Question

When 1 mol of a gas burns at constant pressure, it produces 2418 \(\mathrm{J}\) of heat and does 5 \(\mathrm{J}\) of work. Determine \(\Delta E, \Delta H, q,\) and \(w\) for the process

Short answer

\(\Delta E = 2413 \, \mathrm{J}\), \(\Delta H = 2418 \, \mathrm{J}\), \(q = 2418 \, \mathrm{J}\), \(w = -5 \, \mathrm{J}\)

Step-by-step solution

Understanding the given quantities

The given values for the problem are the heat produced, which is 2418 J, and the work done on the surroundings, which is 5 J.

Determining \(q\)

The heat (\(q\)) given off during the burning process at constant pressure is already provided as 2418 J. Therefore, \(q = 2418 \, \mathrm{J}\).

Determining \(w\)

The work (\(w\)) done by the system on the surroundings is also provided as 5 J. However, in thermodynamics, when work is done by the system, it is considered negative, so \(w = -5 \, \mathrm{J}\).

Determining \(\Delta E\) using the first law of thermodynamics

The change in internal energy \(\Delta E\) of the system can be found using the first law of thermodynamics, which states that \(\Delta E = q + w\). Substituting the values for \(q\) and \(w\), we get \(\Delta E = 2418 \, \mathrm{J} + (-5 \, \mathrm{J}) = 2413 \, \mathrm{J}\).

Determining \(\Delta H\) using the definition at constant pressure

The enthalpy change \(\Delta H\) at constant pressure is equal to the heat \(q_p\) exchanged in the process. Therefore, \(\Delta H = q = 2418 \, \mathrm{J}\).

Key concepts

First Law of Thermodynamics

The first law of thermodynamics is an expression of the conservation of energy principle. It asserts that the energy of an isolated system is constant. Energy can be transformed from one form to another, but it cannot be created or destroyed.

In the context of chemistry and the problem given, it relates to the internal energy changes occurring in a chemical reaction. When a mole of gas burns, it involves the conversion of chemical energy in the substance into thermal energy (heat) and work done on the surroundings. The first law can be mathematically represented as: \[ \triangle E = q + w \]
where \( \triangle E \) is the change in internal energy, \( q \) is the heat exchanged, and \( w \) is the work done by the system. When work is done by the system, like expansion against atmospheric pressure, it is considered negative because energy is being given out by the system.

Enthalpy Change

Enthalpy change, represented as \(\triangle H\), is a thermodynamic property that measures the heat exchanged at constant pressure. It is an important concept in chemistry as it corresponds to the heat absorbed or released during a chemical reaction taking place at constant pressure.

For reactions occurring at constant pressure, the heat exchanged (\(q_p\)) is exactly the enthalpy change. The enthalpy change is given by:\[ \triangle H = q_p \]
In the exercise, since the gas burns at constant pressure, the heat produced directly provides us with the value for \(\triangle H\). This simplicity allows scientists and engineers to use enthalpy as a useful indicator for the heat of reaction.

Internal Energy

Internal energy, denoted as \( U \) or \( E \), is the total of all forms of energy within a system. It includes kinetic energy due to the motion of atoms and molecules, potential energy arising from the forces between them, and chemical energy associated with the electron configurations and chemical bonds.

In thermodynamics, we are often interested in the change in internal energy (\( \triangle E \)) rather than its absolute value. The change tells us whether energy is absorbed by the system (\( \triangle E > 0 \)) or released (\( \triangle E < 0 \)). In the exercise provided, the change in internal energy for the burning gas includes the heat released and the work done on or by the system; thus, when calculating \( \triangle E \), the positive value indicates the system has overall lost energy to the surroundings.