CHEM 031 Class Notes

CHEM 031 Class Notes - Intermolecular Forces(Chapter 10...

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Intermolecular Forces (Chapter 10) Phase - homogenous portion of a system separated from other portions by a well-defined phase boundary Solid & Liquid – condensed phases; held together by Intermolecular Forces Gas Intermolecular Forces 1. Ion-Ion Interactions Coulomb’s Law: F  q 1 q 2 /r 2 Maximize force by decreasing distance / increasing charge 2. Dipole-Dipole Forces Electrostatic attraction between polar molecules Arrange with dipoles in the same direction 3. Hydrogen Bonds Strong example of dipole-dipole Interaction represented by ----- line H – X; where X is replaced by N, O or F 4. Ion-Dipole Forces Electrostatic interaction (attraction) Interaction between ion and dipole Induced Dipole Not permanent Instantaneous (transient) dipole arising from fluctuations in electron cloud Polarizability – ease with which flucuations occur o Increases with number of electrons 1. Ion – Induced Dipole Forces 2. Dipole-Induced Dipole Forces 3. London Dispersion Forces Transient dipole, Ubiquitous reaction Falls off 1/r 6 Increases with size – more electrons further from nucleus means more polarizable Dependent on shape Properties of Liquids Surface tension – the amount of energy it takes to create a unit area of new surface o Unbalanced intermolecular forces at surface o Increases with strong intermolecular forces Viscosity – resistance of liquids to flow Special Properties of Water 1
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High surface tension and viscosity High specific heat – energy need to raise average kinetic energy Ice floats – solid is less dense than liquid Phase Transitions Vapor pressure – pressure exerted by gas in equilibrium with its liquid H vap – energy required to evaporate 1 mol of liquid Boiling point – temperature at which vapor pressure of the liquid equals the atmospheric pressure o Clausius-Clapeyron equation: ln(P) = - H vap /RT + C o ln(P 1 / P 2 ) = ( H vap / R)((T 1 – T 2 )/(T 1 T 2 )) Normal melting point – solid and liquid coexists (equilibrium); 1 atm H fus = energy required to melt 1 mol of solid Hess’s Law - H is only phase dependent (not pathway) H sub = H fus + H vap Phase Diagram Triple point – solid, liquid, gas coexist at equilibrium Lines represent where two phases exist at equilibrium o The solid-liquid line for water has a negative slope due to the density of ice Critical point – critical temperature above which substance cannot condense Solutions (Chapter 10) H sol = H lattice + H solvation Lattice energy represented by ‘u’ H solvation = o Cavitation energy – energy required to break solvent-solvent interactions – endothermic 2
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o Solute-solvent interaction – exothermic Born-Haber Cycle – application of Hess’s Law o H f o = H sub + H BE + IE + EA + U o Used to calculate lattice energy Concentrations 1. Molarity (M) = mol solute / L solution 2. Molality (m) = mol solution / kg solvent 3.
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