Problem 36 - Chemistry The Central Science In Si Units

Question

Suppose that the gas-phase reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow\) \(2 \mathrm{NO}_{2}(g)\) were carried out in a constant-volume container at constant temperature. (a) Would the measured heat change represent \(\Delta H\) or \(\Delta E ?(\mathbf{b})\) If there is a difference, which quantity is larger for this reaction? (c) Explain your answer to part (b).

Step-by-Step Solution

Answer

(a) The heat change represents \(\Delta E\). (b) \(\Delta E\) is larger. (c) \(\Delta H = \Delta E - RT\), hence \(\Delta H < \Delta E\).

Step 1: Understanding the Reaction Setup

The reaction occurs in a gas system at constant volume and constant temperature. Under constant volume, any heat change of the system is equal to the change in internal energy, \( \Delta E \).

Step 2: Distinguish Between \( \Delta H \) and \( \Delta E \)

The enthalpy change, \( \Delta H \), is related to the internal energy change via the equation \( \Delta H = \Delta E + \Delta (PV) \). Since the reaction is at constant temperature, \( \Delta (PV) \) relates to the change in moles of gas (\( nRT \)).

Step 3: Determine the Change in Gas Moles

Calculate the change in the number of gas moles: Before reaction: 2 moles NO + 1 mole O\(_2\) = 3 moles. After reaction: 2 moles NO\(_2\) = 2 moles. The change in moles, \( \Delta n \), is -1.

Step 4: Calculate \( \Delta (PV) \) Contribution

Since \( \Delta (PV) = \Delta nRT \) and \( \Delta n = -1 \), \( \Delta (PV) = -RT \). Hence, \( \Delta H = \Delta E - RT \).

Step 5: Compare \( \Delta H \) and \( \Delta E \)

From \( \Delta H = \Delta E - RT \), it follows that \( \Delta H < \Delta E \) for this exothermic reaction carried out at constant volume.

Key Concepts

Enthalpy Change (ΔH)

In thermochemistry, enthalpy change, represented by \(\Delta H\), describes the heat exchange in a system at constant pressure. It is essential when predicting how energy in the form of heat changes during chemical reactions. However, in reactions occurring at constant volume, like the one described in the exercise, calculating \(\Delta H\) requires adjustments due to not maintaining constant pressure.

The relationship between enthalpy \(\Delta H\) and internal energy \(\Delta E\) can be shown with the equation:

• \( \Delta H = \Delta E + \Delta (PV) \)

Since pressure and volume changes reflect in the \( \Delta (PV) \) part of the equation, getting \(\Delta H\) at constant volume involves determining how these factors interplay with the change in the number of moles of gas, \(n\).

Remember, a decrease in moles of gas typically leads to a decrease in \(PV\), meaning for a reaction where gas moles decrease, \(\Delta H\) often turns out lower than \(\Delta E\) at constant volume.

Internal Energy Change (ΔE)

Internal energy change, \( \Delta E \), signifies the total energy change within a system as it undergoes a chemical reaction. In a constant volume setting, like the one discussed here, \( \Delta E \) equates directly to the heat exchanged because no work is done from volume changes.

For a gas-phase reaction in a closed container, altered moles of gas during reactions cause changes in pressure and volume. However, under constant volume, any heat input or output directly changes \( \Delta E \). Thus at constant volume, the heat measured is a reflection of \( \Delta E \).

More succinctly, in thermochemistry, \( \Delta E \) comprises both \( q \), the heat exchanged, and \( w \), the work done by or on the system:

• \( \Delta E = q + w \)

But under constant volume conditions, no volume work occurs, simplifying to:

• \( \Delta E = q_{v} \)

where \( q_{v} \) is the heat at constant volume.

Constant Volume Process

In a constant volume process, also known as an isochoric process, the volume of the system doesn’t change, even though reactions occur. It means that any energy change merely involves changes in heat, not in volume or mechanical work.

For a gas-phase reaction, such as the one being discussed, the constant volume process impacts the energy calculation by focusing entirely on the internal energy change \( \Delta E \). Unlike reactions where volume can change, leading to pressure and volume work, here all the energy exchange is in the form of heat alone.

This simplifies calculations because:

• The total heat change observed is equivalent to the internal energy change, \( \Delta E \).
• No work is done as there's no change in volume, \( w=0 \).
• Thus, \( \Delta E = q_{v} \) is the core equation during such a process.

Such a setup provides a clear picture of internal energy shifts, revealing the nature of the substance's energy transformation free from the confounding factors of changing volumes.