Question

The specific heat of $I_2$ in vapour and solid state are 0.031 and 0.055 cal/g respectively. The heat of sublimation of iodine at $200$ is 6.096 kcal ${}^{-1}$. The heat of sublimation of iodine at $250$ will be:

• A. $3.8kcal{}^{-1}$
• B. $4.8kcal{}^{-1}$
• C. $2.28kcal{}^{-1}$
• D. $5.8kcal{}^{-1}$

Answer 1

The correct option is D $5.8kcal{}^{-1}$

Solution:- (D) $5.8kcal/mol$

${I_2}_{\left(s\right)}⟶{I_2}_{\left(v\right)}$

$\Delta H=6.0696kcal/mol=6069.6kcal/mol$ at $200℃$

$\Delta C_p={C_p}_{product}0{C_p}_{reactant}$

$\Rightarrow \Delta C_p=0.031-0.055=-0.024cal/g$

${C_p}_{\left(\text{per mole}\right)}={C_p}_{\left(\text{per gm}\right)}\times \text{Mol. wt.}$

$\therefore {C_p}_{\left(\text{per mole}\right)}=-0.024\times 253.8=-6.09cal/mol$

Now as we know that,

$\Delta H=\Delta C_p\Delta T$

$H_2-H_1=-6.09(T_2-T_1)$

$H_2-6069.6=-6.09(250-200)$

$H_2=6069.6-304.5=5765.1cal/g\approx 5.8kcal/mol$

Hence the heat of sublimation of iodine at $250℃$ will be $5.8kcal/mol$.