Ionic Crystals: How Do We Measure the
Stability of the Compound?
In the gaseous phase, use this formula


( Z e)( Z e)
Ek
d
Energy of the formation of an
ionic bond in the _________
phase
More energy is released,
this released energy is called
enthalpy of lattice
formation, _______
Lattice energy= -ΔHLF,
is a measure of the
stability in solid phase.
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What Can the Coulomb’s Law Tell Us?


( Z e)( Z e)
Ek
d
 Generally the attractive force between a pair of oppositely
charged ions increases with increased charge on the ions and
with decreased ionic sizes.
 Coulomb’s Law can be used as a rough estimate for lattice
energy.
2
Practice
Which of the following would you expect to have the largest lattice
energy?
1. MgCl2
2. MgBr2
3. NaBr
4. CaCl2
5. CaBr2
H LF 

3
cation ch arg e  anion ch arg e
r  r
Practice
Which of the following ionic compounds would you expect to
require more energy to decompose to gaseous ions than CaS(s)?
1. NaF(s)
2. SrS(s)
3. AlN(s)
4. MgBr2(s)
5. SrSe(s)
4
What Can ΔHLF Tell Us?
 ΔHLF is important because it affects such properties as
__________________________________________:
 If the cation and anion have the same charge and size, they
will pack well and have high ΔHLF. As such, it will require a
lot of energy to break up the lattice and they tend to be
insoluble. Ionic compounds with mismatched ions tend to
have low ΔHLF and are more soluble.
Generally higher lattice energy means higher
boiling point, higher melting point, and lower
solubility
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What Can ΔHLF Tell Us?
 Carbonates of alkaline earth metals decompose to the metal
oxide and carbon dioxide:
MgCO3 undergoes this reaction at 300 °C while CaCO3
requires heating to 840 °C. Because Mg2+ is smaller than
Ca2+, it is a better match for the small spherical O2- than the
larger nonspherical CO32Reactions occur to form
more stable products.
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Practice
Which of the following ionic compounds would you expect to have
the lowest boiling point?
1. KF
2. BeO
3. BN
4. KBr
5. NaCl
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What Happens When Lattice Energy Is Low?
 Compounds such as Mg(ClO4)2 and CuSO4 have small cations relative
to their anions. As such, they can literally pull water out of the air
(and are therefore useful as drying agents) to surround the small cation
with the relatively small water molecules. This gives a hydrated
solid:
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So How Do We Find ΔHLF?
 ΔHLF is a measure of how stable an ionic compound is in lattice form:
The enthalpy of lattice formation will always be a negative number. The
enthalpy of lattice decomposition will always be a positive number. It requires
input of energy to break bonds!
 In a lattice, every ion is attracted to all ions of opposite charge and repelled by
all ions with the same charge. Knowing the lattice structure, we could
calculate the lattice enthalpy based on these attractions and repulsions;
however, we will not be doing that in Chemistry 1000. Instead, we will
calculate lattice enthalpy using a Born-Fajans-Haber cycle.
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Standard States and Standard Enthalpy
Changes
Define a particular state as a standard state.
Standard enthalpy of reaction, H°
 The enthalpy change of a reaction in which all
reactants and products are in their standard states.
Standard State
 The pure element or compound at a pressure of
1 bar and at the temperature of interest.
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Hess’s Law
 Hess’s law of constant heat summation
 If a process occurs in stages or steps (even hypothetically),
the enthalpy change for the overall process is the sum of
the enthalpy changes for the individual steps.
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½N2(g) + ½O2(g) → NO(g)
H = +90.25 kJ
NO(g) + ½O2(g) → NO2(g)
H = -57.07 kJ
½N2(g) + O2(g) → NO2(g)
H = +33.18 kJ
What is the Born-Fajans-Haber cycle?
 A Born-Fajans-Haber cycle breaks down an overall
reaction into steps for each of which the enthalpy is known
or calculatable.
 Based on Hess’s Law the enthalpies of the steps should
add up to be the same as the enthalpy of the overall
reaction.
 In order to calculate the lattice energy, the following step
should be part of the cycle.
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Example
 Given the following data, calculate ΔHLF for NaCl(s)





13
Enthalpy of sublimation of Na = +108 kJ/mol
First ionization energy for Na = +496 kJ/mol
Enthalpy of bond dissociation for Cl2(g) = +243 kJ/mol
Enthalpy of electronic attraction for Cl = -349 kJ/mol
Enthalpy of formation for NaCl = -411 kJ/mol
Example
 Using Hess’s Law, we have
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Na(s) → Na(g)
H = +108
Na(g) → Na+(g) + e
1/2Cl2(g) → Cl(g)
Cl(g) + e → Cl-(g)
H = +496
H = +122
H = -349
Na+(g) + Cl-(g) → NaCl(s)
HLF = ?
½Cl2(g) + Na(s) → NaCl(s)
H = -411
Energy Diagram of the Born-Fajans-Haber
Cycle for the Formation of NaCl (s)
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One More Example
 Given the following data, calculate ΔHLF for LiF(s)





16
Enthalpy of sublimation of Li = +155.2 kJ/mol
Enthalpy of bond dissociation for F2(g) = +150.6 kJ/mol
First ionization energy for Li = + 520 kJ/mol
Enthalpy of electronic attraction for F = -328 kJ/mol
Enthalpy of formation for LiF = -594.1 kJ/mol
LiF Example
 Using Hess’s Law, we have
Li(s) → Li(g)
H = +155.2
1/2F2(g) → F(g)
Li(g) → Li+(g) + e
H = +150.6/2=+75.3
H = +520
F(g) + e → F-(g)
H = -328
½F2(g) + Li(s) → LiF(s)
17
H = -594.1
Energy Diagram of the Born-Fajans-Haber
Cycle for the Formation of LiF (s)
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Recap
•
Enthalpy
•
Standard state
•
Standard enthalpy change
•
Hess’s Law
•
Standard enthalpy of formation
•
Calculating enthalpy of reaction from enthalpies of formation
•
Lattice energy and ΔHLF
•
Born-Fajans-Haber cycle and the use of the cycle to calculate unknown
energies
•
Using lattice energy or ΔHLF to justify ionic formulas
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Readings for Next Class
21.2, 21.3
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