8.4 Gibbs and Helmholtz Free Energy, Spontaneity, and Fundamental Equations

Gibbs and Helmholtz Free Energy, Spontaneity, and Fundamental Equations

To predict the spontaneity and equilibrium state of a chemical system without evaluating the entropy changes of the entire universe, thermodynamics defines auxiliary state functions: the Helmholtz energy ($A$) and the Gibbs energy ($G$). These potentials encapsulate both the First and Second Laws into compact state variables whose differentials govern equilibrium and maximum work output.

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1. Helmholtz Free Energy ($A$) and the Maximum Work Theorem

The Helmholtz free energy is defined mathematically as:

$A \equiv U - TS$

Criterion of Spontaneity at Constant Temperature and Volume

From the Clausius inequality, for any process: $dS \ge \frac{\delta q}{T} \implies T\,dS - \delta q \ge 0$. At constant volume ($dV = 0$) with $PV$ work only, the First Law gives $\delta q_v = dU$. Substituting yields: $T\,dS - dU \ge 0 \implies -(dU - T\,dS) \ge 0$ For an isothermal process ($T = \text{const}$), $d(TS) = T\,dS$. Therefore: $-d(U - TS)_{T, V} \ge 0 \implies (dA)_{T, V} \le 0$

• $(dA)_{T,V} < 0$: Spontaneous process at constant $T$ and $V$.
• $(dA)_{T,V} = 0$: Thermodynamic equilibrium at constant $T$ and $V$.

The Maximum Total Work Theorem

Consider a system exchanging heat and performing both expansion and non-expansion work: $dU = \delta q + \delta w_{total}$. By the Clausius inequality, $\delta q \le T\,dS$. Substituting into the First Law: $dU \le T\,dS + \delta w_{total} \implies \delta w_{total} \ge dU - T\,dS$ Under isothermal conditions ($T = \text{const}$), $dU - T\,dS = dA_T$. Therefore: $\delta w_{total} \ge dA_T$ Recalling the IUPAC convention where work delivered to the surroundings is $-w_{total}$:

$-w_{total, max} = -\Delta A_T$ The decrease in Helmholtz energy ($-\Delta A$) represents the maximum total work (expansion plus non-expansion work) that a closed isothermal system can deliver to its surroundings.

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2. Gibbs Free Energy ($G$) and Maximum Non-Expansion Work

The Gibbs free energy is defined mathematically as:

$G \equiv H - TS = U + PV - TS = A + PV$

Criterion of Spontaneity at Constant Temperature and Pressure

At constant pressure ($dP = 0$), heat absorbed under mechanical equilibrium equals enthalpy change: $\delta q_p = dH$. Substituting into the Clausius inequality ($T\,dS \ge \delta q$): $T\,dS - dH \ge 0 \implies -(dH - T\,dS) \ge 0$ Under isothermal conditions ($T = \text{const}$), $dH - T\,dS = d(H - TS) = dG$. Thus:

$(dG)_{T, P} \le 0$

• $(dG)_{T,P} < 0$: Spontaneous process at constant $T$ and $P$.
• $(dG)_{T,P} = 0$: Chemical equilibrium at constant $T$ and $P$.
• $(dG)_{T,P} > 0$: Non-spontaneous process (the reverse process is spontaneous).

The Maximum Non-Expansion Work Theorem

Dividing total work into $PV$ boundary work and non-expansion work (such as electrical work $\delta w_{elec}$ in a battery or fuel cell): $\delta w_{total} = -P\,dV + \delta w_{non-PV}$ From the First Law and Clausius inequality: $dU = \delta q - P\,dV + \delta w_{non-PV} \le T\,dS - P\,dV + \delta w_{non-PV}$ $\delta w_{non-PV} \ge dU + P\,dV - T\,dS = d(U + PV - TS)_{T,P} = dG_{T,P}$ Therefore, the maximum non-expansion work obtainable by the surroundings is:

$-w_{non-PV, max} = -\Delta G_{T, P}$ In electrochemistry this gives $\Delta G = -nFE_{cell}$, the relationship from which the Nernst equation is derived.

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3. The Four Cases of Spontaneity: Interplay of $\Delta H$ and $\Delta S$

At constant temperature and pressure, the change in Gibbs free energy is: $\Delta G = \Delta H - T\Delta S$ Because $T$ is in Kelvin ($T > 0$), the signs of $\Delta H$ (enthalpic driving force) and $\Delta S$ (entropic driving force) define four distinct thermodynamic regimes:

Quantitative Crossover Temperature Calculation

At the crossover temperature ($T_{cross}$), $\Delta G^\circ = 0$: $\Delta H^\circ - T_{cross}\Delta S^\circ = 0 \implies T_{cross} = \frac{\Delta H^\circ}{\Delta S^\circ}$

Worked Step-by-Step Example: Decomposition of Limestone $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ Given Data at $298.15\text{ K}$:

• $\Delta H^\circ = +178.3\text{ kJ/mol} = +178300\text{ J/mol}$
• $\Delta S^\circ = +160.5\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$

Step 1: Calculate $\Delta G^\circ$ at $298.15\text{ K}$: $\Delta G^\circ_{298} = 178300\text{ J/mol} - (298.15\text{ K})(160.5\text{ J/mol}\cdot\text{K}) = 178300 - 47853 = +130447\text{ J/mol} \approx +130.4\text{ kJ/mol}$ Because $\Delta G^\circ > 0$, limestone is completely stable at room temperature against decomposition.

Step 2: Calculate the crossover decomposition temperature ($T_{cross}$): $T_{cross} = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{178300\text{ J/mol}}{160.5\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}} = 1110.9\text{ K}$ $T_{cross}(^\circ\text{C}) = 1110.9 - 273.15 = 837.8^\circ\text{C} \approx 838^\circ\text{C}$ Conclusion: At temperatures exceeding $1111\text{ K}$ ($838^\circ\text{C}$), $-T\Delta S^\circ$ becomes sufficiently negative to overcome the endothermic enthalpy barrier, rendering calcination spontaneous at $1\text{ bar}$ of $\text{CO}_2$.

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4. The Gibbs-Helmholtz Equation

The Gibbs-Helmholtz equation governs how free energy varies with temperature at constant pressure.

Mathematical Derivation

Applying the calculus quotient rule to $(G/T)$ with respect to $T$: $\left(\frac{\partial (G/T)}{\partial T}\right)_P = \frac{1}{T}\left(\frac{\partial G}{\partial T}\right)_P - \frac{G}{T^2}$ From the fundamental equation $dG = -S\,dT + V\,dP$, the partial derivative is $(\frac{\partial G}{\partial T})_P = -S$. Substituting gives: $\left(\frac{\partial (G/T)}{\partial T}\right)_P = -\frac{S}{T} - \frac{G}{T^2} = -\frac{TS + G}{T^2} = -\frac{H}{T^2}$

Expressed in terms of reaction changes:

$\left(\frac{\partial (\Delta G / T)}{\partial T}\right)_P = -\frac{\Delta H}{T^2}$

Using the chain rule with respect to $(1/T)$: $\left(\frac{\partial (\Delta G / T)}{\partial (1/T)}\right)_P = \Delta H$

Connection to the van 't Hoff Equation

Because $\Delta G^\circ = -RT \ln K_{eq} \implies \frac{\Delta G^\circ}{T} = -R \ln K_{eq}$, substituting into Gibbs-Helmholtz yields: $\frac{d \ln K_{eq}}{dT} = \frac{\Delta H^\circ}{RT^2} \iff \frac{d \ln K_{eq}}{d(1/T)} = -\frac{\Delta H^\circ}{R}$ Integration from $T_1$ to $T_2$ yields the standard two-point van 't Hoff equation: $\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

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5. The Four Fundamental Thermodynamic Equations

For a closed system of constant composition undergoing reversible processes, combining the First and Second Laws yields four master differential relations:

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6. Maxwell Relations and Thermodynamic Equations of State

Because $U, H, A,$ and $G$ are thermodynamic state functions, their mixed second partial derivatives are invariant under order of differentiation (Euler's reciprocity criterion). This generates the four Maxwell relations:

1. From $dU = T\,dS - P\,dV$: $\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V$
2. From $dH = T\,dS + V\,dP$: $\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P$
3. From $dA = -S\,dT - P\,dV$: $\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$
4. From $dG = -S\,dT + V\,dP$: $\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P$

The First Thermodynamic Equation of State (Internal Pressure $\pi_T$)

The dependence of internal energy on volume at constant temperature is termed the internal pressure: $\pi_T \equiv \left(\frac{\partial U}{\partial V}\right)_T$ Dividing $dU = T\,dS - P\,dV$ by $dV$ at constant $T$: $\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial S}{\partial V}\right)_T - P$ Substituting Maxwell relation 3: $(\frac{\partial S}{\partial V})_T = (\frac{\partial P}{\partial T})_V$ yields:

$\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P$

Application to Ideal and Real Gases:

• Ideal Gas: $P = \frac{nRT}{V} \implies (\frac{\partial P}{\partial T})_V = \frac{nR}{V}$. Substituting gives: $\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{nR}{V}\right) - P = P - P = 0$ This provides the rigorous thermodynamic proof of Joule's second law: internal energy of an ideal gas is completely independent of volume and pressure, depending solely on temperature.
• van der Waals Gas: $P = \frac{RT}{V_m - b} - \frac{a}{V_m^2} \implies (\frac{\partial P}{\partial T})_V = \frac{R}{V_m - b}$. Substituting gives: $\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{R}{V_m - b}\right) - \left[\frac{RT}{V_m - b} - \frac{a}{V_m^2}\right] = \frac{a}{V_m^2}$ The internal pressure is strictly positive and directly measures intermolecular attractive forces ($a$) that pull molecules together.

The Second Thermodynamic Equation of State

$\left(\frac{\partial H}{\partial P}\right)_T = V - T\left(\frac{\partial V}{\partial T}\right)_P$ For an ideal gas, $(\frac{\partial H}{\partial P})_T = 0$, confirming that enthalpy of an ideal gas is also purely a function of temperature.