JEE Main 2026 — Apr 4, Shift 2 — Chemistry solution

JEE Main2026Apr 4, Shift 2Chemistry

For the following reaction at 50°C and 2 atm pressure,

$2 N_{2} O_{5} (g) ⇌ 2 N_{2} O_{4} (g) + O_{2} (g)$

N $_{2}$ O $_{5}$ 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 mol $^{- 1}$ K $^{- 1}$ , log 2 = 0.30, log 3 = 0.48, ln 10 = 2.303, °C + 273 = K

Your answer

Solution

Take initial moles of N $_{2}$ O $_{5}$ = 1. With 50% dissociation, x = 0.5 (in the stoichiometric variable), so:

Species	N $_{2}$ O $_{5}$	N $_{2}$ O $_{4}$	O $_{2}$
moles at eqm	0.5	0.5	0.25

Total moles = 1.25. Mole fractions: $y_{N_{2} O_{5}} = 0.4$ , $y_{N_{2} O_{4}} = 0.4$ , $y_{O_{2}} = 0.2$ .

Partial pressures at P $_{t o t a l}$ = 2 atm: $P_{N_{2} O_{5}} = 0.8$ , $P_{N_{2} O_{4}} = 0.8$ , $P_{O_{2}} = 0.4$ atm.

$K_{p} = \frac{P _{N_{2} O_{4}}^{2} \cdot P _{O_{2}}}{P _{N_{2} O_{5}}^{2}} = \frac{( 0.8 ) ^{2} ( 0.4 )}{( 0.8 ) ^{2}} = 0.4$ .

$Δ G ° = - 2.303 R T lo g K_{p} = - 2.303 \times 8.314 \times 323 \times lo g (0.4)$ .

$lo g 0.4 = lo g (4/10) = 2 lo g 2 - 1 = 0.60 - 1 = - 0.40$ .

$Δ G ° = - 2.303 \times 8.314 \times 323 \times (- 0.40) = + 2473.8$ J/mol.

Magnitude: $∣Δ G °∣ \approx 2474$ J/mol.

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