Spontaneous change and entropy

Changes that happen on their own

A drop of ink spreads through a glass of water. Hot tea cools in a cold room. Iron left in damp air rusts. Nobody has to push these changes along. Each is a spontaneous change: once it can start, it goes on by itself.

The reverse changes never happen by themselves. The ink does not gather back into a drop, and the tea does not warm up again. A spontaneous change has a direction.

"Spontaneous" says nothing about speed. Diamond turning into graphite is spontaneous, but far too slow to notice.

Is giving out heat enough?

Many spontaneous reactions are exothermic: burning, rusting, a metal dissolving in acid. It is tempting to think that a fall in enthalpy, $\Delta H < 0$, drives every spontaneous change.

But look at these:

• Ice melts in a warm room, although melting takes in heat.
• Ammonium chloride dissolves in water and the beaker turns cold. The change is endothermic, yet it happens on its own.
• Two gases in connected flasks mix completely, although $\Delta H$ is almost zero.

So a fall in enthalpy helps, but it cannot be the whole answer. Something else must count.

Entropy

That something is entropy, $S$. Entropy measures how spread out, or disordered, the particles and their energy are. The more ways the particles can be arranged, the higher the entropy. Like enthalpy, entropy is a state function: its change depends only on the start and the end.

You can often predict the sign of $\Delta S$:

• Melting and boiling: solid to liquid to gas, the particles move ever more freely. $\Delta S$ is positive.
• Dissolving a solid in a liquid: the ions or molecules spread through the solvent. $\Delta S$ is usually positive.
• A reaction that makes more moles of gas has $\Delta S > 0$. One that uses up gas has $\Delta S < 0$.
• Heating a substance raises its entropy, because the particles move faster in more ways.

Measuring an entropy change

When heat $q_{rev}$ flows into a system reversibly at temperature $T$, the entropy change is

• $\Delta S = \dfrac{q_{rev}}{T}$

The unit is J/(K mol). The same heat makes a bigger difference to a cold system than to a hot one.

Melting and boiling take place reversibly at a fixed temperature. So $q_{rev}$ is just the enthalpy change.

• Ice melts at $273\ \text{K}$ with $\Delta H_{fus} = 6.01\ \text{kJ/mol}$: $\Delta S = \dfrac{6010\ \text{J/mol}}{273\ \text{K}} = 22.0\ \text{J/(K mol)}$.
• Water boils at $373\ \text{K}$ with $\Delta H_{vap} = 40.7\ \text{kJ/mol}$: $\Delta S = \dfrac{40\,700}{373} = 109\ \text{J/(K mol)}$.

Boiling raises the entropy far more than melting. A gas is far more spread out than a liquid.

The second law

The entropy of the system can fall in a spontaneous change. Water freezes in a freezer, and ice is more ordered than water. What rises is the total entropy, of the system and its surroundings together:

• $\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0$ for every spontaneous change.

This is the second law of thermodynamics. At equilibrium, $\Delta S_{total} = 0$.

The surroundings receive the heat that the system gives out. At constant pressure that heat is $-\Delta H$, so

• $\Delta S_{surroundings} = \dfrac{-\Delta H}{T}$

Freezing water gives out heat, which raises the entropy of the surroundings. Below $273\ \text{K}$, that gain is larger than the loss inside the water, so freezing is spontaneous.

The third law and absolute entropies

As a pure crystal is cooled, its particles settle into one perfect arrangement. The third law of thermodynamics says that the entropy of a perfectly crystalline substance is zero at 0 K. This gives a starting point, so absolute entropies can be measured.

Tables list the standard molar entropy, $S^\circ$, of each substance at $298\ \text{K}$ and $1$ bar. Unlike $\Delta_f H^\circ$, $S^\circ$ is not zero for elements. For a reaction,

• $\Delta_r S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$

For $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$, with $S^\circ$ of $191.6$, $130.7$ and $192.5\ \text{J/(K mol)}$:

• $\Delta_r S^\circ = 2(192.5) - [191.6 + 3(130.7)] = 385.0 - 583.7 = -198.7\ \text{J/(K mol)}$

Four moles of gas become two, so the entropy falls, as expected.

Practise with new numbers

Try it in Python

The program works out the entropy change of melting ice and of boiling water from their enthalpy changes.

Common mistakes

• Thinking only exothermic changes are spontaneous. Melting ice and dissolving ammonium chloride take in heat and still happen.
• Mixing kJ and J. Enthalpies are usually in kJ/mol and entropies in J/(K mol). Change one before you combine them.
• Using degrees Celsius in $\Delta S = q_{rev}/T$. The temperature must be in kelvin.
• Looking only at the system. The second law is about the total entropy, system and surroundings together.
• Taking $S^\circ$ of an element as zero. That is true of $\Delta_f H^\circ$, not of the standard entropy.

Remember these

• Spontaneous: happens on its own; it says nothing about speed.
• Entropy measures how spread out matter and energy are. Gas $>$ liquid $>$ solid.
• $\Delta S = q_{rev}/T$; for melting and boiling, $\Delta S = \Delta H / T$.
• Second law: $\Delta S_{total} > 0$ for a spontaneous change, with $\Delta S_{surr} = -\Delta H/T$.
• Third law: a perfect crystal has $S = 0$ at $0\ \text{K}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Is $\Delta S$ positive or negative when steam condenses on a cold mirror?

2. Predict the sign of $\Delta S$ for $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$.

3. Why is boiling water's entropy change about five times its entropy change of melting?

4. Water freezing has $\Delta S_{system} < 0$. How can it be spontaneous at $263\ \text{K}$?

5. Why can absolute entropies be listed in tables, but not absolute enthalpies?