Post on 02-Jan-2016
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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 Born–Haber cycles to calculate latticeenthalpies from experimental data.
In pairs recap the following definitions and illustrate them with examples:
Standard enthalpy of formation
IE
EA
Mean bond enthalpy
Second IE
Second EA
Standard enthalpy of atomisation, is the enthalpy change which accompanies the formation of one mole of gaseous atoms form the elements in its standard state under standard conditions.
Use iodine as an example
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)
Lattice Enthalpy Definition(s)Lattice Enthalpy Definition(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
1. Lattice Formation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice is formed from its isolated gaseous ions.’
Values highly EXOTHERMICEXOTHERMIC 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)
Lattice Enthalpy Definition(s)Lattice Enthalpy Definition(s)
NaCl(s)
Na+(g) + Cl–(g)
Lattice Enthalpy Definition(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 highly ENDOTHERMICENDOTHERMIC 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)
Calculating Lattice EnthalpyCalculating Lattice Enthalpy
SPECIAL POINTS
you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY
it is CALCULATED USING A BORN-HABER CYCLE
Calculating Lattice EnthalpyCalculating Lattice Enthalpy
SPECIAL POINTS
you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY
it is CALCULATED USING A BORN-HABER CYCLE
greater chargedensities of ions = greater attraction
= larger lattice enthalpy
Calculating Lattice EnthalpyCalculating Lattice Enthalpy
SPECIAL POINTS
you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY
it is CALCULATED USING A BORN-HABER CYCLE
greater chargedensities 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 relativevalues of Lattice and Hydration Enthalpies
Cl¯ Br¯ F¯ O2-
Na+ -780 -742 -918 -2478
K+ -711 -679 -817 -2232
Rb+ -685 -656 -783
Mg2+ -2256 -3791
Ca2+ -2259
Lattice Enthalpy ValuesLattice Enthalpy Values
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.
Units: kJ mol-1
Cl¯ Br¯ F¯ O2-
Na+ -780 -742 -918 -2478
K+ -711 -679 -817 -2232
Rb+ -685 -656 -783
Mg2+ -2256 -3791
Ca2+ -2259
Lattice Enthalpy ValuesLattice Enthalpy Values
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.
Cl¯Na+ 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.
K+
Born-Haber Cycle For Sodium ChlorideBorn-Haber Cycle For Sodium Chloride
kJ mol-1
Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) – 411
Enthalpy of atomisation of sodium Na(s) ——> Na(g) + 108
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
Lattice Enthalpy of NaCl Na+(g) + Cl¯(g) ——> NaCl(s) ?
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1This 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
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
Na(s) + ½Cl2(g)
NaCl(s)
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——> NaCl(s)
1
1
2
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
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
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——> NaCl(s)
Enthalpy of atomisation of sodium
Na(s) ——> Na(g)
1
2
1
3
2
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
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
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——> NaCl(s)
Enthalpy of atomisation of sodium
Na(s) ——> Na(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g)
1
2
3
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1
4
3
2
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
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
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Na+(g) + Cl(g)
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——> NaCl(s)
Enthalpy of atomisation 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
2
3
4
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1
54
3
2
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
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
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Na+(g) + Cl(g)
Na+(g) + Cl–(g)
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——> NaCl(s)
Enthalpy of atomisation 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
2
3
4
5
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1
6
54
3
2
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Na+(g) + Cl(g)
Na+(g) + Cl–(g)
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——> NaCl(s)
Enthalpy of atomisation 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
2
3
4
5
6
Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other.
Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other.
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1
6
54
3
2
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Na+(g) + Cl(g)
Na+(g) + Cl–(g)
CALCULATING THE LATTICE ENTHALPY CALCULATING THE LATTICE ENTHALPY
Apply Hess’s Law
16 5 4 3 2 = - - - - +
The minus shows you are going in the opposite direction to the definition
= - (-364) - (+500) - (+121) - (+108) + (-411)= - 776 kJ mol-1
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1
6
54
3
2
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Na+(g) + Cl(g)
Na+(g) + Cl–(g)
CALCULATING THE LATTICE ENTHALPY CALCULATING THE LATTICE ENTHALPY
Apply Hess’s Law
16 5 4 3 2 = - - - - +
The minus shows you are going in the opposite direction to the definition
= - (-364) - (+500) - (+121) - (+108) + (-411)= - 776 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)= - 776 kJ mol-1
16 5 4 3 2
H
CaO (s)
Ca2+ (g) + O2- (g)
H lattice energy
of formation
Ca (s) + ½ O2 (g)
H formation
H atomisation(s)
Ca (g) + O (g)
Ca2+ (g) + 2 e- + O (g)
H ionisation energy/iesH electron affinity/ies
CaO
193
248
590
1150
–142
+844
?
– 635 = 193 + 248 + 590 + 1150 – 142 + 844 + Hlattice
Hlattice = – 635 – 193 – 248 – 590 – 1150 + 142 – 844 = – 3518 kJ mol-1
–635
1
65
4
3
2
Mg(s) + Cl2(g)
MgCl2(s)
Mg(g) + Cl2(g)
Mg(g) + 2Cl(g) Mg2+(g) + 2Cl–(g)
7
Mg+(g) + 2Cl(g)
Mg2+(g) + 2Cl(g)
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
2
3
4
5
7
6
Born-Haber Cycle - MgCBorn-Haber Cycle - MgCll22
H
CoCl3 (s)
Co3+ (g) + 3 Cl- (g)
H lattice energy
of formation
Co (s) + 3/2 Cl2 (g)
H formation
H atomisation(s)
Co (g) + 3 Cl(g)
Co3+ (g) + 3 e- + 2 Cl (g)
H ionisation energy/iesH electron affinity/ies
CoCl3
427
3(121)
757
1640
3230
3(–364)
–5350
?
Hformation = 427 + 3(121) + 757 + 1640 + 3230 – 3(364) – 5350 = – 25 kJ mol-1
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 Born–Haber cycles to calculate latticeenthalpies from experimental data.