14.4 Entropy (ΔS), Gibbs Free Energy (ΔG) & Spontaneity

14.4 Entropy (ΔS), Gibbs Free Energy (ΔG) & Spontaneity

Quick Summary: A spontaneous process proceeds naturally under specified conditions without continuous external energy input. The Second Law of Thermodynamics asserts that all spontaneous processes increase universal entropy: $\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$. Microscopic entropy reflects microstate dispersal ($S = k_B \ln W$). The Third Law establishes that a pure, perfect crystal at $0\text{ K}$ has zero entropy ($S = 0$), allowing absolute standard molar entropies ($S^\circ$) to be measured. Gibbs Free Energy ($\Delta G = \Delta H - T\Delta S$) serves as the operational criterion for spontaneity at constant $T$ and $P$: $\Delta G < 0$ is spontaneous (exergonic), $\Delta G = 0$ is dynamic equilibrium, and $\Delta G > 0$ is nonspontaneous (endergonic). The four-quadrant matrix governs temperature dependence and crossover points ($T = \Delta H / \Delta S$), while free energy connects directly to the equilibrium constant via $\Delta G^\circ = -RT \ln K$.

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1. Spontaneity & the Second Law of Thermodynamics

A spontaneous process occurs without ongoing external driving force (e.g., heat flowing from hot to cold):

• Spontaneity vs. Kinetics: Spontaneity defines thermodynamic favorability, not reaction rate. Diamond conversion to graphite is spontaneous at $298\text{ K}$ ($\Delta G^\circ = -2.9\text{ kJ/mol}$), but imperceptibly slow due to high activation energy.

The Second Law of Thermodynamics

Spontaneity requires universal entropy increase: $\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$ Surroundings entropy changes via heat transfer at temperature $T$: $\Delta S_{\text{surroundings}} = -\frac{\Delta H_{\text{system}}}{T}$

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2. Statistical Entropy & the Third Law

Boltzmann defined entropy through accessible microstates ($W$): $S = k_B \ln W$ where $k_B = 1.381 \times 10^{-23}\text{ J/K}$. Entropy measures matter and energy dispersal.

The Third Law & Standard Entropies ($S^\circ$)

The Third Law of Thermodynamics states that the entropy of a pure, perfectly crystalline substance at absolute zero ($0\text{ K}$) is exactly zero ($S = 0\text{ J/(mol}\cdot\text{K)}$). Because $W = 1$ at $0\text{ K}$, all substances have positive absolute standard molar entropies ($S^\circ > 0$) at $T > 0\text{ K}$: $\Delta S^\circ_{\text{rxn}} = \sum n S^\circ(\text{products}) - \sum m S^\circ(\text{reactants})$

Predicting the Sign of $\Delta S_{\text{system}}$

1. Phases: $S^\circ(\text{solid}) < S^\circ(\text{liquid}) \ll S^\circ(\text{gas})$.
2. Change in Gas Moles ($\Delta n_{\text{gas}}$): If $\Delta n_{\text{gas}} > 0 \implies \Delta S > 0$; if $\Delta n_{\text{gas}} < 0 \implies \Delta S < 0$.
3. Dissolution: Dissolving crystalline solids in liquids typically increases entropy ($\Delta S > 0$).
4. Temperature: Higher temperature populates higher energy levels ($\Delta S > 0$).

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3. Gibbs Free Energy ($\Delta G$) & Spontaneity Criteria

Multiplying $\Delta S_{\text{univ}} = \Delta S_{\text{sys}} - \Delta H_{\text{sys}}/T$ by $-T$ defines Gibbs Free Energy ($G = H - TS$): $\Delta G = \Delta H - T\Delta S \quad (\text{at constant } T, P)$

Spontaneity Criteria

• $\Delta G < 0$ (Exergonic): Spontaneous in the forward direction.
• $\Delta G = 0$ (Equilibrium): System is at dynamic chemical equilibrium.
• $\Delta G > 0$ (Endergonic): Nonspontaneous forward; spontaneous in reverse.

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4. Four-Quadrant Spontaneity Matrix & Crossover Temperature

Enthalpy and entropy compete to determine the sign of $\Delta G$:

Four-Quadrant Spontaneity Matrix

Crossover Temperature Calculation

In Quadrants 3 and 4, setting $\Delta G = 0$ gives the threshold temperature: $T_{\text{crossover}} = \frac{\Delta H}{\Delta S}$ (Note: convert $\Delta H$ to J to match $\Delta S$.)

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5. Standard Free Energy of Formation & Free Energy of Reaction

The standard free energy of formation ($\Delta G^\circ_f$) is the free-energy change when one mole of a compound forms from its elements in their standard states; tables usually list values at $298.15\text{ K}$. As with $\Delta H^\circ_f$, $\Delta G^\circ_f = 0$ for any element in its standard state, such as $\text{O}_2(g)$, $\text{C(graphite)}$, or $\text{Br}_2(l)$.

The standard free energy of reaction can be found in two equivalent ways:

1. From formation data: $\Delta G^\circ_{\text{rxn}} = \sum n\,\Delta G^\circ_f(\text{products}) - \sum m\,\Delta G^\circ_f(\text{reactants})$
2. From enthalpy and entropy: $\Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T\Delta S^\circ_{\text{rxn}}$

A negative $\Delta G^\circ_f$ means the compound is thermodynamically stable relative to its elements at 298 K. Nitric oxide, NO, has a positive $\Delta G^\circ_f$ (about $+87\text{ kJ/mol}$), so it is unstable with respect to $\text{N}_2$ and $\text{O}_2$ yet persists because its decomposition is slow.

Worked Example: Methane Combustion from Formation Data

$\text{CH}_4(g) + 2\,\text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2\,\text{H}_2\text{O}(l)$ $\Delta G^\circ = [-394.4 + 2(-237.1)] - [-50.5 + 2(0)] = -868.6 + 50.5 = -818.1\text{ kJ}$

Worked Example: Ammonia Synthesis Two Ways

For $\text{N}_2(g) + 3\,\text{H}_2(g) \longrightarrow 2\,\text{NH}_3(g)$:

• Formation data: $\Delta G^\circ = 2(-16.4) - 0 = -32.8\text{ kJ}$
• Enthalpy and entropy: with $\Delta H^\circ = -92.2\text{ kJ}$ and $\Delta S^\circ = -198.7\text{ J/K}$, $\Delta G^\circ = -92.2 - (298.15)(-0.1987) = -33.0\text{ kJ}$, which agrees within rounding.
• Equilibrium constant: $K = e^{-\Delta G^\circ/RT} = e^{32{,}800/(8.314 \times 298.15)} = e^{13.23} \approx 5.6 \times 10^5$ at 298 K.

Formation values apply only at the table temperature. To estimate $\Delta G^\circ$ at another temperature, use $\Delta H^\circ - T\Delta S^\circ$ with $\Delta H^\circ$ and $\Delta S^\circ$ treated as roughly constant, which is how the crossover temperature in the four-quadrant discussion above is found.

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6. Free Energy, Equilibrium & Reaction Coupling

Under non-standard conditions: $\Delta G = \Delta G^\circ + RT \ln Q$ At dynamic equilibrium ($\Delta G = 0, Q = K$): $\Delta G^\circ = -RT \ln K \implies K = e^{-\frac{\Delta G^\circ}{RT}}$

• $\Delta G^\circ < 0 \implies K > 1$ (products favored at standard equilibrium).
• $\Delta G^\circ > 0 \implies K < 1$ (reactants favored at standard equilibrium).

The van 't Hoff Relationship

$\ln K = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T}\right) + \frac{\Delta S^\circ}{R}$ Plotting $\ln K$ vs $1/T$ yields slope $= -\Delta H^\circ / R$.

Reaction Coupling

A nonspontaneous reaction ($\Delta G_1 > 0$) can be driven by coupling to an exergonic reaction ($\Delta G_2 \ll 0$) through a shared intermediate ($\Delta G_{\text{net}} = \Delta G_1 + \Delta G_2 < 0$), such as coupling phosphorylation to ATP hydrolysis ($\Delta G^\circ = -30.5\text{ kJ/mol}$).

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7. Worked Quantitative Thermodynamic Examples

Example 1: Crossover Temperature for Decomposition

Problem: For $\text{CaCO}_3(s) \longrightarrow \text{CaO}(s) + \text{CO}_2(g)$, $\Delta H^\circ = +178.3\text{ kJ/mol}$ and $\Delta S^\circ = +160.5\text{ J/(mol}\cdot\text{K)}$. Find the temperature above which the reaction becomes spontaneous under $1\text{ atm}$. $T_{\text{crossover}} = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{178,300\text{ J/mol}}{160.5\text{ J/(mol}\cdot\text{K)}} = 1111\text{ K} \implies 838\ ^\circ\text{C}$ Conclusion: Because both $\Delta H$ and $\Delta S$ are positive (Quadrant 4), decomposition is spontaneous above $1111\text{ K}$ ($838\ ^\circ\text{C}$).

Example 2: Calculating $K$ from Standard Free Energy

Problem: If $\Delta G^\circ = -17.1\text{ kJ/mol}$ at $298\text{ K}$, calculate $K$. $\ln K = -\frac{-17,100\text{ J/mol}}{(8.314\text{ J/(mol}\cdot\text{K)})(298\text{ K})} = +6.902 \implies K = e^{6.902} = 1.0 \times 10^3$