Born-Haber Cycles and Lattice Enthalpies
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Transcript Born-Haber Cycles and Lattice Enthalpies
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
2xH
H
Mg
θ
1st +2nd IE
θ
EA
Cl
LATTICE
H
MgCl
ENTHALPY
θ
Mg (g)
+
Hθat Mg
2Cl (g)
LAT
2xHθat 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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