MODULE 4
 

4.1      ENERGETICS

 

Simple treatment (including calculations) of the Born-Haber cycle for the halides of Group I and II. 

(Lattice enthalpy will be regarded as the enthalpy of lattice breaking) 

 

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

 

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.)

 

 

M

X

Compound

ΔHFθ

kJmol-1

ΔHat.

kJmol-1

ΔHI.E.

kJmol-1

ΔHat.

kJmol-1

ΔHe.a.

kJmol-1

KBr

-392

+89

+420

+112

-342

BaCl2

-860

+175

(1st) +500

(2nd) +1000

+121

-364

 

Use of Lattice Energies

 

·        The melting points of ionic solids depend on their lattice enthalpies.. The greater the lattice energy the higher the melting point of the compound.

 

 

NaF

NaCl

NaBr

NaI

ΔHlatt. kJ mol-1

915

781

743

699

m. pt.  oC

995

808

750

662

 

·        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.

 

Compound

Theoretical lattice energy

kJ mol-1

Calculated lattice energy (via Born-Haber cycle)

kJ mol-1

NaCl

766

776

NaBr

731

742

NaI

686

699

AgCl

768

890

AgBr

759

877

AgI

736

867

 

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.

 

 

Group I

Group II

Li

Na

K

Rb

Cs

Be

Mg

Ca

Sr

Ba

F

1029

915

813

779

735

3456

2883

2583

2427

2289

Cl

849

776

710

685

676

2983

2489

2197

2109

2049

Br

804

742

671

656

644

2895

2416

2125

2046

1937

I

753

699

643

624

607

2803

2314

2038

1954

1841

 

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-.

 

 

Li+F-

Mg2+O2-

K+Cl-

Ba2+S2-

Interionic distance (nm)

 

0.204

 

0.206

 

0.314

 

0.318