•Lattice energy•Born-haber cycle

A2 – CHEMICAL ENERGETICS

Lattice energyLattice 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 < 0Provide a measure of strength of ionic bond

between ions in the crystal lattice.Higher lattice energy (more exothermic),

stronger bond.

Lattice energyFactors affecting magnitude of lattice energies:a)Ionic charge increase, more exothermic, higher

lattice energy.b)Ionic radius decrease, more exothermic, higher

lattice energy.c)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

Electron affinityElectron 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

Born-Haber CycleLattice energy cannot be determined directly

– use Born-Haber Cycle to determine.Born-Haber Cycle for formation of NaCl :

Born-Haber CycleAlternatively energy level diagram can be

constructed.Constructing Born-Haber cycle (energy level

diagram) of NaCl:Stage 1 : Formation of NaClStage 2 : Atomisation of sodiumStage 3 : Ionisation of sodiumStage 4 : Atomisation of chlorineStage 5 : Formation of chloride ion (electron

affinity)H < 0 , arrows pointing down.H < 0 , arrows pointing up.

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 .

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:1)Crystal lattice breaks down forming isolated

gaseous ions.Energy (equal to lattice energy) absorbed to break

ionic bonds and pull ions apart.2)Hydration of gaseous ions.Energy (equal to hydration energy) released when

bonds formed between H2O molecules and ions.

3 methods to calculate ∆Hsol.

1.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

3. Born-Haber cylcle.Lattice energy of NaCl = -776 kJmol-1

Enthalpy change of hydration/kJmol-1: Na+ = -390; Cl- = -381

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