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Using Born Haber Cycles to Determine Lattice Enthalpies 15.2.3
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Watch this video 1. Enthalpy level diagrams.mp4 Enthalpy level diagrams.mp4 http://youtu.be/b7GAPUgvRLk
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Watch this video 2. Hess Cycles & Enthalpy.mp4 Hess Cycles & Enthalpy.mp4 http://youtu.be/e9G2q9Ea6tQ
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Review of ΔH variables ΔH˚ = ? ΔH c ˚ = ? ΔH f ˚ = ? ΔH lat ˚ = ? ΔH i ˚ = ? ΔH e ˚ = ?
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15.2.3 - 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.
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Watch this video 3. Born Haber Cycle (Finding Lattice Energy).mp4 Born Haber Cycle (Finding Lattice Energy).mp4 http://youtu.be/BbTZoJ_K_l4
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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)
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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)
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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 -
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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)
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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)
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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
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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 ?
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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
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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 http://youtu.be/Y-lSATTDMcU 5. Born Haber (Hf of magnesium oxide).mp4 5. Born Haber (Hf of magnesium oxide).mp4 http://youtu.be/2JtHvspX7HE
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We’ll spend the rest of the period trying out some old IB Born-Haber problems.
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