The shape of the lesson
By the end of the lesson, learners can
- Predicting the sign: Compare the before and after particle pictures in the simulator and watch how freely the particles move.
- Calculating Δ S ° from data: Work through these with the standard entropy values given.
- Solid < liquid < gas for entropy.
- More moles of gas: ΔS positive.
- ΔS° = ΣS°(products) − ΣS°(reactants), times coefficients.
- Units: J K−1 mol−1; elements are not zero.
One picture every idea returns to
Entropy, S, is a measure of the dispersal of energy in a system: the more ways the particles and their energy can be arranged, the greater the entropy. Its units are J K−1 mol−1, not kJ. Entropy increases from solid to liquid to gas (ice 48, water 70, steam 189), when temperature rises, when a solid dissolves, and when a reaction produces more moles of gas. A positive ΔS means energy becomes more dispersed. The standard entropy change of a reaction is ΔS° = ΣS°(products) − ΣS°(reactants), using the stoichiometric coefficients; for CaCO3 → CaO + CO2 it is 40 + 214 − 93 = +161 J K−1 mol−1.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Energy likes to spread out.. Entropy measures how many ways the particles and their energy can be arranged. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Predicting the sign | Compare the before and after particle pictures in the simulator and watch how freely the particles move. Project tab 5 of the lab (Chemical energetics (A Level) lab, entropy, in particles) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Calculating Δ S ° from data | Work through these with the standard entropy values given. Project tab 5 of the lab (Chemical energetics (A Level) lab, entropy, in particles) and let a student drive it while the class predicts. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Calculating Δ S ° from data, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Entropy values are in J K −1 mol −1, not kJ. | Elements do not have zero entropy, so O 2, H 2 and N 2 must be included. Multiply by the coefficients, and always do products minus reactants. The state symbol matters: H 2 O(l) is 70 but H 2 O(g) is 189. |
Entropy change: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
State what is meant by entropy, and give its units.
Predict the sign of ΔS for each change and explain: (a) I2(s) → I2(g); (b) 2SO2(g) + O2(g) → 2SO3(g); (c) KNO3(s) dissolving in water; (d) cooling water from 50 °C to 20 °C.
Calculate ΔS° for N2(g) + 3H2(g) → 2NH3(g). S° / J K−1 mol−1: N2 192, H2 131, NH3 193.
Calculate ΔS° for 2Mg(s) + O2(g) → 2MgO(s). S° / J K−1 mol−1: Mg 33, O2 205, MgO 27.
The entropy change for C(graphite) + O2(g) → CO2(g) is only +3.3 J K−1 mol−1. Explain why it is so small.
The standard entropy of H2O(l) is 70 and of H2O(g) is 189 J K−1 mol−1. Calculate ΔS for boiling 1 mol of water and explain the sign.
Answer key and marking notes
- Entropy is a measure of the dispersal of energy in a system: the number of ways the particles and their energy can be arranged.
- Units: J K−1 mol−1.
- (a) Positive: solid to gas, far more freedom of movement.
- (b) Negative: 3 mol of gas become 2 mol of gas.
- (c) Positive: ions leave the ordered lattice and spread through the solution.
- (d) Negative: at lower temperature the energy is spread over fewer energy levels.
- ΣS°(products) = 2 × 193 = 386; ΣS°(reactants) = 192 + 3 × 131 = 585.
- ΔS° = 386 − 585 = −199 J K−1 mol−1.
- Products: 2 × 27 = 54; reactants: 2 × 33 + 205 = 271.
- ΔS° = 54 − 271 = −217 J K−1 mol−1.
- There is 1 mol of gas on each side, so the number of moles of gas does not change.
- Graphite has a very small entropy (5.7), so the only change is between two similar gases (214 − 205).
- ΔS = 189 − 70 = +119 J K−1 mol−1.
- Positive: gas particles move freely and randomly in a far larger volume, so energy is much more dispersed.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What are the units of standard entropy, S°?
- AJ K−1 mol−1
- BkJ mol−1
- CJ mol−1
- DkJ K−1
Which reaction has a positive entropy change?
- ACaCO3(s) → CaO(s) + CO2(g)
- BN2(g) + 3H2(g) → 2NH3(g)
- C2Mg(s) + O2(g) → 2MgO(s)
- DH2O(g) → H2O(l)
Which list puts these substances in order of INCREASING standard entropy?
- AC(diamond), H2O(l), CO2(g)
- BCO2(g), H2O(l), C(diamond)
- CH2O(l), C(diamond), CO2(g)
- DC(diamond), CO2(g), H2O(l)
Calculate ΔS° for 2SO2(g) + O2(g) → 2SO3(g). S° / J K−1 mol−1: SO2 248, O2 205, SO3 257.
- A−187 J K−1 mol−1
- B−196 J K−1 mol−1
- C+187 J K−1 mol−1
- D+18 J K−1 mol−1
Which statement about entropy is correct?
- Athe entropy of a substance increases as its temperature rises
- Belements in their standard states have zero standard entropy
- Ca solid has a higher entropy than the same substance as a gas
- Dstandard entropies are measured in kJ mol−1
Calculate ΔS° for 2H2O2(l) → 2H2O(l) + O2(g). S° / J K−1 mol−1: H2O2(l) 110, H2O(l) 70, O2(g) 205.
- A+125 J K−1 mol−1
- B+165 J K−1 mol−1
- C−125 J K−1 mol−1
- D+363 J K−1 mol−1
Which reaction would have an entropy change closest to zero?
- AH2(g) + Cl2(g) → 2HCl(g)
- B2H2(g) + O2(g) → 2H2O(l)
- CNH4Cl(s) → NH3(g) + HCl(g)
- DC2H4(g) + H2(g) → C2H6(g)
For N2(g) + O2(g) → 2NO(g), ΔS° = +25 J K−1 mol−1. S°(N2) = 192 and S°(O2) = 205 J K−1 mol−1. What is S°(NO)?
- A211 J K−1 mol−1
- B422 J K−1 mol−1
- C186 J K−1 mol−1
- D372 J K−1 mol−1
Quiz key
Standard entropies are quoted in J K−1 mol−1, which is why they must be divided by 1000 before use with enthalpy changes in kJ. kJ mol−1 is the unit of enthalpy change. J mol−1 leaves out the per-kelvin part. kJ K−1 uses the wrong energy unit and leaves out per mole.
Decomposing CaCO3 makes a gas from a solid (0 to 1 mol of gas), so energy becomes more dispersed and ΔS = +161 J K−1 mol−1. The Haber reaction turns 4 mol of gas into 2 mol, so ΔS is negative (−199). Burning magnesium uses up a mole of gas and forms only solid (−217). Condensing steam to liquid water lowers the entropy (−119).
Diamond is a rigid giant structure with very few ways to arrange its energy (S° = 2.4), liquid water is higher (70) and gaseous CO2 highest (214) J K−1 mol−1. The second list is decreasing order. The third puts a liquid below a solid. The fourth puts the gas below the liquid.
ΔS° = 2(257) − [2(248) + 205] = 514 − 701 = −187 J K−1 mol−1; negative because 3 mol of gas become 2 mol. −196 ignores the coefficients (257 − 248 − 205). +187 is reactants minus products. +18 leaves out O2 (514 − 496), as if elements had zero entropy.
Raising the temperature gives the particles more energy, spread over more energy levels, so entropy increases. Elements have positive standard entropies (O2 is 205 J K−1 mol−1); it is ΔHf that is zero for elements. Gases have far higher entropy than solids. Entropy units are J K−1 mol−1.
ΔS° = [2(70) + 205] − 2(110) = 345 − 220 = +125 J K−1 mol−1; positive because a gas forms from liquids. +165 ignores the coefficients (70 + 205 − 110). −125 is reactants minus products. +363 uses the entropy of steam (189) instead of liquid water.
H2 + Cl2 → 2HCl has 2 mol of gas on each side, so ΔS is small (about +20 J K−1 mol−1). Forming liquid water from 3 mol of gas is strongly negative. Decomposing a solid into 2 mol of gas is strongly positive. Adding hydrogen to ethene turns 2 mol of gas into 1 mol, so ΔS is clearly negative.
2S°(NO) = 25 + 192 + 205 = 422, so S°(NO) = 211 J K−1 mol−1. 422 forgets to divide by the coefficient 2. 186 subtracts ΔS instead of adding it ((397 − 25) ÷ 2). 372 subtracts ΔS and also forgets to halve.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.