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Exam 1 Study Guide

by: Emily Grace

Exam 1 Study Guide Chem 032

Emily Grace
GPA 3.85

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About this Document

Study Guide 1 covers chapters 12 & 13
General Chemistry 2
Professor Ruggles
Study Guide
50 ?




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This 4 page Study Guide was uploaded by Emily Grace on Friday February 5, 2016. The Study Guide belongs to Chem 032 at University of Vermont taught by Professor Ruggles in Spring 2016. Since its upload, it has received 142 views. For similar materials see General Chemistry 2 in Chemistry at University of Vermont.


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Date Created: 02/05/16
Chem 32: Exam 1 Study Guide/ Outline (Ch. 12, 13) Predicting Solubility General Forces: Solute/Solvent Forces > Solute/Solute Forces > Solvent/ Solvent Forces Forces of Attraction: Electrostatic (ionic) > Network Covalent > H-bonding > Dipole/Dipole > Dispersion  Ionic/ Electrostatic: o Charge: greater charge = stronger force o Size: smaller ions = stronger force  Network Covalent: generally 500g/mol or larger; covalent  Hydrogen Bonding o –OH o –NH 2 o –CO H2  Dipole/Dipole: no symmetry  Dispersion: symmetry **LIKE DISSOLVES LIKE** Solute: what’s being dissolved (s,l,g) Solvent: dissolving agent (l) What matters when predicting solubility? 1. Phase- liquids are more soluble than solid/gas 2. Solids are more soluble than gasses (exception: O gas2soluble in H O) 2 3. Liquid v. Liquid Solute a. Identify whether the solvent is polar, non polar, or H-bonding (which forces are present?) b. Identify the solute’s forces of attraction c. Match the forces for best solubility  Solutes with –OH, -NH , 2CO H,2water is the best choice for solvent (or some other h-bonding solvent)  Solutes without H-Bonding: If dipole-dipole polar solvent; If dispersion non-polar solvent  As we increase # of Carbons, we observe less solubility in H O 2 4. Solid v. Solid Solute a. If Ionic system, want H 2 as solvent b. Covalent or molecular system, want molecular solvent (dipole, dispersion) c. weaker forces in solute aids solubility 5. Gas v. Gas Solute a. gas solute with strongest forces of attraction aids solubility Temperature Effects:  With increase in temperature, we observe greater solubility in solid and liquid solutes. We observe decrease in solubility in gas solutes Definitions:  Miscible: (liq solute) good mixing; homogeneous  Non-miscible: (liq solute) no mixing; heterogeneous  Non-saturated solution: (dilute solution) more solute can dissolve in solvent  Saturated solution: (concentrated) no more solute can dissolve (T=25C)  Super-saturated: past saturation due to increase in temperature Units:  % Volume: (vol solute/ vol solution) x 100%  % Mass: (mass solute/ mass solution) x 100%  ppt- parts per thousand  ppm- parts per million  ppb- parts per billion  M (molarity): mols solute/ L solution  m (molality): mols solute/ kg solvent  N (normality): mols H / L solution Gas Solutes = Henry’s Law Sgas= K H gas(S= molarity); (K H Henry’s constant, gas specific) Pgas partial or pure  Partial = (gas(Ptotal **GREATER PRESSURE = GREATER SOLUBILITY** Liquid Solutes = Raout’s Law Psolution( liquid1liquid1 (liquid2Pliquid2 Vapor pressure = Atmospheric Pressure = Boiling point **INCREASE IN TEMP = INCREASE IN VAPOR PRESSURE SOLID SOLUTES = ENERGETICS OF SOLUTION H soln H soluteH solventH mix H soln -H latticeH hydration H solute-H lattice Vapor Pressure Lowering Psoln= (solventPsolvent P = ( solutePsolvent **physical properties of solution depend on how much solute is in solution** Van’t Hoff Factor (i) Covalent (molecular): ipredicted 1 Ionic: ipredicted 1 Percent Ionization: ( i obsipred x 100% Boiling Point Elevation T = (i)(K b(m)  K =mblal boiling point constant (C x Kg/ mol) Freezing Point Depression T = (i)(K F(m)  K =mFlal boiling point constant (C x Kg/ mol) Osmotic Pressure  = (i)(M)(R)(T) M= mol/L R= 0.082 (L x atm)/(mol x K) T= Kelvin CHEMICAL KINETICS (CH13) Rate =  [substance] / T [concentration in Molarity] Instantaneous Rates:  In regards to starting material, must add “-” sign  Stoichiometry is important, must divide by respective stoich Rate Expressions: AB m Rate = k [A]  m = rxn order  k = rate constant  rate = always in terms of starting materials Zero-order: m=0 (horizontal line) First-order: m=1 (linear) Second-order: m=2 (exponential) More than one starting material: A + B C Rate = k [A] [B] n  m= individual rxn order [A]  n= individual rxn order [B]  overall order = m+n Integrated Rate Laws: AB Zero-Order: [A] =t-kt + [A] o 1/2= [A] 0 2k First-Order: ln[A] = -tt + ln[A] o t1/2= .0693/ k Second- Order: 1/[A] = kt +t1/[A] t o 1/2= 1 / k[A] 0 t1/2= half life Activation Energy: the energy required to initiate a rxn ln(k 2k 1 = -E / A (1/T – 1/2 ) 1  EA= activation energy  T = kelvin


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