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.
The lattice energy for an ionic compound, ΔlattH⦵, is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions (298K, 101 kPa).
Lattice energies by this definition (see note at the end of this post) are always exothermic as this the lattice energy for the formation of an ionic lattice. The more exothermic, the stronger the electrostatic attractions / ionic bonds between the oppositely charged ions in the lattice.
e.g. Mg2+(g) + O2-(g) ⇾ MgO(s) ΔlattH⦵ = -3842 kJ mol-1
A Born-Haber cycle combines a number of experimental values to determine the theoretical value for the lattice energy for an ionic compound, since a lattice energy cannot be measured directly itself.
We can break the Born-Haber cycle down into a number of steps although it makes more sense to think of this cycle as an energy level diagram:
- Atomise the metal element in the ionic compound
Mg(s) ⇾ Mg(g) ΔatH⦵ = +148 kJ mol-1
The enthalpy of atomisation, ΔatH⦵, is the enthalpy change when 1 mole of gaseous atoms is formed from the element under standard conditions.
- Ionise the gaseous metal atoms to form cations – this is done one electron at a time for ions with a 2+ or 3+ charge.
Mg(g) ⇾ Mg+(g) + e– 1st ionisation enthalpy, ΔieH⦵ = +736 kJ mol-1
Mg+(g) ⇾ Mg2+(g) + e– 2nd ionisation enthalpy, ΔieH⦵ = +1450 kJ mol-1
- Atomise the non-metal element to form gaseous atoms – be careful to follow the definition when writing equations, we are forming 1 mole of atoms, NOT starting with 1 mole of the element.
½O2(g) ⇾ O(g) ΔatH⦵ = +249 kJ mol-1
- Ionise the non-metal gaseous atoms to form anions – once again we add one electron at a time from anions with 2- or 3- charge.
O(g) + e– ⇾ O–(g) 1st electron affinity, ΔeaH⦵ = -141 kJ mol-1
O–(g) + e– ⇾ O2-(g) 2nd electron affinity, ΔeaH⦵ = +798 kJ mol-1
The first electron affinity, ΔeaH⦵, is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous ions to from 1 mole of gaseous ions with a single negative charge under c standard conditions
- Find a value for the enthalpy of formation of our ionic compound.
Mg(s) + ½O2(g) ⇾ MgO(s) ΔfH⦵ = -602 kJ mol-1
- Pull all of the above into an energy level diagram and calculate the value of the lattice energy of formation of the 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.
VERY IMPORTANT NOTE:
You should be aware that technically the lattice energy or lattice enthalpy is both the energy needed to form gaseous ions from one mole of a solid ionic compound / ionic lattice AND the energy released when one mole of a solid ionic compound / ionic lattice is formed from its ions in the gaseous state.
And herein lies much confusion … text books can use the term to mean both!
Clearly forming a lattice is highly exothermic as millions of strong electrostatic attractions pull the gaseous ions into a regular crystalline arrangement. Breaking up a lattice to from gaseous ions must be an highly endothermic process.
Na+(g) + Cl–(g) ⇾ NaCl(s) ΔH⦵ = -787 kJ mol-1
NaCl(s) ⇾ Na+(g) + Cl– (g) ΔH⦵ = +787 kJ mol-1
How do we tell which version of lattice energy a question or text book is talking about?
- Check the sign – if it is negative then we are forming a lattice and if it is positive we are breaking it apart
- Text books and exam questions often talk in terms of ‘enthalpy of formation of a lattice’ which equates to the exothermic process
It doesn’t actually make any difference to our explanations or our calculations as long as we all agree on a definition before beginning. Just be aware!
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.Â
    (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
1.
2.
3.
4.
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