0
$\begingroup$

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$\ce{J/mol}$.

Can someone please help me solve this question?

$\endgroup$
  • $\begingroup$ I get the same answer as you do. $\endgroup$ – Chester Miller Jul 17 '18 at 13:03

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.