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Standard Gibbs energy change from the degree of dissociation at equilibrium

This type of question, asked 2 times: 14 Jun 2022, 4 Apr 2026 S2. All 2 questions below, oldest first.

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  1. Question 1

    14 Jun 2022

    52
    2O3(g)⇌3O2(g)2O_{3}(g) \rightleftharpoons 3O_{2}(g)
    At 300 K300\text{ }K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm1\text{ }atm pressure is (−)000‾ J mol−1(-)\underline{\phantom{000}}\ J\ mol^{-1} (Nearest integer) [Given: ln⁡1.35=0.3\ln 1.35 = 0.3 and R=8.3 J K−1 mol−1R = 8.3\text{ }J\text{ }K^{-1}\text{ }mol^{-1}]

  2. Question 2

    4 Apr 2026 S2

    47
    For the following reaction at 50∘C50^{\circ}C and 2 atm pressure, 2 N2O5( g)⇌2 N2O4( g)+O2( g)2{\text{ }N}_{2}O_{5}(\text{ }g) \rightleftharpoons 2{\text{ }N}_{2}O_{4}(\text{ }g) + O_{2}\left( \text{ }g \right), N2O5N_{2}O_{5} is 50%50\% dissociated. The magnitude of standard free energy change at this temperature is x . x=x = ____\_\_\_\_ Jmol−1J{mol}^{- 1} [Nearest integer] Given : R=8.314 J mol−1 K−1,log⁡2=0.30R = 8.314\text{ }J{\text{ }mol}^{- 1}{\text{ }K}^{- 1},\log2 = 0.30,log⁡3=0.48,ln⁡10=2.303, ∘C+273=K\log3 = 0.48,\ln10 = 2.303,\ ^{\circ}C + 273 = K