energi kisi dan born haber(1)
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Copyright © 2007 Pearson Education, Inc., publishing as Benjamin Cummings
Lattice Energies in Ionic Solids
Copyright McGraw-Hill 2009 2 Copyright McGraw-Hill 2009
Ionic Bonding
Na• Cl• ••
••
•• + Na+ Cl ••
••
••
••
+
IE1 + EA1 = 496 kJ/mol 349 kJ/mol = 147 kJ/mol
fH = – 410.9 kJ/mol m.p. = 801oC
• Ionic bond: electrostatic force that holds
oppositely charge particles together
• Formed between cations and anions
• Example
Copyright McGraw-Hill 2009 3 Copyright McGraw-Hill 2009
Ionic Bonding
Na• Cl• ••
••
•• + Na+ Cl ••
••
••
••
+
IE1 + EA1 = 496 kJ/mol 349 kJ/mol = 147 kJ/mol
fH = – 410.9 kJ/mol m.p. = 801oC
• 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?
Copyright McGraw-Hill 2009 4
Microscopic View of NaCl Formation
Copyright McGraw-Hill 2009 5 Copyright McGraw-Hill 2009
NaCl(s) Na+(g) + Cl(g) Hlattice = -788 kJ/mol
+
-
-
-
-
-
-
-
- +
+
+ +
+
+
+
• 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)
Copyright McGraw-Hill 2009 6 Copyright McGraw-Hill 2009
Q = amount of charge
d = distance of separation
d
Q1 Q2
• Coulombic attraction:
2
21
d
QQF
• 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
Copyright McGraw-Hill 2009 8
Lattice energies of alkali metal iodides
Copyright McGraw-Hill 2009 9 Copyright McGraw-Hill 2009
The ionic radii sums for LiF and MgO are 2.01 and 2.06 Å,
respectively, yet their lattice energies are 1030 and 3795 kJ/mol. Why
is the lattice energy of MgO nearly four times that of LiF?
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 Hess’s law, via a
Born-Haber Cycle.
11
The lattice enthalpy change is the standard
molar enthalpy change for the following process:
M+
(gas) + X
-
(gas) MX
(solid)
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.
ΔH0
L
ΔH0
L
Lattice enthalpy
Copyright McGraw-Hill 2009 12 Copyright McGraw-Hill 2009
• Born-Haber cycle: A method to determine
lattice energies
Born-Haber Cycles
enthalpy H
eg for sodium chloride:
NaCl (s)
Na+ (g) + Cl- (g)
H lattice enthalpy Na (s) + ½ Cl2 (g)
H formation
H atomisation
Na (g) + ½ Cl2 (g)
H atomisation
Na (g) + Cl (g)
Na+ (g) + e- + Cl (g)
H first ionisation energy H first electron affinity
Born-Haber Cycles
There are two routes from
enthalpy H
NaCl (s)
Na+ (g) + Cl- (g)
H lattice association Na (s) + ½ Cl2 (g)
H formation
H atomisation
Na (g) + ½ Cl2 (g)
H atomisation
Na (g) + Cl (g)
Na+ (g) + e- + Cl (g)
H first ionisation energy H first electron affinity
elements to ionic compound
HatmNa + HatmCl + H1st IE + H1st EA + Hlattice = Hformation
Apply Hess’s Law:
: applying Hess’s Law
Born-Haber Cycles: applying Hess’s 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.
Chem 59-250 Ionic Bonding
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
NaCl(s)
Na(s) Na(g) Na+(g)
½ Cl2(g) Cl(g) Cl-(g)
H°ea H°d
H°ie H°sub
H°f
Lattice Energy, Hlattice
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.
Chem 59-250
NaCl(s)
Na(s) Na(g) Na+(g)
½ Cl2(g) Cl(g) Cl-(g)
H°ea H°d
H°ie H°sub
H°f
Lattice Energy, Hlattice
Ionic Bonding
Born-Haber cycle
H°f = H°sub + H°ie + 1/2 H°d + H°ea + 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
Chem 59-250
NaCl2(s)
Na(s) Na(g) Na+2(g)
Cl2(g) 2 Cl(g) 2Cl-(g)
H°ea H°d
(H°ie1 + H°ie2)
H°sub
H°f
Lattice Energy, Hlattice
Ionic Bonding
If we can predict the lattice energy, a Born-Haber cycle analysis can tell us
why certain compounds do not form. E.g. NaCl2
H°f = H°sub + H°ie1 + H°ie2 + H°d + H°ea + Hlattice
H°f = 109 + 496 + 4562 + 242 + 2*(-349) + -2180
H°f = +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
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 Å / 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 Avogadro’s 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 Å) = r+ + r-
M is a Madelung constant
d* = exponential scaling factor for the repulsive term = 0.345 Å
n = the number of ions in the formula unit
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.
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(H1–5)
• Lattice Energy = Hf – (H1 + H2 + H3 + H4)
Periodic Trends in Lattice Energy
Coulomb’s Law
charge A X charge B electrostatic force a
distance2
(since energy = force X distance)
charge A X charge B or, electrostatic energy a
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
35
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
Copyright © 2007 Pearson Education, Inc., publishing as Benjamin Cummings
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
Copyright © 2007 Pearson Education, Inc., publishing as Benjamin Cummings
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
The Relationship between Lattice Energies and
Physical Properties
8–38
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.
8–39
– The formation of an ionic compound will be exothermic (enthalpy is negative) if and only if the enthalpy of step 5 in the Born–Haber 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.
Predicting the Stability of Ionic Compounds
8–40
John A. Schreifels
Chemistry 212
Chapter 12-40
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.
8–41
Kelarutan ion
-Entalpi kisi
• AgCl(s) → Ag+ (g) + Cl-(g) ∆H = 917 kj/mol
solvasi
• Ag+ (g) = H2O → Ag+ (aq) ∆H= -475 kj/mol
solvasi
• Cl- (g) = H2O → Cl-(aq) ∆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
Chapter 12-41
8–43
untuk reaksi dibawah ini
Cs(s) + ½ 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
tersebut
John A. Schreifels
Chemistry 212
Chapter 12-43
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