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Standard Gibbs energy change from a given equilibrium constant

This type of question, asked 3 times: 29 Jan 2023, 31 Jan 2024, 1 Feb 2024. All 3 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    29 Jan 2023

    56
    Consider the following reaction approaching equilibrium at 27∘C27^{\circ}C and 1 atm1\text{ }atm pressure
    A+B⇌kf=103kr=102C+DA + B\underset{k_{r}=10^{2}}{\overset{k_{f}=10^{3}}{\rightleftharpoons}}C + D
    The standard Gibbs energy change (ΔrGθ)\left( \Delta_{r}G^{\theta} \right) at 27∘C27^{\circ}C is (−)( - ) ____\_\_\_\_ kJ mol−1kJ\,mol^{- 1} (Nearest integer). (Given: R=8.3 J K−1 mol−1R = 8.3\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} and ln⁡10=2.3\ln10 = 2.3 )

  2. Question 2

    31 Jan 2024

    58
    Consider the following reaction at 298 K298\text{ K}. 32O2(g)⇌O3(g)\dfrac{3}{2}O_{2(g)} \rightleftharpoons O_{3(g)}; Kp=2.47×10−29K_{p} = 2.47 \times 10^{-29}. ΔrG⊖\Delta_{r}G^{\ominus} for the reaction is ______ kJkJ. (Given R=8.314 J K−1 mol−1R = 8.314\,J\,K^{-1}\,mol^{-1})

  3. Question 3

    1 Feb 2024

    148
    For a certain reaction at 300 K, K=10300\text{ }K,\text{ }K = 10, then ΔG∘\Delta G^{\circ} for the same reaction is - _______ ×10−1 kJ mol−1\times10^{- 1}\text{ }kJ{\text{ }mol}^{- 1}. (Given R=8.314JK−1 mol−1R = 8.314JK^{- 1}{\text{ }mol}^{- 1} )