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Clausius-clapeyron And Phase Transitions

Mediumchemistry

The enthalpy of vaporisation of benzene is 30.8 kJ/mol at its boiling point of 353 K. What is the entropy change during the vaporisation of 1 mole of benzene at its boiling point?

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About This Question

Subject
chemistry
Chapter
chemical thermodynamics
Topic
clausius-clapeyron and phase transitions
Difficulty
Medium
Year
2025
Tags
entropy of vaporisationphase transitionTrouton's rulereversible processboiling point thermodynamics

Solution

Correct Answer:

At the boiling point, vaporisation is a reversible phase transition occurring at constant temperature and pressure. For a reversible process at constant temperature, the entropy change is: (\Delta S = \frac{\Delta H_{vap}}{T}). This follows from the definition (dS = \delta q_{rev}/T) and the fact that the heat exchanged at the boiling point under equilibrium conditions is exactly (\Delta H_{vap}). Given (\Delta H_{vap} = 30{,}800) J/mol and (T = 353) K: (\Delta S = \frac{30{,}800}{353} = 87.3) J/mol·K. This is consistent with Trouton's Rule, which states that for many non-associated liquids, (\Delta S_{vap} \approx 87{-}88) J/mol·K — a notable regulariy. Option 95.4 J/mol·K uses a lower temperature of 322 K incorrectly. Option 74.1 J/mol·K results from dividing by 415 K. Option 108.6 J/mol·K uses an incorrect (\Delta H) of 38.3 kJ/mol. This is covered in NCERT under entropy changes during phase transitions. Plausibility check: benzene is a non-associated, non-polar liquid, so Trouton's Rule applies and the calculated value of (\approx 87) J/mol·K is precisely on target.

This medium difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of clausius-clapeyron and phase transitions. It appeared in the 2025 exam.

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