Question

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

• A. $704 \,J K^{-1}\,mol^{-1}$
• B. $80 \,JK^{-1}\,mol^{-1}$
• C. $24.64 \,JK^{-1}\,mol^{-1}$
• D. $70 \,JK^{-1}\,mol^{-1}$

Answer 1

Answer: (B)

Entropy of vaporisation $(\Delta S)$
$=\frac{\text { enthalpy of vaporisation }(\Delta H)}{\text { boiling point (in } K )}$
Given, $\Delta H=24.64 \,kJ \,mol ^{-1} $ and
boiling point $=35+273=308\, K $
$\therefore \Delta S=\frac{24.64 \times 10^{3} \,J \,mol ^{-1}}{308 \,K }$
$=80\, JK ^{-1} \,mol ^{-1}$