Free Energy and Equilibrium

Core Concept

At equilibrium ΔG = 0 and Q = K. Substituting these into ΔG = ΔG° + RT ln Q gives the master link between thermodynamics and chemical equilibrium: ΔG° = −RT ln K (equivalently K = e^(−ΔG°/RT)). This single equation says a reaction's equilibrium constant is fixed by its standard free-energy change: a more negative ΔG° means a larger K and more products at equilibrium. It is how we predict equilibrium position from tables of thermodynamic data, with no experiment required.

Key Ideas

• At equilibrium, ΔG = 0. The forward and reverse free-energy driving forces exactly cancel.
• ΔG° is not zero at equilibrium. ΔG° is a fixed standard-state constant; only the actual ΔG is zero.
• Exponential sensitivity. Because K = e^(−ΔG°/RT), a modest change in ΔG° produces a huge change in K (at 298 K, a 5.7 kJ/mol shift changes K by a factor of 10).
• Sign → size. ΔG° < 0 ⇒ K > 1 (products favored); ΔG° > 0 ⇒ K < 1 (reactants favored); ΔG° = 0 ⇒ K = 1.
• Temperature dependence. Combined with ΔG° = ΔH° − TΔS°, this predicts how K changes with T (van 't Hoff).

Equations and Variables

ΔG° = −RT ln K

How It Works

1. Compute ΔG° from ΔG°f tables (or from ΔH° and ΔS°).
2. Convert ΔG° to J/mol.
3. Solve ln K = −ΔG°/RT, then K = e^(ln K).
4. Interpret K: >1 product-favored, <1 reactant-favored, ≈1 balanced.

Worked Example

Calculate K at 25 °C for N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), given ΔG° = −33.0 kJ/mol.

• Convert: ΔG° = −33.0 kJ/mol = −33,000 J/mol
• ln K = −ΔG°/(RT) = 33,000 / (8.314 × 298) = 33,000 / 2477.6 = 13.3
• K = e^13.3 = 6.1 × 10⁵

The large K reflects a strongly product-favored equilibrium at room temperature — consistent with the industrial Haber process value (though industrially it runs hot, where K is smaller, and relies on high pressure and catalysis).

How it works

1. Compute ΔG° from ΔG°f tables (or from ΔH° and ΔS°).
2. Convert ΔG° to J/mol.
3. Solve ln K = −ΔG°/RT, then K = e^(ln K).
4. Interpret K: >1 product-favored, <1 reactant-favored, ≈1 balanced.

Common confusions

• "ΔG° = 0 at equilibrium." — Wrong. ΔG = 0 at equilibrium; ΔG° is a constant that is zero only if K happens to equal 1.
• "K = 1 whenever ΔG° = 0." — Correct as stated, but students often mistake any equilibrium for ΔG° = 0. Equilibrium ⇒ ΔG = 0, not ΔG° = 0.
• "Bigger ΔG° means bigger K." — Wrong. More negative ΔG° means bigger K (K = e^(−ΔG°/RT)).
• "K can be computed in °C." — Wrong. T must be in Kelvin.
• "Units of ΔG° don't matter." — Wrong. Use J/mol with R = 8.314 J/(mol·K).

Quick review

• ΔG° = −RT ln K; K = e^(−ΔG°/RT).
• At equilibrium ΔG = 0 and Q = K.
• More negative ΔG° ⇒ larger K (product-favored).
• K > 1 ⇔ ΔG° < 0; K < 1 ⇔ ΔG° > 0.
• Use Kelvin and J/mol.