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Thermodynamics question

2010 · Q67
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Thermodynamics question

2010 · Q67

NEETChemistryThermodynamicsMCQ+4 / −1
For vaporization of water at 1 atmospheric pressure, the values of Δ\DeltaΔH and Δ\DeltaΔS are 40.63 kJ mol−-−1 and 108.8 J K−-−1 mol−-−1, respectively. The temperature when Gibb's energy change (Δ\DeltaΔG) for this transformation will be zero, is
  1. A
    273.4 K
  2. B
    393.4 K
  3. C
    373.4 K
  4. D
    293.4 K
View written solutionFree

Correct answer: C

We know, from Gibb's equation,

Δ\Delta ΔG = Δ\Delta ΔH – TΔ\Delta ΔS

When Δ\Delta ΔG = 0, Δ\Delta ΔH = TΔ\Delta ΔS

∴\therefore∴ T=ΔHΔST = {{\Delta H} \over {\Delta S}}T=ΔSΔH​ = 40.63×103108.8{{40.63 \times {{10}^3}} \over {108.8}}108.840.63×103​ = 373.4 K

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