Problem 34 - Chemistry: The Central Science

Question

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant- pressure process, the system releases heat to the surroundings. Does the enthalpy of the system increase or decrease during the process? (c) In a constant-pressure process, \(\Delta H=0\). What can you conclude about \(\Delta E, q\), and \(w\) ?

Step-by-Step Solution

Answer

(a) The enthalpy change of a process will equal the amount of heat transferred into or out of the system under constant pressure conditions, i.e., \(\Delta H = q_p\). (b) During a constant-pressure process when the system releases heat to the surroundings, the enthalpy of the system decreases since \(q_p < 0\). (c) In a constant-pressure process with \(\Delta H=0\), we can conclude that \(\Delta E=-P\Delta V\), \(q=0\), and \(w=-P\Delta V\).

Step 1: Part a: Condition when enthalpy change equals heat transferred

At constant pressure, the enthalpy change (\(\Delta H\)) of a process is given by \(\Delta H = q_p\), where \(q_p\) represents heat transferred into or out of the system at constant pressure. The condition we seek is the one where \(\Delta H = q_p\) always holds true.

Step 2: Part b: Change in enthalpy during a constant-pressure process

Given that heat is released to the surroundings during the constant-pressure process, it implies that \(q_p < 0\). Since \(\Delta H = q_p\) at constant pressure, the enthalpy of the system must decrease during the process as \(\Delta H < 0\).

Step 3: Part c: Inference about \(\Delta E\), \(q\), and \(w\) at constant pressure when \(\Delta H = 0\)

In a constant-pressure process, the enthalpy change (\(\Delta H\)) is related to the change in internal energy (\(\Delta E\)), heat transfer (\(q\)), and work (\(w\)) performed by the system as follows: \[\Delta H = \Delta E + P\Delta V = q_p\] Given that \(\Delta H = 0\), we have: \[0 = \Delta E + P\Delta V\] We can rearrange the equation to obtain the relationship between \(\Delta E\) and \(P\Delta V\): \[\Delta E = -P\Delta V\] Since the process occurs at constant pressure, heat transfer (\(q\)) and work (\(w\)) are related as: \[q = \Delta E + w\] Substituting \(\Delta E = -P\Delta V\) and the work is given by \(w = -P\Delta V\) (since work is done by the system): \[q = -P\Delta V - (-P\Delta V) = 0\] Thus, at constant pressure when \(\Delta H = 0\), we can conclude that \(\Delta E = -P\Delta V\), \(q = 0\), and \(w = -P\Delta V\).

Key Concepts

constant-pressure process

In a constant-pressure process, the pressure of the system remains unchanged throughout the entire process. This consistency in pressure is a common scenario in many real-world applications. Imagine boiling water in an open pot - the pressure above the water is constantly atmospheric pressure.
At constant pressure, any change in the system is described by the behavior of its properties rather than changes in pressure. One key relationship that emerges is that the change in enthalpy (\(\Delta H\)) of the system equals the heat exchanged (\(q_p\)) between the system and its surroundings. This is represented mathematically as:\[\Delta H = q_p\]

• Heat added to the system leads to an increase in enthalpy.
• Heat lost from the system results in a decrease in enthalpy.

Under these conditions, the system's enthalpy becomes the straightforward measure of the heat exchanged, simplifying calculations and insights into the system's thermal dynamics.

heat transfer

Heat transfer refers to the movement of thermal energy from one part of a system to another, or to the surrounding environment. It's a key driver of change in a thermodynamic system, directing the flow of energy.
During constant-pressure processes, where pressure remains the same, heat transfer can significantly influence the system. The direction and magnitude of heat transfer determine how the system's enthalpy changes:

• If the system gains heat (positive heat transfer), enthalpy increases.
• If the system loses heat (negative heat transfer), enthalpy decreases.

This relationship is crucial because it allows us to predict and analyze the behavior of a system undergoing thermal changes without varying its pressure. Mastering the concept of heat transfer in these processes enables us to predict system states and transformations effectively.

internal energy

Internal energy (\(\Delta E\)) is a measure of the total energy contained within a system, encompassing both the kinetic and potential energies of its particles. In thermodynamics, understanding internal energy is essential for analyzing how systems exchange energy through work and heat.
In constant-pressure processes, the relationship between internal energy change and other forms of energy exchange is captured by the equation:\[\Delta E = q + w\]where \(q\) is the heat transfer and \(w\) is the work done by the system. When enthalpy (\(\Delta H\)) is zero, it indicates a balance where the system's pressure-volume work (\(P\Delta V\)) aligns with any internal changes:

• If \(\Delta H = 0\), then \(\Delta E = -P\Delta V\)
• This implies that the energy lost in the form of expansion work is compensated by the internal energy change, with no net heat transfer (\(q = 0\)).

This balance is a fascinating aspect of thermodynamics, revealing how internal adjustments can occur while maintaining overall energy equilibrium.