Topic 12 · page 4 of 17

Why entropy changes

Entropy rises with changes of state to a gas, when an ionic solid dissolves, and when a reaction makes more moles of gas. Count the moles of gas first to predict the sign of the entropy change of the system.

12 min read12.5(i)12.5(ii)12.5(iii)

Read first: Entropy of substances

The system is the chemicals that are reacting or changing. Its entropy change is written $\Delta S_\text{system}$ . This page is about predicting whether $\Delta S_\text{system}$ is positive or negative, without a calculator. Page 5 shows how to calculate it.

Changes of state 12.5(i)

Going from solid to liquid to gas, the particles become freer to move and spread out, so $\Delta S_\text{system}$ is positive. The reverse changes, freezing and condensing, have negative $\Delta S_\text{system}$ .

\ce{H2O(l) -> H2O(g)} \qquad S^\ominus: 70 \to 189\ \text{J K}^{-1}\text{ mol}^{-1}

Boiling gives a much bigger increase than melting. Page 3 gave the reason.

Dissolving an ionic solid 12.5(ii)

Two things happen when an ionic solid dissolves in water, and they pull in opposite directions.

Ions leave an ordered lattice and disperse through the solution. This increases entropy.
Water molecules arrange themselves round each ion in hydration shells (see page 14). That ordering of water molecules decreases entropy.

For most salts the first effect wins and $\Delta S_\text{system}$ is positive. For ions that are small and highly charged, such as Mg²⁺ and Al³⁺, the ions hold their water molecules very tightly. The ordering of the water is so great that $\Delta S_\text{system}$ can be small or even negative.

Reactions with a change in moles 12.5(iii)

More moles of products than reactants means more particles to arrange, and more ways to arrange them. Gases count most, because gas particles are so much more dispersed than particles in solids or liquids.

Reaction	Moles of gas	$\Delta S_\text{system}$
$\ce{CaCO3(s) -> CaO(s) + CO2(g)}$	0 → 1	positive
$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$	4 → 2	negative
$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$	1 → 0	negative
$\ce{2H2(g) + O2(g) -> 2H2O(l)}$	3 → 0	negative
$\ce{N2O4(g) -> 2NO2(g)}$	1 → 2	positive

Moles of solid or liquid hardly matter beside moles of gas. When the number of moles of gas does not change, or there is no gas, look at changes of state, dissolving, and the total number of particles.

MethodPredicting the sign of the entropy change of the system

Count the moles of gas on each side. More gas in the products means positive. Less means negative.
If the moles of gas are equal or there is no gas, look for changes of state (solid to liquid to gas), dissolving, or a change in the number of particles.
Write the sign with its reason, in terms of dispersal: "more moles of gas, so more ways of arranging the particles and energy".

Check yourself

Give the sign of $\Delta S_\text{system}$ for each change, with a reason.

$\ce{H2O(l) -> H2O(g)}$
$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$
$\ce{NH4NO3(s) + aq -> NH4+(aq) + NO3-(aq)}$
$\ce{2H2(g) + O2(g) -> 2H2O(l)}$
$\ce{N2O4(g) -> 2NO2(g)}$
Suggest why $\Delta S_\text{system}$ for dissolving a salt of small, highly charged ions can be negative.

Answers cover these until you have tried the questions

Positive. Liquid to gas: the particles become much more dispersed.
Negative. 1 mol of gas becomes no gas, so the particles are less free and there are fewer ways of arranging them.
Positive. The ordered lattice breaks up and the ions disperse through the solution.
Negative. 3 mol of gas become liquid: far fewer ways of arranging the particles.
Positive. 1 mol of gas becomes 2 mol of gas, so more particles with more ways of arranging them and the energy.
Small, highly charged ions attract water molecules very strongly, so the water molecules become tightly ordered around the ions. This loss of entropy of the water can outweigh the gain in entropy of the ions leaving the lattice, so the overall $\Delta S_\text{system}$ can be negative.