# How to calculate the enthalpy of fusion at a different temperature?

The heat capacity of liquid iodine is $80.7$ $\ce{J*K^{-1}*mol^{-1}}$, and its enthalpy of vaporization is $41.96$ kJ $\ce{*mol^{-1}}$ at its boiling point ($184.3$ $\ce{^oC}$). Furthermore, the enthalpy of formation of gaseous iodine is $+62.44$ kJ$\ce{*mol^{-1}}$, and the molar heat capacity of gaseous iodine is $36.90$ J$\ce{*K^{-1}*mol^{-1}}$. Calculate the enthalpy of fusion of iodine at $25^o\ce{C}$.

I have given this question a lot of thought, but I am still not getting the correct answer.

My approach is as follows:

At $184.3^o$ $\ce{C}$, $\ce{I_2_{(l)}\rightarrow I_2_{(g)}}$, $\Delta H_{vap} = 41.96$ k$\ce{J/mol}$.

At $25^o$ $\ce{C}$, $\ce{I_2_{(l)}\rightarrow I_2_{(g)}}$, $\Delta H_{vap} = 41.96 + (25 - 184.3)*(36.90-80.7)/1000 = 48.94$ kilojoules per mole. This was determined using Kirchhoff's law for enthalpies: $\Delta H(T_2) = \Delta H(T_1) + (T_2-T_1)*\Delta C_p$.

At $25^o$ $\ce{C}$, $\ce{I_2_{(s)}\rightarrow I_2_{(g)}}$, $\Delta H_{sub} = 62.44$ k$\ce{J/mol}$.

Since enthalpy is a state function, at $25^o$ $\ce{C}$, $\Delta H_{sub} = \Delta H_{vap} + \Delta H_{fus}$. Plugging in the values, we get: $62.44 = 48.94 + \Delta H_{fus}$, or $\Delta H_{fus} = 13.50$ k$\ce{J/mol}$.

However, the textbook gives $12.71$ k$\ce{J/mol}$.