1. Born-Haber Cycle

2. Lattice Energy
We recall that the lattice energy is the energy required to completely separate one mole of a solid ionic compound into its gaseous ions
Ex: NaCl(s)  Na+(g) + Cl-(g)
We cannot measure the lattice energy directly, we can use either Coulomb’s law (using charges and radius) or the Born Haber Cycle

3. 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

5. 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
Let’s consider: Li(s) + ½ F2(g)  LiF (s)
DH = kJ/mol

6. Step 1: Convert from Solid to Gas
The standard enthalpy change of atomisation is the ΔH required to produce one mole of gaseous atoms
Li (s)  Li (g) ΔHoat = kJmol-1

7. NOTE: for diatomic gaseous elements, F2, ΔHoat is equal to half the bond energy (enthalpy)
½ F2(g)  F(g) ΔHoat = ½ E (F-F)
ΔHoat = ½ ( )
ΔHoat = kJmol-1

8. 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.
F(g) + e-  F-(g) ΔHo = -328 kJmol-1
For most atoms = exothermic, but gaining a 2nd electron is endothermic due to the repulsion between the anion and the electron

9. Becoming cations Ionization energy
Enthalpy change for one mole of a gaseous element or cation to lose electrons to form a mole of positively charged gaseous ions
Li(g)  Li+(g) + e- IE1= +520 kJmol-1

10. Lattice Enthalpy
Energy required to convert one mole of the solid compound into gaseous ions.
LiF(s)  Li+(g) + F-(g) ΔHolat = ?
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
We will use the reverse of this lattice energy equation to find the lattice energy

12. Calculations ½ F2(g)  F2(g) + 75.3 Li(s) + ½ F2(g)  LiF (s) -594.1
Calculate the lattice energy of LiF(s) using the following: (kJmol-1)
Li(s)  Li(g)
½ F2(g)  F2(g)
Li(g)  Li+(g) + e
F(g) + e-  F-(g) -328
Li+(g) + F-(g)  LiF(g) ???
Li(s) + ½ F2(g)  LiF (s)
Step 5: kJ/mol
We reverse this to get the lattice energy, 1017 kJ/mol

13. Magnitude of Lattice enthalpy
Lattice energy is a measure of stability and we can use this to explain the formation of compounds
For ex: Mg+2 and Cl- form MgCl2, not MgCl
1st ionization energy of Mg is half as much as 2nd, so it would make sense for Mg to only lose 1 electron, BUT
1. MgCl2 fulfills octet rule, but the stability gained through this does NOT overweigh the energy input through 1st ionization
2. Answer lies in lattice energy  enough to compensate for 1st ionization

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

15. Lattice Energy Strength
Greater the lattice energy, the higher the melting point is (more stronger)
Higher the charge, the higher the lattice energy
Ex: Mg+2 > Li+

16. 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.

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

18. 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