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JEE Main 2022 · 24 Jun · Shift 1, Chemistry question 82

2O3(g)⇌3O2(g)\mathrm{2O_3(g)\rightleftharpoons3O_2(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^{-1}. (Nearest integer)

[Given : ln⁡1.35=0.3\ln1.35=0.3 and R=8.3 J K−1mol−1R=8.3\ \mathrm{J\,K^{-1}mol^{-1}}]

Answer and worked solution

Official answer: 747

Equilibrium. Start with 1 mol O3\mathrm{O_3}; half dissociates: O3\mathrm{O_3} 0.5 mol, O2\mathrm{O_2} 0.75 mol, total 1.25 mol.

Partial pressures at 1 atm. pO3=0.4p_{\mathrm{O_3}}=0.4, pO2=0.6p_{\mathrm{O_2}}=0.6.

Constant. Kp=0.630.42=0.2160.16=1.35K_p=\dfrac{0.6^3}{0.4^2}=\dfrac{0.216}{0.16}=1.35.

Free energy. ΔG∘=−RTln⁡K=−8.3×300×0.3=−747\Delta G^\circ=-RT\ln K=-8.3\times300\times0.3=-747 J mol−1^{-1}.

Answer: 747

Topics:Equilibrium and Gibbs energyLaw of mass action and Kc, Kp
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