BORN-HABER
CYCLES
A guide for A level students
KNOCKHARDY PUBLISHING
2008
SPECIFICATIONS
BORN-HABER CYCLES
INTRODUCTION
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understand selected topics at AS and A2 level Chemistry. It is based on
the requirements of the AQA and OCR specifications but is suitable for
other examination boards.
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it may be used for classroom teaching using an interactive white board.
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www.knockhardy.org.uk/sci.htm
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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
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)
+
Na+ Cl¯(s)
Cl¯(g)
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
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
Example
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.
Na+(g)
+
Cl¯(g)
Na+(g) + Cl–(g)
NaCl(s)
Na+ Cl¯(s)
Lattice Enthalpy Definition(s)
2. Lattice Dissociation Enthalpy
‘The enthalpy change when ONE MOLE of an ionic lattice
dissociates into isolated gaseous ions.’
Values
Example
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.
Na+ Cl¯(s)
Na+(g)
Na+(g) + Cl–(g)
NaCl(s)
+
Cl¯(g)
Calculating Lattice Enthalpy
SPECIAL POINTS
you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY
it is CALCULATED USING A BORN-HABER CYCLE
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
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
Lattice Enthalpy Values
Cl¯
Br¯
F¯
O2-
Na+
-780
-742
-918
-2478
K+
-711
-679
-817
-2232
Rb+
-685
-656
-783
Mg2+
-2256
Ca2+
-2259
-3791
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.
Lattice Enthalpy Values
Cl¯
Br¯
F¯
O2-
Na+
-780
-742
-918
-2478
K+
-711
-679
-817
-2232
Rb+
-685
-656
-783
Mg2+
-2256
Ca2+
-2259
-3791
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.
Na+
Cl¯
K+
Cl¯
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.
Born-Haber Cycle For Sodium Chloride
kJ mol-1
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——> NaCl(s)
Enthalpy of sublimation of sodium
Na(s)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g)
+ 121
Ist Ionisation Energy of sodium
Na(g) ——> Na+(g) + e¯
+ 500
Electron Affinity of chlorine
Cl(g) + e¯ ——> Cl¯(g)
– 364
Na+(g) + Cl¯(g) ——> NaCl(s)
?
Lattice Enthalpy of NaCl
——> Na(g)
– 411
+ 108
Born-Haber Cycle - NaCl
1
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——>
NaCl(s)
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 = - 411 kJ mol-1
1
NaCl(s)
Born-Haber Cycle - NaCl
1
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——>
2
NaCl(s)
Enthalpy of sublimation of sodium
Na(s)
——>
Na(g)
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 = + 108 kJ mol-1
1
NaCl(s)
Born-Haber Cycle - NaCl
1
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——>
2
Enthalpy of sublimation of sodium
Na(s)
3
NaCl(s)
——>
Na(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——>
Cl(g)
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 = + 121 kJ mol-1
1
NaCl(s)
Born-Haber Cycle - NaCl
1
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——>
2
Na+(g) + Cl(g)
Enthalpy of sublimation of sodium
Na(s)
3
NaCl(s)
——>
Na(g)
4
Enthalpy of atomisation of chlorine
½Cl2(g) ——>
Cl(g)
Na(g) + Cl(g)
4
Ist Ionisation Energy of sodium
Na(g) ——> Na+(g) + e¯
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 = + 500 kJ mol-1
1
NaCl(s)
Born-Haber Cycle - NaCl
1
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——>
2
Na+(g) + Cl(g)
Enthalpy of sublimation of sodium
Na(s)
3
NaCl(s)
——>
5
Na(g)
4
Enthalpy of atomisation of chlorine
½Cl2(g) ——>
Na+(g) + Cl–(g)
Cl(g)
Na(g) + Cl(g)
4
Ist Ionisation Energy of sodium
Na(g) ——> Na+(g) + e¯
5
Electron Affinity of chlorine
Cl(g) + e¯
——>
3
Na(g) + ½Cl2(g)
Cl¯(g)
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 = - 364 kJ mol-1
1
NaCl(s)
Born-Haber Cycle - NaCl
1
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——>
2
Na+(g) + Cl(g)
Enthalpy of sublimation of sodium
Na(s)
3
NaCl(s)
——>
5
Na(g)
4
Enthalpy of atomisation of chlorine
½Cl2(g) ——>
Na+(g) + Cl–(g)
Cl(g)
Na(g) + Cl(g)
4
Ist Ionisation Energy of sodium
Na(g) ——> Na+(g) + e¯
5
Electron Affinity of chlorine
Cl(g) + e¯
6
——>
3
Na(g) + ½Cl2(g)
Cl¯(g)
Lattice Enthalpy of NaCl
2
Na(s) + ½Cl2(g)
Na+(g) + Cl¯(g) ——> NaCl(s)
Lattice Enthalpy is exothermic.
Oppositely charged ions are
attracted to each other.
6
1
NaCl(s)
Born-Haber Cycle - NaCl
CALCULATING THE LATTICE ENTHALPY
Na+(g) + Cl(g)
Apply Hess’s Law
6
= -
5
-
4
-
3
-
5
2 + 1
4
The minus shows you are going in the
opposite direction to the definition
= - (-364) - (+500) - (+121) - (+108) + (-411)
= - 776 kJ mol-1
Na+(g) + Cl–(g)
Na(g) + Cl(g)
3
Na(g) + ½Cl2(g)
2
6
Na(s) + ½Cl2(g)
1
NaCl(s)
Born-Haber Cycle - NaCl
CALCULATING THE LATTICE ENTHALPY
Na+(g) + Cl(g)
Apply Hess’s Law
6
= -
-
5
-
4
-
3
5
2 + 1
4
The minus shows you are going in the
opposite direction to the definition
= - (-364) - (+500) - (+121) - (+108) + (-411)
= - 776 kJ mol-1
Na+(g) + Cl–(g)
Na(g) + Cl(g)
3
Na(g) + ½Cl2(g)
OR…
2
Ignore the signs and just use the values;
If you go up you add, if you come down
you subtract the value
6
=
5
-
4
-
3
-
2
-
Na(s) + ½Cl2(g)
1
1
= (364) - (500) - (121) - (108) - (411)
= - 776 kJ mol-1
6
NaCl(s)
Born-Haber Cycle - MgCl2
1
2
Enthalpy of formation of MgCl2
Mg(s) + Cl2(g) ——> MgCl2(s)
Mg2+(g) + 2Cl(g)
Enthalpy of sublimation of magnesium
Mg(s) ——> Mg(g)
5
6
3
Enthalpy of atomisation of chlorine
½Cl2(g) ——>
Mg+(g)
+ 2Cl(g)
x2
Cl(g)
4
4
5
6
Ist Ionisation Energy of magnesium
Mg(g) ——> Mg+(g) + e¯
2nd Ionisation Energy of magnesium
Mg+(g) ——> Mg2+(g) + e¯
7
3
Mg(g) + Cl2(g)
2
Electron Affinity of chlorine
Cl(g) + e¯
——>
Cl¯(g)
Mg2+(g) + 2Cl–(g)
Mg(g) + 2Cl(g)
x2
Lattice Enthalpy of MgCl2
Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)
7
Mg(s) + Cl2(g)
1
MgCl2(s)
BORN-HABER
CYCLES
THE END
KNOCKHARDY PUBLISHING