Clausius-clapeyron And Phase Transitions

Question

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?

RankGuru · Accepted 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.

Clausius-clapeyron And Phase Transitions

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