Problem 43 - Chemistry The Molecular Science

Question

Calculate how much energy must be transferred to vaporize \(125 \mathrm{~g}\) benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) at its boiling point, \(80.1{ }^{\circ} \mathrm{C}\). (The enthalpy of vaporization of benzene is \(30.8 \mathrm{~kJ} / \mathrm{mol}\).)

Step-by-Step Solution

Answer

49.28 kJ of energy is required to vaporize 125 g of benzene at its boiling point.

Step 1: Determine Molar Mass of Benzene

To start, we first find the molar mass of benzene (C extsubscript{6}H extsubscript{6}). Carbon (C) has a molar mass of 12.01 g/mol and hydrogen (H) has a molar mass of 1.01 g/mol. Therefore, the molar mass of benzene is: \[6(12.01) + 6(1.01) = 72.06 + 6.06 = 78.12\text{ g/mol}.\]

Step 2: Convert Mass to Moles

Next, convert the given mass of benzene (125 g) into moles using the molar mass from Step 1: \[\text{moles of C}_6\text{H}_6 = \frac{125 \text{ g}}{78.12 \text{ g/mol}} \approx 1.60 \text{ mol}.\]

Step 3: Calculate Energy for Vaporization

Now, use the enthalpy of vaporization to calculate the total energy needed to vaporize the benzene. With an enthalpy of vaporization of 30.8 kJ/mol, the energy required is: \[\text{Energy} = 1.60 \text{ mol} \times 30.8 \text{ kJ/mol} \approx 49.28 \text{ kJ}.\]

Key Concepts

Molar Mass Calculation

Before diving into the enthalpy of vaporization, it's essential to understand the molar mass calculation. Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). For a compound like benzene (C_6H_6), calculating its molar mass involves looking at each component atom's molar mass and multiplying it by the number of atoms present.
Let's break it down:

• Carbon (C) has a molar mass of 12.01 g/mol, and there are six carbon atoms in benzene.
• Hydrogen (H) has a molar mass of 1.01 g/mol, with six hydrogen atoms in benzene.

To find the total molar mass of benzene, multiply the number of carbon atoms by the molar mass of carbon, add it to the product of the number of hydrogen atoms and their respective molar mass:\[6 \times 12.01 ext{ g/mol} + 6 \times 1.01 ext{ g/mol} = 72.06 + 6.06 = 78.12 ext{ g/mol}\]This gives us a molar mass of benzene as 78.12 g/mol. Understanding how to calculate this is crucial when converting from a mass in grams to moles in chemical calculations.

Boiling Point

The boiling point is the temperature where a liquid turns into vapor, becoming equal to the atmospheric pressure surrounding it. For benzene, the boiling point is quite specific—80.1°C. This distinctive point is essential because it's when benzene has enough energy to overcome intermolecular forces that keep it in a liquid state.
Why is this important? At its boiling point, the liquid requires energy to change state instead of changing temperature. All energy goes into breaking intermolecular bonds rather than increasing kinetic energy (temperature). This concept is related to understanding enthalpy changes or how much energy is needed in boiling or condensation processes.
For students, grasping why this is important helps in comprehending broader topics like energy changes in chemical reactions and processes. Concepts of boiling points offer insight into real-world applications, such as refining processes or the work of cooling systems.

Energy Calculation

In chemistry, energy calculations become pivotal, especially when dealing with phase changes, such as vaporization. Here, our focus is on calculating the energy required to vaporize benzene. This involves using the enthalpy of vaporization, which is the energy needed to convert one mole of liquid into gas at constant temperature and pressure.For benzene, the enthalpy of vaporization is given as 30.8 kJ/mol. After finding the number of moles in 125 g of benzene (approximately 1.60 moles), we can calculate the total energy required. The formula used is straightforward: multiply the number of moles by the enthalpy of vaporization:\[ ext{Energy} = ext{moles of C}_6 ext{H}_6 \times ext{enthalpy of vaporization}\]So, calculate it as:\[1.60 ext{ mol} \times 30.8 ext{ kJ/mol} \approx 49.28 ext{ kJ}\]This energy, approximately 49.28 kJ, is how much needs to be transferred to vaporize the benzene. This concept underlines many processes in thermodynamics and chemistry, helping to foster deeper understanding of how heat or energy relates to changes in material states.