1. Advanced Higher Chemistry Unit 2
Applications of Hess’s Law 1
Born – Haber Cycles

2. Born - Haber Cycles
The enthalpy of formation is the energy change when 1 mole of a substance is formed from its constituent elements at standard conditions.
e.g. Mg(s) Cl2(g)  MgCl2(s) H = X ?
These cycles are a means of showing clearly all of the processes going on in the formation of 1 mole of an ionic compound.
It is nothing more than the application of Hess’s Law to find the enthalpy change of a particular step that is difficult to measure directly.

3. There are several component steps in making an ionic compound
Enthalpy of atomisation - breaking up the metallic lattice - this is ENDOTHERMIC
Enthalpy of atomisation - breaking up the non-metal molecules - this is ENDOTHERMIC
First/ Second etc ionisation enthalpy - turning the metal atoms into positively charged ions -this is ENDOTHERMIC
Electron gain enthalpy - turning the non-metal atoms into negatively charged ions - this can be EXO or ENDOTHERMIC
Lattice enthalpy - forming bond between the ions to form the lattice - this EXOTHERMIC

4. The Enthalpy changes needed to make ionic compounds
What it means
The metallic lattice is broken up into individual atoms
1. Enthalpy of sublimation (or atomisation)
The non-metal molecules or structure is broken up into atoms
2. Enthalpy of dissociation(or atomisation)
The requisite number of outer electron(s) are removed
3. The appropriate number of ionisation energies of the metal
The non-metal atoms gain electron(s)
4. The electron gain enthalpy
5. Enthalpy of lattice-forming of the ionic substance
The ionic bonds are formed
The overall enthalpy change for the reaction
6. Enthalpy of formation of the substance  H

5. The net result of all on these energy steps is the enthalpy of formation for the substance H.
Let us look at an example, the formation of 1 mole of magnesium chloride MgCl2(s)

6. Electron gain enthalpy = 2 x - 348.7kJ mol-1
Mg2+(g) + 2Cl(g)
Electron gain enthalpy = 2 x kJ mol-1
Second ionisation enthalpy of Mg = kJmol-1
Mg+(g) + 2Cl(g)
Mg2+(g) + 2Cl-(g)
First ionisation enthalpy of Mg = kJmol-1
Mg(g) + 2Cl(g)
Lattice enthalpy = kJmol-1
Enthalpy of atomistaion
of chlorine = kJmol-1
Mg(g) + Cl2(g)
Enthalpy of atomistaion of Mg = kJmol-1
Mg(s) + Cl2(g)
H Enthalpy of formation
Mg2+(Cl-)2(s)

7. H0r = Ho for all of the stages involved
H (kJ)
Enthalpy Change
+ 147
1. Enthalpy of atomisation of Mg
+ 243
2. Enthalpy of atomisation of Cl
3. First ionisation energy of Mg
+ 744
4. Second ionisation energy of Mg
+ 1460
5. Electron affinity of Cl
2 x – 348.7
6. Lattice enthalpy of MgCl2
- 2326
7. Enthalpy of formation of MgCl2

8. The Enthalpy of solution of substance is the energy change
when 1 mole completely dissolves in water
When an ionic compound dissolves in water, two processes occur:
1. The lattice breaks down releasing ions
2. The ions then become hydrated by forming bonds with
water molecules
The enthalpy of solution of a salt, just like the Born – Haber cycle
for the formation of an ionic compound, can be regarded as an
energy balance between the endothermic process of lattice breaking
and the exothermic hydration of ions.

9. Enthalpy change What it means 1. Enthalpy of lattice breaking
The ionic lattice is broken up
into gaseous ions
2. Enthalpy of hydration of
metal ion
Bonds being formed between
the metal ions and the water
molecules
3. Enthalpy of hydration of
non-metal ion
Bonds being formed between
the non-metal ions and the
water molecules
4. Enthalpy of solution
Overall enthalpy change

10. _______________________________
Na+ (g) + Cl(g)
_______________________________
Enthalpy of
hydration of sodium
ions
Lattice
enthalpy
405(H2)
+771(H1)
Na+Cl(s)
______________
Na+ (g) + (aq)  Na+ (aq)
__________________
Enthalpy of solution
of sodium chloride
Enthalpy of hydration
of chloride ions
 362(H3)
+ 4(H4)
Cl(g) (aq)  Cl(aq)
Na+Cl(s) +(aq)  Na+ (aq) + Cl(aq)
___________________
___________________________

11. Enthalpy change described
Equation
H kJmol-1
Enthalpy of lattice-breaking of
sodium chloride
Na+Cl(s)  Na+(g) + Cl(g)
+771(H1)
Enthalpy of hydration of sodium
ions
405(H2)
Na+(g) (aq)  Na+(aq)
Enthalpy of hydration of chloride
ions
Cl(g) (aq)  Cl(aq)
362(H3)
Na+Cl(s) + (aq) 
Na+(aq) + Cl(aq)
Enthalpy of solution of sodium
chloride
+ 4(H4)

12. H4 = H1 + H2 + H3 = 771 + (- 404) + (-362) = + 4kJ mol-1
On applying Hess’s law , the enthalpy of solution can be calculated
thus,
H4 = H H H3
= (- 404) + (-362) = + 4kJ mol-1