Question

When 1g liquid naphthalene (${{C}_{10}}{{H}_{8}}$ ) solidifies, 149 J of heat is evolved. The enthalpy of fusion of naphthalene is:
A. 14532 J
B. 15882 J
C. 19892 J
D. 19072 J

Answer 1

Calculate the molecular weight of naphthalene first. Heat evolved by 1g of naphthalene is given. Therefore heat evolved by 1 mole of naphthalene can be calculated. This is the enthalpy of solidification. Now note that, the process of solidification and fusion are the exact opposite of each other. Therefore the magnitude of enthalpy of fusion is same as the numerical value for enthalpy of solidification, but opposite in sign, i.e.
$\Delta {H_{solidification}} = - \Delta {H_{fusion}}$
Now find the enthalpy of fusion.

Formula used: $\Delta {H_{solidification}} = - \Delta {H_{fusion}}$

Complete step by step answer:
Molecular weight of naphthalene ((${{\text{C}}_{{\text{10}}}}{{\text{H}}_{\text{8}}}$ )) is given by =$(10 \times 12) + (8 \times 1)$= (120+8) = 128 g
Now, heat evolved by 1 g of Naphthalene while solidifying is given as 149 J
Therefore, heat evolved by 128 g of Naphthalene is (149×128) = 19072 J
This is the enthalpy of solidification.
Now, since the process of solidification and fusion are the exact opposite to each other, therefore the magnitude of enthalpy of fusion is same as the numerical value for enthalpy of solidification, but opposite in sign, i.e.
$\Delta {H_{solidification}} = - \Delta {H_{fusion}}$
Therefore, heat absorbed for the fusion of 1 mole of naphthalene will be = 19072 J
$\Delta {H_f} = + 19072{\text{ J}}$

Hence, option D is the correct answer.

Additional information:
The molar heat of fusion of a substance is the heat absorbed by 1 mole of that substance while being converted from solid to liquid state. Since the melting of a substance absorbs heat, it follows that the freezing of a substance releases heat. So the molar heat of solidification is the heat released by 1 mole of that substance as while solidifying.
Since the process of solidification and fusion are the exact opposite to each other, therefore the magnitude of enthalpy of fusion is same as the numerical value for enthalpy of solidification, but opposite in sign, i.e.
$\Delta {H_{solidification}} = - \Delta {H_{fusion}}$

Note: The process of solidification and fusion are the exact opposite to each other, therefore the magnitude of enthalpy of fusion is same as the numerical value for enthalpy of solidification, but opposite in sign, i.e.
$\Delta {H_{solidification}} = - \Delta {H_{fusion}}$
While using this concept, be careful about the signs.