1. Why do molecules form the way they do? Bond Enthalpies, Hess’ Law, The Born-Haber Cycle, and Heats of Reaction Textbook Reference: Chapter 6 with parts from Chapter 9

2. Molecular Compounds Why does oxygen form O 2 rather than O 8 (more accurately 4O 2 rather than O 8 )?  We know that oxygen is a diatomic, but this is not a reason this is merely an observation of trend.  We need to consider  H BDE (Bond Dissociation Energy) which is the energy required to cleave a covalent bond. BDE O 2 = 498 kJ/mol BDE 4O 2 = 4 x 498 kJ = 1992 kJ Meaning 1992 kJ is required to break 4 moles of O 2 OR 1992 kJ of energy is given off when we form 4 moles of O 2 from O atoms. BDE O—O = 146 kJ/mol BDE O 8 = 8 x 146 kJ = 1168 kJ Meaning only 1168 kJ is given off when we form 8 O—O single bonds in O 8. O 2 is energetically favored.

3. Some other elements to consider Why is phosphorus P 4 rather than 2P 2 ?  P 4 (white phosphorus) is tetrahedral  P—P BDE = 209 kJ  P≡P BDE = 490 kJ Why is sulfur S 8 rather than S 2 ?  This is the converse of oxygen which prefers O 2.  S—S BDE = 266 kJ  S=S BDE = 427 kJ

4. The Ionic Lattice... One More Time The find the lattice energy (  H latt )of an ionic compound we can use the following formula, known as the Born-Lande Equation  H latt = (-LA)(z + )(z - )(e 2 )(1 – 1/n) 4  r Where: L = 6.022 x 10 23 A = Madelung Constant z = summation of charges on the ions e = electron charge = 1.6 x 10 -19 C  = permittivity in a vacuum = 8 x 10 -12 F/m r = distance between the ions n = Born constant

5. Lattice Energy there has to be an easier way... (this would be a pretty lousy slide if there wasn’t) We use what’s called the Born-Haber cycle, which makes use of some specific heats of reaction (  H rxn ).  H f ° ≡ the standard heat of formation of a compound from its elements  H sub ≡ the heat of sublimation (solid  gas)  H BDE ≡ the Bond Dissociation Energy for a covalent bond  H I1 ≡ first ionization energy (neutral atom losing an e -, always positive)  H I2 ≡ second ionization energy (+1 to +2, large and positive)  H EA ≡ electron affinity (always a negative term except Be and N)  H latt ≡ lattice energy (always negative, usually quite large)

6. Formation of NaCl (s) Na (s) + ½ Cl 2(g)  H sub = 107 kJ Na (g) + ½ Cl 2(g)  H BDE = 122 kJ* * BDE Cl 2(g) = 244 kJ, so ½ (244 kJ) = 122 kJ Na (g) + Cl (g)  H I1 = 496 kJ Na + (g) + Cl (g) + e -  H EA = -349 kJ Na + (g) + Cl - (g)  H latt = -787 kJ NaCl (s)  H f ° = ???

7. How do we calculate  H f ° from the Born-Haber Cycle? From our work with Hess’ Law we know that energies are additive.  Therefore we can add up all of the components from the cycle which yield the overall formation reaction (from the elements).  H f NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl

8. Formation of NaCl (s) Na (s) + ½ Cl 2(g)  H sub = 107 kJ Na (g) + ½ Cl 2(g)  H BDE = 122 kJ* Na (g) + Cl (g)  H I1 = 496 kJ Na + (g) + Cl (g) + e -  H EA = -349 kJ Na + (g) + Cl - (g)  H latt = -787 kJ NaCl (s)  H f ° = -411 kJ/mol  H f NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl

9.  H sub Mg (s)  Mg (g) = 146 kJ/mol  H I1 Mg (g)  Mg + (g) = 738 kJ/mol  H I2 Mg + (g)  Mg 2+ (g) = 1451 kJ/mol  H BDE F 2(g) = 159 kJ/mol of F 2  H EA F = -328 kJ/mol of F  H form MgF 2(s) = -1124 kJ/mol (this is a  H° f ) Determine the lattice energy of MgF 2(s)

10. Lattice Energy of MgF 2(s) Mg (s) + F 2(g)  H sub = 146 kJ Mg (g) + F 2(g)  H BDE = 159 kJ Mg (g) + 2F (g)  H I1 = 738 kJ Mg 2+ (g) + 2F (g) + 2e -  H EA = -656 kJ (2 x -328 kJ) Mg 2+ (g) + 2F - (g)  H latt = ? MgF 2(s)  H f ° = -1124 kJ/mol  H I2 = 1451 kJ Mg + (g) + 2F (g) + e -  H latt =  H f – (  H sub +  H BDE +  H I1 +  H I2 +  H EA ) = -2962 kJ/mol MgF 2(s)

11. Lets leave it here as far as new material...

12. Your Assignment (and no not if you choose to accept it, just accept it) 1)Using your notes and the textbook suggest possible reasons why some reactions are exothermic and some are endothermic (5.4.2) in terms of average bond energy/enthalpy. 2)The combustion of methane is represented by the equation CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l) + 890.3 kJ. a) what mass of CH 4(g) must be burned to give off 1.00 x 10 5 kJ of heat? b) how much heat is produced when 2.78 moles of CO 2(g) are generated? 3) Using standard enthalpies of formation from Appendix B in your textbook calculate the standard enthalpy change for the following reactions: a) NH 3(g) + HCl (g)  NH 4 Cl (s) b) 3C 2 H 2(g)  C 6 H 6(l) c) FeO (s) + CO (g)  Fe (s) + CO 2(g)

13. 4)When burning a Dorito you find that the temperature of 150 g of water in an aluminum (mass 12 g) can is raised by 64 K. What amount of energy was released by the Dorito? You may assume that no heat was lost to the surrounding and it was completely transferred to the can and water. 5)Use the following 2 reactions calculate the  H rxn for 2NO 2(g)  N 2 O 4(g). N 2(g) + 2O 2(g)  N 2 O 4(g) ;  H = 9.2 kJ and N 2(g) + 2O 2(g)  2NO 2(g) ;  H = 33.2 kJ 6)Calculate the enthalpy of reaction: BrCl (g)  Br (g) + Cl (g)  H = ? Using the following data: Br 2(l)  Br 2(g)  H = 30.91 kJ Br 2(g)  2Br (g)  H = 192.9 kJ Cl 2(g)  2Cl (g)  H = 243.4 kJ Br 2(l) + Cl 2(g)  2BrCl (g)  H = 29.2 kJ 7) Question 9.33 from your textbook. Using a Born-Haber cycle for KF to calculate  H EA of fluorine.