Why Ice Melts on Its Own Even Though Melting Absorbs Heat

Water flows downhill, a hot cup of chai cools, and an ink drop spreads through a glass of water — all on their own. It is tempting to think a process runs by itself only if it gives out heat, but melting ice absorbs heat and still happens every summer day.

The missing ingredient is entropy, and the quantity that combines it with enthalpy is Gibbs energy.

This part covers spontaneity, entropy and the second law, Gibbs energy, and the link between Gibbs energy and the equilibrium constant. Take $R=8.314$ J/mol K.

**A spontaneous process is one that can proceed on its own without continuous outside help; many spontaneous processes are exothermic, but some are endothermic, so a negative $\Delta H$ cannot by itself decide spontaneity.

Endothermic yet spontaneous:

- Melting ice** above $0$ °C, $\Delta H=+6.01$ kJ/mol
- Evaporation of water from a wet cloth
- Dissolving ammonium nitrate in water, which makes the solution colder

Worked example. One mole of ice melting at $25$ °C absorbs $6.01$ kJ from its surroundings — energy goes in, not out — yet the ice melts without any help.

Each of these endothermic processes produces more disorder: rigid ice becomes free-moving water, liquid becomes gas, and an ordered crystal breaks into dispersed ions.

An everyday example. Water poured on a hot terrace in summer evaporates on its own, absorbing heat from the floor.

The substance. Spontaneous does not mean fast — a spontaneous reaction may be far too slow to notice.

**Entropy $S$ measures the randomness or disorder of a system; for heat absorbed reversibly at temperature $T$, $\Delta S=\frac{q_{rev}}{T}$, and the second law states that the total entropy of the system plus surroundings increases in every spontaneous process.**

$$\Delta S_{total}=\Delta S_{system}+\Delta S_{surroundings}>0\text{(spontaneous)}$$

Trends: $S_{solid}<S_{liquid}<S_{gas}$; reactions that increase the number of gas molecules raise entropy.

**Worked example 1 — melting ice at $273$ K.**

$$\Delta S=\frac{6010}{273}\approx 22.0\text{ J/K mol}$$

**Worked example 2 — boiling water at $373$ K** ($\Delta H=40.66$ kJ/mol):

$$\Delta S=\frac{40660}{373}\approx 109.0\text{ J/K mol}$$

Worked example 3 — heat flowing downhill. $1000$ J passes from a body at $400$ K to one at $300$ K:

$$\Delta S_{total}=-\frac{1000}{400}+\frac{1000}{300}=-2.5+3.33\approx +0.83\text{ J/K}$$

Total entropy rises, so heat flows spontaneously from hot to cold.

An everyday example. Rangoli powder scattered by a gust of wind never gathers itself back into the pattern — disorder grows on its own.

The substance. A system's entropy can fall, as when water freezes below $0$ °C, provided the surroundings gain even more entropy.

**Gibbs energy is $G=H-TS$, and at constant temperature and pressure $\Delta G=\Delta H-T\Delta S$; a process is spontaneous when $\Delta G<0$, at equilibrium when $\Delta G=0$, and non-spontaneous when $\Delta G>0$, so a reaction changes over at $T=\frac{\Delta H}{\Delta S}$.

The four cases:

- $\Delta H<0$, $\Delta S>0$ — spontaneous at all temperatures
- $\Delta H>0$, $\Delta S<0$ — never spontaneous
- $\Delta H<0$, $\Delta S<0$ — spontaneous at low temperatures
- $\Delta H>0$, $\Delta S>0$ — spontaneous at high temperatures

Worked example 1 — limestone.** CaCO${}_3$ $\to$ CaO + CO${}_2$ has $\Delta H=+178.3$ kJ and $\Delta S=+160.4$ J/K.

$$\text{At }298\text{ K}:\Delta G=178.3-298\times 0.1604\approx +130.5\text{ kJ}\text{(not spontaneous)}$$

$$T=\frac{178.3\times 10^3}{160.4}\approx 1112\text{ K}$$

Above about $1112$ K, the decomposition becomes spontaneous.

Worked example 2 — ice. With $\Delta H=6010$ J/mol and $\Delta S=22.0$ J/K mol:

$$\Delta G_{298}=6010-298\times 22.0=-546\text{ J}\Delta G_{263}=6010-263\times 22.0=+224\text{ J}$$

Ice melts at $25$ °C but not at $-10$ °C.

An everyday example. A lime kiln roasts limestone at a very high temperature to make quicklime (chuna), because the reaction is spontaneous only when hot.

The substance. **Use the same energy unit for $\Delta H$ and $T\Delta S$** — converting $\Delta S$ to kJ/K is where most errors happen.

**The standard Gibbs energy change and the equilibrium constant are linked by $\Delta G^{\circ }=-RT\ln K=-2.303RT\log K$, so a negative $\Delta G^{\circ }$ means $K>1$ and products are favoured, while a positive $\Delta G^{\circ }$ means $K<1$.**

At $298$ K, $2.303RT=2.303\times 8.314\times 298\approx 5706$ J/mol.

**Worked example 1 — from $K$ to $\Delta G^{\circ }$.** $K=10$ at $298$ K:

$$\Delta G^{\circ }=-5706\times \log 10\approx -5.71\text{ kJ/mol}$$

**Worked example 2 — from $\Delta G^{\circ }$ to $K$.** $\Delta G^{\circ }=+10$ kJ/mol at $298$ K:

$$\log K=-\frac{10000}{5706}\approx -1.75\Rightarrow K\approx 0.018$$

Worked example 3 — ammonia. N${}_2$ + 3H${}_2$ $\rightleftharpoons$ 2NH${}_3$ has $\Delta G^{\circ }=-32.9$ kJ at $298$ K:

$$\log K=\frac{32900}{5706}\approx 5.77\Rightarrow K\approx 5.8\times 10^5$$

An everyday example. Fertiliser plants making ammonia rely on an equilibrium that thermodynamics says favours the product at room temperature, even though the reaction there is extremely slow.

The substance. **A large $K$ says nothing about speed** — catalysts and higher temperatures are needed to reach equilibrium quickly.