1. Lattice Energies
AP Material

2. Lattice Energy
Lattice energy: The energy required to completely separate a mole of a solid ionic compound into its elements in their standard states
It is a measure of just how much stabilization results from arranging oppositely charged ions in an ionic solid

3. The principal reason that ionic compounds are stable is the attraction between oppositely charged ions
This attraction draws the ions together, releasing energy and causing ions to form a solid array, or lattice

4. The formation of ionic compounds is highly exothermic (releasing heat and light):
Na(s) + ½ Cl2 (g) → NaCl(s) ∆ H = -788kJ/mol
The reverse reaction is highly endothermic:
NaCl(s) → Na(s) + ½ Cl2(g) ∆ H = 788kJ/mol

5. Calculating Lattice Energy
Eel = кQ1Q2 d
where,
К is a constant (8.99 x 109 J-m/C2)
Q1 and Q2 are the charges on the ions
d is the distance between their centers

6. Therefore, the magnitude of the lattice energy is dependant upon:
the charges on the particles (Q1 and Q2)
the distance between the nuclei

7. Summary
For a given arrangement of ions, the lattice energy increases as the charges on the ions increase and as their radii decrease

8. Example: Arrange in order of increasing lattice energy
a) NaF, CsI and CaO
CsI < NaF < CaO
b) ZrO2,MgF2, CaF2 and
CaF2 < MgF2 < ZrO2
c) LiF, NaF, CaF2, AlF3
NaF < LiF < CaF2 < AlF3
d) CaCl2, LiCl, Al2O3, NaCl
NaCl < LiCl < CaCl2 < Al2O3

9. The Born-Haber Cycle Lattice energy cannot be determined by experiment
It is calculated by envisioning the formation of an ionic compound occurring in steps (Hess’s Law)
Called the Born-Haber Cycle after Max Born and Fritz Haber (German scientists)

11. Steps in the formation of NaCl
1: Na(s) + ½ Cl2(g) → NaCl(s) ∆Hf = -411kJ
2: Na(s) → Na(g) ∆H°f = 108 kJ
3: 1/2Cl2(g) → Cl(g) ∆H°f = 122 kJ
4: Na(g) → Na+(g) + e I1(Na) = 496 kJ
5: Cl(g) + e → Cl-(g) E(Cl) = -396 kJ
6. Overall : ΔH6= Σ Δ H1 to Δ H5