Energi Kisi Dan Born Haber(1)

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


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


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)


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


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


• 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


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


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


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

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

Energi Kisi Dan Born Haber(1)

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