The molar heats of fusion and vapourisation of benzene are 10.9 and $31.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The changes in entropy for the solid $\rightarrow$ liquid and liquid $\rightarrow$ vapour transitions for benzene are $x$ and $y . \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$,…

Question

The molar heats of fusion and vapourisation of benzene are 10.9 and $31.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The changes in entropy for the solid $\rightarrow$ liquid and liquid $\rightarrow$ vapour transitions for benzene are $x$ and $y . \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. The value of $(y-x)\left(\right.$ in $\left.\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$ is (At 1 atm, benzene melts at $5.5^{\circ} \mathrm{C}$ and boils at $80^{\circ} \mathrm{C}$ )

Aicharya · Accepted Answer

$\begin{aligned} & \quad \Delta \mathrm{H}_{	ext {fus }}\left(\mathrm{C}_6 \mathrm{H}_6ight)=10.9 \mathrm{~kJ} \mathrm{~mol}^{-1} \ & \Delta \mathrm{H}_{	ext {vap }}\left(\mathrm{C}_6 \mathrm{H}_6ight)=31.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \ & 	ext { Melting point of benzene }=5.5^{\circ} \mathrm{C} \ & =(5.5+273.15) \mathrm{K} \ & =278.65 \mathrm{~K} \ & 	ext { Boiling point of benzene }=80^{\circ} \mathrm{C} \ & =(80+273.15) \mathrm{K} \ & =353.15 \mathrm{~K} \ & \mathrm{y}-\mathrm{x}=\frac{\Delta \mathrm{H}_{	ext {vap }}}{\mathrm{T}_{	ext {b.p }}}-\frac{\Delta \mathrm{H}_{	ext {fus }}}{\mathrm{T}_{\mathrm{m} . \mathrm{p}}}\end{aligned}$ $\begin{aligned} & =\frac{31 	imes 1000}{353.15}-\frac{10.9 	imes 1000}{278.65} \ & =48.66 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\end{aligned}$

The molar heats of fusion and vapourisation of benzene are 10.9 and $31.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$…

Solution