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Using Born Haber Cycles to Determine Lattice Enthalpies 15.2.3.

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Presentation on theme: "Using Born Haber Cycles to Determine Lattice Enthalpies 15.2.3."— Presentation transcript:

1 Using Born Haber Cycles to Determine Lattice Enthalpies

2 Watch this video 1. Enthalpy level diagrams.mp4 Enthalpy level diagrams.mp4

3 Watch this video 2. Hess Cycles & Enthalpy.mp4 Hess Cycles & Enthalpy.mp4

4 Review of ΔH variables  ΔH˚ = ?  ΔH c ˚ = ?  ΔH f ˚ = ?  ΔH lat ˚ = ?  ΔH i ˚ = ?  ΔH e ˚ = ?

5 Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides and use it to calculate the enthalpy change  Experimental lattice energies can NOT be determined directly.  The formation of an ionic compound is supposed to take place in steps.  Q – What do we know about thermodynamic values and stepwise reactions?  An energy cycle based on Hess’s Law, known as the Born-Haber cycle can be used instead.

6 Watch this video 3. Born Haber Cycle (Finding Lattice Energy).mp4 Born Haber Cycle (Finding Lattice Energy).mp4

7 Born-Haber Cycle Steps  Elements (standard state) converted into gaseous atoms  Losing or gaining electrons to form cations and anions  Combining gaseous anions and cations to form a solid ionic compound We will start with the following reaction Na(s) + 1/2Cl 2 (g)  NaCl(s)

8 Steps 1 & 2: Atomisation  The standard enthalpy change of atomisation is the ΔH required to produce one mole of gaseous atoms  Na(s)  Na(g) ΔH o atom = +107 kJmol -1  NOTE: for diatomic gaseous elements, Cl 2, ΔH o atom is equal to half the bond energy (enthalpy)  Cl 2 (g)  Cl(g) ΔH o atom = ½ E (Cl-Cl) ΔH o atom = ½ (+242 ) ΔH o atom = +121 kJmol -1 Na(s) + 1/2Cl 2 (g)  NaCl(s) Na(s) + 1/2Cl 2 (g)  Na(g) + Cl(g)

9 Step 3: Forming gaseous cations  Ionisation energy Enthalpy change for one mole of a gaseous element or cation to lose electrons to form a mole of positively charged gaseous ions  Na(g)  Na + (g) + e- IE 1 = +494 kJmol -1 Na(g) + Cl(g)  Na + (g) + Cl(g) + e -

10 Step 4: Forming gaseous anions  Electron Affinity Enthalpy change when one mole of gaseous atoms or anions gains electrons to form a mole of negatively charged gaseous ions.  Cl(g) + e-  Cl - (g) ΔH o = -364 kJmol -1 For most atoms = exothermic, but gaining a 2 nd electron is endothermic due to the repulsion between the anion and the electron Na + (g) + Cl(g) + e -  Na + (g) + Cl - (g)

11 Step 5: Combination of gaseous ions to form a solid crystal ionic compound.  Lattice Enthalpy This step is highly exothermic because you are going from high energy reactants to a lower energy product. The term lattice enthalpy (ΔH˚ lat ) by definition applies to the reverse of this process (ending with gaseous ions) which is why it’s value is always > 0. Na + (g) + Cl - (g)  NaCl (s)

12 Born-Haber Cycles enthalpy H eg for sodium chloride: NaCl (s) Na + (g) + Cl - (g) H lattice enthalpy Na (s) + ½ Cl 2 (g) H formation H atomisation Na (g) + ½ Cl 2 (g) H atomisation Na (g) + Cl (g) Na + (g) + e - + Cl (g) H first ionisation energy H e ˚ first electron affinity According to Hess’s Law (and the 1 st law of Thermodynamics), the sum of all these processes is zero. Write an expression that would = ΔH˚ f

13 Born-Haber Cycles enthalpy H eg for sodium chloride: NaCl (s) Na + (g) + Cl - (g) H lattice enthalpy Na (s) + ½ Cl 2 (g) H formation H atomisation Na (g) + ½ Cl 2 (g) H atomisation Na (g) + Cl (g) Na + (g) + e - + Cl (g) H first ionisation energy H e ˚ first electron affinity H f = H atom (for both Na and Cl)+ H i + H e - H lat Q-How would you solve for H lat ?

14 Born-Haber Cycles enthalpy H eg for sodium chloride: NaCl (s) Na + (g) + Cl - (g) H lattice enthalpy Na (s) + ½ Cl 2 (g) H formation H atomisation Na (g) + ½ Cl 2 (g) H atomisation Na (g) + Cl (g) Na + (g) + e - + Cl (g) H first ionisation energy H e ˚ first electron affinity H lat = H atom (for both Na and Cl)+ H i + H e - H f

15 So that was emotional. Let’s watch some videos of problems being worked.  4. Born Haber (Hf of sodium bromide).mp4 4. Born Haber (Hf of sodium bromide).mp4  5. Born Haber (Hf of magnesium oxide).mp4 5. Born Haber (Hf of magnesium oxide).mp4

16 We’ll spend the rest of the period trying out some old IB Born-Haber problems.


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