Enthalpy of sublimation of iodine is 24 cal g$$^{-1}$$ at 200°C. If specific heat of I$$_2$$(s) and I$$_2$$(vap) are 0.055 and 0.031 cal g$$^{-1}$$ K$$^{-1}$$ respectively, then enthalpy of sublimation of iodine at 250°C in cal g$$^{-1}$$ is:

We are given that the enthalpy of sublimation of iodine at $$200^{\circ}\text{C}$$ (which is $$T_1 = 200 + 273 = 473\ \text{K}$$) is $$\Delta H_{\text{sub},\,1}=24\ \text{cal g}^{-1}$$. We want the same quantity at $$250^{\circ}\text{C}$$, i.e. at $$T_2 = 250 + 273 = 523\ \text{K}$$.

The temperature dependence of an enthalpy change is obtained from Kirchhoff’s relation, which in its integral form reads

$$\Delta H_{2}=\Delta H_{1}+\int_{T_1}^{T_2}\Delta C_p\,dT,$$

where $$\Delta C_p=C_{p,\text{vap}}-C_{p,\text{solid}}$$ is the difference in specific heats between the two phases involved in the process.

From the data, we have

$$C_{p,\text{vap}}=0.031\ \text{cal g}^{-1}\text{K}^{-1},\qquad C_{p,\text{solid}}=0.055\ \text{cal g}^{-1}\text{K}^{-1}.$$

So the specific-heat difference is

$$\Delta C_p=C_{p,\text{vap}}-C_{p,\text{solid}} =0.031-0.055 =-0.024\ \text{cal g}^{-1}\text{K}^{-1}.$$

Because both specific heats are given as constants over the temperature range, the integral simplifies to multiplication:

$$\int_{T_1}^{T_2}\Delta C_p\,dT =\Delta C_p\,(T_2-T_1).$$

The temperature rise is

$$T_2-T_1 = 523\ \text{K}-473\ \text{K}=50\ \text{K}.$$

Therefore, the change in enthalpy due to heating is

$$\Delta C_p\,(T_2-T_1)=(-0.024)\times50=-1.2\ \text{cal g}^{-1}.$$

Adding this correction to the original enthalpy of sublimation, we get

$$\Delta H_{\text{sub},\,2} =\Delta H_{\text{sub},\,1}+(-1.2) =24-1.2 =22.8\ \text{cal g}^{-1}.$$

Hence, the correct answer is Option C.