born-haber cycles

Born-Haber cycles

L.O.:Define and apply the terms enthalpy of formation, ionisation enthalpy, enthalpy of atomisation of an element and of a compound, bond dissociation enthalpy, electron affinity, lattice enthalpy (defined as either lattice dissociation or lattice formation), enthalpy of hydration and enthalpy of solution.

Construct Born–Haber cycles to calculate latticeenthalpies from experimental data.

In pairs recap the following definitions and illustrate them with examples:

Standard enthalpy of formation

IE

EA

Mean bond enthalpy

Second IE

Second EA

Standard enthalpy of atomisation, is the enthalpy change which accompanies the formation of one mole of gaseous atoms form the elements in its standard state under standard conditions.

Use iodine as an example

THERE ARE TWO DEFINITIONS OF LATTICE ENTHALPY

1. Lattice Formation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice is formed from its isolated gaseous ions.’

Example Na+(g) + Cl¯(g) Na+ Cl¯(s)

Lattice Enthalpy Definition(s)Lattice Enthalpy Definition(s)

2. Lattice Dissociation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice dissociates into isolated gaseous ions.’

Example Na+ Cl¯(s) Na+(g) + Cl¯(g)

MAKE SURE YOU CHECK WHICH IS BEING USED

1. Lattice Formation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice is formed from its isolated gaseous ions.’

Values highly EXOTHERMICEXOTHERMIC strong electrostatic attraction between oppositely charged ions a lot of energy is released as the bond is formed

relative values are governed by the charge density of the ions.

Example Na+(g) + Cl¯(g) Na+ Cl¯(s)


NaCl(s)

Na+(g) + Cl–(g)


2. Lattice Dissociation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice dissociates into isolated gaseous ions.’

Values highly ENDOTHERMICENDOTHERMIC strong electrostatic attraction between oppositely charged ions a lot of energy must be put in to overcome the attraction relative values are governed by the charge density of the ions.

Example Na+ Cl¯(s) Na+(g) + Cl¯(g)

NaCl(s)

Na+(g) + Cl–(g)

Calculating Lattice EnthalpyCalculating Lattice Enthalpy

SPECIAL POINTS

you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY

it is CALCULATED USING A BORN-HABER CYCLE


SPECIAL POINTS



greater chargedensities of ions = greater attraction

= larger lattice enthalpy


SPECIAL POINTS



greater chargedensities of ions = greater attraction

= larger lattice enthalpy

Effects

Melting point the higher the lattice enthalpy,the higher the melting point of an ionic compound

Solubility solubility of ionic compounds is affected by the relativevalues of Lattice and Hydration Enthalpies

Cl¯ Br¯ F¯ O2-

Na+ -780 -742 -918 -2478

K+ -711 -679 -817 -2232

Rb+ -685 -656 -783

Mg2+ -2256 -3791

Ca2+ -2259

Lattice Enthalpy ValuesLattice Enthalpy Values

Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions.

Units: kJ mol-1

Cl¯ Br¯ F¯ O2-

Na+ -780 -742 -918 -2478

K+ -711 -679 -817 -2232

Rb+ -685 -656 -783

Mg2+ -2256 -3791

Ca2+ -2259

Lattice Enthalpy ValuesLattice Enthalpy Values

Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions.

Cl¯Na+ Cl¯

The sodium ion has the same charge as a potassium ion but is smaller. It has a higher charge density so will have a more effective attraction for the chloride ion. More energy will be released when they come together.

K+

Born-Haber Cycle For Sodium ChlorideBorn-Haber Cycle For Sodium Chloride

kJ mol-1

Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) – 411

Enthalpy of atomisation of sodium Na(s) ——> Na(g) + 108

Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) + 121

Ist Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯ + 500

Electron Affinity of chlorine Cl(g) + e¯ ——> Cl¯(g) – 364

Lattice Enthalpy of NaCl Na+(g) + Cl¯(g) ——> NaCl(s) ?

Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll

1This is an exothermic process so energy is released. Sodium chloride has a lower enthalpy than the elements which made it.

VALUE = - 411 kJ mol-1

This is an exothermic process so energy is released. Sodium chloride has a lower enthalpy than the elements which made it.


Na(s) + ½Cl2(g)

NaCl(s)

Enthalpy of formation of NaCl

Na(s) + ½Cl2(g) ——> NaCl(s)

1

1

2

This is an endothermic process. Energy is needed to separate the atoms. Sublimation involves going directly from solid to gas.

VALUE = + 108 kJ mol-1

This is an endothermic process. Energy is needed to separate the atoms. Sublimation involves going directly from solid to gas.



Na(s) + ½Cl2(g)

NaCl(s)

Na(g) + ½Cl2(g)



Enthalpy of atomisation of sodium

Na(s) ——> Na(g)

1

2

1

3

2

Breaking covalent bonds is an endothermic process. Energy is needed to overcome the attraction the atomic nuclei have for the shared pair of electrons.


Breaking covalent bonds is an endothermic process. Energy is needed to overcome the attraction the atomic nuclei have for the shared pair of electrons.



Na(s) + ½Cl2(g)

NaCl(s)

Na(g) + ½Cl2(g)

Na(g) + Cl(g)




Na(s) ——> Na(g)

Enthalpy of atomisation of chlorine

½Cl2(g) ——> Cl(g)

1

2

3


1

4

3

2

All Ionisation Energies are endothermic. Energy is needed to overcome the attraction the protons in the nucleus have for the electron being removed.


All Ionisation Energies are endothermic. Energy is needed to overcome the attraction the protons in the nucleus have for the electron being removed.


Na(s) + ½Cl2(g)

NaCl(s)

Na(g) + ½Cl2(g)

Na(g) + Cl(g)

Na+(g) + Cl(g)




Na(s) ——> Na(g)


½Cl2(g) ——> Cl(g)

Ist Ionisation Energy of sodium

Na(g) ——> Na+(g) + e¯

1

2

3

4


1

54

3

2

Electron affinity is exothermic. Energy is released as the nucleus attracts an electron to the outer shell of a chlorine atom.


Electron affinity is exothermic. Energy is released as the nucleus attracts an electron to the outer shell of a chlorine atom.


Na(s) + ½Cl2(g)

NaCl(s)

Na(g) + ½Cl2(g)

Na(g) + Cl(g)

Na+(g) + Cl(g)

Na+(g) + Cl–(g)




Na(s) ——> Na(g)


½Cl2(g) ——> Cl(g)


Na(g) ——> Na+(g) + e¯

Electron Affinity of chlorine

Cl(g) + e¯ ——> Cl¯(g)

1

2

3

4

5


1

6

54

3

2

Na(s) + ½Cl2(g)

NaCl(s)

Na(g) + ½Cl2(g)

Na(g) + Cl(g)

Na+(g) + Cl(g)

Na+(g) + Cl–(g)




Na(s) ——> Na(g)


½Cl2(g) ——> Cl(g)


Na(g) ——> Na+(g) + e¯


Cl(g) + e¯ ——> Cl¯(g)

Lattice Enthalpy of NaCl

Na+(g) + Cl¯(g) ——> NaCl(s)

1

2

3

4

5

6

Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other.

Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other.


1

6

54

3

2

Na(s) + ½Cl2(g)

NaCl(s)

Na(g) + ½Cl2(g)

Na(g) + Cl(g)

Na+(g) + Cl(g)

Na+(g) + Cl–(g)

CALCULATING THE LATTICE ENTHALPY CALCULATING THE LATTICE ENTHALPY

Apply Hess’s Law

16 5 4 3 2 = - - - - +

The minus shows you are going in the opposite direction to the definition

= - (-364) - (+500) - (+121) - (+108) + (-411)= - 776 kJ mol-1


1

6

54

3

2

Na(s) + ½Cl2(g)

NaCl(s)

Na(g) + ½Cl2(g)

Na(g) + Cl(g)

Na+(g) + Cl(g)

Na+(g) + Cl–(g)

CALCULATING THE LATTICE ENTHALPY CALCULATING THE LATTICE ENTHALPY

Apply Hess’s Law

16 5 4 3 2 = - - - - +

The minus shows you are going in the opposite direction to the definition

= - (-364) - (+500) - (+121) - (+108) + (-411)= - 776 kJ mol-1

OR…

Ignore the signs and just use the values;

If you go up you add, if you come down you subtract the value

= - - - -

= (364) - (500) - (121) - (108) - (411)= - 776 kJ mol-1

16 5 4 3 2

Construct a Born-Haber cycle for KCl

H

CaO (s)

Ca2+ (g) + O2- (g)

H lattice energy

of formation

Ca (s) + ½ O2 (g)

H formation

H atomisation(s)

Ca (g) + O (g)

Ca2+ (g) + 2 e- + O (g)

H ionisation energy/iesH electron affinity/ies

CaO

193

248

590

1150

–142

+844

?

– 635 = 193 + 248 + 590 + 1150 – 142 + 844 + Hlattice

Hlattice = – 635 – 193 – 248 – 590 – 1150 + 142 – 844 = – 3518 kJ mol-1

–635

1

65

4

3

2

Mg(s) + Cl2(g)

MgCl2(s)

Mg(g) + Cl2(g)

Mg(g) + 2Cl(g) Mg2+(g) + 2Cl–(g)

7

Mg+(g) + 2Cl(g)

Mg2+(g) + 2Cl(g)

Enthalpy of formation of MgCl2

Mg(s) + Cl2(g) ——> MgCl2(s)

Enthalpy of sublimation of magnesium

Mg(s) ——> Mg(g)


½Cl2(g) ——> Cl(g) x2

Ist Ionisation Energy of magnesium

Mg(g) ——> Mg+(g) + e¯

2nd Ionisation Energy of magnesium

Mg+(g) ——> Mg2+(g) + e¯


Cl(g) + e¯ ——> Cl¯(g) x2

Lattice Enthalpy of MgCl2

Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)

1

2

3

4

5

7

6

Born-Haber Cycle - MgCBorn-Haber Cycle - MgCll22

Construct a Born-Haber cycle for CoCl3

H

CoCl3 (s)

Co3+ (g) + 3 Cl- (g)

H lattice energy

of formation

Co (s) + 3/2 Cl2 (g)

H formation

H atomisation(s)

Co (g) + 3 Cl(g)

Co3+ (g) + 3 e- + 2 Cl (g)

H ionisation energy/iesH electron affinity/ies

CoCl3

427

3(121)

757

1640

3230

3(–364)

–5350

?

Hformation = 427 + 3(121) + 757 + 1640 + 3230 – 3(364) – 5350 = – 25 kJ mol-1

Born-Haber cycles

L.O.:Define and apply the terms enthalpy of formation, ionisation enthalpy, enthalpy of atomisation of an element and of a compound, bond dissociation enthalpy, electron affinity, lattice enthalpy (defined as either lattice dissociation or lattice formation), enthalpy of hydration and enthalpy of solution.

Construct Born–Haber cycles to calculate latticeenthalpies from experimental data.

born-haber cycles

Documents