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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]
2 O3( g)1−x ⇌ 3 O2( g)3 x2\underset{1-x}{2 \mathrm{O}_3(\mathrm{~g})} \rightleftharpoons \underset{\frac{3 \mathrm{x}}{2}}{3 \mathrm{O}_2(\mathrm{~g})}1−x2 O3( g) ⇌ 23 x3 O2( g)
Given, x=0.5x=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=[2]3 ×(0.5)2 × 1.25[3(0.5)]3 × 1
∴ kp=278 × 0.51.25=1.35\therefore \mathrm{k}_{\mathrm{p}}=\frac{27}{8} \times \frac{0.5}{1.25}=1.35∴ kp=827 × 1.250.5=1.35
Δ G∘ =−2.303 RT log kp =−2.303 × 8.3 × 300 log 1.35 =−8.3 × 300 ln (1.35) =−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} Δ G∘ =−2.303 RT log kp =−2.303 × 8.3 × 300 log 1.35 =−8.3 × 300 ln (1.35) =−747 J mol−1
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