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CHM 142 (1)
Chapter 11

# CHM 142--Chapter 11.docx

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School
Department
Chemistry & Biochemistry
Course
CHM 142
Professor
Dr.James D.Garrity
Semester
Spring

Description
11.1 Solution-homogenous mixture Solvent-present in the greater number of moles Solute (can be more than 1)-being dissolved in solvent. Present in lesser number of moles Dissolution of ionic solids in a polar solvent Enthalpy of solution (energy change that happens during process) depends on:  Energies holding solute ions in crystal lattice (ΔH{ion-ion})  Attractive forces holding solvent molecules together (ΔH{dipole- dipole})  Energy that results from interaction of solute and solvent (ΔH{ion- dipole})  ΔH{soln} = ΔH{ion-ion} + ΔH{dipole-dipole} + ΔH{ion-dipole} When the solvent is water  ΔH{dipole-dipole} + ΔH{ion-dipole} = ΔH{hydration} Lattice Energy (U) –the energy released when one mole of the ionic compound (MX) is formed from its free ions in the gas phase.  M(g) + X(g)  MX(s)  U = k(Q 1Q2) / d o k is a proportionality constant depending on the lattice structure For compounds with similar lattice structure (k is the same), lattice energy depends on ionic charge and ionic radius. Lattice energy (U)- energy released when crystal lattice is formed ΔH{ion-ion} – energy required to remove ions from crystal lattice ΔH{ion-ion} = -U ΔH{soln] = ΔH{hydration} – U Born-Haber Cycle –Algebraic sum of enthalpy changes associated with formation of 1 mole of ionic solid from constituent elements  Na(s) + ½ Cl 2(g)  NaCl (s)  ΔH ° = -411.2kJ o sublimation of 1 mole Na(s)  Na(g) = ΔH sub o breaking bonds of ½ mole Cl 2(g) = ½ ΔH BE o ionization of 1 mole of Na(g) atoms = IE 1(ionization energy. 1 because it’s first ionization energy because you remove 1 e) o ionization of 1 mole Cl (g) atoms = EA 1 (electron affinity) o Figure 11.2 in book o ΔH ° = ΔH sub+ ½ ΔH BE + IE1+ EA 1+ U  you know (or can look up) all of these numbers except lattice energy, so solve for U 11.2 Vapor pressure –pressure exerted by a gas in equilibrium with its liquid Equilibrium –rates of 2 opposing processes (evaporation and condensation) are equal.  No net change (of # of molecules in liquid phase and # in gas phase)  Dynamic process. (Evaporation and condensation haven’t stopped) 3 things affect vapor pressure  temperature o higher temperature = higher kinetic energy (E k); more molecules with sufficient Ek to overcome attractive forces in liquid phase  Surface Area  Intermolecular forces of attraction In general, VP of solution is lower than VP of pure solvent. Raoult’s Law –VP of solution is proportional to mole fraction of solvent (with a non-volatile solute)  A non-volatile solute has basically no vapor pressure  P soluti= X solve* Psolvent ° o X is mole fraction o P is vapor pressure o P solventis the vapor pressure of the solvent by itself at 1 atm VP is a colligative property. The identity of the solute doesn’t matter, only the amount Ideal solutions obey Raoult’s law  Ideal solution means the strength of interaction between solvent and solute are similar to those between solvent and solvent. 11.3 Normal boiling point—temperature at which P=1 atm Plot of ln(P) vs. 1/T yields a line  M=- ΔH(vap) / R o R=8.314 J/mol*K o ΔH(vap)=energy required for a liquid to become vapor at its
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