Question

The enthalpy of vaporization of substance is $840\, J - mol ^{-1}$ and its boiling point is $-173^{\circ} C$. Its entropy of vaporization is:

• A. $42\, J \,mol ^{-1} K^{-1}$
• B. $21 \,J \,mol ^{-1}\, K ^{-1}$
• C. $84 \,J \,mol ^{-1}\, K^{-1}$
• D. $8.4 \,J \,mol^{-1} \,K^{-1}$

Answer 1

Answer: (D)

$\Delta S=\frac{\Delta H}{T}$
where $\Delta S=$ entropy of vaporisation.
$\Delta H=$ enthalpyof vaporisation $=840\, J / mol$
$T=-173^{\circ} C=-173+273=100\, K$
$\therefore \Delta S=\frac{840}{100}=8.4 \,J\,mol ^{-1} K ^{-1}$