Gibbs energy, spontaneity and equilibrium

Gibbs energy gives a useful constrained criterion

Gibbs energy is G = H − TS. For a change evaluated at the same fixed temperature, ΔG = ΔH − TΔS. Under the usual closed-system fixed-temperature and fixed-pressure setting without additional imposed non-PV work, a spontaneous direction lowers G; equilibrium minimizes it subject to the system’s constraints.

Here spontaneous means thermodynamically favoured under the stated conditions, not instantaneous. A large kinetic barrier can make a favoured process slow. An externally driven process needs its coupling and transfers included in the broader description rather than being judged from the uncoupled part alone.

Convert units before subtracting TΔS

For the next model, molar reaction changes ΔᵣH, ΔᵣS and ΔᵣG are defined per mole of the stipulated reaction amount on its written stoichiometric basis. An original hypothetical model has ΔᵣH = +20.0 kJ mol⁻¹ and ΔᵣS = +50.0 J mol⁻¹ K⁻¹, treated as approximately constant over 300–500 K within one stated phase and composition scope. Use ΔᵣG = ΔᵣH − TΔᵣS and convert ΔᵣS to +0.0500 kJ mol⁻¹ K⁻¹. At 300 K the model gives ΔᵣG = +5.0 kJ mol⁻¹; at 500 K it gives −5.0 kJ mol⁻¹.

The model crosses zero at T = ΔᵣH/ΔᵣS = 400 K. This is a fictional constant-parameter example, not a real substance’s transition temperature. Real molar reaction changes ΔᵣH and ΔᵣS can vary with temperature, and a phase change can invalidate the extrapolation.

The actual reaction depends on its current composition

For a consistently defined reaction, ΔᵣG = ΔᵣG° + RT ln Q. The subscript r means reaction; the degree symbol denotes the standard-state value. R is the molar gas constant, T is absolute temperature, and Q is the reaction quotient formed from dimensionless activities with the written stoichiometric powers.

A logarithm needs a dimensionless argument, so bare dimensional concentrations cannot simply be inserted into ln Q. A dilute approximation uses appropriate ratios to reference concentrations or pressures. Changing composition can change the actual ΔᵣG even while ΔᵣG° at the same temperature stays fixed.

At equilibrium the actual reaction Gibbs energy is zero

At equilibrium Q = K and ΔᵣG = 0 for the allowed reaction direction, giving ΔᵣG° = −RT ln K with the same dimensionless convention. The standard value generally does not become zero merely because the mixture has reached equilibrium.

An original model stipulates ΔᵣG° = −5.00 kJ mol⁻¹, T = 298 K, dimensionless Q = 10 and a chosen rounded calculation value R = 8.314 J mol⁻¹ K⁻¹. Treat these as exact numerical inputs for this fictional calculation; R here is not asserted to be the exact SI gas constant. The chosen inputs give K ≈ 7.52 and actual ΔᵣG ≈ +0.705 kJ mol⁻¹. Since Q > K, the reverse direction is favoured in this model, despite the negative standard-state value. The later equilibrium chapter develops activity quotients in more detail.

Worked examples

Try it yourself

1. Does negative ΔG alone establish that a reaction is fast?

   Show the answer and reasoning

   No.

   The Gibbs criterion concerns thermodynamic direction, while barriers and kinetics govern rate.

2. Convert 50.0 J mol⁻¹ K⁻¹ to kJ mol⁻¹ K⁻¹.

   Show the answer and reasoning

   0.0500 kJ mol⁻¹ K⁻¹.

   Divide by 1000 before combining with an enthalpy expressed in kJ.

3. At equilibrium, which is zero: actual ΔᵣG or necessarily ΔᵣG°?

   Show the answer and reasoning

   Actual ΔᵣG.

   The standard-state value obeys −RT ln K and need not be zero.

4. What kind of argument is required inside ln Q?

   Show the answer and reasoning

   A dimensionless quotient.

   Use dimensionless activities or explicitly defined reference ratios.

What to carry forward

• State temperature, pressure and work constraints for the criterion.
• Use matching units in the enthalpy–entropy balance.
• Actual reaction Gibbs energy depends on composition through Q.
• Equilibrium sets actual ΔᵣG to zero; it does not set every standard value to zero.