The molar heat of fusion of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(9.92 \mathrm{kJ} / \mathrm{mol}\). Its molar heat of vaporization is \(30.7 \mathrm{kJ} / \mathrm{mol}\). 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?

Short Answer

Expert verified
The heat required to melt 8.25 g benzene at its normal melting point is approximately 1.05 kJ. The heat required to vaporize 8.25 g benzene at its normal boiling point is approximately 3.24 kJ. The heat of vaporization is more than three times the heat of fusion because vaporization requires more energy to completely break the molecular bonds, whereas fusion only requires weakening the bonds.

Step by step solution

01

Calculate the molar mass of benzene

To find the molar mass of benzene, we can simply add up the molar masses of the individual elements making up the molecule. For benzene, this means adding together the masses of six carbon atoms and six hydrogen atoms: Molar mass of benzene = (6 × molar mass of carbon) + (6 × molar mass of hydrogen) Molar mass of benzene ≈ (6 × 12.01 g/mol) + (6 × 1.01 g/mol) = 72.06 + 6.06 ≈ 78.12 g/mol
02

Find the moles of benzene

Now that we've determined the molar mass of benzene, we can find the moles of benzene: Moles = mass / molar mass Moles ≈ 8.25 g / 78.12 g/mol ≈ 0.1056 mol
03

Calculate the heat required to melt the benzene

We can now calculate the heat required to melt the benzene using the molar heat of fusion: Heat = Molar heat × Moles Heat = 9.92 kJ/mol × 0.1056 mol ≈ 1.05 kJ So, the heat required to melt 8.25 g benzene at its normal melting point is approximately 1.05 kJ.
04

Calculate the heat required to vaporize the benzene

Similarly, we can calculate the heat required to vaporize the benzene using the molar heat of vaporization: Heat = Molar heat × Moles Heat = 30.7 kJ/mol × 0.1056 mol ≈ 3.24 kJ So, the heat required to vaporize 8.25 g benzene at its normal boiling point is approximately 3.24 kJ.
05

Explain why the heat of vaporization is more than three times the heat of fusion

The heat of vaporization is more than three times the heat of fusion because the process of vaporization requires more energy compared to the process of fusion. When a substance undergoes fusion, its molecules change from a solid phase to a liquid phase, meaning the molecular bonds become slightly weaker during this phase transition. This requires some amount of heat, which is the heat of fusion. On the other hand, when a substance undergoes vaporization, its molecules are changing from a liquid phase to a gaseous phase. This requires enough energy to completely break the molecular bonds within the substance so that they are no longer attracted to each other and can move independently of one another. The energy needed to break these bonds is much larger than the energy needed to merely weaken them, as in the case of fusion. That's why the heat of vaporization is significantly higher than the heat of fusion.

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