log 2 = 0.30, log⁡3=0.48\log 3 = 0.48, ln⁡10=2.303\ln 10 = 2.303, ∘^\circC + 273 = K

We corrected a printing error in the original paper

Source printed “R = 8.314 mol−1^{-1} K−1^{-1}” (unit J missing).

Video solution coming soon

We're animating every PYQ step by step. Subscribe to get them first.

Subscribe

See the full 4 Apr 2026, Shift 2 paper

More on Gibbs energy and equilibrium constant