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BORN-HABER CYCLES

A guide for A level students

KNOCKHARDY PUBLISHING2008

SPECIFICATIONS

BORN-HABER CYCLESINTRODUCTION

This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.

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Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...

www.knockhardy.org.uk/sci.htm

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BORN-HABER CYCLES

CONTENTS• Lattice Enthalpy

• Definition of enthalpy changes

• Born-Haber cycle for sodium chloride

• Calculation of Lattice Enthalpy

• Born-Haber cycle for magnesium chloride

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)

2. Lattice Dissociation Enthalpy‘The enthalpy change when ONE MOLE of an ionic latticedissociates 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 EXOTHERMICstrong electrostatic attraction between oppositely charged ionsa lot of energy is released as the bond is formedrelative 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 latticedissociates into isolated gaseous ions.’

Values highly ENDOTHERMICstrong electrostatic attraction between oppositely charged ionsa lot of energy must be put in to overcome the attractionrelative 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 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 -2478K+ -711 -679 -817 -2232Rb+ -685 -656 -783Mg2+ -2256 -3791Ca2+ -2259

Lattice 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 -2478K+ -711 -679 -817 -2232Rb+ -685 -656 -783Mg2+ -2256 -3791Ca2+ -2259

Lattice 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 Chloride

kJ mol-1

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

Enthalpy of sublimation 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 - NaCl

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 NaClNa(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 sublimation of sodiumNa(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)



Enthalpy of atomisation of chlorine½Cl2(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)Enthalpy of formation of NaClNa(s) + ½Cl2(g) ——> NaCl(s)



Ist Ionisation Energy of sodiumNa(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)





Electron Affinity of chlorineCl(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)





Electron Affinity of chlorineCl(g) + e¯ ——> Cl¯(g)

Lattice Enthalpy of NaClNa+(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

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

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

1

65

4

3

2Mg(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 MgCl2Mg(s) + Cl2(g) ——> MgCl2(s)

Enthalpy of sublimation of magnesiumMg(s) ——> Mg(g)

Enthalpy of atomisation of chlorine½Cl2(g) ——> Cl(g) x2

Ist Ionisation Energy of magnesiumMg(g) ——> Mg+(g) + e¯

2nd Ionisation Energy of magnesiumMg+(g) ——> Mg2+(g) + e¯

Electron Affinity of chlorineCl(g) + e¯ ——> Cl¯(g) x2

Lattice Enthalpy of MgCl2Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)

1

2

3

4

5

7

6

Born-Haber Cycle - MgCl2

BORN-HABER CYCLES

THE END

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