JEE Main 2026 · 4 Apr · Shift 2 · Chemistry Q72

ChemistryInteger+4 / -1Question 72 of 75

For the following reaction at 50 °C and at 2 atm pressure,
$\ce{2N2O5(g) <=> 2N2O4(g) + O2(g)}$
$\ce{N2O5}$ 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, ln10 = 2.303, °C + 273 = K

Your answer

Answer and worked solution

Official answer: 2474

Moles at equilibrium. From 2 mol $\ce{N2O5}$ : 1 mol $\ce{N2O5}$ , 1 mol $\ce{N2O4}$ , 0.5 mol $\ce{O2}$ ; 2.5 mol in all.

Partial pressures (total 2 atm): 0.8, 0.8 and 0.4 atm, so $K_p=\frac{0.8^2(0.4)}{0.8^2}=0.4$ .

Free energy. $\log0.4=2(0.30)-1=-0.40$ , so
$|\Delta G^\circ|=2.303RT(0.40)=2.303(8.314)(323)(0.40)=2474\text{ J mol}^{-1}.$

Answer: 2474

JEE Main 2026 · 4 Apr · Shift 2, Chemistry question 72