TOPIC 2 THERMOCHEMISTRY 20222023

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-1

101

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

102

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) = -2489kJ

Mg+(g) 2Cl-(g) 103

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

104

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) 105

ΔHa(Mg) = +149 kJ

Mg(s) + Cl2(g)

ΔHf = ?

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

106

ExampElex4ercise 3

Construct the Born-Haber cycle for CaO to
calculate the first electron affinity of oxygen.

Data :

Enthalpy of formation of CaO = -635 kJ mol-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

107

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) + 10
8

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

109

ii. Energy level diagram of CaO

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

IE2 = +1150 kJ EA1 = ? 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) 110

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

111

• 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)

112

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.

113

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

114

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) = -2489kJ

Na+(g) 2Cl-(g) 115

IE2(Na)

= +4562 kJ

Na2+(g) +

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

116

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) 117

ΔHa(Cl) = +121 kJ x 2

Na(g) + Cl2(g)

ΔHa(Na) = +107 kJ

Na(s) + Cl2(g)

ΔHf = ?

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
Ionic compound NaCl2 cannot form under standard
conditions because it is unstable (highly
endothermic)

118