CHAPTER 11
Solutions and
Their Colligative
Properties
Enthalpy of Solution
• Dissolution of Ionic Solids:
– Enthalpy of solution (ΔHsoln) depends on:
• Energies holding solute ions in crystal lattice.
• Attractive force holding solvent molecules together.
• Interactions between solute ions and solvent
molecules.
– ΔHsoln = ΔHion-ion + ΔHdipole-dipole + ΔHion-dipole
– When solvent is water:
• ΔHsoln = ΔHion-ion + ΔHhydration
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Calculating Lattice Energies Using the Born-Haber Cycle
Lattice energy (U) is the energy required to completely separate one mole of
a solid ionic compound into gaseous ions. It is always endothermic.
e.g. MgF2(s)  Mg2+(g) + 2F-(g)
Q Q
Uk  
d
Q+ is the charge on the cation
Lattice energy (E) increases as Q
increases and/or as r decreases.
Compoun
d
Lattice
Energy (U)
MgF2
2957
Q= +2,-1
MgO
3938
Q= +2,-2
LiF
LiCl
1036
853
radius F <
radius Cl
Q- is the charge on the anion
d is the distance between the ions
k is a proportionality constant
that depends on lattice structure
Hess’ Law is used to calculate Ulattice
The calculation is broken down into a series of steps (cycles)
Steps:
1. sublimation of 1 mole of the metal
tricky 2. if present, breaking bond of a diatomic
= ΔHsub
= ½ ΔHBE
3. ionization of the metal(g) atom
= IE1 + IE2 etc
4. electron affinity of the nonmetal atom
= EA1 + EA2 etc
5. formation of 1 mole of the salt from ions(g)
= ΔHlast = -Ulattice
2
Tricky part
M(s) + ½ X2(g)  MX(s)
M+(g) + X(g)
ΔHf°
M+(g) + X(g)
EA1 + EA2 etc
IE1 + IE2 etc
M+(g) + X-(g)
M(g) + X(g)
tricky
M(g) + ½ X2(g)
ΔHsub
½ ΔHBE
ΔHlast = -Ulattice
M(s) + ½ X2(g)
ΔHf
MX(s)
Calculating Ulattice
Tricky part
ΔHf° =
ΔHsub + ½ ΔHBE +  IE +  EA + ΔHlast
Rearranging and solving for ΔHlast -
ΔHlast = ΔHf − ΔHsub − ½ ΔHBE −  IE −  EA
Ulattice = -ΔHlast
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Calculating Ulattice for NaCl(s)
Na+(g) + Cl(g)
Na+(g) + Cl(g)
EA1
IE1
Na+(g) + Cl-(g)
Na(g) + Cl(g)
Na(g) + ½ Cl2(g)
ΔHsub
½ ΔHBE
ΔHlast = -Ulattice
Na(s) + ½ Cl2(g)
ΔHf
NaCl(s)
ΔHlast = ΔHf − ΔHsub − ½ ΔHBE − IE1 − EA1
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Sample Exercise 11.2
In Sample Exercise 10.1, we predicted that the ion–
ion attraction in CaF2 is greater than in NaF. Confirm
this prediction by calculating the lattice energies of
(a) NaF and (b) CaF2 given the following
information:
Na(s) + ½ F2(g)  NaF2(s)
Ca(s) + F2(g)  CaF2(s)
Tricky part
tricky
Sample Exercise 11.2
tricky
Hlast =
-U
Hlast = - U
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