Enthalpy of sublimation of Iodine is 24 Cal/g at ${200}^{0}C$   If specific heat of  $I_{2\left(s\right)}$  and  $I_{2\left(vap\right)}$  are 0.055 and  $0.031Cal{g}^{-1}{K}^{-1}$ respectively, then enthalpy of iodine at ${250}^{0}C$   in  $Cal{g}^{-1}$ is.

 $I_{2\left(s\right)}\to I_{2\left(g\right)}$
Heat of reaction depends on temperature
$$\begin{gathered} \therefore \Delta H_{T_2}=\Delta H_{T_1}+{\int }_{T_1}^{T_2}\Delta c_{p}dT \\ \Delta c_p=c_{p}ofproduct-c_{p}ofreactats \\ =0.031-0.055 =-0.024cal/g \\ \therefore \Delta H_{250}=\Delta H_{200}+\Delta C_P(T_2-T_1) \\ =24-50\times 0.024 =22.8cal/g \end{gathered}$$