7.3 Chemical potential and equilibrium constants

Chemical potential connects Gibbs free energy to the composition of a system, telling you how the energy changes when you add or remove a small amount of a particular species. This is the central concept for understanding why reactions proceed in a given direction and where equilibrium lies. It bridges the gap between bulk thermodynamic quantities and the molecular-level changes that drive reactions.

Chemical Potential and Gibbs Free Energy

Definition and Relationship

Chemical potential ($\mu_i$) is defined as the partial molar Gibbs free energy of component $i$:

$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{T, P, n_{j \neq i}}$

In plain terms, it answers the question: "If you add an infinitesimal amount of species $i$ to the mixture, holding temperature, pressure, and all other amounts fixed, how much does the total Gibbs energy change?"

A few important consequences follow from this definition:

• In a system at equilibrium, the chemical potential of each component is the same in every phase it occupies. If it weren't, material would spontaneously transfer from the high-$\mu$ phase to the low-$\mu$ phase until the potentials equalize.
• The total Gibbs energy of a mixture can be written as $G = \sum_i n_i \mu_i$.
• The Gibbs-Duhem equation constrains how chemical potentials in a mixture can change simultaneously: $\sum_i n_i \, d\mu_i = 0$ at constant $T$ and $P$. This means the chemical potentials of different components in a mixture are not independent of one another.

Expressions for Chemical Potential

The general form relates chemical potential to activity ($a_i$):

$\mu_i = \mu_i^\circ + RT \ln a_i$

where $\mu_i^\circ$ is the standard chemical potential (the value at a defined reference state). The activity takes different forms depending on the system:

• Ideal gas: $a_i = p_i / p^\circ$, so $\mu_i = \mu_i^\circ + RT \ln(p_i / p^\circ)$, where $p_i$ is the partial pressure and $p^\circ = 1$ bar.
• Ideal solution: $a_i = x_i$, so $\mu_i = \mu_i^\circ + RT \ln x_i$, where $x_i$ is the mole fraction.
• Non-ideal systems use fugacity (for gases) or activity coefficients (for solutions) to correct for deviations from ideal behavior. That's beyond the scope here, but the general $\mu = \mu^\circ + RT \ln a$ framework still holds.

Notice that because $\ln x_i$ and $\ln(p_i/p^\circ)$ are negative when $x_i < 1$ or $p_i < p^\circ$, the chemical potential of a species in a mixture is always lower than its pure-component standard value. This is why mixing is thermodynamically favorable for ideal systems.

Equilibrium Constants and Chemical Potentials

Deriving the Equilibrium Constant

For a general reaction $\sum_i \nu_i A_i = 0$ (using the convention that stoichiometric coefficients $\nu_i$ are positive for products and negative for reactants), the Gibbs energy change is:

$\Delta G = \sum_i \nu_i \mu_i$

Substituting $\mu_i = \mu_i^\circ + RT \ln a_i$:

$\Delta G = \underbrace{\sum_i \nu_i \mu_i^\circ}_{\Delta G^\circ} + RT \ln \underbrace{\prod_i a_i^{\nu_i}}_{Q}$

This gives the fundamental equation:

$\Delta G = \Delta G^\circ + RT \ln Q$

where $Q$ is the reaction quotient. At equilibrium, $\Delta G = 0$ and $Q = K$, so:

$0 = \Delta G^\circ + RT \ln K$

Rearranging:

$\Delta G^\circ = -RT \ln K \qquad \text{or equivalently} \qquad K = \exp\!\left(-\frac{\Delta G^\circ}{RT}\right)$

This is the central result connecting chemical potentials (through $\Delta G^\circ$) to the equilibrium constant.

The van 't Hoff Equation

Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, you can write:

$\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$

This is the van 't Hoff equation in its integrated form (assuming $\Delta H^\circ$ and $\Delta S^\circ$ are approximately temperature-independent). It shows that a plot of $\ln K$ vs. $1/T$ gives a straight line with slope $-\Delta H^\circ / R$ and intercept $\Delta S^\circ / R$. The differential form is:

$\frac{d \ln K}{d(1/T)} = -\frac{\Delta H^\circ}{R}$

Calculating Equilibrium Constants

From Standard Chemical Potentials

1. Look up (or calculate) the standard chemical potential $\mu_i^\circ$ for each reactant and product. For pure substances, $\mu^\circ$ equals the standard molar Gibbs energy, which in turn equals $\Delta G_f^\circ$ (the standard Gibbs energy of formation).

2. Compute the standard Gibbs energy change: $\Delta G^\circ = \sum \nu_{\text{products}} \mu_{\text{products}}^\circ - \sum \nu_{\text{reactants}} \mu_{\text{reactants}}^\circ$

3. Calculate $K$: $K = \exp\!\left(-\frac{\Delta G^\circ}{RT}\right)$

Worked example: For $\ce{A + B <=> C}$ with $\mu_A^\circ = -20$ kJ/mol, $\mu_B^\circ = -30$ kJ/mol, $\mu_C^\circ = -60$ kJ/mol at $T = 298$ K:

$\Delta G^\circ = (-60) - (-20 + (-30)) = -10 \text{ kJ/mol}$

Convert to J: $\Delta G^\circ = -10{,}000$ J/mol.

$K = \exp\!\left(\frac{10{,}000}{8.314 \times 298}\right) = \exp(4.04) \approx 57$

From Standard Formation Gibbs Energies

The procedure is identical, since $\mu_i^\circ$ for a pure substance equals $\Delta G_{f,i}^\circ$:

$\Delta G^\circ = \sum \nu_{\text{products}} \Delta G_{f,\text{products}}^\circ - \sum \nu_{\text{reactants}} \Delta G_{f,\text{reactants}}^\circ$

Worked example: For $\ce{2A + B <=> C}$ with $\Delta G_f^\circ(A) = -20$ kJ/mol, $\Delta G_f^\circ(B) = -30$ kJ/mol, $\Delta G_f^\circ(C) = -100$ kJ/mol at 298 K:

$\Delta G^\circ = (-100) - [2(-20) + (-30)] = -100 - (-70) = -30 \text{ kJ/mol}$

$K = \exp\!\left(\frac{30{,}000}{8.314 \times 298}\right) = \exp(12.11) \approx 1.8 \times 10^5$

Predicting Reaction Direction

Using Chemical Potentials

The system always evolves to lower its total Gibbs energy. Compare the sum of chemical potentials on each side of the reaction:

• If $\sum \nu_{\text{products}} \mu_{\text{products}} < \sum \nu_{\text{reactants}} \mu_{\text{reactants}}$, the forward reaction lowers $G$, so the reaction proceeds toward products.
• If the inequality is reversed, the reverse reaction is spontaneous.
• When the two sums are equal, the system is at equilibrium.

Using the Reaction Quotient

Equivalently, compare $Q$ to $K$:

• $Q < K$: The system has too little product relative to equilibrium. The reaction proceeds forward ($\Delta G < 0$).
• $Q > K$: The system has too much product. The reaction proceeds in reverse.
• $Q = K$: Equilibrium. No net change.

The value of $K$ itself tells you about the equilibrium position:

• $K \gg 1$: Products strongly favored at equilibrium.
• $K \ll 1$: Reactants strongly favored.
• $K \approx 1$: Comparable amounts of reactants and products.

Effects on Chemical Potentials and Equilibrium Constants

Temperature Effects

Temperature influences chemical potentials through the Gibbs-Helmholtz relation:

$\left(\frac{\partial(\mu_i / T)}{\partial(1/T)}\right)_P = H_i$

where $H_i$ is the partial molar enthalpy of component $i$. The practical consequence for equilibrium constants comes from the van 't Hoff equation:

$\frac{d \ln K}{d(1/T)} = -\frac{\Delta H^\circ}{R}$

• Exothermic reactions ($\Delta H^\circ < 0$): Increasing $T$ decreases $K$. The equilibrium shifts toward reactants.
• Endothermic reactions ($\Delta H^\circ > 0$): Increasing $T$ increases $K$. The equilibrium shifts toward products.

This is the quantitative basis for Le Chatelier's principle with respect to temperature.

Pressure Effects

Pressure affects chemical potentials through:

$\left(\frac{\partial \mu_i}{\partial P}\right)_T = V_i$

where $V_i$ is the partial molar volume. For reactions involving gases, the key factor is the change in the total number of moles of gas, $\Delta n_{\text{gas}}$:

• If $\Delta n_{\text{gas}} < 0$ (fewer moles of gas in products), increasing pressure shifts equilibrium toward products.
• If $\Delta n_{\text{gas}} > 0$ (more moles of gas in products), increasing pressure shifts equilibrium toward reactants.

An important subtlety: the thermodynamic equilibrium constant $K$ (defined in terms of activities) does not change with pressure at constant temperature. What changes is the composition at equilibrium. For ideal gases, $K_p$ is pressure-independent, but $K_x$ (expressed in mole fractions) depends on total pressure when $\Delta n_{\text{gas}} \neq 0$.

Composition Effects

Changing the amounts of reactants or products alters their activities and therefore their chemical potentials via $\mu_i = \mu_i^\circ + RT \ln a_i$.

• Adding more reactant increases $Q$ relative to its current value... actually, adding reactant decreases $Q$ (since reactant activities appear in the denominator of $Q$). This makes $Q < K$, so the reaction shifts forward to re-establish equilibrium.
• Removing product has the same effect: $Q$ decreases, driving the forward reaction.

The equilibrium constant $K$ itself does not change with composition at fixed $T$. Only $Q$ changes, and the system adjusts until $Q = K$ again.