AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/9/9-1)
Unit 9 · Topic 9.1
9.1 Introduction to Entropy
Entropy, S, measures how spread out matter and energy are. You can predict the sign of an entropy change, ΔS, by asking whether particles get more freedom and space (melting, boiling, a gas expanding, more moles of gas) or whether energy is spread over a wider range of motions (higher temperature).
Key terms
- entropy (S)
- ΔS
- dispersal of matter
- dispersal of energy
- moles of gas
What entropy measures
Entropy describes how dispersed, or spread out, matter and energy are in a system. The more ways the particles and their energy can be arranged, the higher the entropy. ΔS is the change in entropy: positive when things become more spread out, negative when they become more concentrated or ordered.
You don't need to calculate entropy from scratch. You need to predict the sign of ΔS for a process and compare roughly how big different changes are.
Dispersal of matter
- Phase changes: solid → liquid → gas increases entropy, because particles become freer to move and usually take up more space. The reverse (condensing, freezing, deposition) decreases entropy.
- Gas expanding: at constant temperature, a gas that spreads into a larger volume has more room to move, so its entropy increases.
- Moles of gas in a reaction: if a reaction makes more moles of gas than it uses up, ΔS is usually positive. If it makes fewer, ΔS is usually negative. This is the most useful rule for reactions.
- Forming a precipitate (dissolved ions → solid) usually decreases entropy.
Dispersal of energy
Temperature matters too. As a substance warms, its particles have a wider range of kinetic energies; the Maxwell–Boltzmann distribution gets broader and flatter (topic 3.5). Energy is spread over more possibilities, so entropy increases with temperature even without a phase change.
Judging relative size
Gases have far more entropy than liquids or solids, so changes in the amount of gas dominate. Vaporizing a liquid or subliming a solid gives a much larger entropy increase than melting a solid, because the particles go from being in contact to being far apart.
When a reaction has the same number of moles of gas on both sides, such as N₂(g) + O₂(g) → 2NO(g), ΔS is small, and you can't reliably predict its sign without data.
Dissolving a solid usually increases entropy, but not always. When small, highly charged ions dissolve, water molecules arrange tightly around them, which can make ΔS negative (topic 9.6).
Quick reference
| Process | Sign of ΔS | Reason |
|---|---|---|
| Ice melting | Positive | Particles become freer to move |
| Steam condensing | Negative | Gas particles come together as a liquid |
| Gas expanding into a vacuum | Positive | More volume to spread into |
| 2H₂(g) + O₂(g) → 2H₂O(l) | Negative | 3 mol gas → 0 mol gas |
| Heating a solid | Positive | Wider range of particle energies |
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Predicting the sign of ΔS
Predict the sign of ΔS for each process and give a reason: (a) H₂O(l) → H₂O(g); (b) 2SO₂(g) + O₂(g) → 2SO₃(g); (c) CaCO₃(s) → CaO(s) + CO₂(g); (d) Ag⁺(aq) + Cl⁻(aq) → AgCl(s); (e) a sample of N₂ gas cooled from 50 °C to 25 °C at constant volume.
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- Step 1: (a) Positive: liquid → gas, so the molecules become far more dispersed.
- Step 2: (b) Negative: 3 mol of gas become 2 mol of gas, so matter becomes less dispersed.
- Step 3: (c) Positive: a solid produces a gas (0 → 1 mol gas).
- Step 4: (d) Negative: freely moving dissolved ions become a solid with fixed positions.
- Step 5: (e) Negative: at a lower temperature, the particles' kinetic energies are spread over a narrower range, so energy is less dispersed.
Answer: (a) +; (b) −; (c) +; (d) −; (e) −
- Example 2
Trap: when gas moles don't change
A student says ΔS for N₂(g) + O₂(g) → 2NO(g) must be exactly zero because there are 2 mol of gas on each side. Evaluate the claim.
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- Step 1: Equal moles of gas means the moles-of-gas rule gives no clear prediction, so ΔS is expected to be small.
- Step 2: Small isn't zero. Different gases have different standard entropies, so the products and reactants won't match exactly.
- Step 3: Using table values, ΔS° = 2(210.8) − 191.6 − 205.2 = +24.8 J/K, a small positive value, much smaller than for reactions that change the number of moles of gas.
Answer: Not exactly zero: ΔS is small (about +25 J/K from table data) and its sign can't be predicted reliably from moles of gas alone.
Common mistakes
- Counting all moles instead of moles of gas. Solids, liquids and dissolved species matter much less.
- Assuming every dissolving process has a positive ΔS.
- Saying a reaction with equal moles of gas on both sides has ΔS = 0. It's small, not zero.
- Forgetting that heating a substance (with no phase change) also increases its entropy.
On the exam
- Expect 'predict the sign of ΔS and justify' questions. A full justification names what becomes more or less dispersed, such as 'the number of moles of gas decreases from 3 to 2'.
- Sign of ΔS is often the first step in a longer question about thermodynamic favorability (topic 9.3).
Connected topics
Videos
Check yourself
4 questions on 9.1 Introduction to Entropy. Pick an answer to see if you got it, and why.
For which of the following processes is ΔS° most clearly positive?
Which of the following best explains why ΔS is positive when liquid bromine evaporates, Br₂(l) → Br₂(g)?
What is the sign of ΔS for CO₂(g) → CO₂(aq), the dissolving of carbon dioxide gas in water, and why?
For which of the following processes is ΔS negative?
0 of 4 answered