Two competing tendencies, folded into one number
Every process in nature is pulled by two tendencies at once: systems tend toward lower energy (favoring negative ΔH, exothermic change) and toward higher disorder (favoring positive ΔS, increasing entropy). These two tendencies don't always agree, and reactions routinely have to trade one off against the other. Josiah Willard Gibbs combined them into a single quantity that settles the argument by weighting entropy's contribution by temperature:
ΔG = ΔH - TΔS ΔG < 0 → spontaneous (thermodynamically favored, proceeds forward) ΔG > 0 → non-spontaneous (favored in reverse) ΔG = 0 → system is at equilibrium
The four classic cases
Because ΔG mixes a temperature-independent term (ΔH) and a temperature-scaled term (TΔS), the sign of ΔG can depend entirely on temperature — and there are exactly four combinations of sign(ΔH) and sign(ΔS) worth memorizing:
ΔH < 0, ΔS > 0 → ΔG always negative ALWAYS spontaneous ΔH < 0, ΔS < 0 → ΔG negative at LOW T spontaneous only when cold ΔH > 0, ΔS > 0 → ΔG negative at HIGH T spontaneous only when hot ΔH > 0, ΔS < 0 → ΔG always positive NEVER spontaneous
Ice melting is the textbook case of row two flipped: ΔH is positive (heat must be absorbed to melt it) and ΔS is positive (a liquid is more disordered than a solid), so it's spontaneous only above the crossover temperature — 0°C at standard pressure, and not a coincidence: it's exactly where ΔG passes through zero.
Finding the crossover temperature
Whenever ΔH and ΔS have the same sign, there's a specific temperature at which ΔG switches sign. Setting ΔG = 0 and solving:
0 = ΔH - T·ΔS T = ΔH / ΔS ← the crossover / equilibrium temperature
This is the general form behind every phase-change temperature you already know — melting points, boiling points — each one is simply the T at which ΔG for that particular phase transition crosses zero.
From ΔG° to the equilibrium constant K
Gibbs free energy's real power in chemistry is that it links directly to how far a reaction proceeds before settling into equilibrium, through the relation ΔG° = -RT ln K, where R is the gas constant and K is the equilibrium constant:
ΔG° very negative → K >> 1 reaction runs essentially to completion ΔG° near zero → K ≈ 1 substantial amounts of both sides present ΔG° very positive → K << 1 reaction barely proceeds forward at all
Spontaneous does not mean fast
A common trap: ΔG < 0 says nothing about how quickly a reaction happens, only that it's thermodynamically favored to happen eventually. That's a separate question, kinetics, governed by activation energy and reaction mechanism — a topic ΔG is completely silent on. Diamond converting to graphite has a comfortably negative ΔG at room temperature and pressure, yet diamonds are, for practical human timescales, forever; the activation barrier is simply enormous.
Frequently asked questions
Does a spontaneous reaction (negative ΔG) always happen fast?
No — spontaneity and rate are unrelated questions. ΔG is a thermodynamic quantity: it tells you the reaction is thermodynamically favored and will eventually proceed toward equilibrium. Rate is a kinetic question governed by activation energy. Diamond turning into graphite has negative ΔG but happens so slowly it's effectively permanent at room temperature.
What does it mean for ΔG to equal exactly zero?
ΔG = 0 marks equilibrium — the forward and reverse processes are equally favored, so there's no net driving force in either direction. Setting ΔG = ΔH - TΔS = 0 and solving gives the crossover temperature T = ΔH/ΔS at which a temperature-dependent reaction switches from spontaneous to non-spontaneous, like ice melting exactly at 0°C.
How is ΔG related to the equilibrium constant K?
Through ΔG° = -RT ln K, connecting standard free energy directly to how far a reaction proceeds before reaching equilibrium. A large negative ΔG° gives a large K (products strongly favored); a large positive ΔG° gives a tiny K (reactants strongly favored); ΔG° = 0 gives K = 1.
Try it live
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