Thermodynamics, Entropy, and Gibbs Free Energy

The relationship between entropy and microstates is defined by the Boltzmann formula: $S = k \ln W$ where $k$ is the Boltzmann constant ($1.38 \times 10^{-23} \text{ J/K}$).

Factors Affecting Entropy

Several factors influence the entropy of a substance:

1. Phase/State: Generally, entropy increases from solid to liquid to gas.
2. Temperature: As temperature increases, molecular disorder increases, leading to higher entropy.
3. Molar Mass: Larger molar masses often correspond to higher standard entropies.
4. Molecular Complexity: More complex molecules have more ways to distribute energy, resulting in higher entropy.
5. Gas Production: In a chemical reaction, if the total number of gas molecules increases, $\Delta S^\circ > 0$. If the number of gas molecules decreases, $\Delta S^\circ < 0$.

Exam Tip: When looking at a chemical equation to determine the sign of $\Delta S$, count the moles of gas on the reactant side versus the product side. A decrease in gas moles usually means a negative $\Delta S$.

The Third Law of Thermodynamics

The Third Law of Thermodynamics states that the absolute entropy ($S^a$) of a perfect crystalline substance is zero at absolute zero ($0 \text{ K}$). Because most substances are not perfect crystals, their absolute entropy is always a positive value.

Enthalpy and the Surroundings

While entropy describes the system, we must also consider the enthalpy change ($\Delta H$), which is the difference in the sum of the internal energy and $PV$ work between reactants and products.

The entropy change of the surroundings ($\Delta S_{\text{surr}}$) is driven by the heat exchanged between the system and its environment. At constant pressure and temperature, this is calculated as: $\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T}$

This relationship explains why exothermic reactions ($\Delta H < 0$) increase the entropy of the surroundings (making $\Delta S_{\text{surr}}$ positive). Even if a reaction causes a decrease in the system's entropy (like water freezing), it can still be spontaneous if the reaction is exothermic enough to create a larger increase in the entropy of the surroundings.

Gibbs Free Energy

To simplify the study of spontaneity without having to calculate the entropy of the entire universe every time, we use Gibbs Free Energy ($G$). Gibbs free energy combines the system's enthalpy and entropy into a single value that predicts spontaneity.

The change in Gibbs free energy is calculated using: $\Delta G = \Delta H - T\Delta S$

The sign of $\Delta G$ tells us the spontaneity of the process at constant temperature and pressure:

• $\Delta G < 0$: Spontaneous (exergonic)
• $\Delta G > 0$: Nonspontaneous (endergonic)
• $\Delta G = 0$: The system is at equilibrium

Standard Free Energy and Nonstandard Conditions

The standard free energy of reaction ($\Delta G^\circ$) refers to the change in free energy when all reactants and products are in their standard states (e.g., $1 \text{ M}$ concentration or $1 \text{ atm}$ pressure).

However, most real-world reactions occur under nonstandard conditions. To find the actual free energy change ($\Delta G$), we use the reaction quotient ($Q$): $\Delta G = \Delta G^\circ + RT \ln Q$

At equilibrium, $\Delta G = 0$, which allows us to relate the standard free energy to the equilibrium constant ($K$): $\Delta G^\circ = -RT \ln K$

Chemical Potential and Equilibrium

The relationship between $\Delta G^\circ$ and $K$ is qualitative:

• If $\Delta G^\circ < 0$, then $K > 1$ (products are favored).
• If $\Delta G^\circ > 0$, then $K < 1$ (reactants are favored).
• If $\Delta G^\circ = 0$, then $K = 1$.

Note on Chemical Potential: Spontaneity is ultimately determined by comparing the chemical potential energy of the system before the reaction with the free energy after the reaction.

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Worked Example: Calculating $\Delta G$

Problem: Calculate $\Delta G$ at $25^\circ\text{C}$ for the reaction: $\text{Al}(s) + \text{Fe}_2\text{O}_{3(s)} \rightarrow \text{Fe}(s) + \text{Al}_2\text{O}_{3(s)}$ Given: $\Delta H = -847.6 \text{ kJ}$ and $\Delta S = -41.3 \text{ J/K}$.

Solution:

1. Convert $\Delta S$ to $\text{kJ/K}$: $\Delta S = -41.3 \text{ J/K} \times \left(\frac{1 \text{ kJ}}{1000 \text{ J}}\right) = -0.0413 \text{ kJ/K}$
2. Convert temperature to Kelvin: $T = 25 + 273 = 298 \text{ K}$
3. Use the Gibbs equation: $\Delta G = \Delta H - T\Delta S$ $\Delta G = (-847.6 \text{ kJ}) - (298 \text{ K})(-0.0413 \text{ kJ/K})$ $\Delta G = -847.6 \text{ kJ} + 12.3094 \text{ kJ}$ $\Delta G = -835.29 \text{ kJ}$ Since $\Delta G$ is negative, the reaction is spontaneous.