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Thermodynamics question

2021 · 20 Jul · Shift 2 · Q17
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  5. /2021 · 20 Jul · Shift 2 · Q17

Thermodynamics question

2021 · 20 Jul · Shift 2 · Q17

JEE MainChemistryThermodynamicsNumerical+4 / −1
For a given chemical reaction A →\to→ B at 300 K the free energy change is −-− 49.4 kJ mol −-− 1 and the enthalpy of reaction is 51.4 kJ mol −-− 1. The entropy change of the reaction is ‾\underline{\hspace{2cm}}​ JK −-− 1 mol −-− 1.
Numerical answer
View written solutionFree

Correct answer: 336

  1. Use the Gibbs free energy relation

    For a reaction, ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

    Rearranging for entropy change: ΔS=ΔH−ΔGT\Delta S = \frac{\Delta H - \Delta G}{T}ΔS=TΔH−ΔG​

  2. Substitute the given values

    Given: ΔG=−49.4 kJ mol−1\Delta G = -49.4\ \text{kJ mol}^{-1}ΔG=−49.4 kJ mol−1 ΔH=51.4 kJ mol−1\Delta H = 51.4\ \text{kJ mol}^{-1}ΔH=51.4 kJ mol−1 T=300 KT = 300\ \text{K}T=300 K

    So, ΔS=51.4−(−49.4)300\Delta S = \frac{51.4 - (-49.4)}{300}ΔS=30051.4−(−49.4)​

    ΔS=100.8300 kJ K−1mol−1\Delta S = \frac{100.8}{300}\ \text{kJ K}^{-1}\text{mol}^{-1}ΔS=300100.8​ kJ K−1mol−1

  3. Calculate

    ΔS=0.336 kJ K−1mol−1\Delta S = 0.336\ \text{kJ K}^{-1}\text{mol}^{-1}ΔS=0.336 kJ K−1mol−1

  4. Convert to J K−1^{-1}−1 mol−1^{-1}−1

    Since 1 kJ=1000 J1\ \text{kJ} = 1000\ \text{J}1 kJ=1000 J, ΔS=0.336×1000=336 J K−1mol−1\Delta S = 0.336 \times 1000 = 336\ \text{J K}^{-1}\text{mol}^{-1}ΔS=0.336×1000=336 J K−1mol−1

  5. Final answer

    336\boxed{336}336​

  6. Comparison with stored answer

    Stored correct answer = 336336336

    This matches the derived answer.

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