Problem 19 - Chemistry and Chemical Reactivity

Question

How much energy 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

49.28 kJ of energy is required.

Step 1: Calculate the number of moles of benzene

First, you need to determine the molar mass of benzene, \ \( \text{C}_6\text{H}_6 \). This is calculated as follows: \( 6 \times 12.01 \text{ g/mol} + 6 \times 1.01 \text{ g/mol} = 78.11 \text{ g/mol} \). Then, use this molar mass to convert the mass of benzene (125 g) to moles: \[ \text{moles of benzene} = \frac{125 \text{ g}}{78.11 \text{ g/mol}} \approx 1.60 \text{ moles} .\]

Step 2: Use the heat of vaporization

The next step involves the use of the heat of vaporization, which is the amount of energy required to vaporize one mole of a substance. For benzene, this value is 30.8 kJ/mol. Multiply the number of moles of benzene by the heat of vaporization:\[ \text{Energy required} = 1.60 \text{ moles} \times 30.8 \text{ kJ/mol} = 49.28 \text{ kJ} .\]

Step 3: Finalize the calculation

Ensure that all calculations are correct and that the units match. The final amount of energy required is 49.28 kJ.

Key Concepts

Understanding Molar Mass

When dealing with chemical calculations, the concept of molar mass is vital. The molar mass of a substance is the mass of one mole of its particles. One mole represents Avogadro's number, approximately \(6.022 \times 10^{23}\) atoms, molecules, or ions. In simpler terms, the molar mass is the sum of the atomic masses of all the atoms in a molecule, expressed in grams per mole \(\text{g/mol}\).
For benzene \(\text{C}_6\text{H}_6\), calculating the molar mass involves adding the atomic masses of carbon (\(12.01 \text{ g/mol}\)) and hydrogen (\(1.01 \text{ g/mol}\)). Here's how:

• Benzene has 6 carbon atoms, so: \(6 \times 12.01 = 72.06 \text{ g/mol}\)
• It also has 6 hydrogen atoms, so: \(6 \times 1.01 = 6.06 \text{ g/mol}\)

Adding these gives the molar mass of benzene: \(72.06 + 6.06 = 78.12 \text{ g/mol}\). Understanding molar mass allows us to convert between mass and moles, a fundamental step in chemical calculations.

Energy Calculation using Heat of Vaporization

The heat of vaporization is the energy required to convert a given amount of a liquid into a gas at a constant temperature. This value is crucial when calculating the energy needed for processes like vaporization. For benzene, the heat of vaporization is given as \(30.8 \text{kJ/mol}\).
To find out how much energy is needed to vaporize 125 grams of benzene, we first convert the mass to moles using its molar mass (as calculated earlier: \(78.11 \text{ g/mol}\)). This step tells us how many moles of benzene we have.

• Number of moles = \(\frac{125 \text{ g}}{78.11 \text{ g/mol}} \approx 1.60 \text{ moles}\)

Next, we multiply the number of moles by the heat of vaporization to find the total energy required:

• Energy required = \(1.60 \text{ moles} \times 30.8 \text{kJ/mol} = 49.28 \text{kJ}\)

This calculation provides us with the energy necessary to vaporize benzene at its boiling point.

Applying Chemical Calculations

Chemical calculations allow us to predict and understand changes in a chemical system. They involve steps often repeated in various contexts across chemistry. The problem presented is an excellent example, as it combines concepts like molar mass and energy calculations.
In this scenario, taking the known quantity of benzene (125 g), we needed to perform calculations in stages:

• First, determine the moles of benzene using molar mass.
• Next, use the heat of vaporization to calculate the required energy.
• Finally, verify calculations for accuracy.

This typical step-by-step approach ensures a clear pathway from the given information to the solution. By breaking it down, chemical calculations empower us to solve complex problems systematically and accurately. Practicing these steps enhances understanding and proficiency in chemistry fundamentals.