Question

Ethanol boils at ${78.4}^{o}C$ and the enthalpy of vaporisation of ethanol is $42.4kJmo{l}^{-1}.$ Calculate the entropy of vaporisation of ethanol.

• A. $101.5J{K}^{-1}mo{l}^{-1}$
• B. $150J{K}^{-1}mo{l}^{-1}$
• C. $120.66J{K}^{-1}mo{l}^{-1}$
• D. $135J{K}^{-1}mo{l}^{-1}$

Answer 1

The correct option is C $120.66J{K}^{-1}mo{l}^{-1}$

We are given that,
$△H_{vap}=42.4kJmo{l}^{-1},T=273+78.4=351.4K$

We know that,
$△S_{vap}=\frac{△H_{vap}}{T}$
putting the values,
$=\frac{42.4\times {10}^3}{351.4}=120.66J{K}^{-1}mo{l}^{-1}$


Theory:

Entropy Calculation with phase transformation:
Phase change occurs at constant Pressure & Temperature and is considered reversible if it occurs at its transition temperature.

Entropy change for Vaporisation:
At constant pressure, $q=q_p$ = $\Delta H$
$\Delta S_{vap}$ = $\frac{\Delta H_{vap}}{T_b}$ where $T_b$ is boiling point of a substance.