Presentation on theme: "Ch. 9: Chemical Bonding I: Lewis Theory (cont’d) Dr. Namphol Sinkaset Chem 200: General Chemistry I."— Presentation transcript:
Ch. 9: Chemical Bonding I: Lewis Theory (cont’d) Dr. Namphol Sinkaset Chem 200: General Chemistry I
I. Chapter Outline I.Introduction II.Energy of Ionic Bond Formation III.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 - IE 1 = 496 kJ/mole Cl (g) + e - Cl - (g) EA = -349 kJ/mole Na (g) + Cl (g) Na + (g) + Cl - (g) IE 1 + 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. Born-Haber Cycle for NaCl
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 CaCl 2 given the following enthalpy values: heat of sublimation for Ca = kJ/mole, Cl 2 bond energy = 243 kJ/mole, 1 st ionization energy Ca = kJ/mole, 2 nd ionization energy Ca = kJ/mole, 3 rd ionization energy Ca = kJ/mole, electron affinity Cl = -349 kJ/mole, heat of formation CaCl 2 = 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. Formation of H 2
III. Bond Energies Breaking bonds require energy… A-B (g) A (g) + B (g) ΔH˚ break = BE A-B Forming bonds releases energy... A (g) + B (g) A-B (g) ΔH˚ form = -BE A-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. Estimating Rxn Enthalpies
III. Calculating w/ BE’s A reaction can be considered a two-step process: 1)Heat absorbed to break bonds 2)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.