One mole of a solid melts at its melting point of 250 K, the liquid is then heated to 350 K and vapourised at 350 K. Given Δ H f u s = 2.0 kJ/mol, Δ H v a p = 35 kJ/mol and the average molar heat

Question

One mole of a solid melts at its melting point of 250 K, the liquid is then heated to 350 K and vapourised at 350 K. Given Δ H f u s ​ = 2.0 kJ/mol, Δ H v a p ​ = 35 kJ/mol and the average molar heat capacity of the liquid = 80 J/K/mol, find the total entropy change in J/K/mol. ( ln 1.4 = 0.34 )

TestHub · Accepted Answer

Fusion at 250 K: 250 2000 ​ = 8.0 J/K. Heating the liquid 250 K to 350 K: 80 × ln 250 350 ​ = 80 × 0.34 = 27.2 J/K. Vaporisation at 350 K: 350 35000 ​ = 100 J/K. Total = 8.0 + 27.2 + 100 = 135.2 ≈ 135 J/K/mol. 127 drops the fusion step, 175 uses 250 K for vaporisation, and 81 inverts the logarithm. Answer: (B).

Chemistry - Thermodynamics - II Question with Solution | TestHub

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