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Thermodynamics question

2023 · 6 Apr · Shift 1 · Q19
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Thermodynamics question

2023 · 6 Apr · Shift 1 · Q19

JEE MainChemistryThermodynamicsNumerical+4 / −1
The value of log⁡K\log \mathrm{K}logK for the reaction A⇌B\mathrm{A} \rightleftharpoons \mathrm{B}A⇌B at 298 K298 \mathrm{~K}298 K is ‾\underline{\hspace{2cm}}​. (Nearest integer) Given: ΔH∘=−54.07 kJ mol−1ΔS∘=10 J K−1 mol−1\Delta \mathrm{H}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}\Delta \mathrm{S}^{\circ}=10 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}ΔH∘=−54.07 kJ mol−1ΔS∘=10 J K−1 mol−1(Take 2.303×8.314×298=57052.303 \times 8.314 \times 298=57052.303×8.314×298=5705 )
Numerical answer
View written solutionFree

Correct answer: 10

  1. Use the relation between ΔG∘\Delta G^\circΔG∘ and equilibrium constant

    ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circΔG∘=ΔH∘−TΔS∘ and ΔG∘=−2.303RTlog⁡K\Delta G^\circ = -2.303RT\log KΔG∘=−2.303RTlogK

  2. Convert units properly

    Given: ΔH∘=−54.07 kJ mol−1=−54070 J mol−1\Delta H^\circ = -54.07\,\text{kJ mol}^{-1} = -54070\,\text{J mol}^{-1}ΔH∘=−54.07kJ mol−1=−54070J mol−1 ΔS∘=10 J K−1mol−1\Delta S^\circ = 10\,\text{J K}^{-1}\text{mol}^{-1}ΔS∘=10J K−1mol−1 T=298 KT=298\,\text{K}T=298K

  3. Calculate ΔG∘\Delta G^\circΔG∘

    ΔG∘=−54070−(298)(10)\Delta G^\circ = -54070 - (298)(10)ΔG∘=−54070−(298)(10) ΔG∘=−54070−2980=−57050 J mol−1\Delta G^\circ = -54070 - 2980 = -57050\,\text{J mol}^{-1}ΔG∘=−54070−2980=−57050J mol−1

  4. Use ΔG∘=−2.303RTlog⁡K\Delta G^\circ = -2.303RT\log KΔG∘=−2.303RTlogK

    −57050=−2.303×8.314×298 log⁡K-57050 = -2.303\times 8.314\times 298\,\log K−57050=−2.303×8.314×298logK

    Given: 2.303×8.314×298=57052.303\times 8.314\times 298 = 57052.303×8.314×298=5705

    So, −57050=−5705log⁡K-57050 = -5705\log K−57050=−5705logK

    log⁡K=570505705=10\log K = \frac{57050}{5705} = 10logK=570557050​=10

  5. Nearest integer

    10\boxed{10}10​

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