Lattice enthalpy

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Lattice enthalpy

• Textbook reference p166-171

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Born-Haber cycles

• L.O.
• Explain and use the term lattice enthalpy.
• Use the lattice enthalpy of a simple ionic solid
  and relevant energy terms to construct BornHaber
  cycles.

3

• In pairs recap the following definitions and
  illustrate them with examples
• Standard enthalpy of formation
• IE
• EA
• Second IE
• Second EA

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The enthalpy change of atomisation is the
enthalpy change that takes place when one mole of
gaseous atoms forms from the elements in its
standard state. Use iodine as an example
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Atomisation enthalpies

• 1. Write out the atomisation enthalpies for the
  following elements
• Hydrogen
• zinc
• Iodine
• Mercury
• nitrogen

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Lattice Enthalpy Definition
1. Lattice 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)
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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)
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Calculating Lattice Enthalpy
SPECIAL POINTS you CANNOT MEASURE LATTICE
ENTHALPY DIRECTLY it is CALCULATED USING A
BORN-HABER CYCLE
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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
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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 so
lubility of ionic compounds is affected by the
relative values of Lattice and Hydration
Enthalpies
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Lattice Enthalpy Values
Cl Br F O2- Na -780 -742 -918 -2478
K -711 -679 -817 -2232 Rb -685 -656 -783 Mg
2 -2256 -3791 Ca2 -2259
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.
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Lattice Enthalpy Values
Cl Br F O2- Na -780 -742 -918 -2478
K -711 -679 -817 -2232 Rb -685 -656 -783 Mg
2 -2256 -3791 Ca2 -2259
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.
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.
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Born-Haber Cycle For Sodium Chloride
kJ mol-1 Enthalpy of formation of NaCl
Na(s) ½Cl2(g) gt NaCl(s)
411 Enthalpy of atomisation of sodium Na(s)
gt Na(g) 108 Enthalpy of atomisation
of chlorine ½Cl2(g) gt Cl(g)
121 Ist Ionisation Energy of sodium Na(g)
gt Na(g) e 500 Electron Affinity
of chlorine Cl(g) e gt Cl(g)
364 Lattice Enthalpy of NaCl Na(g)
Cl(g) gt NaCl(s) ?
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Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl Na(s) ½Cl2(g)
gt 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 - 411 kJ
mol-1
1
NaCl(s)
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Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl Na(s) ½Cl2(g)
gt NaCl(s) Enthalpy of atomisation of
sodium Na(s) gt 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 108
kJ mol-1
1
NaCl(s)
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Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl Na(s) ½Cl2(g)
gt NaCl(s) Enthalpy of atomisation of
sodium Na(s) gt Na(g) Enthalpy of
atomisation of chlorine ½Cl2(g) gt 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 121 kJ mol-1
1
NaCl(s)
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Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl Na(s) ½Cl2(g)
gt NaCl(s) Enthalpy of atomisation of
sodium Na(s) gt Na(g) Enthalpy of
atomisation of chlorine ½Cl2(g) gt
Cl(g) Ist Ionisation Energy of sodium Na(g)
gt 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 500 kJ mol-1
1
NaCl(s)
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Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl Na(s) ½Cl2(g)
gt NaCl(s) Enthalpy of atomisation of
sodium Na(s) gt Na(g) Enthalpy of
atomisation of chlorine ½Cl2(g) gt
Cl(g) Ist Ionisation Energy of sodium Na(g)
gt Na(g) e Electron Affinity of
chlorine Cl(g) e gt 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 - 364
kJ mol-1
1
NaCl(s)
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Born-Haber Cycle - NaCl
Enthalpy of formation of NaCl Na(s) ½Cl2(g)
gt NaCl(s) Enthalpy of atomisation of
sodium Na(s) gt Na(g) Enthalpy of
atomisation of chlorine ½Cl2(g) gt
Cl(g) Ist Ionisation Energy of sodium Na(g)
gt Na(g) e Electron Affinity of
chlorine Cl(g) e gt
Cl(g) Lattice Enthalpy of NaCl Na(g)
Cl(g) gt 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
6
2
Na(s) ½Cl2(g)
6
1
Lattice Enthalpy is exothermic. Oppositely
charged ions are attracted to each other.
NaCl(s)
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Born-Haber Cycle - NaCl
CALCULATING THE LATTICE ENTHALPY Apply Hesss Law
Na(g) Cl(g)
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- - - - The
minus shows you are going in the opposite
direction to the definition - (-364) -
(500) - (121) - (108) (-411) - 776 kJ
mol-1
1
5
6
4
3
2
4
Na(g) Cl(g)
Na(g) Cl(g)
3
Na(g) ½Cl2(g)
6
2
Na(s) ½Cl2(g)
1
NaCl(s)
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Born-Haber Cycle - NaCl
CALCULATING THE LATTICE ENTHALPY Apply Hesss Law
Na(g) Cl(g)
5
4
1 2 3456
Na(g) Cl(g)
Na(g) Cl(g)
3
Na(g) ½Cl2(g)
6
2
Na(s) ½Cl2(g)
1
NaCl(s)
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Construct a Born-Haber cycle for KCl
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Find easy exam qs,
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Born-Haber Cycle - MgCl2
Enthalpy of formation of MgCl2 Mg(s) Cl2(g)
gt MgCl2(s) Enthalpy of sublimation of
magnesium Mg(s) gt Mg(g) Enthalpy of
atomisation of chlorine ½Cl2(g) gt
Cl(g) x2 Ist Ionisation Energy of
magnesium Mg(g) gt Mg(g) e 2nd
Ionisation Energy of magnesium Mg(g) gt
Mg2(g) e Electron Affinity of
chlorine Cl(g) e gt Cl(g)
x2 Lattice Enthalpy of MgCl2 Mg2(g) 2Cl(g)
gt MgCl2(s)
1
Mg2(g) 2Cl(g)
2
5
6
Mg(g) 2Cl(g)
3
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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CaO
Ca2 (g) 2 e- O (g)
590 1150
142 844
193 248
?
?H formation
635
635 193 248 590 1150 142 844
?Hlattice ?Hlattice 635 193 248 590
1150 142 844 3518 kJ mol-1
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Construct a Born-Haber cycle for CoCl3
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CoCl3
Co3 (g) 3 e- 2 Cl (g)
757 1640 3230
3(364)
427 3(121)
5350
?H formation
?
?Hformation 427 3(121) 757 1640 3230
3(364) 5350 25 kJ mol-1
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Born-Haber cycles

• L.O.
• Define and apply the terms enthalpy of formation,
  ionisation enthalpy, enthalpy of atomisation of
  an element and of a compound, bond dissociation
  enthalpy, electron affinity, lattice enthalpy
  (defined as either lattice dissociation or
  lattice formation), enthalpy of hydration and
  enthalpy of solution.
• Construct BornHaber cycles to calculate lattice
• enthalpies from experimental data.