Energi Kisi Dan Born Haber

Lattice Energies in Ionic
Solids
Copyright 2007 Pearson Education, Inc., publishing as Benjamin Cummings
Ionic Bonding
Ionic bond: electrostatic force that holds
oppositely charge particles together
Formed between cations and anions
Example
Na+

Cl
Na
Cl
IE1 + EA1 = 496 kJ/mol 349 kJ/mol = 147 kJ/mol

m.p. =
801oC
H = 410.9 kJ/mol
Copyright McGraw-Hill 2009
Ionic Bonding
Energy must be input to remove an electron from a metal (IE is

positive). Generally, the energy released when the non metal
accepts an electron (EA) does not compensate for the IE.
So why do ionic compounds form?
m.p. = 801oC
+
Na + Cl
Na +
Cl
IE1 + EA1 = 496 kJ/mol 349 kJ/mol = 147 kJ/mol
H f = 410.9 kJ/mol
Microscopic View of NaCl Formation
Lattice energy = the energy required to
completely separate one mole of a solid

ioniccompound into gaseous ions (energi yang
dilepaskan ketika ion-ion bergabung

membentuk kristal)
+
-
+
+ -
NaCl(s) Na+(g) + Cl(g)
+
-
Hlattice = -788 kJ/mol
Coulombic attraction:
Q1 Q2
F
d2
Q1
Q = amount of charge
d = distance of separation
Q2
Lattice energy (like a coulombic force) depends on
Magnitude of charges
Distance between the charges
Calculating Lattice Energy

Lattice energy
1. Directly related to the product of the ion charges
and inversely related to the internuclear distance
2. Depends on the product of the charges of the ions
3. Inversely related to the internuclear distance, ro ,
and is inversely proportional to the size of the ions
Lattice energies of alkali metal iodides
The ionic radii sums for LiF and MgO are 2.01 and 2.06 ,
respectively, yet their lattice energies are 1030 and 3795 kJ/mol. Why
is the lattice energy of MgO nearly four times that of LiF?
Copyright
2009
Lattice Energy
Remember, IE and EA are for adding/removing an

electron to/from an atom in the gaseous state.
Ionic compounds are usually solids. The release of

energy on forming the solid, called the lattice energy is
the driving force for the formation of ionic compounds.
Because of high lattice energies, ionic solids tend to be

hard and have high melting points. Ionic compounds are
insulators in the solid state, because electrons are
localized on the ions, but conduct when molten or in
solution, due to flow of ions (not electrons).
Lattice energies can be calculated using Hesss law, via a

Born-Haber Cycle.
Lattice enthalpy
0
The lattice enthalpy change HL is the standard

molar enthalpy change for the following process:
M+(gas) + X-(gas) MX(solid)
HL
Because the formation of a solid from a gas of ions is

always exothermic, lattice enthalpies (defined in this
way !!) are usually negative numbers.
If entropy considerations are neglected the most
stable crystal structure of a given compound is the one
with the highest lattice enthalpy.
11
Born-Haber cycle: A method to determine

lattice energies
Copyright
2009
12
Born-Haber Cycles
eg for sodium chloride:
enthalpy H
Na+ (g) + e- + Cl (g)

H
first ionisation energy
Na (g) + Cl (g)
H
atomisation
first electron affinity
Na+ (g) + Cl- (g)
Na (g) + Cl2 (g)

H atomisation
Na (s) + Cl2 (g)
H
formation
NaCl (s)
H lattice
enthalpy
Born-Haber Cycles : applying Hesss Law

There are two routes from elements to ionic compound
enthalpy H
Na+ (g) + e- + Cl (g)

H
first ionisation energy
Na (g) + Cl (g)
H
atomisation
first electron affinity
Na+ (g) + Cl- (g)
Na (g) + Cl2 (g)
Apply
Hesss
Law:
H atomisation
Na (s) + Cl2 (g)
formation
H lattice
association
NaCl (s)
HatmNa + HatmCl + H1st IE + H1st EA + Hlattice = Hformation
Born-Haber Cycles: applying Hesss Law

HatmNa + HatmCl + H1st IE + H1st EA + Hlattice = Hformation
Rearrange to find the lattice energy:
Hlattice = Hformation - (HatmNa + HatmCl + H1st IE + H1st
EA)
So Born-Haber cycles can be used to calculate a measure of
ionic bond strength based on experimental data.
Ionic Bonding
Chem 59-250
The energy that holds the arrangement of ions together is called the lattice energy,
Hlattice , and this may be determined experimentally or calculated.
Uo is a measure of the energy released as the gas phase ions are assembled into a
crystalline lattice. A lattice energy must always be exothermic.
E.g.: Na+(g) + Cl-(g) NaCl(s) Hlattice = -788 kJ/mol
Lattice energies are determined experimentally using a Born-Haber cycle

such as this one for NaCl. This approach is based on Hess law and can be used to
determine the unknown lattice energy from known thermodynamic values.
Hsub
Hie
Na(s) Na(g) Na+(g)
Cl2(g) Cl(g) Cl-(g)

Hd
Hea
Hf
Lattice Energy, Hlattice

NaCl(s)
Ionic Bonding
Chem 59-250
Born-Haber cycle
Hsub
Hie
Na(s) Na(g) Na+(g)
Cl2(g) Cl(g) Cl-(g)

Hd
Hea
Hf

NaCl(s)
Hf = Hsub + Hie + 1/2 Hd + Hea + Hlattice

-411 = 109 + 496 + 1/2 (242) + (-349) + Hlattice
Hlattice = -788 kJ/mol
You must use the correct stoichiometry and signs to obtain the correct lattice energy.
Practice Born-Haber cycle analyses at: http://chemistry2.csudh.edu/lecture_help/bornhaber.html
Ionic Bonding
Chem 59-250
If we can predict the lattice energy, a Born-Haber cycle analysis can tell us
why certain compounds do not form. E.g. NaCl2
(Hie1 + Hie2)
Na(s) Na(g) Na+2(g)
Hsub
Cl2(g) 2 Cl(g) 2Cl-(g)
Hd Hea
Hf

NaCl2(s)
Hf = Hsub + Hie1 + Hie2 + Hd + Hea + Hlattice
Hf = 109 + 496 + 4562 + 242 + 2*(-349) + -2180

Hf = +2531 kJ/mol
This shows us that the formation of NaCl2 would be highly endothermic and very
unfavourable. Being able to predict lattice energies can help us to solve many
problems so we must learn some simple ways to do this.
Chem 59-250
Ionic Bonding
The equations that we will use to predict lattice energies for crystalline solids are
the Born-Mayer equation and the Kapustinskii equation, which are very similar to
one another. These equations are simple models that calculate the attraction and
repulsion for a given arrangement of ions.
Born-Mayer Equation:
Hlattice = (e2 / 4 e0) * (N zA zB / d0) * M * (1 (d* / d0))
Hlattice = 1390 (zA zB / d0) * M * (1 (d* / d0)) in kJ/mol
Kapustinskii equation :
Hlattice = (1210 kJ / mol) * (n zA zB / d0) * (1 (d* / d0))
Where:
e is the charge of the electron, 1,602 x 10-19 C
e0 is the permittivity of a vacuum 1,11x10-10 C2J-1m-1
N is Avogadros number, (e2 / 4 e0) = 2,307 x 10-28 J m
zA is the charge on ion A, zB is the charge on ion B
d0 is the distance between the cations and anions (in ) = r+ + rM is a Madelung constant
d* = exponential scaling factor for the repulsive term = 0.345
n = the number of ions in the formula unit
Figure 9.6
The Born-Haber cycle for lithium fluoride
Calculating Lattice Energy
Step 1: Convert elements to atoms in the gas state

e.g. for Li, Li (s) Li (g)
H1 = Hatomization
for F, 1/2 F2 (g) F (g)
H2 = 1/2 (Bond Energy)
Step 2: Electron transfer to form (isolated) ions

Li (g) Li+ (g) + e
H3 = IE1
F (g) + e F (g)
H4 = EA1
Step 3: Ions come together to form solid

Li+ (g) + F (g) LiF (s)
H5 = Lattice Energy
Overall: Li (s) + 1/2 F2 (g) LiF (s)
H = Hf = S(H15)
Lattice Energy = Hf (H1 + H2 + H3 + H4)
Periodic Trends in Lattice Energy

Coulombs Law
electrostatic force a
charge A X charge B
distance2
or, electrostatic energy a
charge A X charge B
distance
(since energy = force X distance)
So, lattice energy increases, as ionic radius decreases

(distance between charges is smaller).
Lattice energy also increases as charge increases.
Figure 9.7
Trends in lattice energy
Application of lattice enthalpy calculations:

(exercises)
Solubility of salts in water

other applications (Born-Haber cycle necessary):
thermal stabilities of ionic solids
stabilities of high oxidation states of cations
calculations of electron affinity data
lattice enthalpies and stabilities of non existent
compounds
35
The Relationship between Lattice Energies and

Physical Properties
1. Melting point
a. Temperature at which the individual ions have
enough kinetic energy to overcome the attractive forces
that hold them in place
b. Temperature at which the ions can move freely and

substance becomes a liquid
c. Varies with lattice energies for ionic substances that
have similar structures
The Relationship between Lattice Energies and

Physical Properties
2. Hardness
a. Resistance to scratching or abrasion
b. Directly related to how tightly the ions are held
together electrostatically
3. Solubility of ionic substances in water:
a. The higher the lattice energy, the less soluble the
compound in water
Predicting the Stability of Ionic Compounds

The most important factor in determining the stability of a
substance is the amount of energy it contains. Highly
energetic substances are generally less stable while
less-energetic substances are generally more stable,
making the lattice energy the most important factor in
determining the stability of an ionic compound, since
lattice energy is the sum of the electrostatic interaction
energies between ions in a crystal.
Positive enthalpy values make the formation of an ionic
substance energetically unfavorable. Negative enthalpy
values make the formation favorable.
838
Predicting the Stability of Ionic Compounds

The formation of an ionic compound will be exothermic
(enthalpy is negative) if and only if the enthalpy of step 5
in the BornHaber cycle, and therefore the lattice energy
(U), is a large negative number.
Compounds that have the same anion but different
cations, the trend for increasing lattice energies parallels
the trend of decreasing cation size
Compounds that have the same cation but different
anions, the trend for increasing lattice energies parallels
the trend of decreasing anion size
Compounds of ions with larger charges have greater
lattice energies than compounds of ions with lower
charges.
839
Ionic Solutions
Solubility affected by:
Energy of attraction (due Ion-dipole force) affects the solubility. Also
called hydration energy,
Lattice energy (energy holding the ions together in the lattice.
Related
to the charge on ions; larger charge means higher lattice energy.
Inversely proportional to the size of the ion; large ions mean
smaller lattice energy.
Solubility increases with increasing ion size, due to decreasing

lattice energy; Mg(OH)2(least soluble), Ca(OH)2, Sr(OH)2,
Ba(OH)2(most soluble) (lattice energy changes dominant).
Energy of hydration increases with for smaller ions than bigger
ones; thus ion size. MgSO4(most soluble),... BaSO4 (least
soluble.) Hydration energy dominant.
John A. Schreifels
Chemistry 212
840
Chapter 12-40
Kelarutan ion
-Entalpi kisi
AgCl(s) Ag+ (g) + Cl-(g)
solvasi
Ag+ (g) = H2O Ag+ (aq)
solvasi
Cl- (g) = H2O Cl-(aq)
H = 917 kj/mol
H= -475 kj/mol
H= -369 kj/mol
Maka kelarutan
AgCl(s) + H2O Ag+ (aq) + Cl-(aq) H = 73 kj/mol
Jika 3 diketahui maka reaksi keempat bisa dihitung,
tapi juga perlu memperhitungkan entropi kelarutan,
HSAB, struktur elektronik, kristal struktur dll.
John A. Schreifels
Chemistry 212
841
Chapter 12-41
untuk reaksi dibawah ini

Cs(s) + F2 (g) CsF(s)
Hf = -553.5 kj/mol
Diketahui (kJ/mol):
Energi atomisasi Cs
+76.5
Energi ionisasi pertama Ce
+375.7
Energy disosiasi ikatan F2
+158.8
afinitas elektron F
-328.2
Entalpi pembentukan standar CsF(s)
-553.5
Gambarkan siklus termodinamika (siklus Born Haber)
dari pembentukan CsF dari reaksi antara Cs(s)
dengan F2(g) berdasarkan nilai enthalpy yang
diberikan (dalam kJ/mol). Beri tanda atau keterangan
untuk tiap langkah dan hitung energy kisi dari reaksi 843
tersebut
John A. Schreifels
Chemistry 212
Chapter 12-43

Energi Kisi Dan Born Haber

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Lattice Energies in Ionic

Copyright 2007 Pearson Education, Inc., publishing as Benjamin Cummings

IE1 + EA1 = 496 kJ/mol 349 kJ/mol = 147 kJ/mol

Energy must be input to remove an electron from a metal (IE is

So why do ionic compounds form?

IE1 + EA1 = 496 kJ/mol 349 kJ/mol = 147 kJ/mol

Copyright McGraw-Hill 2009

Microscopic View of NaCl Formation

Copyright McGraw-Hill 2009

Lattice energy = the energy required to

completely separate one mole of a solid

dilepaskan ketika ion-ion bergabung

NaCl(s) Na+(g) + Cl(g)

Hlattice = -788 kJ/mol

Copyright McGraw-Hill 2009

Lattice energy (like a coulombic force) depends on

Copyright McGraw-Hill 2009

Calculating Lattice Energy

Lattice energies of alkali metal iodides

Copyright McGraw-Hill 2009

Remember, IE and EA are for adding/removing an

Ionic compounds are usually solids. The release of

Because of high lattice energies, ionic solids tend to be

Lattice energies can be calculated using Hesss law, via a

The lattice enthalpy change HL is the standard

M+(gas) + X-(gas) MX(solid)

Because the formation of a solid from a gas of ions is

Born-Haber cycle: A method to determine

Na+ (g) + e- + Cl (g)

first ionisation energy

first electron affinity

Na+ (g) + Cl- (g)

Na (g) + Cl2 (g)

Born-Haber Cycles : applying Hesss Law

Na+ (g) + e- + Cl (g)

first ionisation energy

first electron affinity

Na+ (g) + Cl- (g)

Na (g) + Cl2 (g)

HatmNa + HatmCl + H1st IE + H1st EA + Hlattice = Hformation

Born-Haber Cycles: applying Hesss Law

Lattice energies are determined experimentally using a Born-Haber cycle

Cl2(g) Cl(g) Cl-(g)

Lattice Energy, Hlattice

Cl2(g) Cl(g) Cl-(g)

Lattice Energy, Hlattice

Hf = Hsub + Hie + 1/2 Hd + Hea + Hlattice

Lattice Energy, Hlattice

Hf = Hsub + Hie1 + Hie2 + Hd + Hea + Hlattice

Hf = 109 + 496 + 4562 + 242 + 2*(-349) + -2180

The Born-Haber cycle for lithium fluoride

Calculating Lattice Energy

Step 1: Convert elements to atoms in the gas state

for F, 1/2 F2 (g) F (g)

H2 = 1/2 (Bond Energy)

Step 2: Electron transfer to form (isolated) ions

Step 3: Ions come together to form solid

Overall: Li (s) + 1/2 F2 (g) LiF (s)

Lattice Energy = Hf (H1 + H2 + H3 + H4)