Question

Enthalpy of sublimation of iodine is 8.3 $kcalmo{l}^{-1}$ at 350 K. If molar heat capacity of $I_{2\left(s\right)}$ and $I_{2\left(vap\right)}$ are 14 and 8 $calmo{l}^{-1}{K}^{-1}$ respectively, then what is entropy of sublimation of $I_{2\left(s\right)}$ at 400 K in $calmo{l}^{-1}{K}^{-1}$? (pressure = 1 atm)

Answer 1

Enthalpy of sublimation at 350 K = 8300 $calmo{l}^{-1}$
Molar heat capacity $\Delta H_{\left(I_2\right(s\left)\right)}=14cal/molK$ and $\Delta H_{\left(I_2\right(Vap\left)\right)}=8cal/molK$
So, to get vapours of iodine at 400 K, we can first sublime it, and then heat the vapours, or first heat the solid and then sublime it, the change in enthalpy will be the same in both:
$\Delta H_{\left(subat350K\right)}+n{C}_{P\left(vap\right)}\times \Delta T=n{C}_{P\left(s\right)}\times \Delta T+\Delta H_{\left(subat400K\right)}$
$\Rightarrow 8300+8\times 50=14\times 50+\Delta H_{\left(subat400K\right)}=8000cal/mol$

So, Entropy of sublimation at 400 K:
$\Delta S_{\left(subat400K\right)}=\frac{\Delta H_{\left(subat400K\right)}}{T_{sub}}=\frac{8000}{400}=20cal/molK$