Question

The enthalpy of vaporisation of a certain liquid at its boiling point of ${35}^{\circ }C$ is $24.64kJ{mol}^{-1}$. The value of change in entropy for the process is

• A. $704J{K}^{-1}{mol}^{-1}$
• B. $80J{K}^{-1}{mol}^{-1}$
• C. $24.64J{K}^{-1}{mol}^{-1}$
• D. $7.04J{K}^{-1}{mol}^{-1}$

Answer 1

The correct option is B $80J{K}^{-1}{mol}^{-1}$

Entropy of vaporisation $\left(\Delta S\right)=\frac{enthalpyofvaporisation\left(\Delta H\right)}{boilingpoint\left(inK\right)}$

Given that,

$\Delta H=24.64kJ{mol}^{-1}$ and boiling point, $T_b=35+273=308K$

$\therefore \Delta S=\frac{24.64\times {10}^{3}J{mol}^{-1}}{308K}$

$=80J{K}^{-1}{mol}^{-1}$