JEE MAIN - Chemistry (2022 - 24th June Morning Shift - No. 13)
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]
Answer
747
Explanation
$\underset{1-x}{2 \mathrm{O}_3(\mathrm{~g})} \rightleftharpoons \underset{\frac{3 \mathrm{x}}{2}}{3 \mathrm{O}_2(\mathrm{~g})}$
Given, $x=0.5$
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{[3(0.5)]^{3} \times 1}{[2]^{3} \times(0.5)^{2} \times 1.25}$
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{27}{8} \times \frac{0.5}{1.25}=1.35$
$$ \begin{aligned} \Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\\\ &=-2.303 \times 8.3 \times 300 \log 1.35 \\\\ &=-8.3 \times 300 \ln (1.35) \\\\ &=-747 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} $$
Given, $x=0.5$
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{[3(0.5)]^{3} \times 1}{[2]^{3} \times(0.5)^{2} \times 1.25}$
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{27}{8} \times \frac{0.5}{1.25}=1.35$
$$ \begin{aligned} \Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\\\ &=-2.303 \times 8.3 \times 300 \log 1.35 \\\\ &=-8.3 \times 300 \ln (1.35) \\\\ &=-747 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} $$
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