1. Title: Lesson 7 Lattice Enthalpies and Enthalpy Change of Solution
Learning Objectives:
Understand the term lattice enthalpy
Identify and explain trends in lattice enthalpy
Explain how lattice enthalpy and enthalpies of hydration are related to enthalpy change of solution

2. Theoretical lattice enthalpies can be calculated from the ionic model
Assumption is the crystal is made up of perfectly spherical ions.
Assumption is the only interaction is due to electrostatic forces between ions.
The energy needed to separate the ions depends on the product of the ionic charges and the sum of the ionic radii. An increase in ionic radius decreases attraction. An increase in ionic charge increases the attraction.
K is a constant that depends on the geometry of the lattice
n and m are the magnitude of the charges on the ions.
Ionic radii can be determined from X-ray diffraction measurements

4. Solutions

6. Lattice enthalpies depend on the size and charge of the ions
What pattern can you see in the table? What is changing as you move down each group?
Lattice enthalpy decreases as the size of the cation or anion increases
Explain the difference in lattice enthalpies for these ionic structures...

7. Comparing Lattice Enthalpies
Study Table 18 in the data booklet.
What is the relationship between lattice enthalpy and the charge on an ion?
Explain why this occurs
Use suitable pairs of lattice enthalpies to illustrate your answer
What is the relationship between lattice enthalpy and the size of an ion:

8. Key Points
Lattice enthalpy is the equivalent of bond enthalpy in ionic compounds
Born-Haber cycles are a specialist type of Hess cycle
Each individual step in the reaction is included
Lattice enthalpy:
Increases with ionic charge
Decreases with ionic size

10. Solutions

12. Enthalpies of solution
Enthalpy of solution can be calculated by measuring the temperature change of the solution (q = mct)
Enthalpy of solution can also be obtained by measuring enthalpy changes for solutions with increasing volumes of water until a limit is reached.
Ionic compounds dissolved readily in water as the ions are strongly attracted to the polar solvent water. (See diagrams below)
Ions separated and surrounded by water are said to be hydrated.
The strength of interaction between the polar water molecules and the separated ions is given by their hydration enthalpies.

13. Enthalpy of hydration
‘The enthalpy of hydration of an ion is the enthalpy change that occurs when one mole of gaseous ions is dissolved to form an infinitely dilute solution of one mole of aqueous ions (under standard temperature and pressure).’
As there is a force of attraction between the ions and the polar water molecules, it is exothermic (negative enthalpy value)
QUESTION: What do we notice about the enthalpies as we move down the groups? (Think attraction and ionic radius)
Think: Solid is sublimed into gaseous ions, then plunged into water!

14. Hydration enthalpies of the ions are inversely proportional to the ionic radii
A = Constant
Down group 1 and 7 exothermic hydration enthalpy decreases
Across period 3 exothermic hydration enthalpy increases due to increased ionic charge and decrease in ionic radii.  Increased attraction to polar water molecules.
This leads to:
B = Constant
n = charge of ion

15. Enthalpy change of solution is related to the lattice enthalpy and hydration enthalpies

17. Solutions

18. Li+(g) + F(g) By Hess’s Law LEdiss = - (-616) Li+(g) + F-(g)
Q1. Use the following data to construct a Born-Haber cycle for the formation of LiF and then calculate a value for its lattice association enthalpy.
Enthalpy of atomisation of Li(s) = kJ mol-1
Enthalpy of atomisation of F2(g) = kJ mol-1
First ionisation energy of Li(g) = kJ mol-1
First electron affinity of F(g) = kJ mol-1
Enthalpy of formation of LiF(s) = kJ mol-1
Li+(g) + F(g)
I1[Li]
EA1[F]
By Hess’s Law
= - 334
= + 520
LEdiss =
- (-616)
Li+(g) + F-(g)
Li(g) + F(g)
+ 520
ΔHØat [Li(s)] + ΔHØat [½F2(l)]
+ (- 334) =
= kJ mole-1
LEdiss
Li(s) + ½F2(l)
ΔHØf [LiF]
= - 616
LiF(s)

19. The Ca2+ ion is smaller and more highly charged than Li+
Q2 The values of the lattice association energies of potassium iodide and calcium iodide experimentally determined from Born-Haber cycles and theoretically calculated from an ionic model are shown below.
Experimental Lattice Dissociation Energy / kJ mole-1
Theoretical Lattice Dissociation Energy / kJ mole-1
KI(s)
-651
-636
CaI2(s)
-2074
-1905
Explain why the experimental lattice energy of potassium iodide is less
exothermic than the experimental lattice energy of calcium iodide.
The Ca2+ ion is smaller and more highly charged than Li+
(i.e. Ca2+ more charge dense than Li+)
forces of attraction between Ca2+ and I- ions > K+ and I- ions
 more energy needed to overcome forces to separate Ca2+ & I-.
Explain why the experimental and theoretical values of the lattice energy are almost the same for potassium iodide, but are significantly different for calcium iodide.
Bonding in KI is pure ionic but
CaI2 shows significant covalency because
the I- ion is significantly polarised by the more charge dense
Ca2+ but not by the low charge density K+ ion.

20. Click Here For DATA M2+(g) + 2Cl(g) By Hess’s Law M2+(g) + 2Cl-(g)
Q3. Use a Born-Haber cycle for the formation of MCl2 to calculate the first electron affinity of chlorine, given that metal M has ΔHat = +150, I1 = +736 and I2 = kJ mole-1 and ΔHf of MCl2 is -642 kJ mole-1 and LEdiss of MCl2 is kJ mole-1.
Click Here For DATA
M2+(g) + 2Cl(g)
2EA1[Cl]
I1[M] + I2[M] =
By Hess’s Law
M2+(g) + 2Cl-(g) =
- (-642)
M(g) + 2Cl(g)
(121)
ΔHØat [M(s)] + 2ΔHØat [½Cl2(g)]
+ 2EA1[Cl]
= (+121)
LEdiss
M(s) + Cl2(g)
EA1[Cl]
= kJ mole-1 =
ΔHØf [MCl2]
= - 642
MCl2(s)

21. DATA : ALL VALUES IN kJ mole-1
ELEMENT
Hoat I1 I2 I3
EA1
EA2
Li(s)
+159
Li(g)
+520
Na(s)
+107
Na(g)
+496
K(s)
+89
K(g)
+419
Cs(s)
+76
Cs(g)
+376
Mg(s)
+148
Mg(g)
+738
+1451
Ca(s)
+178
Ca(g)
+590
+1145
Al(s)
+326
Al(g)
+578
+1817
+2745
1/2F2(g)
+79
F(g)
-328
1/2Cl2(g)
+121
Cl(g)
-349
1/2Br2(g)
+112
Br(g)
-325
1/2I2(g)
I(g)
-295
1/2O2(g)
+249
O(g)
-141
+798
1/2S8(g)
+279
S(g)
-200
+640