The Equilibrium Constant & ΔG°

ΔG° sets the equilibrium constant: ΔG° = −RT ln K, so K is the thermodynamic 'target' composition at a given temperature.

Material & Energy BalancesChemical Engineering Year 1Free preview
⏱️ About 16 min

How does a single number K encode what composition a reaction mixture will settle at?

💡
The big idea: At a fixed temperature, the equilibrium constant K is determined by ΔG° through ΔG° = −RT ln K.
🎯 By the end, you'll be able to
  • State the relationship between ΔG° and the equilibrium constant K
  • Interpret K as a ratio of product/reactant activities at equilibrium
  • Compute K from a given ΔG° and temperature
  • Compute ΔG° from a given K and temperature
📎 Helpful to know first

What the Equilibrium Constant Represents

For a reaction written with stoichiometric coefficients νi, the equilibrium constant K is built from activities (ideal-gas approximations often use partial pressures or mole fractions):

At equilibrium, K equals the value of the product of activities raised to stoichiometric powers.

Physically: a large K means the equilibrium composition is product-favored; a small K means reactant-favored.

\[ \Delta G^\circ = -RT\ln K \]
Core thermodynamic link between standard Gibbs energy change and equilibrium constant at temperature T.

Rearranging: Solving for K or for ΔG°

  • Given ΔG° and T: K = exp(−ΔG°/(RT))
  • Given K and T: ΔG° = −RT ln K

Use R = 8.314 J/(mol·K) and keep units consistent (ΔG° in J/mol if you use R in J/(mol·K)).

⚠️ Pitfall: K is dimensionless (activities), and the sign matters

K is built from activities, so it is dimensionless. In idealized problems you may see Kp or Kx constructed from ratios of pressures or mole fractions; these are stand-ins for activities.

Also: if ΔG° < 0 then ln K > 0 and K > 1 (product-favored). If ΔG° > 0 then K < 1.

📝 Worked example: At T = 298 K, a reaction has ΔG° = −5.708 kJ/mol. Using R = 8.314 J/(mol·K), compute K.
  1. Convert ΔG° to J/mol: ΔG° = −5.708 kJ/mol = −5708 J/mol
  2. Compute −ΔG°/(RT): −ΔG°/(RT) = −(−5708) / (8.314×298) = 5708 / 2477.572
  3. 5708 / 2477.572 = 2.304 (dimensionless)
  4. Compute K = exp(2.304): K ≈ 10.01 ≈ 10.0
✓ K ≈ 10.0 at 298 K.
✏️ Practice: At T = 300 K, K = 2.00 for a reaction. Using R = 8.314 J/(mol·K), compute ΔG° (J/mol).
J/mol
Solution
  1. Use ΔG° = −RT ln K
  2. ln(2.00) = 0.6931
  3. ΔG° = −(8.314)(300)(0.6931) J/mol
  4. (8.314)(300) = 2494.2
  5. ΔG° = −2494.2×0.6931 = −1728.4 J/mol (to 5 sig figs)

Check your understanding

1. If ΔG° is negative at a given temperature, the equilibrium constant K is:
ΔG° = −RT ln K. If ΔG° < 0 then ln K > 0 so K > 1.
2. In ΔG° = −RT ln K, if you use R = 8.314 J/(mol·K), then ΔG° must be in:
Units must match: R in J/(mol·K) requires ΔG° in J/mol to keep the exponent dimensionless.
🔎 Related lessons
The van't Hoff Equation & Le Chatelier's Principle · soon
✅ Key takeaways
  • At fixed T: ΔG° = −RT ln K
  • K is built from equilibrium activities (dimensionless); large K means product-favored equilibrium
  • Compute K from ΔG° by exponentiating: K = exp(−ΔG°/(RT))
  • Compute ΔG° from K by taking a logarithm: ΔG° = −RT ln K
➡️ K is temperature-dependent. Next we quantify how K changes with T using the van't Hoff equation, and connect the sign of ΔH° to the qualitative direction predicted by Le Chatelier's principle.
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