1. A guide for A level students KNOCKHARDY PUBLISHING
BORN-HABER CYCLES
A guide for A level students
2008 SPECIFICATIONS
KNOCKHARDY PUBLISHING

2. BORN-HABER CYCLES
INTRODUCTION
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3. BORN-HABER CYCLES CONTENTS Lattice Enthalpy
Definition of enthalpy changes
Born-Haber cycle for sodium chloride
Calculation of Lattice Enthalpy
Born-Haber cycle for magnesium chloride

4. Lattice Enthalpy Definition(s)
THERE ARE TWO DEFINITIONS OF LATTICE ENTHALPY
1. Lattice Formation Enthalpy
‘The enthalpy change when ONE MOLE of an ionic lattice
is formed from its isolated gaseous ions.’
Example Na+(g) Cl¯(g) Na+ Cl¯(s)
2. Lattice Dissociation Enthalpy
‘The enthalpy change when ONE MOLE of an ionic lattice
dissociates into isolated gaseous ions.’
Example Na+ Cl¯(s) Na+(g) Cl¯(g)
MAKE SURE YOU CHECK WHICH IS BEING USED

5. Lattice Enthalpy Definition(s)
1. Lattice Formation Enthalpy
‘The enthalpy change when ONE MOLE of an ionic lattice
is formed from its isolated gaseous ions.’
Values highly EXOTHERMIC
strong electrostatic attraction between oppositely charged ions
a lot of energy is released as the bond is formed
relative values are governed by the charge density of the ions.
Example Na+(g) Cl¯(g) Na+ Cl¯(s)
NaCl(s)
Na+(g) + Cl–(g)

6. Lattice Enthalpy Definition(s)
2. Lattice Dissociation Enthalpy
‘The enthalpy change when ONE MOLE of an ionic lattice
dissociates into isolated gaseous ions.’
Values highly ENDOTHERMIC
strong electrostatic attraction between oppositely charged ions
a lot of energy must be put in to overcome the attraction
relative values are governed by the charge density of the ions.
Example Na+ Cl¯(s) Na+(g) Cl¯(g)
NaCl(s)
Na+(g) + Cl–(g)

7. Calculating Lattice Enthalpy
SPECIAL POINTS
you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY
it is CALCULATED USING A BORN-HABER CYCLE

8. Calculating Lattice Enthalpy
SPECIAL POINTS
you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY
it is CALCULATED USING A BORN-HABER CYCLE
greater charge
densities of ions = greater attraction
= larger lattice enthalpy

9. Calculating Lattice Enthalpy
SPECIAL POINTS
you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY
it is CALCULATED USING A BORN-HABER CYCLE
greater charge
densities of ions = greater attraction
= larger lattice enthalpy
Effects
Melting point the higher the lattice enthalpy,
the higher the melting point of an ionic compound
Solubility solubility of ionic compounds is affected by the relative
values of Lattice and Hydration Enthalpies

10. Lattice Enthalpy Values
Cl¯ Br¯ F¯ O2- Na K Rb Mg Ca
Units: kJ mol-1
Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions.

11. Lattice Enthalpy Values
Cl¯ Br¯ F¯ O2- Na K Rb Mg Ca
Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions. K+
Cl¯
Cl¯
Na+
The sodium ion has the same charge as a potassium ion but is smaller. It has a higher charge density so will have a more effective attraction for the chloride ion. More energy will be released when they come together.

12. Born-Haber Cycle For Sodium Chloride
kJ mol-1
Enthalpy of formation of NaCl Na(s) ½Cl2(g) ——> NaCl(s) – 411
Enthalpy of sublimation of sodium Na(s) ——> Na(g)
Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g)
Ist Ionisation Energy of sodium Na(g) ——> Na+(g) e¯
Electron Affinity of chlorine Cl(g) e¯ ——> Cl¯(g) – 364
Lattice Enthalpy of NaCl Na+(g) + Cl¯(g) ——> NaCl(s) ?

13. Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl
Na(s) ½Cl2(g) ——> NaCl(s) 1
Na(s) + ½Cl2(g)
This is an exothermic process so energy is released. Sodium chloride has a lower enthalpy than the elements which made it.
VALUE = kJ mol-1 1
NaCl(s)

14. Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl
Na(s) ½Cl2(g) ——> NaCl(s)
Enthalpy of sublimation of sodium
Na(s) ——> Na(g) 1 2
Na(g) + ½Cl2(g) 2
Na(s) + ½Cl2(g)
This is an endothermic process. Energy is needed to separate the atoms. Sublimation involves going directly from solid to gas.
VALUE = kJ mol-1 1
NaCl(s)

15. Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl
Na(s) ½Cl2(g) ——> NaCl(s)
Enthalpy of sublimation of sodium
Na(s) ——> Na(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g) 1 2 3
Na(g) + Cl(g) 3
Na(g) + ½Cl2(g) 2
Na(s) + ½Cl2(g)
Breaking covalent bonds is an endothermic process. Energy is needed to overcome the attraction the atomic nuclei have for the shared pair of electrons.
VALUE = kJ mol-1 1
NaCl(s)

16. Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl
Na(s) ½Cl2(g) ——> NaCl(s)
Enthalpy of sublimation of sodium
Na(s) ——> Na(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g)
Ist Ionisation Energy of sodium
Na(g) ——> Na+(g) e¯ 1
Na+(g) + Cl(g) 2 4 3
Na(g) + Cl(g) 4 3
Na(g) + ½Cl2(g) 2
Na(s) + ½Cl2(g)
All Ionisation Energies are endothermic. Energy is needed to overcome the attraction the protons in the nucleus have for the electron being removed.
VALUE = kJ mol-1 1
NaCl(s)

17. Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl
Na(s) ½Cl2(g) ——> NaCl(s)
Enthalpy of sublimation of sodium
Na(s) ——> Na(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g)
Ist Ionisation Energy of sodium
Na(g) ——> Na+(g) e¯
Electron Affinity of chlorine
Cl(g) e¯ ——> Cl¯(g) 1
Na+(g) + Cl(g) 2 5 4 3
Na+(g) + Cl–(g)
Na(g) + Cl(g) 4 3
Na(g) + ½Cl2(g) 5 2
Na(s) + ½Cl2(g)
Electron affinity is exothermic. Energy is released as the nucleus attracts an electron to the outer shell of a chlorine atom.
VALUE = kJ mol-1 1
NaCl(s)

18. Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl
Na(s) ½Cl2(g) ——> NaCl(s)
Enthalpy of sublimation of sodium
Na(s) ——> Na(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g)
Ist Ionisation Energy of sodium
Na(g) ——> Na+(g) e¯
Electron Affinity of chlorine
Cl(g) e¯ ——> Cl¯(g)
Lattice Enthalpy of NaCl
Na+(g) + Cl¯(g) ——> NaCl(s) 1
Na+(g) + Cl(g) 2 5 4 3
Na+(g) + Cl–(g)
Na(g) + Cl(g) 4 3
Na(g) + ½Cl2(g) 5 2 6
Na(s) + ½Cl2(g) 6 1
Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other.
NaCl(s)

19. Born-Haber Cycle - NaCl
CALCULATING THE LATTICE ENTHALPY
Apply Hess’s Law
Na+(g) + Cl(g) =
The minus shows you are going in the opposite direction to the definition
= - (-364) - (+500) - (+121) - (+108) + (-411)
= kJ mol-1 5 6 5 4 3 2 1 4
Na+(g) + Cl–(g)
Na(g) + Cl(g) 3
Na(g) + ½Cl2(g) 2 6
Na(s) + ½Cl2(g) 1
NaCl(s)

20. Born-Haber Cycle - NaCl
CALCULATING THE LATTICE ENTHALPY
Apply Hess’s Law
Na+(g) + Cl(g) =
The minus shows you are going in the opposite direction to the definition
= - (-364) - (+500) - (+121) - (+108) + (-411)
= kJ mol-1
OR…
Ignore the signs and just use the values;
If you go up you add, if you come down you subtract the value =
= (364) - (500) - (121) - (108) - (411) 5 6 5 4 3 2 1 4
Na+(g) + Cl–(g)
Na(g) + Cl(g) 3
Na(g) + ½Cl2(g) 2 6
Na(s) + ½Cl2(g) 1 6 5 4 3 2 1
NaCl(s)

21. Born-Haber Cycle - MgCl2
Enthalpy of formation of MgCl2
Mg(s) Cl2(g) ——> MgCl2(s)
Enthalpy of sublimation of magnesium
Mg(s) ——> Mg(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g) x2
Ist Ionisation Energy of magnesium
Mg(g) ——> Mg+(g) e¯
2nd Ionisation Energy of magnesium
Mg+(g) ——> Mg2+(g) e¯
Electron Affinity of chlorine
Cl(g) e¯ ——> Cl¯(g) x2
Lattice Enthalpy of MgCl2
Mg2+(g) + 2Cl¯(g) ——> MgCl2(s) 1
Mg2+(g) + 2Cl(g) 2 5 6 3
Mg+(g) + 2Cl(g) 4 4
Mg2+(g) + 2Cl–(g)
Mg(g) + 2Cl(g) 3 5
Mg(g) + Cl2(g) 2 7 6
Mg(s) + Cl2(g) 1 7
MgCl2(s)

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BORN-HABER CYCLES
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