The molar heat of fusion of benzene ‍ is kJ/mol. Its molar heat of vaporization is kJ/mol. Calculate the heat required to melt g benzene at its normal melting point. Calculate the heat required to vaporize g benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?

Question;

The molar heat of fusion of benzene ‍ is 9.92 kJ/mol. Its molar heat of vaporization is 30.7 kJ/mol. Calculate the heat required to melt 8.25 g benzene at its normal melting point. Calculate the heat required to vaporize 8.25 g benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?

The answers to the question are;

a. The heat required to melt g benzene at its normal melting point is equal to 1.048 kJ.

b. The heat required to vaporize g benzene at its normal boiling point is  3.24 kJ.

c. Energy is absorbed to increase entropy of the system.

Explanation:

We note that the molar mass of benzene gas C₆H₆ is 78.11 g/mol

Therefore 8.25 g contains

Number of moles =   which is

Number of moles of C₆H₆ = = 0.1056 moles

Energy required to melt one mole of benzene, C₆H₆  = 9.92 kJ/mol. Therefore, energy required to melt 0.1056 moles =

9.92 kJ/mol × 0.1056 moles = 1.048 kJ

b. The heat of vaporization of benzene, C₆H₆ is given as 30.7 kJ/mol.

That is the energy required to vaporize 1 mole of benzene, C₆H₆  is equal to 30.7 kJ/mol.

Therefore, energy required to vaporize 0.1056 moles =

30.7 kJ/mol × 0.1056 moles = 3.24 kJ

c. The reason if because of the increased entropy of the vaporized C₆H₆, requires more energy to completely free the molecules from the inter-molecular bonds so they can drift to fill the entire container in which they are placed

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