Question

If the enthalpy of vaporization of water is $186.5 \,kJ \, mol^{-1}$, the entropy if its vaporization will be :

• A. $0.5\,k \, JK^{-1} \, mol^{-1}$
• B. $1.0 \,k \, JK^{-1} \, mol^{-1}$
• C. $1.5 k \, JK^{-1} \, mol^{-1}$
• D. $2.0\,k \, JK^{-1} \, mol^{-1}$

Answer 1

Answer: (A)

Given enthalpy of vaporization,
$\Delta H =186 \cdot 5\, kJ\, mol ^{-1}$
Boiling point of water
$=100^{\circ} C =100+273=373\, K$
Entropy change,
$\Delta S =\frac{\Delta H }{ T }=\frac{186 \cdot 5\, kJ\, mol ^{-1}}{373\, k }$
$=0 \cdot 5\, kJ\, mol ^{-1} K ^{-1}$