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Equilibrium constant (log K) from dH and dS, or dS from dH and K

This type of question, asked 4 times: 6 Apr 2023, 27 Jan 2024, 21 Jan 2026 S1, 6 Apr 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    6 Apr 2023

    55
    The value of log⁡K\log K for the reaction A⇋BA \leftrightharpoons B at 298 K298\text{ }K is (Nearest integer) Given: ΔH0=−54.07 kJ mol−1\Delta H^{0}= - 54.07\text{ }kJ{\text{ }mol}^{- 1}
    ΔS∘=10JK−1 mol−1\Delta S^{\circ}= 10JK^{- 1}{\text{ }mol}^{- 1}
    (( Take 2.303×8.314×298=5705)2.303 \times 8.314 \times 298 = 5705)

  2. Question 2

    27 Jan 2024

    145
    For a certain thermochemical reaction M→NM \rightarrow N at T=400 K,ΔH⊖=77.2 kJ mol−1,ΔS=122JK−1T = 400\text{ }K,\Delta H^{\ominus}= 77.2\text{ }kJ{\text{ }mol}^{- 1},\Delta S = 122JK^{- 1}, log⁡\log equilibrium constant (log⁡K)(\log K) is - _______ ×10−1\times10^{- 1}.

  3. Question 3

    21 Jan 2026 S1

    48
    Use the following data:
    SubstanceΔfH⊖(500 K)/kJ mol−1\Delta_{f}H^{\ominus}(500\text{ }K)/kJ\,mol^{-1}S⊖(500 K)/J K−1mol−1S^{\ominus}(500\text{ }K)/J\,K^{-1}mol^{-1}
    AB(g)AB(g)32222
    A2(g)A_{2}(g)6146
    B2(g)B_{2}(g)x280
    One mole each of A2(g)A_{2}(g) and B2(g)B_{2}(g) are taken in a 1L closed flask and allowed to establish the equilibrium at 500 K .
    A2( g)+B2( g)⇌2AB(g)A_{2}(\text{ }g) + B_{2}(\text{ }g) \rightleftharpoons 2AB(g)
    The value of xx (in kJ mol−1kJ\,mol^{-1}) is ____\_\_\_\_ (Nearest integer) (Given: log⁡K=2.2, R=8.3 JK−1mol−1\log K = 2.2,\ R = 8.3\ JK^{- 1}mol^{- 1} )

  4. Question 4

    6 Apr 2026 S1

    50
    Consider the reaction X⇌YX \rightleftharpoons Y at 300 K . If ΔHθ\Delta H^{\theta} and K are 28.40 kJ mol−128.40\text{ }kJ{\text{ }mol}^{- 1} and 1.8×10−71.8 \times 10^{- 7} at the same temperature, then the magnitude of ΔSθ\Delta S^{\theta} for the reaction in JK−1 mol−1JK^{- 1}{\text{ }mol}^{- 1} is ____\_\_\_\_ .(Nearest Integer). (Given : R=8.3JK−1 mol−1,ln⁡10=2.3R = 8.3{JK}^{- 1}{\text{ }mol}^{- 1},\ln10 = 2.3, log⁡3=0.48,log⁡2=0.30\log3 = 0.48,\log2 = 0.30