Free Energy, Spontaneity & K

You’ll be able to

Predict the temperature dependence of spontaneity from the signs of ΔH and ΔS
Relate the sign of ΔG° to whether a reaction is thermodynamically favored
Connect ΔG° to the equilibrium constant K through ΔG° = −RT ln K

Gibbs free energy decides who wins

A reaction is thermodynamically favored (spontaneous) when it releases free energy — when ΔG < 0. Gibbs combined the two competing drives, enthalpy and entropy, into one number: ΔG = ΔH − TΔS. Enthalpy (ΔH) favors reactions that release heat; entropy (ΔS) favors reactions that spread energy and matter out. Temperature (T, always in kelvin) sets how loudly the entropy term speaks.

Gibbs free energy

ΔG = ΔH − TΔS

ΔG < 0 favored, ΔG > 0 not favored, ΔG = 0 at equilibrium. T is in kelvin, so the TΔS term always grows with temperature.

Reading the four sign combinations

Because T is always positive, the signs of ΔH and ΔS tell you when a reaction turns favorable. ΔH < 0, ΔS > 0: favored at all temperatures. ΔH > 0, ΔS < 0: favored at no temperature. ΔH < 0, ΔS < 0: favored only at low T (before the −TΔS penalty overwhelms the released heat). ΔH > 0, ΔS > 0: favored only at high T (once TΔS grows large enough to pay the enthalpy cost).

Thermodynamics vs kinetics

A favored ΔG says a reaction can release free energy — it does not say it will happen fast. Diamond turning to graphite has ΔG < 0, yet the activation energy is so high it never visibly proceeds. Thermodynamic control asks "which way does free energy point?"; kinetic control asks "is there a low-energy path to get there?" A reaction needs both a favorable ΔG and an accessible pathway to actually run.

Free energy and the equilibrium constant

ΔG° = −RT ln K

R = 8.314 J·mol⁻¹·K⁻¹, T in kelvin. K > 1 ⇒ ln K > 0 ⇒ ΔG° < 0 (products favored). K < 1 ⇒ ΔG° > 0 (reactants favored).

Worked example

A reaction has ΔH° = +30.0 kJ·mol⁻¹ and ΔS° = +100.0 J·mol⁻¹·K⁻¹. Above what temperature does it become thermodynamically favored?

1.A reaction becomes favored the moment ΔG turns negative, and it crosses through ΔG = 0 at that threshold temperature.
2.Set ΔG = 0: 0 = ΔH − TΔS, so T = ΔH ÷ ΔS.
3.Match units — convert ΔH to joules: 30.0 kJ = 30 000 J·mol⁻¹.
4.T = 30 000 J·mol⁻¹ ÷ 100.0 J·mol⁻¹·K⁻¹ = 300 K.
5.Both ΔH and ΔS are positive, so this is the "high-T favored" case: above 300 K, TΔS beats ΔH and ΔG < 0.

Answer: Favored above 300 K (about 27 °C).

Checkpoint

A reaction has ΔH < 0 and ΔS < 0. At which temperatures is it thermodynamically favored?

Checkpoint

A reaction at equilibrium has K = 1.0 × 10⁻⁵ at 298 K. What is true of its standard free energy change?

Watch out

Watch your units. ΔH° is almost always tabulated in kJ·mol⁻¹ but ΔS° in J·mol⁻¹·K⁻¹. Convert one so both use the same energy unit before combining them, or your crossover temperature will be off by a factor of 1000.

On the exam

Do not confuse ΔG° with ΔG. ΔG° = −RT ln K compares the standard state to equilibrium; a large positive ΔG° just means a small K. The actual ΔG (which is 0 at equilibrium) depends on the real concentrations through ΔG = ΔG° + RT ln Q.

Answer the 2 checkpoints as you read.