1. Title: Lesson 6 Born-Haber Cycles and Lattice Enthalpies Learning Objectives: – Understand the term lattice enthalpy – Use Born-Haber cycles to calculate lattice enthalpy – Identify and explain trends in lattice enthalpy

2. Main Menu Refresh  The standard enthalpy change of three combustion reactions is given below in kJ. 2C 2 H 6 (g) + 7O 2 (g) → 4CO 2 (g) + 6H 2 O(l)∆H o = –3120 2H 2 (g) + O 2 (g) → 2H 2 O(l)∆H o = –572 C 2 H 4 (g) + 3O 2 (g) → 2CO 2 (g) + 2H 2 O(l) Δ H o = –1411  Based on the above information, calculate the standard change in enthalpy, ∆H o, for the following reaction. C 2 H 6 (g) → C 2 H 4 (g) + H 2 (g)

3. Main Menu Bond Enthalpies  Look at the covalent bond enthalpies on Table 10 in the Data Booklet.  Why do you think there are no ionic bond enthalpies?

4. Main Menu Recap of First Ionisation Energies and Electron Affinities  We know that metals loses electrons and non-metals gain electrons. We can use Ionisation Energies and Electron Affinities to work out the enthalpy changes within an ionic compound.  The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms:  Sodium (on the left of the periodic table) has a relatively low ionisation energy.  The first electron affinity is the enthalpy change when one mole of gaseous atoms attracts one mole of electrons. Values can be found in section 7 of the IB data booklet.  As the electron is attracted to the positively charged nucleus of the Cl atom, the process is EXOTHERMIC

5. Main Menu Lattice Enthalpies  Add the equations for the first ionisation energy and first electron affinity.  The process is ENDOTHERMIC overall. This is energetically UNFAVOURABLE (despite the fact it leads to the formation of ions with stable noble gas configurations).  Oppositely charged ions come together to form an ionic lattice. The strong attraction between the oppositely charged ions means its very EXOTHERMIC.  Lattice Enthalpy  H θ lat expresses this enthalpy change in terms of the reverse ENDOTHERMIC process.  The lattice enthalpy relates to the ‘formation of gaseous ions from one mole of a solid crystal breaking into gaseous ions’. (As seen above)

6. Main Menu  The lattice enthalpy relates to the enthalpy change of ‘formation of gaseous ions from one mole of a solid crystal breaking into gaseous ions’.  Think of lattice enthalpy as ‘lattice disassociation enthalpy’...

7. Main Menu Lattice Enthalpy,  H lat  This is the equivalent of ‘bond strength’ for ionic compounds  It is the enthalpy change when one mole of an ionic compound is converted to gaseous ions.  This is an endothermic process, requiring energy to be put in. MX(s)  M + (g) + X - (g) Compound Lattice Enthalpy kJ mol-1 LiF1049 LiBr820 KF829 CaF 2 2651 Note: Most places define lattice enthalpy the opposite way round, i.e: M + (g) + X - (g)  MX(s) The values would be the same magnitude, but with a negative sign to show they are exothermic. It is just a strange quirk of the IB that they do it this way round….I think so that it fits with average bond enthalpies, which are also represent bond breaking

8. H θ EA ClH θ 1st IE Na H θ at Cl H θ f NaCl Born-Haber Cycle eg for sodium chloride: NaCl (s) Cl - (g) H θ LAT NaCl Na (s) + ½ Cl 2 (g) Na (g) Cl (g) + Na + (g) + e - + NB: H θ f = formation H θ at = atomisation H θ IE = ionisation H θ EA = electron affinity H θ LAT = lattice enthalpy Using the ‘FAIL’ technique F = formation A = atomisation I = ionisation L = lattice enthalpy FORMATION IONISATION Construction of a H θ at Na ATOMISATION LATTICE ENTHALPY

9. Born-Haber Cycle There are two routes fromelements to ionic compound H at Na + H at Cl + H 1st IE Na + H EA Cl - H LAT NaCl = H f NaCl Apply Hess’s Law: : Applying Hess’s Law Clockwise Anti-clockwise = The Indirect route and the Direct route Clockwise arrows must equal the Anti-clockwise arrows H θ EA Cl H θ 1st IE Na H θ at Cl H θ f NaCl NaCl (s) Cl - (g) H θ LAT NaCl Na (s) + ½ Cl 2 (g) Na (g) Cl (g) + Na + (g) + e - + H θ at Na

10. Born-Haber Cycle: Rearrange to find the lattice energy: So Born-Haber cycles can be used to calculate a measure of ionic bond strength based on experimental data. H at Na + H at Cl + H 1st IE Na + H EA Cl - H LAT NaCl = H f NaCl If you want to calculate H LAT NaCl, it is as follows:- H LAT NaCl = [H at Na + H at Cl + H 1st IE Na + H EA Cl] - H f NaCl Calculation Values: H at Na = 107 H at Cl = 121 H 1st IE Na = 496 (kJmol -1 ) H EA Cl = -349 H f NaCl = -411 H LAT NaCl = [(107) + (121) + (496) + (-349)] - (-411) H LAT NaCl = 786 kJmol -1 It’s important to keep all numbers in ( ), whether +ve or –ve, when entering information into your calculator.

11. 2xH θ EA Cl H θ 1st +2nd IE Mg 2xH θ at Cl H θ f MgCl 2 Born-Haber Cycle eg for magnesium chloride: MgCl 2 (s) 2Cl - (g) H θ LAT MgCl 2 Mg (s) + Cl 2 (g) Mg (g) 2Cl (g) + Mg 2+ (g) + 2e - + NB: H θ f = formation H θ at = atomisation H θ IE = ionisation H θ EA = electron affinity H θ LAT = lattice enthalpy Using the ‘FAIL’ technique F = formation A = atomisation I = ionisation L = lattice enthalpy FORMATION IONISATION Construction of a H θ at Mg ATOMISATION LATTICE ENTHALPY

13. Main Menu Born-Haber Cycles  Lattice enthalpies are difficult to measure directly, so we use Born-Haber cycles  These are just a specialised type of Hess cycle  To calculate  H lat :  Start at the bottom and work round clockwise, adding and subtracting according to the arrows. solid ionic compound elements in their standard states metal and atomised non-metal atomised metal and atomised non-metal ionised metal and atomised non-metal ionised metal and ionised non-metal Enthalpy of atomisation (metal) Ionisation energy/s (Metal) Enthalpy of atomisation (non-metal) Enthalpy of formation  H lat = ? Electron affinity/s (Non-metal)

14. Main Menu Building Born-Haber Cycles  Work through the activity herehere  Cut out the equations first and arrange them sensibly on your desk.  We will work through LiF together.  Hint: Don’t forget to include both ionisations for Mg/Ca and both electron affinities for O

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18. Solutions

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