Born-Haber cycles are simply a fancy Hess cycle transposed onto an energy level diagram and we can use them to calculate the lattice energy for an ionic compound.
They might seem a little complicated when you first meet them but if you employ your logic skills, you will have them mastered in no time at all.
Practice questions
- Draw a Born-Haber cycle for potassium bromide and use it to calculate the lattice energy, ΔLEH⦵, for the process KBr(s)  ⇾ K+(g)  +  Br–(g)  .
| ΔatH⦵ /kJ mol-1 | ΔIEH⦵ /kJ mol-1 | ΔEAH⦵ /kJ mol-1 | ΔfH⦵ /kJ mol-1 | |
| Potassium | +89 | +420 | – | – |
| Bromine | +112 | – | -342 | – |
| Potassium bromide | – | – | – | -392 |
2. Draw a Born-Haber cycle for sodium hydride and use it to calculate the first electron affinity for hydrogen.
| ΔatH⦵ /kJ mol-1 | ΔIEH⦵ /kJ mol-1 | ΔfH⦵ /kJ mol-1 | ΔLEH⦵ /kJ mol-1 | |
| Sodium | +108 | +500 | – | – |
| Hydrogen | +218 | – | – | – |
| Sodium hydride | – | – | -57 | +811 |
3. Draw a Born-Haber cycle for barium chloride and use it to calculate the enthalpy of lattice formation for the compound.Â
| ΔatH⦵ /kJ mol-1 | 1st ΔIEH⦵ /kJ mol-1 | 2nd ΔIEH⦵ /kJ mol-1 | ΔEAH⦵ /kJ mol-1 | ΔfH⦵ /kJ mol-1 | |
| Barium | +175 | +500 | +1000 | – | – |
| Chlorine | +112 | – | – | -364 | – |
| Barium chloride | – | – | – | – | -860 |
4. Complete a Born-Haber cycle for strontium oxide using the template below, and use it to calculate the 2nd electron affinity for oxygen.
| ΔatH⦵ /kJ mol-1 | 1st ΔIEH⦵ /kJ mol-1 | 2nd ΔIEH⦵ /kJ mol-1 | 1st ΔEAH⦵ /kJ mol-1 | ΔfH⦵ / kJ mol-1 | ΔLEH⦵ /kJ mol-1 | |
| Strontium | +164 | +500 | +1100 | – | – | – |
| Oxygen | +249 | – | – | -141 | – | – |
| Strontium oxide | – | – | – | – | -590 | +3303 |
5. (a) Define the term ‘enthalpy of lattice formation’ and write an equation to represent the enthalpy of lattice formation of silver iodide.
(b) The lattice formation energy for silver iodide is -869 kJ mol-1 which is determined using experimental data. A calculation for the same term based on a perfect ionic lattice model gives a less exothermic value. Explain this difference.
Answers
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5. (a) Â This is the enthalpy change when one mole of a solid ionic compound / ionic lattice is formed from its gaseous ions.Â
Ag+(g)  +  I–(g)  ⇾  AgI(s)
    (b)  The experimental value is more exothermic indicating that in reality the bonds in silver iodide are stronger than the model predicts , which is due to significant covalent character in the bonding.Â