Question

(a) Calculate the \(P V\) work in kilojoules done during the combustion of \(1 \mathrm{~mol}\) of ethanol in which the volume contracts from \(73.5 \mathrm{~L}\) to \(49.0 \mathrm{~L}\) at a constant external pressure of 1 atm. In which direction does the work energy flow? \((1 \mathrm{~L} \cdot \mathrm{atm}=101 \mathrm{~J} .)\) (b) If \(\Delta H^{\circ}=-1367 \mathrm{~kJ} / \mathrm{mol}\), calculate the change in internal energy \((\Delta E)\) for the combustion of \(1 \mathrm{~mol}\) of ethanol.

Short answer

Work done is 2.4745 kJ; energy flows into the system. \(\Delta E = -1369.47 \, \text{kJ/mol}\).

Step-by-step solution

Calculate Work Done (W) in L·atm

Work done by the system during volume change at constant pressure is given by the formula: \[ W = -P_{ext}(V_f - V_i) \]Given:- Initial volume, \(V_i = 73.5\) L- Final volume, \(V_f = 49.0\) L- External pressure, \(P_{ext} = 1\) atmPlug these values into the formula:\[ W = -1(49.0 - 73.5) = -1(-24.5) = 24.5 \text{ L·atm} \]

Convert L·atm to Joules

We need to convert the work from L·atm to Joules using the conversion factor \(1 \, \text{L·atm} = 101 \, \text{J}\):\[ W = 24.5 \, \text{L·atm} \times 101 \, \text{J/L·atm} = 2474.5 \, \text{J} \]Convert Joules to kilojoules:\[ W = \frac{2474.5}{1000} = 2.4745 \, \text{kJ} \]

Determine Direction of Energy Flow

Since the work calculated is positive, it implies that work is done on the system. Therefore, energy flows into the system.

Use the Relationship Between ΔH and ΔE

We use the relationship between the change in enthalpy (ΔH), change in internal energy (ΔE), and work done at constant pressure (W):\[ \Delta H = \Delta E + W \]Rearrange the equation to solve for \(\Delta E\):\[ \Delta E = \Delta H - W \]

Calculate ΔE

Substitute the given values and previously calculated work:- \( \Delta H^{\circ} = -1367 \, \text{kJ/mol} \)- \( W = 2.4745 \, \text{kJ} \)\[ \Delta E = -1367 \, \text{kJ/mol} - 2.4745 \, \text{kJ} = -1369.4745 \, \text{kJ/mol} \]

Conclusion

The change in internal energy (\( \Delta E \)) for the combustion of 1 mol of ethanol is approximately \(-1369.47 \, \text{kJ/mol}\). The positive work indicates energy flows into the system.

Key concepts

Enthalpy Change

Enthalpy change \( (\Delta H) \) is an important concept in thermodynamics that measures the heat change in a system at constant pressure. It helps in understanding how energy is absorbed or released during a reaction. When the reaction involves combustion, like in our exercise, enthalpy change tells us about the energy needed to break bonds and form new ones.
In this exercise, the enthalpy change \( (\Delta H^\circ) \) for the combustion of ethanol is given as \(-1367 \text{ kJ/mol} \). The negative sign indicates that the reaction releases energy, making it exothermic. This means energy is released into the surroundings as heat.
Understanding enthalpy change allows us to predict the behavior of the reaction concerning energy. This is crucial for designing processes and systems in which heat energy needs to be managed effectively.

• Exothermic reactions have negative \( \Delta H \), indicating heat release.
• Endothermic reactions have positive \( \Delta H \), meaning they absorb heat.

Keep in mind that \( \Delta H \) is always measured under constant pressure, which provides a more predictable measure of heat change in chemical processes.

Work Energy

Work energy \( (W) \) in thermodynamics refers to the energy transferred when a system goes through a volume change under pressure. In this context, it is calculated using the formula \( W = -P_{ext}(V_f - V_i) \). This is a fundamental principle, as work done on or by the system influences the overall energy balance.
In our problem, as the volume of ethanol combusts and decreases from 73.5 L to 49.0 L, work is done on the system by the surrounding. The work here is calculated to be positive \(2.4745 \text{ kJ}\), indicating that the energy flows into the system. It occurs because the system undergoes a contraction.
Understanding work energy helps us identify whether a system is gaining or losing energy due to volume changes. Here are some key points:

• Positive work means energy is added to the system, as the system contracts.
• Negative work would imply that energy is leaving the system, usually during expansion.

This concept is crucial for systems like engines or refrigeration, where changes in volume are associated with work and energy transfer.

Internal Energy

Internal energy \( (\Delta E) \) is the total energy stored within a system, comprising kinetic and potential energy of molecules. It is a key marker of a system's energy state changes involved in chemical reactions.
For any reaction at constant pressure, the relationship \( \Delta H = \Delta E + W \) helps us find the internal energy change by rearranging to \( \Delta E = \Delta H - W \). This equation shows how enthalpy change and work contribute to the change in internal energy.
In our provided exercise, the internal energy change \( \Delta E \) for ethanol combustion was calculated as \(-1369.47 \text{ kJ/mol} \). This negative value signifies an overall loss in energy within the system. The work done (positive \(2.4745 \text{ kJ}\)) partially offsets the large energy release shown by \( \Delta H \).
This concept outlines how those energies interact:

• Internal energy change reflects total energy absorbed or released.
• Positive \( \Delta E \) indicates energy gained by the system, negative signifies loss.

Thus, understanding \( \Delta E \) helps evaluate how a reaction progresses energetically, ensuring energy transformations align with the process goals.