Question

Enthalpy of sublimation of iodine is $24cal{g}^{-1}at{200}^{0}C$. If specific heat $I_2\left(s\right)and{I}_2\left(vap\right)$ are$0.055and0.031cal{g}^{-1}{K}^{-1}$ respectively, then enthalpy of sublimation of iodine ${250}^{0}Cincal{g}^{-1}$ is :

• A. 2.85
• B. 5.7
• C. 22.8
• D. 11.4

Answer 1

The correct option is C 22.8

$I_2\left(s\right)\to I_2\left(g\right)$

Heat of reaction depend upon temperature i.e., it varies with temperature, as given by Kirchoff's equation,
$\Delta H_{T_2}=\Delta H_{T_1}+\underset{T_1}{\overset{T_2}{\int }}\Delta C_{p}dT$

where $\Delta C_p=C_p$ of product $-C_p$ of reactant

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

Now, $\Delta H_{T_2}-\Delta H_{T_1}=\Delta C_p(T_2-T_1)$

$\Delta H_{\left(250\right)}-\Delta H_{\left(200\right)}=-0.024(523-473)$

$\Delta H_{\left(250\right)}=24-50\times 0.024=22.8cal/g$