1. A2 – CHEMICAL ENERGETICS
Lattice energy
Born-haber cycle

2. Lattice energy
Lattice energy of an ionic crystal Hlatt heat energy evolved when 1 mole of crystalline solid is formed from its separate gaseous ions under standard condition (298K and 1 atm).
E.g : Na+(g) + Cl-(g)  Na+Cl-(s)
Hlatt < 0
Provide a measure of strength of ionic bond between ions in the crystal lattice.
Higher lattice energy (more exothermic), stronger bond.

3. Lattice energy Factors affecting magnitude of lattice energies:
Ionic charge increase, more exothermic, higher lattice energy.
Ionic radius decrease, more exothermic, higher lattice energy.
Crystal structure
Lattice energy  (q+ x q-)
(r+ + r-)
Compound
Lattice energy / kJmol-1
NaF
-915
NaCl
-776
NaBr
-742
NaI
-699
MgCl2
-2489
MgO
-3933

4. Electron affinity
Electron affinity the enthalpy change when 1 mole of electrons is added to one mole of atoms or ions in the gaseous state.
Is the measure of attraction of the atom or ion for the extra electron.
First electron affinity enthalpy change when 1 mole of electron is added to 1 mole of gaseous atoms to form singly- charged negative ions.
X(g) + e-  X-(g) H = 1st electron affinity < 0
Second electron affinity is endothermic:
X-(g) + e-  X2-(g) H = 2nd electron affinity > 0

5. Born-Haber Cycle
Lattice energy cannot be determined directly – use Born-Haber Cycle to determine.
Born-Haber Cycle for formation of NaCl :

6. Born-Haber Cycle H < 0 , arrows pointing down.
Alternatively energy level diagram can be constructed.
Constructing Born-Haber cycle (energy level diagram) of NaCl:
Stage 1 : Formation of NaCl
Stage 2 : Atomisation of sodium
Stage 3 : Ionisation of sodium
Stage 4 : Atomisation of chlorine
Stage 5 : Formation of chloride ion (electron affinity)
H < 0 , arrows pointing down.
H < 0 , arrows pointing up.

7. Born-Haber cycle (energy level diagram) of NaCl
By Hess Law:
(-364) + ∆Hlatt [Na+Cl-(s)] kJ mol-1= (-121) + (-500) + (-108) + (-411) kJ mol-1.
*Make sure all arrows are pointing downwards when calculating ∆Hlatt .

8. Aqueous Solution of Ionic Crystals.
When ionic crystals dissolve in water :
M+X-(s) + aq  M+(aq) + X-(aq)
Dissolution of ionic solid (e.g. NaCl) in water occur in 2 imaginary steps:
Crystal lattice breaks down  forming isolated gaseous ions.
Energy (equal to lattice energy) absorbed to break ionic bonds and pull ions apart.
Hydration of gaseous ions.
Energy (equal to hydration energy) released when bonds formed between H2O molecules and ions.

9. 3 methods to calculate ∆Hsol.
Equation method:
Step 1 : Na+Cl-(s)  Na+(g) + Cl-(g), ∆Hlatt
Step 2 : Na+(g) + Cl-(g) + aq  Na+(aq) + Cl-(aq),∆Hhyd
Na+Cl-(s)  Na+(aq) + Cl-(aq) ∆Hsol
If ∆Hhyd > ∆Hlatt  ∆Hsol < 0  salt soluble in H2O.
If ∆Hhyd < ∆Hlatt  ∆Hsol > 0  salt insoluble in H2O.
Enthalpy cycle.
∆Hsol = ∆Hhyd - ∆Hlatt

10. Born-Haber cylcle. Lattice energy of NaCl = -776 kJmol-1
Enthalpy change of hydration/kJmol-1: Na+ = -390; Cl- = -381

11. Exercise :
Using the Born-Haber cycle method, calculate the enthalpy change of solution, Hsol, of LiCl. Predict its solubility of in water.
Lattice energy of LiCl = -848 kJmol-1
Enthalpy change of hydration/kJmol-1: Li+ = -499; Cl- = -381