Question

The relation between the enthalpy of fusion of KCl and naphthalene is:
A. KCl $ < $ naphthalene
B. KCl $ = $ naphthalene
C. KCl $ > $ naphthalene
D. none

Answer 1

Hint: We can define enthalpy of fusion as the heat evolved in the reaction when an element changes its state from one state to another. The units for this are (KJ/mol), or sometimes (J/g) units are also used.

Complete answer:
- We can see that if there is any element ${{A}_{\left( s \right)}}$ and we are changing it in liquid state ${{A}_{\left( l \right)}}$ at constant pressure, then the heat will be evolved in this reaction, and we call this heat evolved as enthalpy of fusion, which is denoted by${{\Delta }_{r}}H$.
${{A}_{\left( s \right)}}\to {{A}_{\left( l \right)}}$
- So, if we convert it from solid to liquid state, we have to apply heat because as we know that solid is present in a very compact form, and we supply heat constantly.
- We know that solid is having compact arrangement of atoms, for example in case of KCl , ${{K}^{+}}$ and \[C{{l}^{-}}\]are very strongly bonded to each other because the electronegativity difference in both of these is high.
- So, KCl is very compact molecule, and if we about compactness or rigidity in KCl or naphthalene,
That is if see whether which one is more rigid than other, then it is found that KCl > naphthalene, because ${{K}^{+}}$ and \[C{{l}^{-}}\] are bonded and are compact as compared to naphthalene.
- The energy which we will provide to KCl in order to dissociate or we can say to change its state from solid to liquid, as compared to energy provided to naphthalene will be more.
- We are giving $\Delta {{H}_{1}}$ heat in KCl and $\Delta {{H}_{2}}$ in naphthalene, so it is clear that we have to give more heat in KCl because we need more heat in order to break bond in KCl
- Hence, we can say that $\Delta {{H}_{1}}$>$\Delta {{H}_{2}}$
Now, we can conclude that the option (c) is correct that is, the relation between the enthalpy of fusion of KCl and naphthalene is KCl >naphthalene.

Note:
- We can say that the stronger the bond present or greater difference in electronegativity between two atoms, more heat energy will be required to break the bond and such molecules will have high enthalpy of fusion.