ENERGETICS - SUMMARYNOTES

1 ENERGETICS

Simple treatment (including calculations) of the Born-Haber cycle for the halides of Group I and
(Lattice enthalpy will be regarded as the enthalpy of lattice breaking)

When you have finished this section you should be able to:

Explain and use the term ‘lattice enthalpy’ as a measure of ionic bond strength.
Construct Born-Haber cycles to calculate the lattice enthalpy of a simple ionic solid

e.g NaCl, MgCl2, using relevant energy terms (enthalpy changes of formation, ionisation energy, enthalpy of atomisation and electron affinity.

Explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of a lattice energy.

Lattice Enthalpy :

Bond enthalpies provide a measure of the strength of covalent bonds. Ionic bonding is an electrostatic attraction between oppositely charged ions.

The attraction acts in all directions, resulting in a giant ionic lattice containing many ions. For ionic compounds the corresponding enthalpy is the lattice enthalpy (also called the lattice energy).

Lattice enthalpy indicates the strength of the ionic bonds in an ionic lattice.

The standard molar lattice enthalpy is the energy required to convert one mole of a solid ionic compound into its constituent gaseous ions under standard conditions.

e.g. NaCl (s) Na+ (g) + Cl- (g)

Born-Haber cycle
This is an application of Hess’s Law and can be used to calculate lattice energies.
Lattice enthalpies cannot be determined directly by experiment and must be calculated indirectly using Hess’s Law and other enthalpy changes that can be found experimentally.

The energy cycle used to calculate a lattice enthalpy is the Born-Haber cycle.
The basis of a Born-Haber is the formation of an ionic lattice from its elements.
In general for an ionic compound a Born-Haber cycle can be written as:

M+ (g) + X- (g)

M+ (g) + X (g) ΔHe.a.

ΔHdiss.
M+ (g) + ½ X2 (g)

ΔHI.E. ΔHlatt.
M (g) + ½ X2 (g)

ΔHsub.
M (s) + ½ X2 (g)

ΔHFθ

M+X- (s)

According to Hess’s Law
ΔHlatt. = (-ΔHFθ) + ΔHsub. + ΔHI.E. + ΔHdiss. + ΔHe.a.

ΔHfθ = enthalpy of formation of MX (s)
ΔHsub. = enthalpy of sublimation of M (s)
ΔHI.E. = ionisation energy
ΔHdiss. = dissociation energy of X2 (g)
ΔHe.a.= electron affinity of X (g)
ΔHlatt. = lattice energy

N.B. The actual figures may be positive or negative and are simply substituted in the above equation.

Consider the reaction between sodium and chlorine to form sodium chloride.
Na (s) + ½ Cl2 (g) NaCl (s)

The reaction can be considered to occur by means of the following steps:

Vapourisation of sodium
Na (s) Na (g) ΔHsub.
The standard enthalpy of sublimation or vaporisation is the enthalpy change when one mole of sodium atoms are vaporised. This is an endothermic process and can be determined experimentally.

Ionisation of sodium
Na (g) Na+ (g) + e- ΔHI.E
The standard enthalpy of ionisation is the energy required to remove one mole of electrons from one mole of gaseous atoms. This is endothermic and can be determined by spectroscopy.

· Dissociation of chlorine molecules
Cl2 (g) 2Cl (g) ΔHdiss.
The standard bond dissociation enthalpy is the energy required to dissociate one mole of chlorine molecules into atoms (i.e. to break one mole of bonds). This is also endothermic and can be determined by spectroscopy.

Ionisation of chlorine atoms

Cl (g) + e- Cl- (g) ΔHe.a.
The electron affinity of chlorine is the energy released when one mole of gaseous chlorine atoms accepts one mole of electrons forming one mole of chloride ions.
Reaction between the ions

Na+ (g) + Cl- (g) NaCl (s) -ΔHlatt.
This is the reverse of the lattice energy. The standard lattice enthalpy is the energy absorbed when one mole of solid sodium chloride is separated into its gaseous ions. It has a positive value and cannot be determined experimentally.


A Born-Haber cycle can be drawn:
Na+ (g) + Cl- (g)

ΔHe.a.
Na+ (g) + Cl (g)

ΔHI.E.
Na (g) + Cl (g) ΔHlatt.

ΔHdiss.
Na (g) + ½ Cl2 (g)

ΔHsub. ΔHFθ
Na (s) + ½ Cl2 (g) NaCl (s)

Applying Hess’s Law
ΔHsub. + ΔHI.E. + ΔHdiss. + ΔHe.a. - ΔHlatt. – ΔHfθ = 0

Calculate the lattice enthalpy of sodium chloride given
ΔHfθ (NaCl) = -411 kJ mol-1
ΔHsub. (Na) = 108.3 kJ mol-1
ΔHI.E. (Na) = 500 kJ mol-1
ΔHdiss. (Cl) = 121 kJ mol-1
ΔHe.a. (Cl) = -364 kJ mol-1
Answer ΔHlatt. = +776 kJ mol-1


Exercise 1
Draw Born-Haber cycles for each of the following ionic compounds and calculate their lattice enthalpies.
(Note : ΔHat. of an element is the energy required to form one mole of gaseous atoms from the element.)


Solubility of ionic compounds is usually governed by its lattice energy. In general the higher the lattice energy the lower the solubility.
(See enthalpy of solution below)
A comparison of calculated and theoretical lattice energies gives an indication of the degree of covalent character in an ionic compound. The greater the difference between the two values the more covalent the compound.

The close agreement between the theoretical and experimental values for the alkali metal halides provides strong evidence that the simple ionic model of a lattice, composed of discrete spherical ions with an even charge distribution, is a very satisfactory one.

For the silver halides the theoretical values are about 15% less than the experimental values based on the Born-Haber cycle. This indicates that the simple ionic model is not very satisfactory.

When there is a large difference in electronegativity between the ions in a crystal , as in the case of the alkali metal halides then the ionic model is satisfactory. However as the difference in electronegativity gets smaller, as in the case of the silver halides, the bonding is stronger than the ionic model predicts.

The bonding in this case is not purely ionic but intermediate in character between ionic and covalent. The ionic bonds have been polarised (Fajans rules) giving some covalent character.


Exercise 2

The figures below give a list of lattice energies in kJ mol-1. Try to find as many patterns and trends in the figures as you can.

RbF -779 CaI2 -2038
BeF2 -3456 CaCl2- 2197
BaI2- 1841 MgCl2- 2489
MgBr2- 2416 KCl -710
CaBr2 -2125 NaF- 915
CsI- 607 LiF- 1029
KBr -671 MgI2 -2314
BaF2 -2289 LiBr- 804
CsBr -644 RbI- 624
LiI -753 SrBr2- 2046
BeI2 -2803 NaBr- 742
LiCl- 849 SrCl2- 2109
NaI -699 BeBr2- 2895
BeCl2- 2983 KF -813
CsCl -676 BaBr2- 1937
KI- 643 CaF2- 2583
MgF2 -2883 NaCl -776
RbCl- 685 SrF2 -2427
SrI2- 1954 RbBr -656
CsF -735 BaCl2- 2049



Trends in lattice enthalpy explained in terms of ionic radius and charge.

Consider the ionisation of an ionic solid MX.
MX (s) Mn+ (g) + Xn- (g)
The ease of separation of the ions and hence the lattice energy is determined by the size of the ions and their charge.


Effect of ionic size
As the ionic radius of both Mn+ and Xn- the lattice energy decreases. The attractive force between the ions decreases and they become easier to separate.
e.g. LiBr 804 kJ mol-1 BeCl2 2983 kJ mol-1
NaBr 742 MgCl2 2489
KBr 671 CaCl2 2197
RbBr 656 SrCl2 2109
CsBr 644 BaCl2 2049

As we descend both Groups I and II the lattice energies become less positive.
For any given metal the lattice energy also decreases in passing from the fluoride to the iodide.
e.g. NaF 915 kJ mol-1 SrF2 2427 kJ mol-1
NaCl 776 SrCl2 2109
NaBr 742 SrBr2 2046
NaI 699 SrI2 1954
This is due to an increase in ionic size from F- to I- which increases the internuclear distance. There is a corresponding decrease in attractive force and hence lattice energy.
When the internuclear distances are about equal, as for RbF and LiI for example, then the lattice energies are almost equal.


Effect of ionic charge
As the charge on Mn+ increases there is a greater attractive force between the ions and lattice energies increase. In addition, the decrease in size of Mn+ with increasing charge increases the attractive force between the ions and also increases the lattice energy.
The ionic radius of the Na+ and Ca2+ ions are very similar. However the lattice energy of CaCl2 is about 3 times that of NaCl.

NaCl 776 kJ mol-1 CaCl2 2197 kJ mol-1

This is due to the increased charge on the metal ion giving greater electrostatic attraction.
In general Group II halides have a lattice energy about three times that of the equivalent Group I halide.
Beryllium halides have considerable covalent character and the lattice energies are bigger than expected.

Exercise 3
What would be the effect on lattice energy of increasing the charge on Xn- ? (i.e. forming a Group VI compound rather than a Group VII compound).
Describe and explain the trends.

For comparable interionic distances the lattice energy would be bigger for X2- ions compared with X- ions. This is because X2- ions exert a stronger electrostatic field compared to X- ions due to the increased charge.
For comparable interionic distances the lattice energy is approximately four times higher when M2+ X2- ions are involved comared to M+ X-.

Exercise 4
Describe and explain the trends in hydration energy.