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 lattice enthalpies from experimental data.

2. In pairs recap the following definitions and illustrate them with examples:  Standard enthalpy of formation  IE  EA  Mean bond enthalpy  Second IE Second EA

3. 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

4. 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 lattice dissociates into isolated gaseous ions.’ Example Na + Cl¯(s) Na + (g) + Cl¯(g) MAKE SURE YOU CHECK WHICH IS BEING USED

5. 1. Lattice Formation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice is formed from its isolated gaseous ions.’ EXOTHERMIC Values highly EXOTHERMIC 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) Lattice Enthalpy Definition(s) NaCl(s) Na + (g) + Cl – (g)

6. Lattice Enthalpy Definition(s) 2. Lattice Dissociation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice dissociates into isolated gaseous ions.’ ENDOTHERMIC Values highly ENDOTHERMIC 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)

7. Calculating Lattice Enthalpy SPECIAL POINTS you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY it is CALCULATED USING A BORN-HABER CYCLE

8. Calculating Lattice Enthalpy SPECIAL POINTS you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY it is CALCULATED USING A BORN-HABER CYCLE greater charge densities of ions = greater attraction = larger lattice enthalpy

9. Calculating Lattice Enthalpy SPECIAL POINTS you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY it is CALCULATED USING A BORN-HABER CYCLE greater charge densities of ions = greater attraction = larger lattice enthalpy Effects Melting pointthe higher the lattice enthalpy, the higher the melting point of an ionic compound Solubilitysolubility of ionic compounds is affected by the relative values of Lattice and Hydration Enthalpies

10. Cl¯ Br¯ F¯ O 2- Na + -780-742-918-2478 K + -711-679-817-2232 Rb + -685-656-783 Mg 2+ -2256-3791 Ca 2+ -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

11. C l ¯ Br¯ F¯ O 2- Na + -780-742-918-2478 K + -711-679-817-2232 Rb + -685-656-783 Mg 2+ -2256-3791 Ca 2+ -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¯Cl¯ Na + Cl¯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+K+

12. Born-Haber Cycle For Sodium Chloride kJ mol -1 Enthalpy of formation of NaC l Na(s) + ½C l 2 (g) ——> NaC l (s) – 411 Enthalpy of atomisation of sodium Na(s) ——> Na(g) + 108 Enthalpy of atomisation of chlorine ½C l 2 (g) ——> C l (g) + 121 Ist Ionisation Energy of sodium Na(g) ——> Na + (g) + e¯ + 500 Electron Affinity of chlorine C l (g) + e¯ ——> C l ¯(g) – 364 Lattice Enthalpy of NaC l Na + (g) + C l ¯(g) ——> NaC l (s) ?

13. Born-Haber Cycle - NaC l 1 This 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. VALUE = - 411 kJ mol -1 Na(s) + ½C l 2 (g) NaC l (s) Enthalpy of formation of NaC l Na(s) + ½C l 2 (g) ——> NaC l (s) 1

14. 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. VALUE = + 108 kJ mol -1 Born-Haber Cycle - NaC l Na(s) + ½C l 2 (g) NaC l (s) Na(g) + ½C l 2 (g) Enthalpy of formation of NaC l Na(s) + ½C l 2 (g) ——> NaC l (s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) 1 2

15. 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. VALUE = + 121 kJ mol -1 Breaking covalent bonds is an endothermic process. Energy is needed to overcome the attraction the atomic nuclei have for the shared pair of electrons. VALUE = + 121 kJ mol -1 Born-Haber Cycle - NaC l Na(s) + ½C l 2 (g) NaC l (s) Na(g) + ½C l 2 (g) Na(g) + C l (g) Enthalpy of formation of NaC l Na(s) + ½C l 2 (g) ——> NaC l (s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½C l 2 (g) ——> C l (g) 1 2 3

16. Born-Haber Cycle - NaC l 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. VALUE = + 500 kJ mol -1 All Ionisation Energies are endothermic. Energy is needed to overcome the attraction the protons in the nucleus have for the electron being removed. VALUE = + 500 kJ mol -1 Na(s) + ½C l 2 (g) NaC l (s) Na(g) + ½C l 2 (g) Na(g) + C l (g) Na + (g) + C l (g) Enthalpy of formation of NaC l Na(s) + ½C l 2 (g) ——> NaC l (s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½C l 2 (g) ——> C l (g) Ist Ionisation Energy of sodium Na(g) ——> Na + (g) + e¯ 1 2 3 4

17. Born-Haber Cycle - NaC l 1 5 4 3 2 Electron affinity is exothermic. Energy is released as the nucleus attracts an electron to the outer shell of a chlorine atom. VALUE = - 364 kJ mol -1 Electron affinity is exothermic. Energy is released as the nucleus attracts an electron to the outer shell of a chlorine atom. VALUE = - 364 kJ mol -1 Na(s) + ½C l 2 (g) NaC l (s) Na(g) + ½C l 2 (g) Na(g) + C l (g) Na + (g) + C l (g) Na + (g) + C l – (g) Enthalpy of formation of NaC l Na(s) + ½C l 2 (g) ——> NaC l (s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½C l 2 (g) ——> C l (g) Ist Ionisation Energy of sodium Na(g) ——> Na + (g) + e¯ Electron Affinity of chlorine C l (g) + e¯ ——> C l ¯(g) 1 2 3 4 5

18. Born-Haber Cycle - NaC l 1 6 5 4 3 2 Na(s) + ½C l 2 (g) NaC l (s) Na(g) + ½C l 2 (g) Na(g) + C l (g) Na + (g) + C l (g) Na + (g) + C l – (g) Enthalpy of formation of NaC l Na(s) + ½C l 2 (g) ——> NaC l (s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½C l 2 (g) ——> C l (g) Ist Ionisation Energy of sodium Na(g) ——> Na + (g) + e¯ Electron Affinity of chlorine C l (g) + e¯ ——> C l ¯(g) Lattice Enthalpy of NaCl Na + (g) + C l ¯(g) ——> NaC l (s) 1 2 3 4 5 6 Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other.

19. Born-Haber Cycle - NaC l 1 6 5 4 3 2 Na(s) + ½C l 2 (g) NaC l (s) Na(g) + ½C l 2 (g) Na(g) + C l (g) Na + (g) + C l (g) Na + (g) + C l – (g) CALCULATING THE LATTICE ENTHALPY Apply Hess’s Law 1 6 5 432 = - - - - + The minus shows you are going in the opposite direction to the definition = - (-364) - (+500) - (+121) - (+108) + (-411) = - 776 kJ mol -1

20. Born-Haber Cycle - NaC l 1 6 5 4 3 2 Na(s) + ½C l 2 (g) NaC l (s) Na(g) + ½C l 2 (g) Na(g) + C l (g) Na + (g) + C l (g) Na + (g) + C l – (g) CALCULATING THE LATTICE ENTHALPY Apply Hess’s Law 1 6 5 432 = - - - - + 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 165432

21. Construct a Born- Haber cycle for KCl

22. H CaO (s) Ca 2+ (g) + O 2 - (g)  H lattice energy of formation Ca (s) + ½ O 2 (g)  H formation  H atomisation(s) Ca (g) + O (g) Ca 2+ (g) + 2 e - + O (g)  H ionisation energy/ies  H electron affinity/ies CaO 193 248 590 1150 –142 +844 ? – 635 = 193 + 248 + 590 + 1150 – 142 + 844 +  H lattice  H lattice = – 635 – 193 – 248 – 590 – 1150 + 142 – 844 = – 3518 kJ mol -1 – 635

23. 1 6 5 4 3 2 Mg(s) + C l 2 (g) MgC l 2 (s) Mg(g) + C l 2 (g) Mg(g) + 2C l (g) Mg 2+ (g) + 2C l – (g) 7 Mg + (g) + 2C l (g) Mg 2+ (g) + 2C l (g) Enthalpy of formation of MgC l 2 Mg(s) + C l 2 (g) ——> MgC l 2 (s) Enthalpy of sublimation of magnesium Mg(s) ——> Mg(g) Enthalpy of atomisation of chlorine ½C l 2 (g) ——> C l (g) x2 Ist Ionisation Energy of magnesium Mg(g) ——> Mg + (g) + e¯ 2nd Ionisation Energy of magnesium Mg + (g) ——> Mg 2+ (g) + e¯ Electron Affinity of chlorine C l (g) + e¯ ——> C l ¯(g) x2 Lattice Enthalpy of MgC l 2 Mg 2+ (g) + 2C l ¯(g) ——> MgC l 2 (s) 1 2 3 4 5 7 6 Born-Haber Cycle - MgC l 2

24. Construct a Born-Haber cycle for CoCl 3

25. H CoCl 3 (s) Co 3+ (g) + 3 Cl - (g)  H lattice energy of formation Co (s) + 3 / 2 Cl 2 (g)  H formation  H atomisation(s) Co (g) + 3 Cl(g) Co 3+ (g) + 3 e - + 2 Cl (g)  H ionisation energy/ies  H electron affinity/ies CoCl 3 427 3(121 ) 757 164 0 323 0 3(– 364) – 5350 ?  H formation = 427 + 3(121) + 757 + 1640 + 3230 – 3(364) – 5350 = – 25 kJ mol -1

26. 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 lattice enthalpies from experimental data.