Question

The specific heats of iodine vapour and solid are $0.031$ and $0.05$ cals/g respectively. If heat of sublimation of iodine is $24$ cals/g at ${200}^{o}C$.
Calculate its value at ${250}^{o}C$.

Answer 1

$C_P=$ molar heat capacity of product $-$ molar heat capacity of reactant

Molar heat capacity $=$ specific heat capacity $\times$ molecular weight

Molecular weight of iodine $=127g$

Given that the specific heats of iodine vapour and solid are $0.031$ and $0.05cal/g$ respectively.

Molar heat capacity of iodine solid $=0.05\times 127=6.35$

Molar heat capacity of iodine vapour $=0.031\times 127=3.9$

$\therefore C_P=6.35-3.9=2.45cal/g$

According to Kirchoff's equation-

$C_p=\frac{{\Delta H}_2-{\Delta H}_1}{T_2-T_1}$

Where as,

${\Delta H}_1=$ Heat of sublimation at temperature $T_1$

${\Delta H}_2=$ Heat of sublimation at temperature $T_2$

$\therefore 2.45=\frac{{\Delta H}_2-24}{250-200}$

$\Rightarrow {\Delta H}_2-24=50\times 2.45$

$\Rightarrow {\Delta H}_2=122.5+24=146.5cal/g$

Hence the required answer is $146.5cal/g$.