Thermodynamics of Equilibrium and Free Energy

Chemical equilibrium is the state of minimum Gibbs free energy of the reaction mixture, where ΔG = 0 and Q = K, and the equilibrium constant is related to the standard free energy change by ΔG° = −RT ln K. In brief: for N₂O₄ ⇌ 2NO₂, ΔG° = +4.8 kJ/mol at 298 K, which corresponds to K = e^−4800/2479 = 0.14, so the equilibrium mixture contains mostly N₂O₄.

Kinetics and Le Châtelier’s principle describe how equilibria behave, but thermodynamics explains why there is a particular equilibrium composition. The central point is that a reaction mixture has a free energy that depends on its composition, and the system moves spontaneously toward the composition with the lowest free energy. This article explains the free energy curve, derives the relation between ΔG° and K, shows how to calculate K from tabulated data and discusses temperature.

Why does equilibrium correspond to a free energy minimum?

Consider a mixture of reactants and products. As the reaction proceeds, the Gibbs free energy G of the mixture changes. Pure reactants have a certain free energy, and pure products have a different one. In between, G is lower than either of the end values, because of the entropy of mixing: a mixture of two substances has a greater entropy than pure ones. As a result, the free energy plotted against the extent of reaction passes through a minimum. This minimum is the equilibrium state. On the left of the minimum the slope of G is negative (ΔG < 0), so the reaction goes forward; on the right it is positive, so the reaction goes in reverse. At the minimum the slope is zero: ΔG = 0.

The position of the minimum depends on ΔG°. If products are much more stable than reactants (ΔG° ≪ 0), the minimum is very close to pure products and K is large. If they are less stable (ΔG° ≫ 0), it is close to pure reactants and K is small. Even for a positive ΔG°, some products form, because mixing always lowers the free energy.

How are ΔG°, ΔG and K related?

For a reaction mixture with reaction quotient Q:

ΔG = ΔG° + RT ln Q

At equilibrium ΔG = 0 and Q = K, so:

ΔG° = −RT ln K, or K = e−ΔG°/RT

Combining the two equations gives ΔG = RT ln(Q/K). The standard free energy change ΔG° is the free energy change for converting pure reactants to pure products in their standard states, and it is a fixed number at a given temperature, whereas ΔG depends on the composition of the mixture. For gases, K is the thermodynamic constant (with partial pressures relative to 1 bar); for solutions, concentrations are relative to 1 mol/L.

How do Q, K and ΔG compare?

Worked examples

Example 1: K from standard free energies of formation

For N₂O₄(g) ⇌ 2NO₂(g), the standard free energies of formation at 298 K are ΔG_f°(NO₂) = +51.3 kJ/mol and ΔG_f°(N₂O₄) = +97.8 kJ/mol.

1. ΔG° = 2(51.3) − 97.8 = +4.8 kJ/mol.
2. ln K = −ΔG° / RT = −4800 / (8.314 × 298.15) = −4800 / 2479 = −1.94.
3. K = e^−1.94 = 0.14.

This agrees with the value of K_p found by direct measurement, about 0.15. Although ΔG° is positive, NO₂ is present in a significant amount.

Example 2: ΔG for a mixture that is not at equilibrium

For the same reaction, K = 0.14. If the partial pressures give Q = 10, then ΔG = ΔG° + RT ln Q = 4.8 + 2.479 × ln 10 = 4.8 + 5.71 = +10.5 kJ/mol. The value is positive, so the reverse reaction is spontaneous: NO₂ combines to form N₂O₄. This is consistent with Q > K.

Example 3: ammonia at different compositions

For N₂ + 3H₂ ⇌ 2NH₃, ΔG° = −33.0 kJ at 298 K, and K = e^33,000/2479 = e^13.3 = 6.0 × 10⁵.

• For Q = 1 (all gases at 1 bar), ΔG = ΔG° = −33.0 kJ: spontaneous forward.
• For Q = 1 × 10⁵, ΔG = −33.0 + 2.479 × ln(10⁵) = −33.0 + 28.5 = −4.5 kJ: still forward.
• For Q = 6.0 × 10⁵, ΔG = −33.0 + 2.479 × 13.3 = 0: equilibrium.

Example 4: finding ΔG° from a measured K

The acid dissociation constant of acetic acid is K_a = 1.8 × 10⁻⁵ at 25 °C. ΔG° = −RT ln K = −2.479 × ln(1.8 × 10⁻⁵) = −2.479 × (−10.92) = +27.1 kJ/mol. The positive value agrees with the small fraction of acid that ionizes.

How does temperature enter?

Since ΔG° = ΔH° − TΔS°, combining with ΔG° = −RT ln K gives ln K = −ΔH°/(RT) + ΔS°/R. The two terms show the two driving forces: an exothermic reaction (ΔH° < 0) raises K, and so does an increase in entropy (ΔS° > 0). The enthalpy term depends on 1/T, so K for an endothermic reaction increases with temperature and for an exothermic reaction decreases. A plot of ln K against 1/T has a slope of −ΔH°/R. For N₂O₄ ⇌ 2NO₂, with ΔH° = +57.2 kJ and ΔS° = +176 J/K, the reaction is driven by entropy at room temperature, since two gas molecules form from one.

How does this link to electrochemistry?

For an electrochemical cell, the maximum electrical work is related to the free energy: ΔG = −nFE. At standard conditions ΔG° = −nFE°, so E° = (RT/nF) ln K. A cell potential can therefore be used to measure an equilibrium constant. A battery that is “dead” has reached equilibrium: ΔG = 0 and E = 0.

Who established these relations?

Josiah Willard Gibbs laid the foundation in 1876–1878 with his work on the equilibrium of heterogeneous substances, in which equilibrium was defined by the minimum of free energy. Jacobus van ‘t Hoff gave the relation between the free energy and the equilibrium constant in 1886 (the van ‘t Hoff isotherm). Gilbert N. Lewis and Merle Randall presented the full treatment, with activities, in their textbook Thermodynamics and the Free Energy of Chemical Substances (1923), which became the standard.

Which mistakes are common?

• Thinking that ΔG° = 0 at equilibrium: it is ΔG that is zero; ΔG° is related to K.
• Using Celsius: T must be in kelvins.
• Unit mismatch: if R = 8.314 J/(mol·K), ΔG° must be in J/mol.
• Assuming that a positive ΔG° means no reaction: it means that K < 1, but some products form.
• Confusing natural and common logarithms: ln K = 2.303 log K.