Chem 106, Week 1 Notes
Chem 106, Week 1 Notes Chem 106 (Dr. Paul Buckley)
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This 8 page Class Notes was uploaded by RebeccaR on Saturday April 2, 2016. The Class Notes belongs to Chem 106 (Dr. Paul Buckley) at Washington State University taught by Dr. Buckley in Spring 2016. Since its upload, it has received 36 views. For similar materials see Principles of Chemistry II (Chem 106, Dr. Paul Buckley) in Chemistry at Washington State University.
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I'm pretty sure these materials are like the Rosetta Stone of note taking. Thanks Rebecca!!!
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Date Created: 04/02/16
12.1-12.2 Monday, January 11, 2016 3:29 PM - Know your Ions - Propertiesof Solutions ○ What is a solution? Homogeneousmixture Composedof a solute dissolved in a solvent ○ How do they form? ○ How do solution properties differ from those displayed by pure substances? ○ What is not a solution? When the individual substances remain distinct, we are dealing with a heterogeneous system □ Example: bowl of M&Ms ○ Types of Solutions Non-ionic vs. ionic Non-ionic □ Sugar dissolved in water Ionic □ Salt dissolved in water Liquid-Solid Liquid-Liquid Liquid-Gas Gas-Gas □ Atmosphere Solute = oxygen, carbon-dioxide Solvent = nitrogen Solid-Solid □ Cu + Zn □ Cu + Sn ○ Solution Themes What drives solution formation? □ Intermolecularforces Dispersion forces ◊ Random motion of e- ◊ Local dipole moments< Hydrogen Bonding □ Thermodynamics Enthalpy (deltaH) Entropy (deltaS) ◊ Tendency towards randomness/chaos Mixing Based Upon Intermolecular Forces □ "Like dissolves Like" □ Ethanol and Water These two are soluble in all proportions(miscible) ….If a solution does not mix in all proportions (imiscible) Solute-Solute IM Forces ◊ Hydrogen Bonding ◊ Dispersion Forces Solvent-SolventIM Forces ◊ Hydrogen Bonding ◊ Dispersion Forces Solvent-SolventIM Forces ◊ Hydrogen Bonding ◊ Dispersion Forces - IM Forces ○ Ion-Ion Gas phase interactions (troposphere) Solutions too ○ Dispersion ○ Dipole-Dipole ○ Hydrogen Bonding ○ Ion-dipole - When a solution forms, solute-soluteand solvent-solventinteractions are replaced by solvent- solute interactions ○ By assessing the thermodynamicfavorabilityof each "step," you can determine whether or not a solution is likely to form - Solution that Does NOT form ○ Example: Octane in Water Solute-Solute (octane) □ Dispersion Forces Solvent-Solvent(water) □ Hydrogen Bonding □ Dispersion Forces □ Water would have to break all of its Hydrogen Bonds in order to solvate the octane, which would take a large amount of energy HYDROGEN BONDING IS STRONGER THAN DISPERSION FORCES In order for this solution to form, energy must be put INTO the system - In order to have a "perfect" solution, the two substances need to have the SAME Intermolecular Forces - Solute is ALWAYS the SMALLER amount (moles) - Solvent is ALWAYS the LARGER amount (moles) 12.3-12.4 Wednesday, January 13, 2016 7:55 AM - Three distinct steps in formationof a solution ○ Solute particles separate deltaH = positive endothermic ○ Solvent particles separate deltaH = positive endothermic ○ Solute and solvent particles mix deltaH = negative Exothermic - Enthalpy of solution ○ deltaH soln = deltaH solute + deltaH solvent + deltaH mixing ○ Magnitude of each deltaH term is determined by the strength of the IM forces - Example: dissolving Octane (solute) in Water (solvent) ○ deltaH solute = positive, small Dispersion forces, weak ○ deltaH solvent = positive, large Hydrogen bonding, strong ○ deltaH mixing = negative, small Dispersion forces only - Example: dissolving NaCl in Water ○ deltaH solute = positive, large Ionic bonding ○ deltaH solvent = positive, large Hydrogen bonding ○ deltaH mixing = negative, large Ion-dipole forces ○ Endothermic reaction ○ Occurs spontaneously ○ The increase in entropy is enough to overcome the 3 kJ gap in enthalpy Allows it to be spontaneous - Dissolving salts in water ○ [ deltaH solvent + deltaH mix ] = deltaH hydration ○ deltaH solution = deltaH solute + deltaH hydration ○ If deltaH solute < deltaH hydration, then deltaH solution = exothermic ○ If deltaH solute > deltaH hydration, then deltaH solution = endothermic ○ deltaH hydration is typically large, exothermic ○ deltaH solute = - (lattice energy) ○ Whether dissolving a salt in water is exothermicor endothermicdepends on the lattice energy ○ A salt with a small lattice energy will have an exothermicdeltaH solution Not a great deal of energy needs to be spent breaking lattice structure Temperature goes up ○ Insoluble salts have relatively large lattice energies Heat supplied from hydrating ions and kinetic energy of water moleculesis not sufficient to disrupt the crystal lattice structure - Other factors affecting solubility ○ Temperature Solids □ As temp increases, so does solubility □ Increase in KE aksjd;lafkjds Gases □ Temp increase, solubility decreases □ Increase in KE allows solute molecules to escape from solute-solventIM forces ○ Pressure Gases □ Higher pressure of gas above the solution, higher the solubility of gasin the solution □ Henry's Law Lectures Page 1 □ Henry's Law Relates concentrationof a dissolved gas to the pressure of the gas above the solution S = kP ◊ S = solubility of dissolved gas in mol/L ◊ K = henry's law constant ◊ P = partial pressure of solute above solution (atm) Example: calculate mass of N2 dissolved at room temperature (25*C) in an 80L aquarium. Total atmospheric pressure is 1.0 atm. Mol fraction N2 = 0.78. Lectures Page 2 Lectures Page 3 12.5-12.6 Friday, January 15, 20168:05 AM - Colligative Properties of Solutions ○ Properties of solution that depend only on the number of solute particles…………….???? Vapor pressure Boilig pts Freezing pts □ How do solutes affect these properties v. the pure solvent? - Changes in vapor pressure ○ Vapor pressure is lower for liquid with solute dissolved in it ○ Nonvolatile solute molecules at the liquid surface ○ Reduces amount of solvent molecules at the surface ○ Reduces the vapor pressure ○ Raoult's Law Vapor Pressure Changes □ Begin with two changes: Non-volatile solid, non-ionic solute Lectures Page 1 Non-volatile solid, non-ionic solute ◊ Example: Dissolving sugar in water Dissolve wax in nonpolar solvent ◊ Psoln = Xsolvent * P^0 solvent ◊ Psoln = [mole fraction of solvent] * [partial pressure of solvent] ◊ Vapor pressure is due to solvent only ◊ Look up other bullet point tbh Example: – Mole fraction of solvent = mole solvent / mole solute + mole solvent - 2 (volatile solute) ○ Ptotal = Pa + Pb Ptotal = XaP Lectures Page 2 - Ideal solutions ○ FILL IN THE REST OF THESE NOTES Lectures Page 3
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