CHEM 108 1st Edition Lecture 8 Current LectureChapter 11 Solutions: Properties and Behavior continued:Enthalpy of Solution:- To make a solution you must:o overcome all attractions between the solute particles (breaking bonds require energy ); therefore ΔHsolute is endothermic o overcome some attractions between solvent molecules (breaking some bonds also require energy);therefore ΔHsolvent is endothermic o form new attractions between solute particles and solvent molecules (forming bonds releases energy); therefore ΔHmix is exothermic o The overall DH for making a solution depends on the relative sizes of the DH for these three processes ΔHsolution= ΔHsolute + ΔHsolvent + ΔHmix - Dissolution of Ionic Solids: Enthalpy of solution (ΔHsolution) depends on: o Energies holding solute ions in crystal lattice (endothermic)o Attractive force holding solvent molecules together (endothermic)o Interactions between solute ions and solvent molecules. (exothermic)o ΔHsolution = ΔHion-ion + ΔHdipole-dipole + ΔHion-dipole ΔHdipole-dipole + ΔHion-dipole = ΔHhydration Lattice energy (U): - The energy released when one mole of the ionic compound forms from its free ions in the gas phase.- Lattice energy depends on: ionic charge and ionic radiuso (U= (Q1Q2)/distance), and Q is charge o So the bigger the charge the more lattice energy, and the smaller the ionic radius the more the lattice energy. - Lattice energy (U)—energy released when crystal lattice is formed. o ΔHion-ion = energy required to remove ions from crystal lattice. o ΔHion-ion = −U And: ΔHsoln = ΔHhydration – U U is negative in this equation because generally U is defined as the energywhen gas-phase ions form bonds, but we are calculating the energy for breaking bonds.Born-Haber Cycle: - Algebraic sum of enthalpy changes associated with formation of ionic solid from constituent elements. o E.g., Na(s) + ½ Cl2 (g) → NaCl(s) ΔHf ° = −411.2 kJ- Steps: o 1. sublimation of 1 mole Na(s) → Na(g) = ΔHsublimation o 2. breaking bonds of ½ mole of Cl2 (g) = ½ ΔHBE o 3. ionization of 1 mole Na(g) atoms = IE1 (Ionization energy)o 4. ionization of 1 mole Cl(g) atoms = EA1 (Electron Affinity)o 5. formation of 1 mole NaCl(s) from ions(g) = Uo ΔHf ° = ΔHsub + ½ ΔHBE + IE1 + EA1 +
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