The shape of the lesson
By the end of the lesson, learners can
- The solution cycle: Set up the triangle in the simulator: solid to gaseous ions (reverse of lattice energy), then gaseous ions to aqueous ions (hydration).
- Charge density and solubility: Switch the solution cycle in the simulator between the Group 2 sulfates and hydroxides, then explain the sulfate trend with the two competing energy terms.
- Hydration: gaseous ion → aqueous ion, always negative.
- Solution: solid → aqueous ions, small, either sign.
- ΔHsol = ΣΔHhyd − LE, counting every ion.
- High charge density (small, highly charged): more exothermic hydration.
One picture every idea returns to
The enthalpy change of hydration is the enthalpy change when 1 mol of gaseous ions dissolves in water, e.g. Na+(g) → Na+(aq); it is always exothermic because water dipoles are attracted to the ion. The enthalpy change of solution is the enthalpy change when 1 mol of an ionic solid dissolves in enough water to give an infinitely dilute solution, e.g. NaCl(s) → Na+(aq) + Cl−(aq). A Hess cycle through the gaseous ions gives LE + ΔHsol = ΣΔHhyd, so for NaCl ΔHsol = (−406 − 377) − (−787) = +4 kJ mol−1. Hydration is more exothermic for ions of higher charge and smaller radius (higher charge density).
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, Pull the lattice apart, let water wrap the ions.. Dissolving an ionic solid is a contest: the lattice energy holds the ions together, the hydration of each ion releases energy as water molecules cluster round it. 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: The solution cycle | Set up the triangle in the simulator: solid to gaseous ions (reverse of lattice energy), then gaseous ions to aqueous ions (hydration). Project tab 4 of the lab (Chemical energetics (A Level) lab, the solution cycle) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Charge density and solubility | Switch the solution cycle in the simulator between the Group 2 sulfates and hydroxides, then explain the sulfate trend with the two competing energy terms. Project tab 4 of the lab (Chemical energetics (A Level) lab, the solution cycle) and let a student drive it while the class predicts. | Round Table |
| 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, 15 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 density and solubility, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Hydration starts from gaseous ions, not the solid. | In the cycle, the solid goes to gaseous ions by the reverse of lattice energy, so Δ H sol = ΣΔ H hyd − LE; adding LE instead gives a hugely negative answer. Multiply each hydration enthalpy by the number of ions in the formula. |
Enthalpies of solution and hydration: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Define (a) enthalpy change of hydration and (b) enthalpy change of solution, with an equation for each using magnesium chloride.
Calculate ΔHsol(LiF). LE(LiF) = −1037; ΔHhyd(Li+) = −519; ΔHhyd(F−) = −506 (all kJ mol−1).
Calculate ΔHsol(MgCl2). LE = −2521; ΔHhyd(Mg2+) = −1920; ΔHhyd(Cl−) = −377 (all kJ mol−1).
NaF has ΔHsol = +1 kJ mol−1. ΔHhyd(Na+) = −406 and ΔHhyd(F−) = −506 kJ mol−1. Calculate the lattice energy of NaF.
Explain why ΔHhyd(Mg2+) is much more exothermic than ΔHhyd(Na+).
Explain, in terms of lattice energy and hydration enthalpy, why BaSO4 is much less soluble than MgSO4.
Answer key and marking notes
- (a) The enthalpy change when 1 mol of gaseous ions dissolves in water to give an infinitely dilute solution: Mg2+(g) → Mg2+(aq).
- (b) The enthalpy change when 1 mol of an ionic solid dissolves in water to give an infinitely dilute solution: MgCl2(s) → Mg2+(aq) + 2Cl−(aq).
- ΣΔHhyd = −519 + (−506) = −1025.
- ΔHsol = −1025 − (−1037) = +12 kJ mol−1.
- Two chloride ions: 2 × (−377) = −754.
- ΣΔHhyd = −1920 − 754 = −2674.
- ΔHsol = −2674 + 2521 = −153 kJ mol−1.
- LE = ΣΔHhyd − ΔHsol.
- LE = (−406 − 506) − 1 = −913 kJ mol−1.
- Mg2+ has twice the charge of Na+.
- Mg2+ also has a smaller ionic radius, so it has a much higher charge density.
- It attracts the δ− oxygen atoms of water molecules more strongly, so more energy is released.
- SO42− is a large ion, so the lattice energy becomes only slightly less exothermic from MgSO4 to BaSO4.
- Ba2+ is much larger than Mg2+, so its hydration enthalpy is much less exothermic.
- So ΔHsol becomes less exothermic (more endothermic) down the group and BaSO4 is less soluble.
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 enthalpy change of solution of potassium chloride?
- AKCl(s) → K+(aq) + Cl−(aq)
- BK+(g) + Cl−(g) → K+(aq) + Cl−(aq)
- CKCl(s) → K+(g) + Cl−(g)
- DK+(g) + Cl−(g) → KCl(s)
Which ion has the most exothermic enthalpy change of hydration?
- AAl3+
- BMg2+
- CNa+
- DK+
LE(NaBr) = −747 kJ mol−1; ΔHhyd(Na+) = −406 kJ mol−1; ΔHhyd(Br−) = −342 kJ mol−1. What is the enthalpy change of solution of NaBr?
- A−1 kJ mol−1
- B+1 kJ mol−1
- C−1495 kJ mol−1
- D+1495 kJ mol−1
Use the energy cycle to calculate the enthalpy change of solution of magnesium chloride.
- A−153 kJ mol−1
- B+224 kJ mol−1
- C+153 kJ mol−1
- D−5195 kJ mol−1
How are water molecules arranged around a hydrated Na+ ion, and why?
- Athe δ− oxygen atoms point towards the ion, because they are attracted to its positive charge
- Bthe δ+ hydrogen atoms point towards the ion, because they are attracted to its positive charge
- Cthe water molecules form covalent bonds to Na+ using their hydrogen atoms
- Dthe water molecules are arranged randomly, because Na+ has no effect on them
ΔHsol(LiCl) = −37 kJ mol−1; ΔHhyd(Li+) = −519 kJ mol−1; ΔHhyd(Cl−) = −377 kJ mol−1. What is the lattice energy of LiCl?
- A−859 kJ mol−1
- B−933 kJ mol−1
- C+859 kJ mol−1
- D−896 kJ mol−1
ΔHsol(MgCl2) = −153 kJ mol−1; LE(MgCl2) = −2521 kJ mol−1; ΔHhyd(Cl−) = −377 kJ mol−1. What is ΔHhyd(Mg2+)?
- A−1920 kJ mol−1
- B−2297 kJ mol−1
- C−2674 kJ mol−1
- D+3122 kJ mol−1
Why does the solubility of the Group 2 sulfates decrease from MgSO4 to BaSO4?
- Athe cation hydration enthalpy becomes much less exothermic, while the lattice energy changes only slightly because the sulfate ion is large
- Bthe lattice energy becomes much more exothermic down the group
- Cthe cation hydration enthalpy becomes more exothermic down the group
- Dthe sulfate ion becomes smaller down the group
Quiz key
The enthalpy change of solution is for 1 mol of the ionic solid dissolving to give an infinitely dilute solution, so it goes from KCl(s) to K+(aq) + Cl−(aq). Gaseous ions going into water is the sum of the two hydration enthalpies. Solid to gaseous ions is the reverse of lattice energy. Gaseous ions to solid is the lattice energy itself.
Al3+ has the highest charge and the smallest radius of these ions, so it has the highest charge density and attracts the δ− oxygen atoms of water most strongly (about −4690 kJ mol−1). Mg2+ (−1920) has a lower charge and is slightly larger. Na+ (−406) and K+ (−317) carry only a 1+ charge and are larger still, K+ being the least exothermic.
The solid is first broken into gaseous ions (+747, the reverse of LE), then the ions are hydrated (−406 − 342 = −748). ΔHsol = −748 − (−747) = −1 kJ mol−1. +1 is LE minus the hydration sum, the wrong way round. −1495 adds LE to the hydration enthalpies instead of subtracting it. +1495 has every sign reversed.
ΣΔHhyd = −1920 + 2(−377) = −2674. ΔHsol = −2674 − (−2521) = −153 kJ mol−1. +224 includes only one chloride ion. +153 has the sign reversed (LE minus the hydration sum). −5195 adds the lattice energy instead of subtracting it.
In water, oxygen is more electronegative, so it carries δ− and the hydrogens carry δ+. A cation attracts the δ− oxygen ends, and these ion-dipole attractions release energy, so hydration is exothermic. The δ+ hydrogen atoms point towards anions such as Cl−, not cations, and like charges would repel. No covalent bonds form through hydrogen. If water molecules were not attracted, there would be no hydration enthalpy.
LE = ΣΔHhyd − ΔHsol = (−519 − 377) − (−37) = −896 + 37 = −859 kJ mol−1. −933 adds ΔHsol instead of subtracting it. +859 has the wrong sign, but lattice energy is always exothermic. −896 is just the hydration sum, which would only equal LE if ΔHsol were zero.
ΣΔHhyd = −2521 + (−153) = −2674. Mg2+ = −2674 − 2(−377) = −2674 + 754 = −1920 kJ mol−1. −2297 subtracts only one chloride. −2674 is the hydration total for all three ions, with the two chlorides not removed. +3122 reverses the lattice energy sign; a hydration enthalpy can never be positive.
Lattice energy depends on the sum of the ionic radii; SO42− is so large that a bigger cation changes it only slightly. Hydration enthalpy depends on the cation alone, and falls sharply as the cation gets bigger. So ΔHsol becomes less exothermic down the group and solubility decreases. Lattice energy actually becomes less exothermic, not more. Larger cations have lower charge density, so hydration becomes less exothermic. The sulfate ion is the same in every compound.
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.