1. 15.2 Born-Haber Cycle
Define and apply the terms lattice enthalpy, and electron affinity
Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds
The relative value of the theoretical lattice enthalpy increases with higher ionic charge and smaller ionic radius due to increased attractive forces
Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides and use it to calculate the enthalpy change
Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character.

2. Born-Haber Cycle
A series of hypothetical steps and their enthalpy changes needed to convert elements to an ionic compound and devised to calculate the lattice energy.
Using Hess’s law as a means to calculate the formation of ionic compounds

4. Born-Haber Cycle Steps
Elements (standard state) converted into gaseous atoms
Losing or gaining electrons to form cations and anions
Combining gaseous anions and cations to form a solid ionic compound

5. Step 1: Atomisation
The standard enthalpy change of atomisation is the ΔH required to produce one mole of gaseous atoms
Na(s)  Na(g) ΔHoat = +109 kJmol-1

6. NOTE: for diatomic gaseous elements, Cl2, ΔHoat is equal to half the bond energy (enthalpy)
Cl2(g)  Cl(g) ΔHoat = ½ E (Cl-Cl)
ΔHoat = ½ (+242 )
ΔHoat = +121 kJmol-1

7. Step 2: Formation of gaseous ions
Electron Affinity
Enthalpy change when one mole of gaseous atoms or anions gains electrons to form a mole of negatively charged gaseous ions.
Cl(g) + e-  Cl-(g) ΔHo = -364 kJmol-1
For most atoms = exothermic, but gaining a 2nd electron is endothermic due to the repulsion between the anion and the electron

8. Becoming cations Ionisation energy
Enthalpy change for one mole of a gaseous element or cation to lose electrons to form a mole of positively charged gaseous ions
Na(g)  Na+(g) + e- IE1= +494 kJmol-1

9. Lattice Enthalpy
Energy required to convert one mole of the solid compound into gaseous ions.
NaCl (s)  Na+(g) + Cl-(g) ΔHolat = +771kJmol-1
It is highly endothermic
We cannot directly calculate ΔHolat , but values are obtained indirectly through Hess’s law for the formation of the ionic compound

11. Calculations
Calculate the lattice energy of NaCl(s) using the following: (kJmol-1)
Enthalpy of formation of NaCl = - 411
Enthalpy of atomisation of Na = +109
Enthalpy of atomisation of Cl = +121
Electron affinity of Cl = - 364
Ionisation energy of Na = + 494
Enthalpy of atomisation + electron affinity + ionisation = enthalpy of formation + lattice energy

12. Magnitude of Lattice enthalpy
The greater the charge on the ions, the greater the electrostatic attraction and hence the greater the lattice enthalpy
Ex: Mg2+ > Na+
The larger the ions, then the greater the separation of the charges and the lower the lattice enthalpy
VICE VERSA

13. Trends ΔHolat Change from NaCl MgO 3889 Increased ionic charge NaCl
771
------
KBr
670
Larger ions

14. Use of Born-Haber Cycles
Empirical value of ΔHolat is found using Born-Haber cycle.
Theoretical value of ΔHolat can be found by summing the electrostatic attractive and repulsive forces between the ions in the crystal lattice.

15. Compound
Empirical value
Theoretical value
NaCl
771
766
KBr
670
667 KI
632
631
AgCl
905
770

16. Agreement
Usually there is good agreement between empirical and theoretical values
If there isn’t good agreement
Implying that the description of the compound as ionic is inappropriate
There could be a significant degree of covalent character in the bonding (EN difference less than 1.7)
Presence of covalent character leads to an increase in ΔHolat