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Why do molecules form the way they do?

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. Molecular Compounds. Why does oxygen form O 2 rather than O 8 (more accurately 4O 2 rather than O 8 )?

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Why do molecules form the way they do?

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  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 O2 rather than O8 (more accurately 4O2 rather than O8)?  We know that oxygen is a diatomic, but this is not a reason this is merely an observation of trend.  We need to consider DHBDE (Bond Dissociation Energy) which is the energy required to cleave a covalent bond. BDE O2 = 498 kJ/mol BDE 4O2 = 4 x 498 kJ = 1992 kJ Meaning 1992 kJ is required to break 4 moles of O2OR 1992 kJ of energy is given off when we form 4 moles of O2 from O atoms. BDE O—O = 146 kJ/mol BDE O8 = 8 x 146 kJ = 1168 kJ Meaning only 1168 kJ is given off when we form 8 O—O single bonds in O8. O2 is energetically favored.

  3. Some other elements to consider Why is phosphorus P4 rather than 2P2?  P4 (white phosphorus) is tetrahedral  P—P BDE = 209 kJ  P≡P BDE = 490 kJ Why is sulfur S8 rather than S2?  This is the converse of oxygen which prefers O2.  S—S BDE = 266 kJ  S=S BDE = 427 kJ

  4. The Ionic Lattice . . . One More Time • The find the lattice energy (DHlatt)of an ionic compound we can use the following formula, known as the Born-Lande Equation • DHlatt = (-LA)(z+)(z-)(e2)(1 – 1/n) 4per • Where: • L = 6.022 x 1023 • A = Madelung Constant • z = summation of charges on the ions • e = electron charge = 1.6 x 10-19 C • e = 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 (DHrxn). DHf° ≡ the standard heat of formation of a compound from its elements DHsub ≡ the heat of sublimation (solid  gas) DHBDE≡ the Bond Dissociation Energy for a covalent bond DHI1 ≡ first ionization energy (neutral atom losing an e-, always positive) DHI2 ≡ second ionization energy (+1 to +2, large and positive) DHEA ≡ electron affinity (always a negative term except Be and N) DHlatt ≡ lattice energy (always negative, usually quite large)

  6. Formation of NaCl(s) Na+(g) + Cl(g) + e- DHEA = -349 kJ DHI1 = 496 kJ Na+(g) + Cl-(g) Na(g) + Cl(g) DHlatt = -787 kJ DHBDE = 122 kJ* Na(g) + ½ Cl2(g) DHsub = 107 kJ Na(s) + ½ Cl2(g) DHf° = ??? NaCl(s) * BDE Cl2(g) = 244 kJ, so ½ (244 kJ) = 122 kJ

  7. How do we calculate DHf° 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). DHf NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl

  8. Formation of NaCl(s) Na+(g) + Cl(g) + e- DHEA = -349 kJ DHI1 = 496 kJ Na+(g) + Cl-(g) Na(g) + Cl(g) DHlatt = -787 kJ DHBDE = 122 kJ* Na(g) + ½ Cl2(g) DHsub = 107 kJ Na(s) + ½ Cl2(g) DHf° = -411 kJ/mol NaCl(s) DHf NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl

  9. Determine the lattice energy of MgF2(s) DHsub Mg(s) Mg(g) = 146 kJ/mol DHI1 Mg(g) Mg+(g) = 738 kJ/mol DHI2 Mg+(g) Mg2+(g) = 1451 kJ/mol DHBDE F2(g) = 159 kJ/mol of F2 DHEA F = -328 kJ/mol of F DHform MgF2(s) = -1124 kJ/mol (this is a DH°f)

  10. Lattice Energy of MgF2(s) Mg2+(g) + 2F(g) + 2e- DHEA = -656 kJ (2 x -328 kJ) DHI2 = 1451 kJ Mg2+(g) + 2F-(g) Mg+(g) + 2F(g) + e- DHI1 = 738 kJ Mg(g) + 2F(g) DHlatt = ? DHBDE = 159 kJ Mg(g) + F2(g) DHsub = 146 kJ Mg(s) + F2(g) DHf° = -1124 kJ/mol MgF2(s) DHlatt = DHf – (DHsub + DHBDE + DHI1 + DHI2 + DHEA) = -2962 kJ/mol MgF2(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) • 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. • The combustion of methane is represented by the equation CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + 890.3 kJ. • a) what mass of CH4(g) must be burned to give off 1.00 x 105 kJ of heat? • b) how much heat is produced when 2.78 moles of CO2(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) NH3(g) + HCl(g)  NH4Cl(s) • b) 3C2H2(g)  C6H6(l) • c) FeO(s) + CO(g)  Fe(s) + CO2(g)

  13. 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. • Use the following 2 reactions calculate the DHrxn for 2NO2(g) N2O4(g). N2(g) + 2O2(g)  N2O4(g); DH = 9.2 kJ and N2(g) + 2O2(g)  2NO2(g); DH = 33.2 kJ • Calculate the enthalpy of reaction: • BrCl(g) Br(g) + Cl(g)DH = ? • Using the following data: • Br2(l) Br2(g)DH = 30.91 kJ • Br2(g)  2Br(g)DH = 192.9 kJ • Cl2(g)  2Cl(g)DH = 243.4 kJ • Br2(l) + Cl2(g)  2BrCl(g)DH = 29.2 kJ • 7) Question 9.33 from your textbook. Using a Born-Haber cycle for KF to calculate DHEA of fluorine.

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