The shape of the lesson
By the end of the lesson, learners can
- The steps, defined: Match each enthalpy term to its equation in the simulator.
- Closing the cycle: Build the Born-Haber cycle step by step in the simulator.
- Charge and size: Compare the lattices in the simulator: smaller, more highly charged ions sit closer together and attract more strongly.
- LE: gaseous ions → 1 mol solid, always negative.
- Atomisation of X2: half the bond energy, per mole of atoms.
- 2+ ions: IE1 + IE2; 2− ions: EA1 + EA2 (EA2 positive).
- LE = ΔHf − (sum of the other steps).
One picture every idea returns to
The lattice energy is the enthalpy change when 1 mol of an ionic compound is formed from its gaseous ions under standard conditions, e.g. Na+(g) + Cl−(g) → NaCl(s); it is always exothermic. A Born-Haber cycle is a Hess cycle linking it to measurable steps: enthalpy change of atomisation of each element, ionisation energies of the metal, electron affinities of the non-metal and the enthalpy change of formation. For NaCl: LE = −411 − (+107 + 496 + 122 − 349) = −787 kJ mol−1. Lattice energy becomes more exothermic when the ions have higher charge and smaller radius, because the electrostatic attraction between them is stronger.
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, Build the lattice from gaseous ions.. Lattice energy cannot be measured directly, but a Born-Haber cycle finds it from steps you can measure: atomising the elements, ionising the metal, adding electrons to the non-metal, and the enthalpy change of formation. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The steps, defined | Match each enthalpy term to its equation in the simulator. Project tab 2 of the lab (Chemical energetics (A Level) lab, the steps, defined) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Closing the cycle | Build the Born-Haber cycle step by step in the simulator. Project tab 1 of the lab (Chemical energetics (A Level) lab, Born-Haber cycle builder) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Charge and size | Compare the lattices in the simulator: smaller, more highly charged ions sit closer together and attract more strongly. Project tab 3 of the lab (Chemical energetics (A Level) lab, charge and size) and let a student drive it while the class predicts. | Corners |
| 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, 16 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 Charge and size, 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.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
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 |
|---|---|
| Lattice energy goes from gaseous ions to the solid, so it is negative; do not reverse it. | Use ½ the bond energy for the atomisation of Cl or O, and remember MgO needs both IE 1 + IE 2 and EA 1 + EA 2, with EA 2 positive. For MgCl 2, the chlorine steps are doubled. |
Lattice energy and Born-Haber cycles: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Define (a) lattice energy, (b) first electron affinity, (c) enthalpy change of atomisation. Give an equation for each, using sodium chloride, chlorine and sodium.
Calculate the lattice energy of NaCl. Data (kJ mol−1): ΔHf(NaCl) −411; ΔHat(Na) +107; IE1(Na) +496; Cl–Cl bond energy +244; EA1(Cl) −349.
Calculate the lattice energy of MgCl2. Data (kJ mol−1): ΔHf(MgCl2) −641; ΔHat(Mg) +148; IE1(Mg) +736; IE2(Mg) +1450; ΔHat(Cl) +122; EA1(Cl) −349.
Explain why the second electron affinity of oxygen is positive, even though the first is negative.
The lattice energies of NaCl and MgO are −787 and −3841 kJ mol−1. Explain the difference.
Place KCl, NaCl and NaBr in order of increasingly exothermic lattice energy and explain.
Answer key and marking notes
- (a) The enthalpy change when 1 mol of an ionic compound is formed from its gaseous ions under standard conditions: Na+(g) + Cl−(g) → NaCl(s).
- (b) The enthalpy change when 1 mol of electrons is added to 1 mol of gaseous atoms to form 1 mol of gaseous 1− ions: Cl(g) + e− → Cl−(g).
- (c) The enthalpy change when 1 mol of gaseous atoms is formed from the element in its standard state: Na(s) → Na(g).
- ΔHat(Cl) = ½ × 244 = +122.
- Sum of the other steps = 107 + 496 + 122 − 349 = +376.
- LE = −411 − 376 = −787 kJ mol−1.
- Two Cl atoms are needed: 2 × 122 = +244 and 2 × (−349) = −698.
- Sum of the other steps = 148 + 736 + 1450 + 244 − 698 = +1880.
- LE = −641 − 1880 = −2521 kJ mol−1.
- The first electron is attracted by the nucleus of the neutral O atom, so energy is released.
- The second electron is added to a negative O− ion and is repelled, so energy must be supplied (+798 kJ mol−1).
- Mg2+ and O2− carry twice the charge of Na+ and Cl−.
- Mg2+ is smaller than Na+ and O2− is smaller than Cl−, so the ions are closer.
- Both give much stronger electrostatic attraction, so far more energy is released when the lattice forms.
- KCl (−718) < NaBr (−747) < NaCl (−787).
- All ions are singly charged, so the order follows ionic size: K+ is larger than Na+ and Br− larger than Cl−, and larger ions attract less strongly.
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.
Which equation represents the lattice energy of magnesium oxide?
- AMg2+(g) + O2−(g) → MgO(s)
- BMg(s) + ½O2(g) → MgO(s)
- CMgO(s) → Mg2+(g) + O2−(g)
- DMg(g) + O(g) → MgO(s)
Which equation represents the first electron affinity of oxygen?
- AO(g) + e− → O−(g)
- B½O2(g) + e− → O−(g)
- CO−(g) + e− → O2−(g)
- DO(g) → O+(g) + e−
Use the Born-Haber cycle for potassium chloride (values in kJ mol−1) to calculate its lattice energy.
- A−718 kJ mol−1
- B−156 kJ mol−1
- C+718 kJ mol−1
- D−1416 kJ mol−1
Which step in the Born-Haber cycle for magnesium oxide is endothermic AND involves gaining an electron?
- Athe second electron affinity of oxygen
- Bthe first electron affinity of oxygen
- Cthe lattice energy of magnesium oxide
- Dthe second ionisation energy of magnesium
Which compound has the LEAST exothermic lattice energy?
- AKBr
- BNaCl
- CLiF
- DMgO
Calculate the lattice energy of MgO from these data (kJ mol−1): ΔHf(MgO) −602; ΔHat(Mg) +148; IE1(Mg) +736; IE2(Mg) +1450; ΔHat(O) +248; EA1(O) −141; EA2(O) +798.
- A−3841 kJ mol−1
- B−3043 kJ mol−1
- C−2391 kJ mol−1
- D−2245 kJ mol−1
For calcium oxide (kJ mol−1): ΔHf −635; ΔHat(Ca) +178; IE1 + IE2(Ca) +1735; ΔHat(O) +248; EA1(O) −141; lattice energy −3453. What is the second electron affinity of oxygen?
- A+798 kJ mol−1
- B−798 kJ mol−1
- C+657 kJ mol−1
- D+550 kJ mol−1
The lattice energies of NaCl and NaBr are −787 and −747 kJ mol−1. Which is the most likely value for the lattice energy of NaF?
- A−923 kJ mol−1
- B−767 kJ mol−1
- C−704 kJ mol−1
- D−3800 kJ mol−1
Quiz key
Lattice energy is the enthalpy change when 1 mol of an ionic compound is formed from its gaseous ions under standard conditions, so it starts from Mg2+(g) and O2−(g) and is exothermic. Starting from Mg(s) and O2(g) is the enthalpy change of formation. Breaking the solid into gaseous ions is the reverse process (lattice dissociation), which is endothermic. Gaseous atoms are not ions, so the last equation leaves out the ionisation and electron affinity steps.
The first electron affinity is the enthalpy change when 1 mol of electrons is added to 1 mol of gaseous atoms to form 1 mol of gaseous 1− ions, so it starts from O(g). Starting from O2(g) would include atomisation as well. Adding an electron to O− is the second electron affinity (+798 kJ mol−1). Removing an electron is the first ionisation energy.
Going round the cycle: ΔHf = 89 + 122 + 419 + (−349) + LE, so LE = −437 − 281 = −718 kJ mol−1. −156 adds all the values together instead of subtracting the other steps from ΔHf. +718 has the wrong sign: forming a lattice from gaseous ions always releases energy. −1416 treats the electron affinity as +349 instead of −349.
O−(g) + e− → O2−(g) is +798 kJ mol−1: the incoming electron is repelled by the negative ion. The first electron affinity gains an electron but is exothermic (−141), because the electron is attracted by the nucleus of a neutral atom. Lattice energy is exothermic and involves no electron transfer. The second ionisation energy is endothermic but removes an electron.
All four are compared by charge and size. KBr has singly charged ions, and K+ and Br− are the largest ions here, so the attraction between them is weakest and the lattice energy is least exothermic. NaCl has smaller ions, LiF smaller still, so both are more exothermic. MgO has 2+ and 2− ions and is by far the most exothermic (about −3800 kJ mol−1).
Sum of the other steps = 148 + 736 + 1450 + 248 − 141 + 798 = +3239, so LE = −602 − 3239 = −3841 kJ mol−1. −3043 leaves out the second electron affinity, as if the oxide ion were O−. −2391 leaves out the second ionisation energy, as if the ion were Mg+. −2245 treats EA2 as −798, but adding an electron to O− is endothermic.
−635 = 178 + 1735 + 248 + (−141) + EA2 + (−3453), so EA2 = −635 − 178 − 1735 − 248 + 141 + 3453 = +798 kJ mol−1. −798 has the sign reversed; EA2 must be endothermic because the electron is repelled by O−. +657 leaves out EA1. +550 uses the whole O=O bond energy (496) instead of half for the atomisation of oxygen.
F− is smaller than Cl−, so the ions in NaF get closer and attract more strongly, so its lattice energy is more exothermic than −787; about −923 kJ mol−1 fits. −767 lies between NaCl and NaBr, which would mean F− is between Cl− and Br− in size. −704 is less exothermic than NaBr, which fits a larger ion such as I−. −3800 is the size expected for a 2+/2− compound like MgO, but NaF has only singly charged ions.
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.