Question

When 1 mole of benzene is vaporized at a constant pressure of 1.00 atm and at its boiling point of \(353.0 \mathrm{K}, 30.79 \mathrm{kJ}\) of energy (heat) is absorbed and the volume change is +28.90 L. What are \(\Delta E\) and \(\Delta H\) for this process?

Short answer

The change in internal energy \(\Delta E\) during the vaporization of 1 mole of benzene is \(\approx 27861.08 \, \mathrm{J}\), and the change in enthalpy \(\Delta H\) is \(\approx 30790 \, \mathrm{J}\).

Step-by-step solution

List down the given information

We know the following information: 1. Heat absorbed during vaporization, \(q = 30.79 \, \mathrm{kJ}\) 2. Volume change, \(\Delta V = 28.90 \, \mathrm{L}\) 3. Constant pressure during vaporization, \(p = 1.00 \, \mathrm{atm}\) 4. Boiling point of benzene, \(T = 353.0 \, \mathrm{K}\) Now, we need to find the work done on the system during vaporization.

Convert the units of given values to SI units

In order to use the given values in our calculations, we need to convert them to SI units. 1. Convert \(\Delta V\) to \(\mathrm{m^3}\): \(\Delta V = 28.90 \, \mathrm{L} \times \frac{1 \, \mathrm{m^3}}{1000 \, \mathrm{L}} = 0.0289 \, \mathrm{m^3}\) 2. Convert the pressure \(p\) to \(\mathrm{Pa}\): \(p = 1.00 \, \mathrm{atm} \times \frac{101325 \, \mathrm{Pa}}{1 \, \mathrm{atm}} = 101325 \, \mathrm{Pa}\)

Calculate the work done on the system

We know that work done on the system under constant pressure can be calculated using the following formula: \(w = -p\Delta V\) Now, we'll substitute the values of \(p\) and \(\Delta V\) to find the work done: \(w = -101325 \, \mathrm{Pa} \times 0.0289 \, \mathrm{m^3} = -2928.9175 \, \mathrm{J}\)

Calculate the change in internal energy

Now we can find the change in internal energy using the formula: \(\Delta E = q + w\) First, convert the heat from kJ to J: \(q = 30.79 \mathrm{kJ} \times \frac{1000 \mathrm{J}}{1 \mathrm{kJ}} = 30790\mathrm{J}\) Next, substitute the values of \(q\) and \(w\) to find \(\Delta E\): \(\Delta E = 30790 \, \mathrm{J} + (-2928.9175 \, \mathrm{J}) = 27861.0825 \, \mathrm{J}\)

Calculate the change in enthalpy

Now we can find the change in enthalpy using the formula: \(\Delta H = \Delta E + p\Delta V\) Substitute the values of \(\Delta E\), \(p\), and \(\Delta V\) to find \(\Delta H\): \(\Delta H = 27861.0825\, \mathrm{J} + 101325 \, \mathrm{Pa} \times 0.0289 \, \mathrm{m^3} = 30790\, \mathrm{J}\)

Presenting the final answer

The change in internal energy \(\Delta E\) during the vaporization of 1 mole of benzene is \(\approx 27861.08 \, \mathrm{J}\), and the change in enthalpy \(\Delta H\) is \(\approx 30790 \, \mathrm{J}\).

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with heat, work, and the forms of energy involved in chemical processes. In the context of the exercise dealing with the vaporization of benzene, thermodynamics principles allow us to understand and quantify energy changes. The internal energy of a system, represented by \( \Delta E \), is the total energy contained within the system, while enthalpy, represented by \( \Delta H \), reflects the heat content under constant pressure. It’s valuable for students to learn how to apply these thermodynamics principles to solve problems involving phase changes, like the vaporization in this exercise, as it builds the conceptual framework needed to tackle more complex systems and reactions in chemistry.

Improving one's grasp of thermodynamics is akin to adding tools to a toolbox; the more tools (knowledge of concepts) you have, the more types of problems (exercises) you can solve confidently. In chemistry education, this is a fundamental step towards developing a robust and versatile understanding of how substances interact and transform.

Vaporization

Vaporization is the phase transition from the liquid state to the gas state. For benzene, this occurs at its boiling point. During vaporization, energy is absorbed to overcome intermolecular forces of attraction within the liquid, which is quantified as the enthalpy change of vaporization.

In chemistry education, exploring vaporization deepens a student's understanding of phase changes and the energy required for those processes. A clear grasp of these concepts enables students to predict and explain the behavior of substances under various temperature and pressure conditions. It's important to convey the practical significance of vaporization, which features in everyday phenomena from cooking to industrial chemical processes.

Chemistry Education

Chemistry education focuses on imparting knowledge about chemical substances and their transformations. Application-based learning, such as calculating the energy change during vaporization, encourages problem-solving skills and a deeper understanding of chemical principles. Exercises like the one provided help students connect textbook theory with practical scenarios.

Effective chemistry education utilises a step-by-step approach, ensuring that students are able to follow the flow of logic and calculations. By breaking down complex reactions into simpler steps, students are more likely to engage with and retain the material. For instance, converting units to standard SI units as an initial step is a fundamental practice that clarifies and standardizes the information before performing calculations.

Using practical examples, such as the vaporization of a common substance like benzene, makes abstract concepts more relatable and easier to understand. This is crucial in chemistry education, as it ties theoretical knowledge to real-world applications.

Energy Changes

Energy changes are central to chemical processes, including phase transitions such as vaporization. The exercise specifically examines the enthalpy change of vaporization, an important concept in thermodynamics and chemistry. The enthalpy change, \( \Delta H \), indicates the amount of heat absorbed or released under constant pressure. On the other hand, the change in internal energy, \( \Delta E \), accounts for the heat and work done on or by the system.

Understanding energy changes enables students to predict the direction of chemical reactions and the stability of products. For example, comprehending that vaporization requires heat can lead to inferences about the cooling effect of evaporation. Additionally, it is fundamental for students to appreciate that energy can be expressed in different units, and that proficiency with unit conversion is necessary for accurate calculations. This bridges the gap between theoretical energy concepts and practical application in diverse chemical contexts.