C9 THERMOCHEMISTRY

Consider the following data:

Ionic Lattice Energy Cation / Anion / Size
Compounds (kJ mol-1)

NaCl 771 Increasing charge on
MgCl2 2493 cation
MgCl2 2493
MgO 3889 Na+ < Mg2+

Increasing charge on
anion
Cl- < O2-

101

Ionisation Energy (IE)

• The energy needed to remove one mole of
electrons from one mole of atoms in gaseous
state to form one mole of gaseous ions.

Example:

First ionisation energy of magnesium
Mg(g)  Mg+(g) + e- IE1 = +740 kJ

Second ionisation energy of magnesium
Mg+(g)  Mg2+(g) + e- IE2 = +1500 kJ

102

Electron Affinity (EA)

• The heat change when one mole of gaseous
atoms gains one mole of electrons to form
one mole of gaseous ions.

Example:
First electron affinity of oxygen
O(g) + e-  Oˉ(g) EA1 = –141.4 kJ

Second electron affinity of oxygen
O–(g) + e-  O2–(g) EA2 = +844 kJ

103

Born-Haber Cycle

 Is an energy cycle that shows the formation
of ionic compounds.

 The process of ionic bond formation occurs in
a few stages which involves the enthalpies
change.

 The Born Haber cycle is often used to
calculate the lattice energy of an ionic
compound.

 It can be constructed as
i. energy cycle diagram or
ii. energy level diagram

104

Example

Construct the Born-Haber cycle for sodium
chloride and calculate the lattice energy of
sodium chloride.

Given;

Enthalpy of formation NaCl = -411 kJ mol-1

Enthalpy of atomisation of Na = +107 kJ mol-1

Ionization energy of Na = +502 kJ mol-1

Enthalpy of atomisation of Cl = +121 kJ mol-1

Electron affinity of Cl = -355 kJ mol-1

Lattice energy of NaCl =?

105

i. Energy cycle diagram of NaCl

Na(s) + ½Cl2(g) ΔHf (NaCl) NaCl(s)

ΔHa (Na) = -411 kJ
= +107 kJ
ΔHa(Cl)
Na(g)
= +121 kJ Hlattice = ?
IE(Na)
= +502 kJ Cl(g)

Na+(g) + EA(Cl)

= -355 kJ

Cl-(g)

106

From Hess’s Law
ΔHf(NaCl)= ΔHa(Na) + ΔHa(Cl) + IE(Na)

+ EA(Cl) + Hlattice
-411 = 107 + 121 + 502 + (-355) + Hlattice
Hlattice = -786 kJ mol-

107

ii. Energy level diagram of NaCl

energy
Na+(g) + e- + Cl(g)

IENa = +502 kJ EACl = -355 kJ

Na(g) + Cl(g) Na+(g) + Cl- (g)

ΔHa(Cl) = +121 kJ

Na(g) + ½ Cl2(g)

ΔHa(Na) = +107 kJ Hlattice

E=0 Na(s) + ½ Cl2(g)

ΔHf (NaCl) = -411 kJ

NaCl(s)

108

From Hess’s Law
ΔHf(NaCl)= ΔHa(Na) + ΔHa(Cl) + IE(Na)

+ EA(Cl) + Hlattice
-411 = 107 + 121 + 502 + (-355) + Hlattice
Hlattice = -786 kJ mol-

109

ExamplEe x3:ercise 2

Construct the Born-Haber cycle for MgCl2 to
calculate the enthalpy of formation of MgCl2.
Data :

Enthalpy of atomisation of Mg = +149 kJ mol–1

Enthalpy of atomisation of Cl2 = +121 kJ mol–1
First ionisation energy of Mg = +740 kJ mol–1

Second ionisation energy of Mg = +1500 kJ mol–1

Electron affinity of Cl = –355 kJ mol–1

Lattice energy of MgCl2 = –2489 kJ mol–1

110

i. Energy cycle diagram of MgCl2

Mg(s) + Cl2(g) ΔHf (MgCl2) = ? MgCl2(s)

ΔHa(Mg) ΔHa(Cl)
= +149 kJ
= 2(+121 kJ)
Mg(g)
2Cl(g)

IE1(Mg) EA(Cl) Hlattice
= +740 kJ =2(-355 kJ) = -2489

Mg+(g) 2Cl-(g) 111

IE2(Mg)
= +1500 kJ

Mg2+(g) +

From Hess’s Law
ΔHf (MgCl2)= ΔHa(Mg) + ΔHa(Cl) + IE1(Mg)+ IE2(Mg)

+ EA(Cl) + Hlattice
= 149 + 2(121) + 740 + 1500

+ 2(-355) + (-2489)
= -568 kJ mol-1

112

ii. Energy level diagram of MgCl2

energy
Mg2+(g) + 2e- + 2Cl(g)

IE2 = +1500 kJ EACl = -355 kJ X 2

Mg+(g) + e- + 2Cl(g) Mg2+(g) + 2Cl- (g)

IE1 = +740 kJ

Mg(g) + 2Cl(g)

E=0 ΔHa(Cl) = +121 kJ X 2 Hlattice = -2489 kJ

Mg(g) + Cl2(g) 113

ΔHa(Mg) = +149 kJ

Mg(s) + Cl2(g)

ΔHf = ? kJ

MgCl2(s)

From Hess’s Law
ΔHf (MgCl2)= ΔHa(Mg) + ΔHa(Cl) + IE1(Mg)+ IE2(Mg)

+ EA(Cl) + Hlattice
= 149 + 2(121) + 740 + 1500

+ 2(-355) + (-2489)
= -568 kJ mol-1

114

ExampElex4ercise 3

Construct the Born-Haber cycle for CaO to
calculate the first electron affinity of oxygen.
Data :
Enthalpy of formation of CaO = -635 kJmol-1
Enthalpy of atomisation of Ca = +193 kJ mol–1
Enthalpy of atomisation of O = +248 kJ mol–1
First ionisation energy of Ca = +590 kJ mol–1
Second ionisation energy of Ca = +1150 kJ mol–1
Second Electron affinity of O = +844 kJ mol–1
Lattice energy of CaO = -3513 kJ mol–1

115

i. Energy cycle diagram of CaO

Ca(s) + ½O2(g) ΔHf (CaO) CaO(s)
= -635 kJ

ΔHa(Ca) ΔHa(O)

= +193 kJ = +248 kJ)

Ca(g) O(g)

IE1(Ca) EA1(O) = ? Hlattice
= +590 kJ = -3513 kJ
O-(g)
Ca+(g)

IE2(Ca) EA2(O)
= +844 kJ
= +1150 kJ
O2-(g)
Ca2+(g) + 116

From Hess’s Law:
ΔHf(CaO) = ΔHa(Ca) + ΔHa(O) + IE1 + IE2 + EA1

+ EA2 + ΔHlattice
-635 = 193 + 248 + 590 + 1150 + EA1

+ 844 + (-3515)
EA1 = -145 kJ mol-1

117

ii. Energy level diagram of CaO

energy Ca2+(g) + O2- (g)
Ca2+(g) + 2e- + O(g)

IE2 = +1150 kJ EA1 = ? kJ EA2 = 844 kJ

Ca+(g) + e- + O(g) Ca2+(g) + e + O- (g)

E=0 IE1 = +590 kJ Hlattice = - 3513 kJ

Ca(g) + O(g) 118

ΔHa(O) = +248 kJ

Ca(g) + ½O2(g)

ΔHa(Ca) = +193 kJ

Ca(s) + ½O2(g)

ΔHf = - 635 kJ

CaO(s)

From Hess’s Law:
ΔHf(CaO) = ΔHa(Ca) + ΔHa(O) + IE1 + IE2 + EA1

+ EA2 + ΔHlattice
-635 = 193 + 248 + 590 + 1150 + EA1

+ 844 + (-3515)
EA1 = -145 kJ mol-1

119

 The Born-Haber cycle is useful for predicting
the stability or the existence of ionic
compounds.

 It can be done by considering:
a) The magnitude of the standard enthalpy

of formation of the ionic compound.
Generally compound with more negative
standard enthalpy of formation values are
more stable than compounds with more
positive (or less negative)

120

b) The lattice energy of the ionic compound.
The larger the lattice energy, the more
stable the solid and the more tightly the
ions are held.

121

Exercise 4

Construct a Born-Haber cycle to explain why ionic
compound NUasCel2thceandnaotat form under standard
conditions. below:

Enthalpy of atomisation of sodium = +107 kJmol-1
First ionization energy of sodium = +502 kJmol-1
Second ionization energy of sodium = +4562 kJmol-1
Enthalpy of atomisation of chlorine = +121kJmol-1
Electron affinity of chlorine = -355 kJmol-1
Lattice energy of NaCl2 = -2489 kJmol-1

122

i. Energy cycle diagram of NaCl2

Na(s) + Cl2(g) ΔHf (NaCl2) = ? NaCl2(s)

ΔHa (Na) ΔHa(Cl)
= +107 kJ
= 2(+121 kJ)
Na(g)
2Cl(g)

IE1(Na) EA(Cl) Hlattice
= +502 kJ =2(-355 kJ) = -2489

Na+(g)

IE2(Na)

= +4562 kJ

Na2+(g) + 2Cl-(g) 123

From Hess’s Law:

ΔHf (NaCl2) = ΔHa(Na) + ΔHa(Cl) + IE1 + IE2
+ EA(Cl) + ΔHlattice

= 107 + 2(121) + 502 + 4562 + 2(-355)
+(-2489)

ΔHf (NaCl2) = +2214 kJ mol-1
Ionic compound NaCl2 cannot form under standard
conditions because it is unstable

124

ii. Energy level diagram of NaCl2

energy
Na2+(g) + 2e- + 2Cl(g)

IE2 = +4562 kJ EACl = -355 kJ X 2

Na+(g) + e- + 2Cl(g) Na2+(g) + 2Cl- (g)

E=0 IE1 = +502 kJ Hlattice = -2489 kJ

Na(g) + 2Cl(g) 125

ΔHa(Cl) = +121 kJ x 2

Na(g) + Cl2(g)

ΔHa(Na) = +107 kJ

Na(s) + Cl2(g)

ΔHf = ? kJ

NaCl2(s)

From Hess’s Law:

ΔHf (NaCl2) = ΔHa(Na) + ΔHa(Cl) + IE1 + IE2
+ EA(Cl) + ΔHlattice

= 107 + 2(121) + 502 + 4562 + 2(-355)
+(-2489)

ΔHf (NaCl2) = +2214 kJ mol-1 126
Ionic compound NaCl2 cannot form under standard
conditions because it is unstable (highly
endothermic)