Problem 21 - Chemistry and Chemical Reactivity

Question

What quantity of heat is required to vaporize \(125 \mathrm{g}\) of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) at its boiling point, \(80.1^{\circ} \mathrm{C} ?\) The heat of vaporization of benzene is \(30.8 \mathrm{kJ} / \mathrm{mol}\).

Step-by-Step Solution

Answer

49.3 kJ of heat is required to vaporize 125 g of benzene.

Step 1: Determine Moles of Benzene

First, find the molar mass of benzene, \(\mathrm{C}_6\mathrm{H}_6\). Carbon (\(\mathrm{C}\)) has an atomic mass of about 12.01 g/mol and hydrogen (\(\mathrm{H}\)) has an atomic mass of about 1.01 g/mol. Thus, the molar mass of benzene is calculated as \(6 \times 12.01 \text{ g/mol} + 6 \times 1.01 \text{ g/mol} \approx 78.11 \text{ g/mol}\). Now, to find the moles of benzene, divide the given mass of benzene by its molar mass:\[\text{Moles of benzene} = \frac{125 \text{ g}}{78.11 \text{ g/mol}} \approx 1.60 \text{ mol}\]

Step 2: Calculate the Heat Required for Vaporization

Using the heat of vaporization, \(30.8 \mathrm{kJ/mol}\), calculate the total heat required to vaporize 1.60 moles of benzene. Use the formula:\[\text{Heat required} = \text{moles} \times \text{heat of vaporization} = 1.60 \text{ mol} \times 30.8 \text{ kJ/mol}\]\[\text{Heat required} \approx 49.3 \text{ kJ}\]

Step 3: Conclusion

Thus, the quantity of heat required to vaporize 125 g of benzene at its boiling point is approximately 49.3 kJ.

Key Concepts

Molar Mass Calculation

Molar mass is an essential concept in chemistry, as it helps in determining the number of moles of a substance. Moles are a way to count particles, and the molar mass acts as a bridge between the mass of a substance and the number of atoms, molecules, or compounds.

• Atomic Mass: Atomic masses are found on the periodic table. For benzene, we consider carbon (C) with an atomic mass of about 12.01 g/mol and hydrogen (H) with an atomic mass of about 1.01 g/mol.
• Molar Mass Calculation: To calculate the molar mass of a compound, multiply the atomic mass of each element by its number of atoms in the molecule and sum these values. For benzene (\(\text{C}_6\text{H}_6\)), the calculation is:\[6 \times 12.01 \text{ g/mol (for C)} + 6 \times 1.01 \text{ g/mol (for H)} \approx 78.11 \text{ g/mol}\]

Understanding molar mass allows us to convert between grams and moles, facilitating various chemical calculations and solutions.

Boiling Point

The boiling point is the temperature at which a liquid turns into a vapor. It’s a critical concept when studying the process of vaporization.
The boiling point of a substance like benzene is determined by its molecular structure and the strength of intermolecular forces.

• Boiling Point of Benzene: For benzene, this is 80.1°C. This means at 80.1°C, benzene changes from a liquid to a gas.
• Significance: Knowing the boiling point helps in calculating the energy needed for vaporization. At the boiling point, all the added heat goes into breaking intermolecular bonds rather than raising the temperature.

This understanding is vital when computing the heat required during the phase change from liquid to vapor.

Moles Calculation

Calculating moles is a fundamental step in chemical quantifications. By converting the mass of a substance to moles, chemists gain a clearer picture of the amount of substance present in terms of its constituent particles.

• Formula: The number of moles (\( n \)) can be calculated using the formula:\[n = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}}\]
• Application to Benzene: For 125 g of benzene:\[\text{Moles of benzene} = \frac{125 \text{ g}}{78.11 \text{ g/mol}} \approx 1.60 \text{ mol}\]

By understanding this calculation, students can apply it to solve further stoichiometric problems and better grasp the nature of substances at the molecular level.