Entropy and spontaneity

A measure of how **dispersed** (spread out) the matter and energy of a system are — more ways to arrange the particles and energy means higher entropy.

When matter/energy become more dispersed: solid → liquid → gas, dissolving, and especially when the **moles of gas increase**.

When the system becomes more ordered: gas → liquid → solid, or when the **moles of gas decrease**.

The change in the **number of moles of gas** — gases have far higher entropy than liquids or solids.

The entropy of **one mole** of a substance under standard conditions; units **J K⁻¹ mol⁻¹**.

**J K⁻¹ mol⁻¹** — note joules, not kilojoules (unlike ΔH).

**gas > liquid > solid** — a gas has the most dispersed matter and energy.

**S = 0** — a single perfectly ordered arrangement with no thermal motion (the only zero-entropy state).

$\Delta S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants})$, each S° × its coefficient.

Ignoring the **stoichiometric coefficients**, and mixing **J** (entropy) with **kJ** (enthalpy).

**Always positive** for any substance above 0 K (entropy is a measure of dispersal, never negative).

Gas particles can occupy far more positions and share energy over more ways of moving, so there are many more arrangements → higher entropy.

$\Delta G = \Delta H - T\,\Delta S$ — it combines the enthalpy change and the entropy change into one test for spontaneity. (Given in the data booklet.)

When **ΔG < 0**. ΔG > 0 is non-spontaneous; ΔG = 0 is at equilibrium.

**No** — spontaneous means thermodynamically **feasible**. Speed is decided by **kinetics** (activation energy), not by ΔG.

ΔH is in **kJ** mol⁻¹ but ΔS is in **J** K⁻¹ mol⁻¹ — convert ΔS to kJ by **dividing by 1000** before combining.

**Kelvin** (K = °C + 273) — never degrees Celsius.

**Always spontaneous** (at every temperature) — ΔG is negative at all T.

**Never spontaneous** — ΔG is positive at all T.

Spontaneous only at **low** temperature (the +TΔS term wins once T is large).

Spontaneous only at **high** temperature (the −TΔS term wins once T is large).

Set **ΔG = 0**, so $T = \dfrac{\Delta H}{\Delta S}$ — use matching units (both in kJ) so T comes out in K.

**Same** signs on ΔH and ΔS ⇒ **temperature decides**. **Opposite** signs ⇒ same answer at every temperature.

A large increase in **disorder** (ΔS > 0) makes −TΔS very negative, so ΔG can be negative even when ΔH > 0 (e.g. dissolving ammonium nitrate).