Gibbs Free Energy

9.3 Gibbs Free Energy

Slide ΔH, ΔS, and T to move a dot along the ΔG vs T line. A 2×2 quadrant grid and competing bar charts show when reactions are spontaneous.

ΔG = ΔH − TΔSSpontaneityEntropyCrossover T

Topic 9.3

Gibbs Free Energy and Thermodynamic Favorability

Explain whether a physical or chemical process is thermodynamically favored based on an evaluation of ΔG°.

The standard Gibbs free energy change (ΔG°) refers to a process in which all reactants and products are present in standard states: pure substances, 1.0 M solutions, or gases at 1.0 atm (or 1.0 bar).

Terminology matters here. EK 9.3.A.2 explains that historically "spontaneous" described processes with ΔG° < 0, but the CED prefers "thermodynamically favored" to avoid the misunderstanding that spontaneous means "suddenly" or "without cause". Use the CED's language on the exam.

Three routes to ΔG°:

1. From free energies of formation, on the equation sheet:

ΔG°_reaction = Σ ΔG°_f(products) − Σ ΔG°_f(reactants)

2. From ΔH° and ΔS°, also on the sheet:

ΔG° = ΔH° − TΔS°

3. From K (topic 9.5): ΔG° = −RT ln K.

EK 9.3.A.4 notes that some cases genuinely require weighing both enthalpy and entropy — the freezing of water and the dissolution of sodium nitrate are the CED's examples.

The sign table is the payoff:

ΔH°	ΔS°	ΔG° < 0 (favored) at:
< 0	> 0	all temperatures
> 0	< 0	no temperature
> 0	> 0	high temperature
< 0	< 0	low temperature

The CED spells out the shortcut: when ΔH° < 0 and ΔS° > 0, no calculation is necessary — the process is favored. When ΔH° > 0 and ΔS° < 0, no calculation is necessary either — it is unfavored.

For the two mixed rows, the crossover temperature where ΔG° = 0 is T = ΔH°/ΔS°.

Key points

ΔG° < 0 means thermodynamically favored. Use that phrase, not “spontaneous”.
ΔH° < 0 with ΔS° > 0 is favored at every temperature; the reverse combination at none.
Mixed signs give a crossover temperature T = ΔH°/ΔS°.
Convert ΔS° from J to kJ before using ΔG° = ΔH° − TΔS°.

Equations

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
on the exam sheetT in kelvin; ΔH° and TΔS° must be in the same energy units.
$Δ G_{rxn}^{\circ} = \sum Δ G_{f, products}^{\circ} - \sum Δ G_{f, reactants}^{\circ}$
on the exam sheet
$T_{crossover} = \frac{Δ H ^{\circ}}{Δ S ^{\circ}}$
not on the sheetThe temperature at which ΔG° = 0.

Common mistakes

Unit mismatch. ΔH° in kJ/mol, ΔS° in J·mol⁻¹·K⁻¹. Convert one.
Celsius in TΔS°. Temperature must be in kelvin.
Saying “spontaneous means fast”. This is exactly the misconception the CED renamed the term to avoid.
Assuming exothermic means favored. A strongly negative ΔS° can override it at high temperature.

Worked example

For the vaporization of water, ΔH° = +40.7 kJ/mol and ΔS° = +109 J·mol⁻¹·K⁻¹. (a) Calculate ΔG° at 25 °C and state whether vaporization is favored. (b) Find the temperature at which ΔG° = 0 and interpret it.

(a) ΔG° at 298 K. Convert entropy first: ΔS° = 0.109 kJ·mol⁻¹·K⁻¹.

ΔG° = ΔH° − TΔS° = 40.7 − (298)(0.109)
ΔG° = 40.7 − 32.5 = +8.2 kJ/mol

Positive, so vaporization is not thermodynamically favored at 25 °C under standard conditions. This makes sense: water at 25 °C and 1 atm is a liquid.

(b) Crossover temperature.
0 = ΔH° − TΔS° → T = ΔH°/ΔS° = 40.7 / 0.109 = 373 K = 100 °C

Interpretation: this is precisely the normal boiling point of water. Below 373 K the enthalpy term dominates and condensation is favored; above 373 K the TΔS° term overtakes it and vaporization becomes favored. At exactly 373 K, ΔG° = 0 and liquid and vapor are in equilibrium — which is the thermodynamic definition of a boiling point.

This reaction sits in the ΔH° > 0, ΔS° > 0 row of the sign table: favored at high temperature.

Full notes for topic 9.3 →