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Gibbs Free Energy: The Sign That Decides Spontaneity

One equation, ΔG = ΔH - TΔS, quietly settles the tug-of-war between a reaction wanting to release energy and wanting to increase disorder.

mysimulator teamUpdated June 2026≈ 8 min read▶ Open the simulation

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
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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.

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