Born-Haber Cycles

A special type of enthalpy cycle used to calculate the Lattice Enthalpy of ionic compounds.

Deep Think Concept

The IB Definition is Endothermic (+):

"The energy required to separate 1 mole of solid crystal into its gaseous ions."

NaCl(s) \rightarrow Na^+(g) + Cl^-(g) \quad \Delta H_{lat} > 0

Note: Some data sources define it as formation (Exothermic). Always check the sign!

We cannot measure Lattice Enthalpy directly. We calculate it using:
Formation = (Sum of all gasification steps) - Lattice Enthalpy

Step Name	Process	Sign
1. Enthalpy of Formation	$Na(s) + \frac{1}{2}Cl_2(g) \rightarrow NaCl(s)$	Exo (-)
2. Atomization of Metal	$Na(s) \rightarrow Na(g)$	Endo (+)
3. Ionization of Metal	$Na(g) \rightarrow Na^+(g) + e^-$	Endo (+)
4. Atomization of Non-Metal	$\frac{1}{2}Cl_2(g) \rightarrow Cl(g)$	Endo (+)
5. Electron Affinity	$Cl(g) + e^- \rightarrow Cl^-(g)$	Exo (-)
6. Lattice Enthalpy	$NaCl(s) \rightarrow Na^+(g) + Cl^-(g)$	Endo (+)

Governed by Coulomb's Law ( $F \propto \frac{q_1q_2}{r ^ 2}$ )

Higher ionic charge = Stronger attraction = Higher Lattice Enthalpy.

MgO (+2/-2) >> NaCl (+1/-1)

Smaller ionic radius = Ions pack closer = Higher Lattice Enthalpy.

LiF > NaCl > KBr

Putting it into Practice

Paper 2 Style

Calculate the Lattice Enthalpy of Potassium Chloride (KCl) using the following data (all kJ/mol):

Formation (KCl): -437

Atomization (K): +89

Ionization (K): +419

Atomization (Cl): +122

Electron Affinity (Cl): -349

[2 Marks]

Suggest whether the Lattice Enthalpy of MgO would be larger or smaller than that of NaCl. Explain your answer.