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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?
Answer:
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