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QJEEM22Ch8b7e: JEE Main 2022 Chemistry - Class 11 JEEM22Ch8b7e Solution: 2O 3 (g) 3O 2 (g) At 300 K, ozone is fifty percent dissociated. …

2O3(g) ⇌\rightleftharpoons 3O2(g)

At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is ( −- ) ____________ J mol −- 1. (Nearest integer)

[Given : ln 1.35 = 0.3 and R = 8.3 J K −- 1 mol −- 1]

Step-by-Step Verified Solution

2O3( g)1−x⇌3O2( g)3x2\underset{1-x}{2 \mathrm{O}_3(\mathrm{~g})} \rightleftharpoons \underset{\frac{3 \mathrm{x}}{2}}{3 \mathrm{O}_2(\mathrm{~g})}

Given, x=0.5x=0.5

∴kp=[3(0.5)]3×1[2]3×(0.5)2×1.25\therefore \mathrm{k}_{\mathrm{p}}=\frac{[3(0.5)]^{3} \times 1}{[2]^{3} \times(0.5)^{2} \times 1.25}

∴kp=278×0.51.25=1.35\therefore \mathrm{k}_{\mathrm{p}}=\frac{27}{8} \times \frac{0.5}{1.25}=1.35

ΔG∘amp;=−2.303RTlog⁡kpamp;=−2.303×8.3×300log⁡1.35amp;=−8.3×300ln⁡(1.35)amp;=−747 J mol−1\begin{aligned}\Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\\\&=-2.303 \times 8.3 \times 300 \log 1.35 \\\\&=-8.3 \times 300 \ln (1.35) \\\\&=-747 \mathrm{~J} \mathrm{~mol}^{-1}\end{aligned}

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