Question

The change in internal energy for the combustion of \(1.0 \mathrm{~mol}\) of octane at a pressure of \(1.0 \mathrm{~atm}\) is \(5084.3 \mathrm{~kJ}\). If the change in enthalpy is 5074.1 kJ, how much work is done during the combustion?

Short answer

-10.2 kJ of work is done during the combustion.

Step-by-step solution

Recall the Relationship between Enthalpy, Internal Energy, and Work

Use the first law of thermodynamics which relates the internal energy change (\(\Delta U\)), the heat added to the system (\(q\)), and the work done by the system (\(w\)): \(\Delta U = q + w\). For a process happening at constant pressure, \(q\) is equal to the change in enthalpy (\(\Delta H\)). Therefore, the equation can be rewritten as \(\Delta U = \Delta H - w\) because work done by the system is negative when the system expands.

Solve for Work

Rearrange the modified first law equation to solve for work: \(w = \Delta H - \Delta U\).

Insert the Values and Calculate Work Done

Substitute the given values for the change in enthalpy (\(\Delta H = 5074.1 \mathrm{~kJ}\)) and the change in internal energy (\(\Delta U = 5084.3 \mathrm{~kJ}\)). Calculate the work done as \(w = 5074.1 \mathrm{~kJ} - 5084.3 \mathrm{~kJ} = -10.2 \mathrm{~kJ}\). The negative sign indicates that the work is done by the system.

Key concepts

First Law of Thermodynamics

The first law of thermodynamics, often referred to as the law of energy conservation, specifies that energy can neither be created nor destroyed; it can only be transformed from one form to another within an isolated system. The principle underlying this law is foundational for understanding how energy is transferred in physical and chemical processes, including combustion.

When considering combustion thermodynamics, the first law provides a way to quantify the energy changes that occur. It equates the change in internal energy \(\Delta U\) of a system to the heat added \(q\) minus the work done by the system \(w\): \[ \Delta U = q - w \]. It’s crucial to note that this formula assumes a sign convention where work done by the system is considered negative because energy is leaving the system.

Internal Energy

Internal energy is the total energy contained within a chemical system and includes kinetic energy from motion at the atomic and molecular levels, as well as potential energy due to molecular structure and intermolecular forces. In thermodynamics, changes in internal energy rather than absolute values are of interest, since those changes can be measured as heat and work.

During a process such as combustion, internal energy changes as chemical bonds are broken and formed. The substances involved undergo transformation, releasing or absorbing energy in the process. The capability to measure these changes in a system’s internal energy is crucial for predicting the direction and spontaneity of chemical reactions.

Enthalpy

Enthalpy is a thermodynamic quantity that represents the total heat content of a system. It is symbolized by \(H\) and is usually expressed in joules or kilojoules per mole. Enthalpy is effectively a measure of the energy that is transferred as heat at constant pressure.

In the context of combustion, the change in enthalpy \(\Delta H\) during a reaction is of particular importance. If the process occurs at constant pressure, the change in enthalpy equals the heat exchanged with the surroundings. A positive \(\Delta H\) indicates an endothermic process, where heat is absorbed from the environment, while a negative value suggests an exothermic process, where heat is released.

Work Done in Thermodynamics

Work in thermodynamics is the energy transfer that occurs when an external effect displaces the system's boundary. For example, during the expansion of gases within an engine, work is done by the system on the surroundings. This is why in our exercise, the work done has a negative value when calculated from the system's perspective; the system is doing work on the environment, using up energy to push outwards.

The amount of work done can be calculated by the integral of pressure over volume change, \(w = -\int PdV\). However, for processes at constant pressure, like the given combustion reaction, the work done is simply the product of pressure and change in volume, \(w = -P\Delta V\). In such scenarios, relating the change in enthalpy to the work done becomes much simpler and is as shown in the provided example.