91Ó°ÊÓ

Enthalpy of formation, often denoted as \( \Delta H_f^0 \), refers to the heat change that occurs when one mole of a compound is formed from its elements in their standard states. This is usually measured under standard conditions of 25°C (298 K) and 1 atmosphere pressure.
In our exercise, we want to determine the Gibbs free energy change for a reaction involving hydrogen and oxygen. We start by identifying the enthalpy of formation values for the substances in the reaction:

• \( \mathrm{H}_2(\mathrm{g}) \) and \( \mathrm{O}_2(\mathrm{g}) \) have \( \Delta H_f^0 \) values of 0 because they are in their standard states.
• \( \mathrm{H}_2\textrm{O}(\mathrm{g}) \) has an enthalpy of formation value of -241.8 kJ/mol.

Using these values, we can calculate the enthalpy change (\( \Delta H \)) for the reaction according to:
\( \Delta H = \sum_{\text{products}} n \Delta H_f^0 - \sum_{\text{reactants}} n \Delta H_f^0 \)
Substituting the respective values and coefficients in the balanced equation, we get:

• \[ \Delta H = [1 \times (-241.8)] - [1 \times 0 + \frac{1}{2} \times 0]\br= -241.8 \textrm{kJ/mol} \]

Another key concept in our problem is entropy change, represented by \( \Delta S \). Entropy is a measure of the randomness or disorder of a system. Each substance has a standard molar entropy, denoted by \( S^0 \).
For the reaction, we use the following standard entropy values at 25°C (298 K) and 1 atm:

• \( \mathrm{H}_2(\textrm{g}) \): 130.68 J/(mol·K)
• \( \mathrm{O}_2(\textrm{g}) \): 205.0 J/(mol·K)
• \( \mathrm{H}_2\textrm{O}(\textrm{g}) \): 188.83 J/(mol·K)

Using these values, we calculate the entropy change of the reaction:
\[ \Delta S = \sum_{\textrm{products}} n S^0 - \sum_{\textrm{reactants}} n S^0 \]
Substituting the values and coefficients, we obtain:

• \[ \Delta S = [1 \times 188.83] - [1 \times 130.68 + \frac{1}{2} \times 205.0]\br= -44.6 \textrm{J/(mol·K)} = -0.0446 \textrm{kJ/(mol·K)} \]

A chemical reaction occurs when substances, known as reactants, transform into different substances called products. This transformation involves breaking old bonds and forming new ones.
In our exercise, the balanced chemical equation is:
\[ \mathrm{H}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} (\mathrm{g}) \]
Here, hydrogen gas (\textrm{H}_2) and oxygen gas (\textrm{O}_2) react to form water vapor (\textrm{H}_2\textrm{O}).
The change in Gibbs free energy (\( \Delta G \)) helps understand the spontaneity of the reaction. We calculate it using Gibbs free energy equation:

• \[ \Delta G = \Delta H - T \Delta S \]

Substituting our calculated values for \( \Delta H \) and \( \Delta S \), and considering T = 298 K:

• \[ \Delta G = -241.8 \textrm{kJ/mol} - (298 \textrm{K}) \times (-0.0446 \textrm{kJ/(mol·K)})\br= -241.8 + 13.3\br= -228.5 \textrm{kJ/mol} \]

This indicates that the reaction is spontaneous under standard conditions.