Title: Lesson 6 Born-Haber Cycles and
Lattice Enthalpies
Learning Objectives:
– Understand the term lattice enthalpy
– Use Born-Haber cycles to calculate lattice enthalpy
– Identify and explain trends in lattice enthalpy
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
The standard enthalpy change of three combustion reactions is given below
in kJ.
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
2H2(g) + O2(g) → 2H2O(l)
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)

∆Ho = –3120
∆Ho = –572
ΔHo = –1411
Based on the above information, calculate the standard change in enthalpy,
∆Ho, for the following reaction.
C2H6(g) → C2H4(g) + H2(g)
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Bond Enthalpies

Look at the covalent bond enthalpies on Table 10 in the
Data Booklet.

Why do you think there are no ionic bond enthalpies?
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Recap of First Ionisation Energies and Electron
Affinities

We know that metals loses electrons and non-metals gain electrons. We can use Ionisation
Energies and Electron Affinities to work out the enthalpy changes within an ionic compound.

The first ionisation energy is the energy needed to remove one mole of electrons
from one mole of gaseous atoms:

Sodium (on the left of the periodic table) has a relatively low ionisation energy.

The first electron affinity
is the enthalpy change when one mole of gaseous
atoms attracts one mole of electrons.Values can be found in section 7 of the IB data
booklet.

As the electron is attracted to the positively charged nucleus of the Cl atom, the process is
EXOTHERMIC
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Lattice Enthalpies

Add the equations for the first ionisation energy and first electron affinity.

The process is ENDOTHERMIC overall. This is energetically UNFAVOURABLE
(despite the fact it leads to the formation of ions with stable noble gas
configurations).

Oppositely charged ions come together to form an ionic lattice. The strong
attraction between the oppositely charged ions means its very EXOTHERMIC.

Lattice Enthalpy Hθlat expresses this enthalpy change in terms of the reverse
ENDOTHERMIC process.

The lattice enthalpy relates to the ‘formation of gaseous ions from one
mole of a solid crystal breaking into gaseous ions’. (As seen above)
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
The lattice enthalpy
relates to the enthalpy
change of ‘formation of
gaseous ions from one
mole of a solid crystal
breaking into gaseous
ions’.

Think of lattice enthalpy
as ‘lattice disassociation
enthalpy’...
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Lattice Enthalpy, Hlat

This is the equivalent of ‘bond strength’ for ionic compounds

It is the enthalpy change when one mole of an ionic
compound is converted to gaseous ions.

This is an endothermic process, requiring energy to be put in.
MX(s) M+(g) + X-(g)
Compound
Lattice Enthalpy
kJ mol-1
LiF
1049
LiBr
820
KF
829
CaF2
2651
Note: Most places define lattice enthalpy the opposite
way round, i.e:
M+(g) + X-(g)  MX(s)
The values would be the same magnitude, but with a
negative sign to show they are exothermic.
It is just a strange quirk of the IB that they do it this way
round….I think so that it fits with average bond
enthalpies, which are also represent bond breaking
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Born-Haber Cycle
Construction
of a
Using the ‘FAIL’ technique
eg for sodium chloride:
NB: Hθ f = formation
HθIE = ionisation
Hθat = atomisation
HθEA = electron affinity
HθLAT = lattice enthalpy
F = formation

A = atomisation

I = ionisation

L = lattice enthalpy

Na+ (g) + e- + Cl- (g)
Hθ1st IE Na
HθEA Cl
IONISATION
Na (g)
+
Hθat Na
LATTICE
H
NaCl
ENTHALPY
θ
Cl (g)
Hθat Cl
ATOMISATION H NaCl
Na (s) + FORMATION
½ Cl (g)
θ
2
f
LAT
NaCl (s)
Born-Haber Cycle
: Applying Hess’s Law
There are two routes from elements to ionic compound
The Indirect route and the Direct route
Clockwise
Na+ (g) + e- + Cl- (g)
Hθ1st IE Na
Na (g)
+
Hθat Na
Na (s)
+
HθEA Cl
Hθ
LAT
NaCl
Cl (g)
Anti-clockwise
Hθat Cl
½ Cl2 (g)
Apply Hess’s Law:
=
Hθ f NaCl
NaCl (s)
Clockwise arrows must equal
the Anti-clockwise arrows
HatNa + HatCl + H 1st IENa + HEACl - HLATNaCl = Hf NaCl
Born-Haber Cycle: Calculation
If you want to calculate HLATNaCl, it is as follows:HatNa + HatCl + H 1st IENa + HEACl - HLATNaCl = Hf NaCl
Rearrange to find the lattice energy:
HLATNaCl = [HatNa + HatCl + H 1st IENa + HEACl] - Hf NaCl
Values:
HatNa = 107
(kJmol-1) HEACl = -349
HatCl = 121
H 1st IENa = 496
Hf NaCl = -411
It’s important to keep all numbers in ( ), whether +ve or –ve,
when entering information into your calculator.
HLATNaCl = [(107) + (121) + (496) + (-349)] - (-411)
HLATNaCl = 786 kJmol-1
So Born-Haber cycles can be used to calculate a measure of
ionic bond strength based on experimental data.
Born-Haber Cycle
Construction
of a
Using the ‘FAIL’ technique
eg for magnesium chloride:
NB: Hθ f = formation
Hθat = atomisation
HθIE = ionisation
HθEA = electron affinity
HθLAT = lattice enthalpy
F = formation

A = atomisation

I = ionisation

L = lattice enthalpy

Mg2+ (g) + 2e- + 2Cl- (g)
IONISATION
2xH
H
Mg
θ
1st +2nd IE
θ
EA
Cl
LATTICE
H
MgCl
ENTHALPY
θ
Mg (g)
+
Hθat Mg
2Cl (g)
LAT
2xHθat Cl
ATOMISATION
H MgCl
Mg (s)
+ Cl (g)
FORMATION
θ
2
f
2
MgCl2 (s)
2
Born-Haber Cycles

Lattice enthalpies are
difficult to measure
directly, so we use
Born-Haber cycles


These are just a
specialised type of Hess
cycle
To calculate Hlat :

Start at the bottom and
work round clockwise,
adding and subtracting
according to the arrows.
ionised metal and atomised non-metal
Ionisation energy/s (Metal)
Electron affinity/s (Non-metal)
atomised metal and atomised
non-metal
Enthalpy of atomisation (metal)
ionised metal and ionised
non-metal
metal and atomised non-metal
Enthalpy of atomisation (non-metal)
elements in their standard
states
Enthalpy of formation
solid ionic compound
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Hlat = ?
Building Born-Haber Cycles

Work through the activity here

Cut out the equations first and arrange them sensibly on
your desk.

We will work through LiF together.

Hint: Don’t forget to include both ionisations for Mg/Ca
and both electron affinities for O
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