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Ch. 9: Chemical Bonding I: Lewis Theory (cont’d). Dr. Namphol Sinkaset Chem 200: General Chemistry I. I. Chapter Outline. Introduction Energy of Ionic Bond Formation Energy of Covalent Bond Formation. I. Energies of Bond Formation.
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Ch. 9: Chemical Bonding I: Lewis Theory (cont’d) Dr. Namphol Sinkaset Chem 200: General Chemistry I
I. Chapter Outline • Introduction • Energy of Ionic Bond Formation • Energy of Covalent Bond Formation
I. Energies of Bond Formation • Previously in this chapter, we talked about how ionic and covalent bonds form. • Now we look at the energies involved. • For ionics, we apply Hess’s Law. • For covalents, we use bond energies (enthalpies).
II. Ionic Bond Formation • Recall that we envision the creation of ionic bond as an e- transfer from the metal to the nonmetal. • However, if we look at the energies involved in these two steps, we see something puzzling.
II. Ion Formation for NaCl • We break up the e- transfer process into two steps and add up the energies. Na(g) Na+(g) + e- IE1 = 496 kJ/mole Cl(g) + e- Cl-(g) EA = -349 kJ/mole Na(g) + Cl(g) Na+(g) + Cl-(g) IE1 + EA = 147 kJ/mole
II. Other Energies • For most ionics, this is typical; ion formation is an endothermic process. • So why do ionics form at all? • There must be a huge exothermic process to offset these endothermic processes. • The strong +/- attractions that are formed are the source of this exo step.
II. Lattice Energy • lattice energy: the energy associated with forming a crystalline lattice of alternating cations and anions from gaseous ions.
II. Born-Haber Cycle • Although lattice energy is a critical component of ionic bond formation, it cannot be measured directly. • We use Hess’s Law in the Born-Haber Cycle.
II. Calculating Lattice Energy • Since enthalpy is independent of path, the following equation applies. ΔH˚f = ΔH˚atom + ΔH˚BE + ΔH˚IE + ΔH˚EA + ΔH˚lattice
II. Sample Problem • Calculate the lattice energy for CaCl2 given the following enthalpy values: heat of sublimation for Ca = 179.3 kJ/mole, Cl2 bond energy = 243 kJ/mole, 1st ionization energy Ca = 589.8 kJ/mole, 2nd ionization energy Ca = 1145.4 kJ/mole, 3rd ionization energy Ca = 4912.4 kJ/mole, electron affinity Cl = -349 kJ/mole, heat of formation CaCl2 = -795.8 kJ/mole.
III. Covalent Bond Strength • The strength of a covalent bond depends on how strongly the e- are held by both nuclei. • The bond energy or bond enthalpy (BE) is the energy needed to overcome this attraction. It is defined as the energy needed to break 1 mole bonds in the gas phase • Another way to envision it…
III. Bond Energies • Breaking bonds require energy… • A-B(g) A(g) + B(g)ΔH˚break = BEA-B • Forming bonds releases energy... • A(g) + B(g) A-B(g)ΔH˚form = -BEA-B • Different bonds have different levels of attraction (have different depths to their wells) and thus different BE’s.
III. BE’s and Chemical Change • In a reaction, bonds are broken and new bonds are formed. • Relative strengths of bonds determine whether rxn is exo or endo. • The energy released or absorbed in a reaction is due to the difference between reactant and product bond energies.
III. Calculating w/ BE’s • A reaction can be considered a two-step process: • Heat absorbed to break bonds • Heat absorbed to make bonds • We can sum these to get the heat of reaction.
III. Sample Problem • Calculate ΔH˚rxn for the reaction shown below. Use the indicated bond enthalpies.