1. Chemsheets AS006 (Electron arrangement)
27/04/2018
BORN-HABER CYCLES

2. At the end of this lesson you will know
BORN-HABER CYCLES
At the end of this lesson you will know
The following terms:
Lattice Enthalpy, 2. Ionization energy, 3. Electron affinity, 4. Atomization
How to construct a Bohr-Haber Diagram
How to calculate any aspect within the diagram
4 minutes on Salt production from the its elements (Formation)
Berkley salt production

3. Standard enthalpy of formation
Definitions
The change in enthalpy when one mole of a substance in its standard state is formed from its elements in their standard states.
Standard enthalpy of formation

4. Formation of Sodium Chloride
kJmol-1
Formation of Sodium Chloride
+800
+800
+800
+700
+700
+700
+600
+600
+600
+500
+500
+500
+400
+400
+400 - +
+300
+300
+300
+200
+200
+200
+100
+100
+100
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100
-400
-300
-200
-100
-400
-300
-200
-100
H = -411kJmol-1 θ F
NaCl(s)

5. I am sure you can see the PROBLEM here , if you look above
Definitions
The energy released when one mole of an ionic compound is formed from its constituent ions in their gaseous states.
Atomization Energy
The energy required to break one mole of bonds between 2 atoms
Is called Bond Energy for Cl2 it is 244 kJ/mol .
Cl2  2 Cl
I am sure you can see the PROBLEM here , if you look above

6. Atomization of Sodium Or Sublimation
kJmol-1
Atomization of Sodium Or Sublimation
+800
+700
+600
+500
+400
+300
+200
Na(g) + 1/2 Cl2(g)
+100
H = +107kJmol-1 θ S
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100

7. Atomization of Chlorine Or Bond Dissociation
kJmol-1
+800
+700
+600
+500
+400
+300
Na(g) + Cl(g)
+200
H = +121kJmol-1 θ D
Na(g) + 1/2 Cl2(g)
+100
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100

8. First ionization energy
Definitions
The energy required to remove the outermost electron of each atom in one mole of an element in its gaseous state.
X (g)  X+ (g) + e −
First ionization energy

9. First Ionization of Sodium
kJmol-1
First Ionization of Sodium
+800
Na+(g) + Cl(g)
+700 e-
+600 e-
+500 e-
H = +502kJmol-1 θ I
+400 e- e-
+300
Na(g) + Cl(g)
+200 +
Na(g) + 1/2 Cl2(g)
+100
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100

10. Definitions
The energy released or absorbed when an electron is added to each atom in one mole of a substance in its gaseous state.
X (g) + e − → X− (g)
Electron affinity

11. Electron Affinity of Chlorine
kJmol-1
Electron Affinity of Chlorine
+800
+800
+800
Na+(g) + Cl(g)
+700
+700
+700
+600
+600
+600
H = -355kJmol-1 θ E
+500
+500
+500
+400
+400
+400
Na+(g) + Cl-(g)
+300
+300
+300
Na(g) + Cl(g)
Na(g) + Cl(g)
Na(g) + Cl(g)
+200
+200
+200 -
Na(g) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g) e-
+100
+100
+100
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100
-400
-300
-200
-100
-400
-300
-200
-100

12. Definition Lattice Energy:
The heat energy (enthalpy) released when one mole of solid in its standard state is formed from its ions in the gaseous state.
Ex: Na+(g) + Cl-(g)  NaCl(s) - + - + - - - + + + - - - + +

13. Born-Haber Cycle for Sodium Chloride
kJmol-1
Born-Haber Cycle for Sodium Chloride
+800
+800
+800
Na+(g) + Cl(g)
+700
+700
+700
+600
+600
+600
+500
+500
+500
+400
+400
+400
Na+(g) + Cl-(g)
+300
+300
+300
Na(g) + Cl(g)
Na(g) + Cl(g)
Na(g) + Cl(g)
Lattice Enthalpy for Sodium Chloride
+200
+200
+200
Na(g) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
+100
+100
+100
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g) - + -
-400
-300
-200
-100
-400
-300
-200
-100
-400
-300
-200
-100 -
H = -786 kJmol-1 θ L + + - - +
NaCl(s)

14. Hformation = sum of all other H’s
Metal ions, e-’s, non-metal atoms (g) H
H ionisation energy
H electron affinity
Gas atoms (g)
Gas ions (g)
Hatomisation(s)
H lattice energy of formation
Elements (std states)
H formation
Ionic compound (s)
Hformation = sum of all other H’s

15. kJ H= +107kJmol-1 Atomisation of sodium +800 +700 +600 +500 +400 +300
+200
Na(g) + 1/2 Cl2(g)
+100
H= +107kJmol-1
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100
Atomisation of sodium

16. Atomisation of chlorinekJ
+800
+700
+600
+500
+400
+300
Na(g) + Cl(g)
+200
H = +121kJmol-1
Na(g) + 1/2 Cl2(g)
+100
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100
Atomisation of chlorine

17. e- leaving e- e- e- leaving e- e- e- + kJ H = +502kJmol-1
+800
Na+(g) + Cl(g)
+700
e- leaving e- e-
+600
e- leaving
+500 e- e-
+400
H = +502kJmol-1 e-
+300
Na(g) + Cl(g)
+200 +
Na(g) + 1/2 Cl2(g)
+100
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100
First Ionisation of sodium

18. First electron affinity of chlorinekJ kJ kJ
+800
+800
+800
Na+(g) + Cl(g)
+700
+700
+700
+600
+600
+600
H = -355kJmol-1
+500
+500
+500 e-
+400
+400
+400
Na+(g) + Cl-(g)
+300
+300
+300
Na(g) + Cl(g)
Na(g) + Cl(g)
Na(g) + Cl(g)
+200
+200
+200 - e-
Na(g) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
+100
+100
+100
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g)
-400
-300
-200
-100
-400
-300
-200
-100
-400
-300
-200
-100
First electron affinity of chlorine

19. H = -411kJ/mol Hformation = 107+121+502+-355+-786
Answer is -411 kJ/mol for heat of formation NaCl kJ kJ kJ
+800
+800
+800
Na+(g) + Cl (g)
+700
+700
+700
+600
+600
+600
+500
+500
+500
+400
+400
+400
Na+(g) + Cl-(g)
+300
+300
+300
Na(g) + Cl(g)
Na(g) + Cl(g) + 502
Na(g) + Cl(g)
+200
+200
+200 - -
Na(g) + 1/2 Cl2(g) + 121
Na(g) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g) +
+100
+100
+100 + - +
Na(s) + 1/2 Cl2(g) + 107
Na(s) + 1/2 Cl2(g)
Na(s) + 1/2 Cl2(g)
H = -786 kJmol-1 - + -
-400
-300
-200
-100
-400
-300
-200
-100
-400
-300
-200
-100
H = -411kJ/mol
Lattice enthalpy for sodium chloride
NaCl(s)

20. Ok, so we just found the heat of formation for NaCl
Using
Born-Haber.
Now lets try another (CaO) and find something different like Lattice Energy.

21. CaO Ca2+(g) + O2– (g) Ca2+(g) + 2e– + O (g) Ca+(g) + e– + O(g)
2nd electron affinity of O
2nd ionisation of Ca
1st electron affinity of O
Ca+(g) + e– + O(g)
1st ionisation of Ca
Ca2+(g) + e– + O – (g)
Ca(g) + O(g)
atomisation of O
Ca(g) + ½O2(g)
Lattice formation of CaO
atomisation of Ca
Ca(s) + ½O2(g)
formation of CaO
CaO (s)

22. CaO Ca2+(g) + O2– (g) Ca2+(g) + 2e– + O (g) Ca+(g) + e– + O(g)
2nd electron affinity of O
2nd ionisation of Ca
1st electron affinity of O
–142
1150
844
Ca+(g) + e– + O(g)
1st ionisation of Ca
Ca2+(g) + e– + O – (g)
590
Ca(g) + O(g)
atomisation of O
248
Ca(g) + ½O2(g)
Lattice formation of CaO
atomisation of Ca X
193
Ca(s) + ½O2(g)
–635
formation of CaO
CaO (s)
Hf CaO= –635;
Atomisation calcium = +193; Atomisation oxygen = +248
1st IE Ca = +590; 2nd IE Ca = +1150
1st e- affinity of oxygen = – nd e- affinity= +844

23. CaO Hlattice = – 635 – 193 – 248 – 590 – 1150 + 142 – 844
Ca2+(g) + O2– (g)
Ca2+(g) + 2e– + O (g)
2nd electron affinity of O
2nd ionisation of Ca
1st electron affinity of O
–142
1150
844
Ca+(g) + e– + O(g)
1st ionisation of Ca
Ca2+(g) + e– + O – (g)
590
Ca(g) + O(g)
atomisation of O
248
Ca(g) + ½O2(g)
Lattice formation of CaO
atomisation of Ca X
193
Ca(s) + ½O2(g)
–635
formation of CaO
CaO (s)
– 635 = – Hlattice
– 635 = – Hlattice
Hlattice = – 635 – 193 – 248 – 590 – – 844
= – 3518 kJ mol-1
© A May-2016

24. Ok, so we just found the heat of Lattice Energy
For CaO
Now try MgBr2 and
Find it’s
Lattice energy
By drawing
and Using
Born-Haber Diagram.

25. MgBr2 Mg2+(g) + 2e– + 2Br (g) Mg+(g) + e– + 2Br(g) Mg2+(g) 2Br – (g)
2nd ionisation of Mg
2 x electron affinity of Br
Mg+(g) + e– + 2Br(g)
Mg2+(g) 2Br – (g)
1st ionisation of Mg
Mg(g) + 2Br(g)
2 x atomisation of Br
Mg(g) + Br2(l)
Lattice formation of MgBr2
atomisation of Mg
Mg(s) + Br2(l)
formation of MgBr2
MgBr2 (s)

26. MgBr2 Hformation = – 518 kJ mol-1 Mg2+(g) + 2e– + 2Br (g)
2nd ionisation of Mg
2 x electron affinity of Br
1450
2(–342)
Mg+(g) + e– + 2Br(g)
Mg2+(g) 2Br – (g)
1st ionisation of Mg
736
Mg(g) + 2Br(g)
2 x atomisation of Br
2(112)
Mg(g) + Br2(l)
Lattice formation of MgBr2
atomisation of Mg
150
–2394
Mg(s) + Br2(l) X
formation of MgBr2
MgBr2 (s)
Hformation = (112) (–342) –2394
Hformation = (112) (–342) –2394
Hformation = – 518 kJ mol-1
© A May-2016

27. Next, find Lattice energy for Na2S

28. Na2S Hlattice = – 370 – 2(107) – 279 – 2(496) + 200 – 649
2Na+(g) + S2–(g)
2nd electron affinity of S
2Na+(g) + 2e– + S(g)
1st electron affinity of S
2 x 1st ionisation of Na
649
–200
2(496)
2Na+(g) + e– +S –(g)
2Na(g) + S(g)
atomisation of S
279
2Na(g) + S(s)
Lattice formation of Na2S
Na2S
2 x atomisation of Na ?
2(107)
2Na(s) + S(s)
formation of Na2S
–370
Na2S(s)
– 370 = 2(107) (496) – Hlattice
Hlattice = – 370 – 2(107) – 279 – 2(496) – 649
= – 2304 kJ mol-1

30. CaBr2 Hformation = 193 + 2(112) + 590 + 1150 – 2(342) – 2125
Ca2+(g) + 2e– + 2Br (g)
2nd ionisation of Ca
2 x electron affinity of Br H
1150
2(–364)
Ca+(g) + e– + 2Br(g)
Ca2+(g) + 2Br – (g)
1st ionisation of Ca
590
Ca(g) + 2Br(g)
2 x atomisation of Ca
2(121)
Ca(g) + Br2(l)
Lattice formation of CaBr2
atomisation of Ca
193
–2125
Ca(s) + Br2(l) ?
formation of CaBr2
CaBr2 (s)
Hformation = (112) – 2(342) – 2125
= – 652 kJ mol-1

31. Al2O3 2Al3+(g) + 6e– + 6O(g) 2Al3+(g) + 6O2–(g) 2Al2+(g) + 4e– + 3O(g)
2 x 3rd ionisation of Al
2(2740)
3 x 1st electron affinity of O
2Al2+(g) + 4e– + 3O(g)
3 x 2nd electron affinity of O
2 x 2nd ionisation of Al
3(–142)
3(844)
2(1820)
2Al+(g) + 2e– + 3O(g)
2Al3+(g) + 3e– + 3O –(g)
2 x 1st ionisation of Al
2(577)
2Al(g) + 3O(g)
Lattice formation of Al2O3
3 x atomisation of O
3(248)
2Al(g) + 3/2O2(g) ?
Al2O3
2 atomisation of Al
2(314)
2Al(s) + 3/2O2(g)
formation of Al2O3
–1669
Al2O3(s)
–1669 = 2(314) + 3(248) + 2(577) + 2(1820) + 2(2740) + 3(-142) + 3(844) + H
H = –1669 – 2(314) – 3(248) – 2(577) – 2(1820) – 2(2740) + 3(142) – 3(844)
= – kJ mol-1

32. CaI2
Ca2+(g) + 2e– + 2I (g)
2nd ionisation of Ca
2 x electron affinity of I H
1150
2 X
Ca+(g) + e– + 2I(g)
Ca2+(g) + 2I – (g)
1st ionisation of Ca
590
Ca(g) + 2I(g)
2 x atomisation of I
2(107)
Ca(g) + I2(s)
Lattice formation of CaI2
atomisation of Ca
193
–2054
Ca(s) + I2(s)
–535
formation of CaI2
CaI2 (s)
– 535 = (107) X – 2054
2X = – 535 – 193 – 2(107) – 590 – 1150 – 2054
2X = X = – 314 kJ mol-1

33. CuO X = – 155 – 248 – 745 – 1960 + 142 – 844 + 4149 = + 339 kJ mol-1
Cu2+(g) + O2–(g)
Cu2+(g) + 2e– + O(g)
2nd electron affinity of O
2nd ionisation of Cu
1st electron affinity of O
1960
844
Cu+(g) + e– + O(g)
–142
1st ionisation of Cu
Cu2+(g) + e– +O –(g)
745
Cu(s) + O(g)
Lattice formation of CuO
atomisation of O
248
Cu(g) + ½O2(g)
–4149
CuO
atomisation of Cu X
Cu(s) + ½O2(g)
formation of CuO
–155
CuO(s)
– 155 = X – – 4149
X = – 155 – 248 – 745 – –
= kJ mol-1

34. AgI
Ag+ (g) + e- + I (g) H
1st ionisation of Ag
–314
1st electron affinity of I
732
Ag (g) + I (g)
Ag+ (g) + I- (g)
Atomisation of I
107
Ag (s) + ½ I2 (s)
Lattice enthalpy of formation of AgI ?
Atomisation of Ag
289
Ag (s) + ½ I2 (s)
–62
Formation of AgI
AgI (s)
–62 = – HLE
HLE = –289 – 107 – – 62
= –876 kJ mol-1

35. BORN-HABER CYCLES You should now know The following terms:
A) Lattice Enthalpy, B) Ionization energy, C) Electron affinity, D) Atomization
How to construct a Bohr-Haber Diagram
How to calculate any aspect within the diagram