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Gibbs free energy and favorability

Enthalpy and entropy each vote on whether a reaction happens, and they can disagree. Gibbs free energy is how the votes are tallied — one number whose sign decides favorability.

§1

ΔG = ΔH − TΔS.

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Gibbs free energy combines enthalpy and entropy into one criterion: ΔG = ΔH − TΔS. Its sign decides favorability — a negative ΔG means the process is thermodynamically favored (spontaneous), and a positive ΔG means it is not.

The two terms can cooperate or compete. When ΔH < 0 and ΔS > 0, ΔG is negative at all temperatures (always favored). When they oppose, temperature decides: the −TΔS term grows with T, so temperature can flip the sign at a crossover.

Watch the signs carefully. An exothermic reaction is not automatically favored — if it lowers entropy enough, the −TΔS term can make ΔG positive at high temperature.

UNIT 9 TOPIC 9.3 • GIBBS FREE ENERGY AND THERMODYNAMIC FAVORABILITY FAVORABILITY MAP AP Chemistry · Unit 9 · Applications of Thermodynamics Combines enthalpy and entropy to predict favorability at constant temperature and pressure. CORE EQUATION ΔG = ΔH − TΔS ΔG Gibbs free energy ΔH enthalpy change T temperature (K) ΔS entropy change WHAT IT MEANS ΔG tells you whether a process is thermodynamically favored (spontaneous) under the given conditions. ΔG < 0 → thermodynamically favored (spontaneous) KEY POINTS ΔH reflects enthalpic effects (bonds, attractions). ΔS reflects disorder / entropy (more disorder is +). T is always in kelvin (K). The larger the magnitude of −TΔS, the more thermodynamically favored the process. STANDARD ΔG OF REACTION ΔG°rxn = ΣΔG°f (products) − ΣΔG°f (reactants) Add standard free energies of formation (ΔG°f) to find ΔG°rxn for the whole reaction. ΔS > 0 entropy increases ΔS < 0 entropy decreases ΔH < 0 enthalpy decreases ΔH > 0 enthalpy increases I. ΔH < 0 , ΔS > 0 Favored at ALL temperatures ΔG < 0 for all T II. ΔH < 0 , ΔS < 0 Favored at LOW temperature only ΔG < 0 at low T III. ΔH > 0 , ΔS > 0 Favored at HIGH temperature only ΔG < 0 at high T IV. ΔH > 0 , ΔS < 0 NEVER favored at any temperature ΔG > 0 for all T SIGN OF ΔG ΔG < 0 → thermodynamically favored (spontaneous) ΔG = 0 → system at equilibrium ΔG > 0 → not favored (nonspontaneous) CED ANCHOR 9.3.A ΔG depends on the balance of enthalpy (ΔH) and entropy (ΔS), and how T scales the −TΔS term.
Fig. 9.3.1 ΔG = ΔH − TΔS combines enthalpy and entropy. A negative ΔG means the process is thermodynamically favored. When ΔH and ΔS oppose, temperature (through the −TΔS term) decides the sign.
§2

Deciding favorability.

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Combine both terms with correct signs.

  1. Get ΔH and ΔS. Note the sign of each; use consistent energy units (ΔS often in J, ΔH in kJ).
  2. Apply the minus sign. ΔG = ΔH − TΔS — the entropy term is subtracted, with T in kelvin.
  3. Read the sign of ΔG. Negative ΔG is favored; positive is not; zero is at equilibrium.
  4. Consider temperature. If ΔH and ΔS oppose, find whether T puts you past the crossover.
§3

The pieces you'll meet.

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One equation, both driving forces.

ΔG
Gibbs free energy
ΔH − TΔS; negative = favored.
ΔH
Enthalpy
Heat term; negative is exothermic.
TΔS
Entropy term
Subtracted; grows with temperature.
favored
Favored
ΔG < 0 (thermodynamically spontaneous).
cross
Crossover
Temperature where ΔG changes sign.
units
Units
Match ΔH (kJ) and ΔS (J) before combining.
§4

Worked example: is it favored?

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Data. A reaction has ΔH = −40 kJ and ΔS = +50 J·K⁻¹. Evaluate ΔG at 298 K.

Entropy term. TΔS = 298 × 50 J = 14,900 J = 14.9 kJ.

Combine. ΔG = ΔH − TΔS = −40 − (+14.9) = −54.9 kJ.

Verdict. ΔG < 0, so the reaction is thermodynamically favored at 298 K. Here both terms help; when they oppose, temperature can flip the outcome.

§5

Mistakes that cost real points.

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Pitfall · 01

"An exothermic reaction is always thermodynamically favored."

Favorability depends on ΔG = ΔH − TΔS, not on ΔH alone. An exothermic reaction that lowers entropy a lot can have a positive ΔG (especially at high temperature) and not be favored. Both terms and the temperature matter.

Fix. Judge favorability from the sign of ΔG, combining ΔH and −TΔS, not from ΔH alone.

Pitfall · 02

"ΔG = ΔH + TΔS."

The entropy term is subtracted: ΔG = ΔH − TΔS. Using a plus sign flips how entropy contributes and can reverse your conclusion about favorability. Keep the minus sign.

Fix. Use ΔG = ΔH − TΔS with the minus sign on the entropy term.

Pitfall · 03

"You can add ΔH in kJ and ΔS in J without converting."

ΔH is usually in kJ and ΔS in J per kelvin, so they must be put in the same units before combining. Mixing them (off by 1000) throws off ΔG by orders of magnitude.

Fix. Convert so ΔH and TΔS share units (e.g. both kJ) before subtracting.

§6

Skill Check.

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Ten scenarios. Pick the chips that match your answer, then check. A scenario marks complete the first time every part is right. Progress saves on this device.

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