With microstates in mind, you can predict whether a process raises or lowers entropy without any numbers. More freedom of motion and more ways to share energy mean more microstates.
Key idea
$\Delta S > 0$ when the moles of gas increase, when a solid melts or a liquid boils, when a solid dissolves (usually), when a substance warms, and when gases or liquids mix. The biggest effect by far is a change in the moles of gas.
Why it works
Gas molecules range over the whole container, so a gas has vastly more positions available than a liquid, and a liquid more than a solid. A warmer substance has more energy to share among its particles, so more ways to share it. Dissolving breaks a crystal's order and lets ions move through the solvent (although small, highly charged ions can order the water around them enough to make $\Delta S$ negative).
Worked examples
Worked example: Count only the gas
Predict the sign of $\Delta S$ for $\mathrm{2H_2(g) + O_2(g) \to 2H_2O(l)}$.
- Moles of gas: 3 on the left, 0 on the right.
- Three moles of gas become a liquid: a large drop in entropy. $\Delta S < 0$.
Answer: Negative.
Worked example: Four quick predictions
Give the sign of $\Delta S$: (a) $\mathrm{CaCO_3(s) \to CaO(s) + CO_2(g)}$; (b) $\mathrm{N_2(g) + 3H_2(g) \to 2NH_3(g)}$; (c) $\mathrm{2H_2O_2(l) \to 2H_2O(l) + O_2(g)}$; (d) $\mathrm{H_2O(l)}$ at 20 °C warmed to 60 °C.
- (a) A gas forms from a solid: positive.
- (b) 4 mol of gas become 2: negative.
- (c) A gas forms from liquids: positive. (d) Warming: positive.
Answer: (a) +, (b) −, (c) +, (d) +.
| Change | Sign of ΔS | Reason |
|---|---|---|
| more moles of gas | + | many more positions |
| solid → liquid → gas | + | more freedom of motion |
| dissolving a solid | usually + | ions or molecules spread out |
| raising T | + | more ways to share energy |
| gas dissolving in water | − | gas loses its freedom |
When the moles of gas do not change, the sign is harder to call by eye, and $\Delta S$ is usually small. Then you need the tabulated values of the next lesson.
Common mistake
Counting all the moles in an equation instead of only the moles of gas. In $\mathrm{C(s) + O_2(g) \to CO_2(g)}$ the moles go from 2 to 1, but the moles of gas stay at 1, and $\Delta S$ is in fact slightly positive. Check: tally (g) species only.