Why ΔG is the deciding number

Enthalpy tells you if a reaction releases heat. Entropy tells you if disorder increases. Neither alone tells you whether the reaction will actually happen. Gibbs free energy combines both into a single criterion: if ΔG < 0 at the temperature you care about, the reaction proceeds in the forward direction. If ΔG > 0, it doesn’t. If ΔG = 0, you’re sitting on the equilibrium line.

The master equation:

ΔG = ΔH − T · ΔS

with T in kelvin, ΔH and ΔS in matching energy units. Decoupling those units is the most common bench-side mistake — more on that in a moment.

The standard workflow

Step 1 — get ΔH for the reaction

ΔH°rxn = Σ n·ΔH°f(products) − Σ n·ΔH°f(reactants)

Elements in their reference states have ΔH°f = 0.

Example: 2 H₂(g) + O₂(g) → 2 H₂O(l)

ΔH = 2(−285.8) − [2(0) + 0] = −571.6 kJ/mol

Step 2 — get ΔS for the reaction

Same pattern with absolute molar entropies:

ΔS°rxn = Σ n·S°(products) − Σ n·S°(reactants)

S°[H₂O(l)] = 69.9, S°[H₂(g)] = 130.7, S°[O₂(g)] = 205.2 J/(mol·K)

ΔS = 2(69.9) − [2(130.7) + 205.2] = 139.8 − 466.6 = −326.8 J/(mol·K)

The negative sign makes physical sense — three moles of gas collapsed into two moles of liquid is a substantial drop in disorder.

Step 3 — fix the units before you plug in

ΔH is in kJ, ΔS is in J. The single most frequent error in this calculation is leaving them mismatched.

ΔS = −326.8 J/(mol·K) = −0.3268 kJ/(mol·K)

Step 4 — apply ΔG = ΔH − TΔS at the temperature of interest

At 298 K:

ΔG = −571.6 − (298)(−0.3268) ΔG = −571.6 − (−97.4) ΔG = −474.2 kJ/mol

Strongly negative. Hydrogen combustion is wildly spontaneous at room temperature, which matches every intuition you have about hydrogen-oxygen mixtures and ignition sources.

Shortcut: ΔG°f directly

When tabulated standard free energies of formation are available, you can skip the ΔH/ΔS round trip:

ΔG°rxn = Σ n·ΔG°f(products) − Σ n·ΔG°f(reactants)

Faster, but you lose the ability to ask “what temperature does this reaction become spontaneous?” That question requires the ΔH and ΔS pieces separately.

Spontaneity vs. temperature — the four cases

The interplay of signs determines how T changes the verdict:

ΔH	ΔS	Behavior
−	+	Spontaneous at all T
−	−	Spontaneous below T = ΔH/ΔS
+	+	Spontaneous above T = ΔH/ΔS
+	−	Never spontaneous

Crossover temperature

Set ΔG = 0 and solve:

T_crossover = ΔH / ΔS

Example — limestone calcination: CaCO₃(s) → CaO(s) + CO₂(g)

ΔH = +178.3 kJ/mol, ΔS = +160.5 J/(mol·K) = 0.1605 kJ/(mol·K)

T = 178.3 / 0.1605 = 1111 K (838 °C)

Below 1111 K the decomposition won’t proceed. Above it, it will. This is exactly why kilns producing quicklime run at red heat — the chemistry doesn’t budge until you get past the crossover.

ΔG° and the equilibrium constant

The bridge between thermodynamics and equilibrium:

ΔG° = −R · T · ln K

with R = 8.314 J/(mol·K). Rearranged:

K = exp(−ΔG° / RT)

Strongly negative ΔG° gives a huge K (products dominate). Strongly positive ΔG° gives a tiny K. ΔG° = 0 means K = 1.

Traps that bite students

Mixing kJ and J. ΔH is usually quoted in kJ, ΔS in J/(mol·K). Convert one before applying the formula. A factor of 1000 in either direction destroys the answer silently — you get a number that looks reasonable but is wrong.

Celsius for T. T in the equation is absolute. Always Kelvin. Add 273.15 to °C.

Standard vs. non-standard conditions. ΔG° applies at standard conditions (1 bar, 1 M, the specified T). Off-standard, use ΔG = ΔG° + RT ln Q.

Confusing spontaneous with fast. ΔG tells you whether a reaction can proceed, not whether it will in finite time. Diamond → graphite has ΔG < 0 at room temperature. Your engagement ring is safe for now.

Practice

Check with the Thermochemistry Calculator:

ΔG at 298 K for ΔH = −125 kJ/mol, ΔS = +45 J/(mol·K).
ΔG at 298 K for ΔH = +50.0 kJ/mol, ΔS = +200 J/(mol·K). Spontaneous?
Crossover T for MgCO₃ decomposition: ΔH = +100.6 kJ/mol, ΔS = +174.8 J/(mol·K).
K from ΔG° = −33.0 kJ/mol at 298 K.
N₂(g) + 3 H₂(g) → 2 NH₃(g): ΔH = −92.2 kJ/mol, ΔS = −198.7 J/(mol·K). Above what T does ammonia synthesis become non-spontaneous?

How to Calculate Gibbs Free Energy