Problem 62 - Chemistry The Science in Context

Question

If a gas-phase reaction mixture contains 1 mole of each reactant and product and has a 1: 1 stoichiometry, is the \(\Delta G_{\text {ran }}^{\circ}\) of the reaction mixture the same as \(\Delta H_{\mathrm{rm}}^{\circ}\) ? Explain why or why not.

Step-by-Step Solution

Answer

Answer: Yes, for this particular reaction, the Gibbs free energy change of the reaction mixture is indeed the same as the enthalpy change.

Step 1: Understand the relationship between Gibbs free energy, enthalpy, and entropy

The equation that relates Gibbs free energy change (\(\Delta G_{\text {ran }}^{\circ}\)), enthalpy change (\(\Delta H_{\mathrm{rm}}^{\circ}\)), and entropy change (\(\Delta S^{\circ}\)) is: \(\Delta G_{\text{ran}}^{\circ} = \Delta H_{\text{rm}}^{\circ} - T\Delta S^{\circ}\) where T is the temperature in Kelvin.

Step 2: Determine the stoichiometry and consider its effect on entropy

The problem states that there is 1 mole of each reactant and product, and the stoichiometry is 1:1. When a reaction has a stoichiometry of 1:1, this causes the entropy change to be \(0\) for ideal gasses because there is no change in the number of molecules present in the reaction mixture.

Step 3: Calculate the Gibbs free energy change

The entropy change is 0, so the equation for Gibbs free energy change can now be written as: \(\Delta G_{\text{ran}}^{\circ} = \Delta H_{\text{rm}}^{\circ} - T\cdot 0\) This simplifies to: \(\Delta G_{\text{ran}}^{\circ} = \Delta H_{\text{rm}}^{\circ}\) Therefore, for this particular reaction with a 1:1 stoichiometry and 1 mole of each reactant and product, the Gibbs free energy change (\(\Delta G_{\text {ran }}^{\circ}\)) of the reaction mixture is indeed the same as the enthalpy change (\(\Delta H_{\mathrm{rm}}^{\circ}\)).

Key Concepts

Gibbs Free Energy

Gibbs free energy, often denoted by \( \Delta G^{\circ} \), measures the maximum reversible work a thermodynamic system can perform at constant temperature and pressure. In simple terms, it's the energy available to do work in a chemical reaction. A negative \( \Delta G \) indicates a spontaneous reaction, while a positive \( \Delta G \) suggests a non-spontaneous reaction.
One can calculate the Gibbs free energy change using the following formula:

• \( \Delta G = \Delta H - T\Delta S \)
Where \( \Delta H \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change. Here, \( T \Delta S \) represents the energy unavailable for doing work. The difference between these, \( \Delta G \), determines the feasibility of the reaction.
In our specific problem, the reaction with a 1:1 stoichiometry in gases means that \( \Delta S \) is zero. As a result, \( \Delta G \) simplifies to \( \Delta H \), making them equal under these conditions.

Enthalpy

Enthalpy, represented by \( \Delta H \), is the heat content of a system at constant pressure. It gives insights into the total energy required or released during a chemical reaction.
When a reaction occurs, \( \Delta H \) indicates if the reaction is endothermic (absorbing heat) or exothermic (releasing heat).

• An exothermic reaction will have \( \Delta H < 0 \), indicating heat release.
• An endothermic reaction reflects \( \Delta H > 0 \), meaning heat absorption.

In the exercise provided, because the entropy change \( \Delta S \) is zero due to 1:1 stoichiometry, \( \Delta G \) entirely depends on \( \Delta H \). This means that understanding \( \Delta H \)'s sign (positive or negative) directly tells us about the nature and feasibility of the reaction.

Entropy

Entropy, denoted by \( \Delta S \), is a measure of disorder or randomness in a system. It describes how energy spreads out and is crucial for determining the feasibility of a process along with enthalpy.
During a reaction, if the system becomes more disordered, \( \Delta S \) is positive. Conversely, if the system becomes more ordered, \( \Delta S \) is negative.

• Positive \( \Delta S \) indicates an increase in randomness.
• Negative \( \Delta S \) shows a decrease in randomness.

In the given exercise, the reaction has equal moles of reactants and products in a 1:1 stoichiometry, leading to \( \Delta S = 0 \). This means there is no change in disorder, simplifying the relation \( \Delta G = \Delta H - T \Delta S \) to \( \Delta G = \Delta H \). This unchanging entropy state is key to understanding why, in this scenario, Gibbs free energy equals the enthalpy.