Problem 98 - Chemistry An Atoms First Approach

Question

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?

Step-by-Step Solution

Answer

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 1: 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

Step 2: 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

Step 3: 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.

Step 4: 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.

Step 5: 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.

Key Concepts

Molar Heat of Fusion

The molar heat of fusion is a measure of the energy required to change a substance from a solid to a liquid at its melting point. This concept is crucial in understanding phase transitions, particularly how solids melt into liquids. The energy that is involved in this process specifically helps to overcome the forces holding the solid together without altering the temperature of the substance.

To calculate the heat absorbed during fusion, we multiply the number of moles by the molar heat of fusion. For benzene, whose molar heat of fusion is 9.92 kJ/mol, and with 0.1056 mol in 8.25 g, the energy needed is approximately 1.05 kJ.

• It indicates how much energy per mole is used to break enough molecular bonds to transition from solid to liquid.
• The heat of fusion value is unique to each substance and depends on the strength of intermolecular forces within that substance.

This energy essentially makes the molecules vibrate more vigorously, weakening the forces holding them in a rigid structure, until it becomes a liquid.

Phase Transitions

Phase transitions involve changes between different states of matter: solid, liquid, and gas. These changes occur without a change in the temperature of the substance, as the energy absorbed or released during the transition is used to alter the bonds between molecules rather than to increase kinetic energy.

During fusion, molecules transition from a solid to a liquid state, which requires overcoming some of the intermolecular forces but not all. In vaporization, however, molecules move from a liquid to a gaseous state, requiring enough energy to break these bonds completely.

• Fusion (melting) requires less energy compared to vaporization due to weaker molecular interaction changes.
• Vaporization involves a complete breaking of interactions, demanding much more energy.

The significant energy requirement for vaporization explains why the heat of vaporization is more than three times greater than the heat of fusion for benzene.

Molar Mass Calculation

Molar mass, expressed in g/mol, is the mass of one mole of a particular substance. Calculating the molar mass of a molecule like benzene involves adding up the atomic masses of its constituent atoms. For benzene \((C_6H_6)\), we find:

\(Molar \ mass = (6 \times 12.01 \, \text{g/mol}) + (6 \times 1.01 \, \text{g/mol}) = 78.12 \, \text{g/mol}\)

• This calculation is fundamental in converting between grams of a substance and moles, which is essential for subsequent calculations of the heat required for phase changes.
• Knowing the exact moles helps in determining the precise amount of energy involved in any given quantity of the substance for fusion or vaporization.

This step sets the foundation for making accurate energy calculations based on the number of moles present.