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Lattice enthalpy

Lattice enthalpy. Textbook reference: p166-171. Born-Haber cycles. L.O.: Explain and use the term: lattice enthalpy . Use the lattice enthalpy of a simple ionic solid and relevant energy terms to construct Born–Haber cycles.

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Lattice enthalpy

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  1. Lattice enthalpy Textbook reference: p166-171

  2. Born-Haber cycles • L.O.: • Explain and use the term: lattice enthalpy. • Use the lattice enthalpy of a simple ionic solid and relevant energy terms to construct Born–Haber cycles.

  3. In pairs recap the following definitions and illustrate them with examples: • Standard enthalpy of formation • IE • EA • Second IE • Second EA

  4. The enthalpy change of atomisation is the enthalpy change that takes place when one mole of gaseous atoms forms from the elements in its standard state. Use iodine as an example

  5. Atomisation enthalpies 1. Write out the atomisation enthalpies for the following elements : • Hydrogen • zinc • Iodine • Mercury • nitrogen

  6. Lattice Enthalpy Definition 1. Lattice 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)

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

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

  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

  10. 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 point the higher the lattice enthalpy, the higher the melting point of an ionic compound Solubility solubility of ionic compounds is affected by the relative values of Lattice and Hydration Enthalpies

  11. Lattice Enthalpy Values Cl¯ Br¯ F¯ O2- Na+-780 -742 -918 -2478 K+ -711 -679 -817 -2232 Rb+ -685 -656 -783 Mg2+ -2256 -3791 Ca2+ -2259 Units: kJ mol-1 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.

  12. Na+ Cl¯ Cl¯ K+ Lattice Enthalpy Values Cl¯ Br¯ F¯ O2- Na+-780 -742 -918 -2478 K+ -711 -679 -817 -2232 Rb+ -685 -656 -783 Mg2+ -2256 -3791 Ca2+ -2259 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. 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.

  13. Born-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) ?

  14. Born-Haber Cycle - NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) 1 Na(s) + ½Cl2(g) 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 1 NaCl(s)

  15. Born-Haber Cycle - NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) 1 2 Na(g) + ½Cl2(g) 2 Na(s) + ½Cl2(g) 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 1 NaCl(s)

  16. Born-Haber Cycle - NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) 1 2 3 Na(g) + Cl(g) 3 Na(g) + ½Cl2(g) 2 Na(s) + ½Cl2(g) 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 1 NaCl(s)

  17. Born-Haber Cycle - NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) Ist Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯ 1 Na+(g) + Cl(g) 2 4 3 Na(g) + Cl(g) 4 3 Na(g) + ½Cl2(g) 2 Na(s) + ½Cl2(g) 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 1 NaCl(s)

  18. Born-Haber Cycle - NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) Ist Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯ Electron Affinity of chlorine Cl(g) + e¯ ——> Cl¯(g) 1 Na+(g) + Cl(g) 2 5 4 3 Na+(g) + Cl–(g) Na(g) + Cl(g) 4 3 Na(g) + ½Cl2(g) 5 2 Na(s) + ½Cl2(g) 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 1 NaCl(s)

  19. Born-Haber Cycle - NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of atomisation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) Ist Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯ Electron Affinity of chlorine Cl(g) + e¯ ——> Cl¯(g) Lattice Enthalpy of NaCl Na+(g) + Cl¯(g) ——> NaCl(s) 1 Na+(g) + Cl(g) 2 5 4 3 Na+(g) + Cl–(g) Na(g) + Cl(g) 4 3 Na(g) + ½Cl2(g) 5 2 6 Na(s) + ½Cl2(g) 6 1 Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other. NaCl(s)

  20. Born-Haber Cycle - NaCl CALCULATING THE LATTICE ENTHALPY Apply Hess’s Law Na+(g) + Cl(g) 5 = - - - - + The minus shows you are going in the opposite direction to the definition = - (-364) - (+500) - (+121) - (+108) + (-411) = - 776 kJ mol-1 5 1 6 4 3 2 4 Na+(g) + Cl–(g) Na(g) + Cl(g) 3 Na(g) + ½Cl2(g) 2 6 Na(s) + ½Cl2(g) 1 NaCl(s)

  21. Born-Haber Cycle - NaCl CALCULATING THE LATTICE ENTHALPY Apply Hess’s Law Na+(g) + Cl(g) 5 4 1 = 2 +3+4+5+6 Na+(g) + Cl–(g) Na(g) + Cl(g) 3 Na(g) + ½Cl2(g) 2 6 Na(s) + ½Cl2(g) 1 NaCl(s)

  22. Construct a Born-Haber cycle for KCl

  23. Find easy exam qs,

  24. Born-Haber Cycle - MgCl2 Enthalpy of formation of MgCl2 Mg(s) + Cl2(g) ——> MgCl2(s) Enthalpy of sublimation of magnesium Mg(s) ——> Mg(g) Enthalpy of atomisation of chlorine 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¯ Electron Affinity of chlorine Cl(g) + e¯ ——> Cl¯(g) x2 Lattice Enthalpy of MgCl2 Mg2+(g) + 2Cl¯(g) ——> MgCl2(s) 1 Mg2+(g) + 2Cl(g) 2 5 6 Mg+(g) + 2Cl(g) 3 4 4 Mg2+(g) + 2Cl–(g) Mg(g) + 2Cl(g) 3 5 Mg(g) + Cl2(g) 2 7 6 Mg(s) + Cl2(g) 1 7 MgCl2(s)

  25. H electron affinity/ies H ionisation energy/ies H Ca2+ (g) + O2- (g) H atomisation(s) Ca (g) + O (g) H lattice energy of formation Ca (s) + ½ O2 (g) CaO (s) CaO Ca2+ (g) + 2 e- + O (g) 590 1150 –142 +844 193 248 ? H formation –635 – 635 = 193 + 248 + 590 + 1150 – 142 + 844 + Hlattice Hlattice = – 635 – 193 – 248 – 590 – 1150 + 142 – 844 = – 3518 kJ mol-1

  26. Construct a Born-Haber cycle for CoCl3

  27. H electron affinity/ies H ionisation energy/ies H Co3+ (g) + 3 Cl- (g) H atomisation(s) Co (g) + 3 Cl(g) H lattice energy of formation Co (s) + 3/2 Cl2 (g) CoCl3 (s) CoCl3 Co3+ (g) + 3 e- + 2 Cl (g) 757 1640 3230 3(–364) 427 3(121) –5350 H formation ? Hformation = 427 + 3(121) + 757 + 1640 + 3230 – 3(364) – 5350 = – 25 kJ mol-1

  28. 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.

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