Entropy, Gibbs Energy & Spontaneity

Why some reactions run on their own and others never do — and the single equation that decides.

High schoolIntro Gen ChemUni Year 1
⏱️ About 22 min

An ice cube melts in your hand even though melting <em>absorbs</em> heat — an endothermic change happening all on its own. If energy release were the whole story, that couldn't happen. Something besides enthalpy is pulling the strings: entropy.

💡
The big idea: Whether a reaction is spontaneous is decided by two things: the enthalpy change (ΔH) and the change in disorder (entropy, ΔS), balanced by temperature. Gibbs free energy combines them: ΔG = ΔH − TΔS. If ΔG is negative, the reaction is spontaneous.
🎯 By the end, you'll be able to
  • Describe entropy as the dispersal of energy and matter (disorder)
  • Predict the sign of ΔS for common changes (melting, boiling, gas produced)
  • Use ΔG = ΔH − TΔS to judge spontaneity, with temperature in kelvin
  • Explain how ΔH and ΔS signs combine to make a reaction spontaneous or not
📎 Helpful to know first

Entropy: nature's tendency to spread out

Entropy (S) measures how spread out the energy and matter of a system are — loosely, its disorder. A tidy crystal has low entropy; the same substance as a gas, molecules flying everywhere, has high entropy. Nature overwhelmingly drifts toward higher entropy, simply because there are vastly more spread-out arrangements than tidy ones.

Entropy generally increases when a solid melts to a liquid, a liquid boils to a gas, a solid dissolves, or a reaction produces more gas molecules than it consumes. Those changes have ΔS > 0.

✨ Spontaneous ≠ fast
A spontaneous change is one that happens on its own without a continuous outside push — it says nothing about speed. Diamond turning to graphite is spontaneous, yet takes longer than the age of the Earth. Spontaneity is about direction (thermodynamics); speed is a separate question (kinetics).

Two drives, one decision

Reactions are pulled by two tendencies: toward lower enthalpy (releasing energy, ΔH < 0) and toward higher entropy (more dispersal, ΔS > 0). Sometimes they agree; often they conflict. Melting ice is uphill in enthalpy (it absorbs heat) but strongly downhill in entropy (liquid is more disordered than ice) — and above 0 °C the entropy drive wins.

Gibbs free energy (G) combines both drives into one number whose sign settles the matter:

\[ \Delta G = \Delta H - T\,\Delta S \]
ΔG = Gibbs free energy change; ΔH = enthalpy change; T = absolute temperature in kelvin (K); ΔS = entropy change. Keep ΔH and TΔS in the same energy units.
🔑 Reading ΔG
ΔG < 0: the reaction is spontaneous in the forward direction. ΔG > 0: non-spontaneous forward (it is spontaneous in reverse). ΔG = 0: the system is at equilibrium — no net drive either way.
⚠️ Watch your units and temperature
T must be in kelvin (K = °C + 273.15) — never Celsius. And ΔS is usually tabulated in J·K⁻¹ while ΔH is in kJ; convert one so both terms share units before subtracting. A stray factor of 1000, or a Celsius temperature, is the most common way this calculation goes wrong.

How the signs combine

Because ΔG = ΔH − TΔS, the four combinations of signs tell a clean story:

  • ΔH < 0, ΔS > 0 → ΔG < 0 always: spontaneous at all temperatures (both drives agree).
  • ΔH > 0, ΔS < 0 → ΔG > 0 always: never spontaneous (both drives oppose).
  • ΔH < 0, ΔS < 0 → spontaneous only at low T (enthalpy wins when T is small).
  • ΔH > 0, ΔS > 0 → spontaneous only at high T (the TΔS term wins when T is large — this is melting ice).
✨ The link to equilibrium (preview)
A negative ΔG means products are favoured, which is exactly what a large equilibrium constant K describes — so ΔG and K point the same way. The precise quantitative bridge between them is developed in the equilibrium module; here, just hold the qualitative link: more negative ΔG ⇒ reaction lies further toward products.
📝 Worked example: For the synthesis of ammonia, ΔH = −92.0 kJ and ΔS = −0.199 kJ·K⁻¹. Is it spontaneous at 298 K?
  1. Both ΔH and ΔS are already in kJ, and T = 298 K — good, no unit surprises.
  2. ΔG = ΔH − TΔS = (−92.0) − (298)(−0.199).
  3. Compute TΔS: (298)(−0.199) = −59.3 kJ, so −TΔS = +59.3 kJ.
  4. ΔG = −92.0 + 59.3 = −32.7 kJ.
✓ ΔG ≈ −32.7 kJ < 0, so the reaction is spontaneous at 298 K. (Because ΔS < 0, higher temperatures would eventually make ΔG positive.)
✏️ Practice: For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ and ΔS = +0.161 kJ·K⁻¹. Calculate ΔG (in kJ) at 298 K.
kJ
Solution
  1. ΔG = ΔH − TΔS = (+178) − (298)(0.161).
  2. TΔS = 298 × 0.161 = 47.98 kJ.
  3. ΔG = 178 − 47.98 = +130.0 kJ.
  4. ΔG > 0, so at 298 K this decomposition is non-spontaneous (limestone is stable at room temperature).
✏️ Practice: For the same reaction (ΔH = +178 kJ, ΔS = +0.161 kJ·K⁻¹), estimate the temperature (in K) at which it becomes spontaneous — i.e. where ΔG = 0.
K
Solution
  1. At the crossover, ΔG = 0, so ΔH − TΔS = 0, which rearranges to T = ΔH ÷ ΔS.
  2. T = 178 ÷ 0.161.
  3. T ≈ 1106 K — above this temperature TΔS outweighs ΔH and the decomposition becomes spontaneous (why kilns run so hot).

Check your understanding

1. A reaction has ΔG = −45 kJ at a given temperature. The reaction is…
A negative ΔG means the forward reaction is spontaneous. ΔG > 0 would be non-spontaneous, and ΔG = 0 would be equilibrium.
2. A reaction with ΔH < 0 and ΔS > 0 is spontaneous at…
With ΔH negative and ΔS positive, ΔG = ΔH − TΔS is negative at every temperature (a negative minus a positive). Both drives agree, so it's always spontaneous.
3. Which change has the largest positive entropy change (ΔS > 0)?
Boiling turns an ordered liquid into a far more dispersed gas — a big increase in entropy. The other three all decrease disorder, so their ΔS is negative.
🔎 Related lessons
Previous: Bond Energies Energy, Heat & Work Collision Theory & Activation Energy
✅ Key takeaways
  • Entropy (S) measures the dispersal of energy and matter; nature drifts toward higher entropy.
  • ΔS > 0 for melting, boiling, dissolving, and reactions that make more gas.
  • Gibbs free energy: ΔG = ΔH − TΔS, with T in kelvin.
  • ΔG < 0 spontaneous; ΔG > 0 non-spontaneous; ΔG = 0 equilibrium.
  • The ΔH and ΔS signs together set whether — and at what temperatures — a reaction runs; a more negative ΔG points further toward products.
➡️ ΔG tells you a reaction <em>can</em> go and how far it settles — but not how fast. A reaction can be strongly spontaneous yet crawl for centuries. Speed is the domain of chemical kinetics: collisions, energy barriers, and catalysts, coming up in the next module.
Want to test yourself on this? Try the Chemistry practice test →