JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Chemical Thermodynamics

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For the following reaction at 50°C and at 2 atm pressure,
2 N₂O₅(g) ⇌ 2 N₂O₄(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⁻¹ K⁻¹, log 2 = 0.30, log 3 = 0.48, ln 10 = 2.303, °C + 273 = K

Answer: 2474

Solution

Taking

1

mol of

N_{2} O_{5}

initially,

50%

dissociation consumes

0.5

mol. From

2 N_{2} O_{5} ⇌ 2 N_{2} O_{4} + O_{2}

this forms

0.5

mol

N_{2} O_{4}

and

0.25

mol

O_{2}

, leaving

0.5

mol

N_{2} O_{5}

, a total of

1.25

mol. The mole fractions are

0.4, 0.4, 0.2

, so at

2

atm the partial pressures are

P_{N_{2} O_{5}} = 0.8

P_{N_{2} O_{4}} = 0.8

and

P_{O_{2}} = 0.4

atm. Hence

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

. With

T = 323 K

Δ G^{\circ} = - 2.303 R T lo g K_{p}

, and since

lo g 0.4 = 2 lo g 2 - 1 = - 0.40

Δ G^{\circ} = - 2.303 (8.314) (323) (- 0.40) = + 2473.8 J mo l^{- 1}

. The magnitude of the standard free-energy change is

2474 J mo l^{- 1}

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Chemical Thermodynamics

Solution

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