Question

# The relation between the enthalpy of fusion of KCl and naphthalene is:A. KCl $<$ naphthaleneB. KCl $=$ naphthaleneC. KCl $>$ naphthaleneD. none

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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.

- 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${{\Delta }_{r}}H$.
${{A}_{\left( s \right)}}\to {{A}_{\left( l \right)}}$
- 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.
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.
- 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 $\Delta {{H}_{1}}$>$\Delta {{H}_{2}}$