InterviewSolution
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InterviewSolution
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Fill up the blanks in the following table related to H2O(l) → H2O(steam) at standard atmospheric pressure.S.No.tºCTKΔH (Enthalpy) change kJ mol-1ΔS (Entrophy change) J mol-1 K-1TΔS kJΔG (Free energy KJ)190363.041.1−−+1.1932110373.040.7109−−3110383.040.1−−− 0.979 |
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Answer» (1) ∆G = ∆H − T∆S + 1.193 = 41.1 − 363 × ∆S 363 × ∆S = 41.1 − 1.93 = 39.907 \(\therefore\) ∆S = \(\frac{39.907s}{363}\) = 0.1099 kJ mol−1 K−1 = 110 J mol−1 K−1 (2) T∆S = 383 × 109 = 41747 = 41.75 kJ mol−1 ∆G = ∆H − T∆S = 40.7 − 41.75 (3) ∆G = ∆H − T∆S − 0.979 = 40.1 − T∆S \(\therefore\) T∆S = 40.1 + 0.979 = 41.079 kJ mol−1 ∆S = \(\frac{T\Delta S}{T}\) = \(\frac{41.079}{383}\) = 0.1072 kJ mol−1 K−1 = 107.2 J mol−1 K−1 . |
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