Thermodynamics and Thermochemistry 2 Question 1
1. Enthalpy of sublimation of iodine is $24 \mathrm{cal} \mathrm{g}^{-1}$ at $200^{\circ} \mathrm{C}$. If specific heat of $\mathrm{I}{2}(s)$ and $\mathrm{I}{2}$ (vap.) are 0.055 and $0.031 \mathrm{cal} \mathrm{g}^{-1} \mathrm{~K}^{-1}$ respectively, then enthalpy of sublimation of iodine at $250^{\circ} \mathrm{C}$ in $\mathrm{cal} \mathrm{g}^{-1}$ is
(2019 Main, 12 April I)
(a) 2.85
(b) 5.7
(c) 22.8
(d) 11.4
Show Answer
Solution:
Key Idea When $q$ is the amount of heat involved in a system then at constant pressure
$$ q=q_{p} \text { and } C_{p} \Delta T=\Delta H $$
Given reaction :
$$ \mathrm{I}{2}(s) \longrightarrow \mathrm{I}{2}(g) $$
Specific heat of $\mathrm{I}_{2}(s)=0.055 \mathrm{cal} \mathrm{g}^{-1} \mathrm{~K}^{-1}$.
Specific heat of $\mathrm{I}_{2}(\mathrm{vap})=0.031 \mathrm{cal} \mathrm{g}^{-1} \mathrm{~K}^{-1}$.
Enthalpy $\left(H_{1}\right)$ of sublimation of iodine $=24 \mathrm{cal} \mathrm{g}^{-1}$
If $q$ is the amount of heat involved in a system then at constant pressure $q=q_{p}$ and
$$ \begin{aligned} \Delta H & =C_{p} \Delta T \ H_{2}-H_{1} & =C_{p}\left(T_{2}-T_{1}\right) \ H_{2} & =H_{1}+\Delta C_{p}\left(T_{2}-T_{1}\right) \ H_{2} & =24+(0.031-0.055)(250-200) \ H_{2} & =24+(-0.024)(50)=24-1.2=22.8 \mathrm{cal} / \mathrm{g} \end{aligned} $$
Thus, the enthalpy of sublimation of iodine at $250^{\circ} \mathrm{C}$ is $22.8 \mathrm{cal} / \mathrm{g}$.
