Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425



Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem

The molar heat of fusion of benzene is 9 . 92 kJ / mol . Its molar heat of vaporization is30.7 kJ/mol. Calculate the heat required to melt 8 . 25 g of benzene at its normal making point. Calculate the hem required to vaporize 3 . 25 g of benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?

Interpretation Introduction


The heat required to melt and heat required to vaporize 8.25g benzene must be calculated and why heat of vaporization is three times more that heat of fusion must also be explained.

Concept Introduction:

During melting and vaporization there is no temperature change as only latent heat is used for phase change.


Given information:

Heat of fusion and heat of vaporization of benzene as 9.92kJ/mol and 30.7kJ/mol respectively.

Heat of fusion of benzene is 9.92kJ/mol

So fusion of 78g benzene needs 9.92kJ heat energy.

Fusion of 1g benzene needs 9.92kJ/78 heat energy

Fusion of 8.25g benzene needs (9.92kJ×8.25)/78 heat energy =(81.84/78)kJ heat energy =1.049kJ

Heat of vaporization of benzene is 30

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