What is the molar entropy of vaporization (ΔSvap) of a liquid that boils at 373 K with an enthalpy of vaporization ΔHvap = 40.66 kJ/mol?
Answer and explanation
Correct answer: B. 109.0 J K^-1 mol^-1
- Step 1: Phase transitions at the normal boiling point occur reversibly at constant temperature and pressure.
- Step 2: The entropy of vaporization is given by: .
- Step 3: Convert enthalpy of vaporization to joules: .
- Step 4: Calculate entropy change: .
- Step 5: (This value aligns closely with Trouton's rule, which predicts for non-associated and polar liquids).
Answer
109.0 J K^-1 mol^-1
Why not the others: '10.90 J K^-1 mol^-1' and '0.109 J K^-1 mol^-1' are off by powers of 10 from arithmetic or conversion errors; '1.52 * 10^4 J K^-1 mol^-1' mistakenly multiplies enthalpy by temperature instead of dividing.
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