# Lattice Enthalpies and Enthalpies of Solution

```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
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.
can be determined from X-ray diffraction measurements


Solutions
Lattice enthalpies depend on the size
and charge of the ions
Explain the difference in lattice
enthalpies for these ionic
structures...
What pattern
can you see in
Lattice
enthalpy
the table?
decreases
What is as the
size
of the cation
changing
as
or you
anion
increases
move
down each
group?
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
What is the relationship between lattice enthalpy and the
size of an ion:


Explain why this occurs
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
Solutions
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.
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).’
Think: Solid is sublimed into
gaseous ions, then plunged into
water!

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
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.


B = Constant
n = charge of ion

Enthalpy change of solution is related to the
lattice enthalpy and hydration enthalpies
Solutions
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) = +160.0 kJ mol-1
Enthalpy of atomisation of F2(g) = +79.0 kJ mol-1
First ionisation energy of Li(g) = +520.0 kJ mol-1
First electron affinity of F(g) = -334.0 kJ mol-1
Enthalpy of formation of LiF(s) = -616.0 kJ mol-1
Li+(g) + F(g)
EA1[F]
I1[Li]
= - 334
= + 520
Li+(g)
Li(g) + F(g)
+
F-(g)
ΔH&Oslash;at [Li(s)] + ΔH&Oslash;at [&frac12;F2(l)]
= + 160 + 79
LEdiss
Li(s) + &frac12;F2(l)
ΔH&Oslash;f [LiF]
= - 616
LiF(s)
By Hess’s Law
LEdiss = - (-616)
+ 160 + 79
+ 520
+ (- 334)
= + 1041 kJ mole-1
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 &gt; K+ and I- ions
 more energy needed to overcome forces to separate Ca2+ &amp; 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.
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 = +1450 kJ mole-1 and
ΔHf of MCl2 is -642 kJ mole-1 and LEdiss of MCl2 is +2526 kJ mole-1.
For DATA
M2+(g) + 2Cl(g)
2EA1[Cl]
I1[M] + I2[M]
= + 736 + 1450
M2+(g) + 2Cl-(g)
+2526 = - (-642)
+ 150 + 2(121)
+ 736 + 1450
+ 2EA1[Cl]
M(g) + 2Cl(g)
ΔH&Oslash;at [M(s)] + 2ΔH&Oslash;at [&frac12;Cl2(g)]
= + 150 + 2(+121)
LEdiss
M(s) + Cl2(g)
= + 2526
ΔH&Oslash;f [MCl2]
= - 642
MCl2(s)
By Hess’s Law
 EA1[Cl]
= - 347 kJ mole-1
DATA :
ALL VALUES IN kJ mole-1
ELEMENT
Hoat
ELEMENT
I1
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
1/2F2(g)
+79
F(g)
-328
1/2Cl2(g)
+121
Cl(g)
-349
1/2Br2(g)
+112
Br(g)
-325
1/2I2(g)
+107
I(g)
-295
1/2O2(g)
+249
O(g)
-141
+798
1/2S8(g)
+279
S(g)
-200
+640
I2
I3
EA1
EA2
+2745
```