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CHEM 112 Test 1 Study Guide

by: Abbey McCoy

CHEM 112 Test 1 Study Guide CHEM 112

Marketplace > University of South Carolina > Art > CHEM 112 > CHEM 112 Test 1 Study Guide
Abbey McCoy
GPA 3.7

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About this Document

This study guide reviews topics that will be covered on the first test.
General Chemistry II
Dr. Lovelace
Study Guide
Chemistry, chemistry 2, Chemistry II, General Chemistry, solubility, henry's law, boiling point, freezing point, concentration, Spontaneity
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This 6 page Study Guide was uploaded by Abbey McCoy on Friday February 5, 2016. The Study Guide belongs to CHEM 112 at University of South Carolina taught by Dr. Lovelace in Winter 2016. Since its upload, it has received 52 views. For similar materials see General Chemistry II in Art at University of South Carolina.


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Date Created: 02/05/16
CHEM 112 Test 1 Study Guide Principles of Solubility:  Dynamic—able to change it o Temperature can change the dynamic process  Why this happens: o Enthalpy (ΔH):  Heat given off or absorbed (endothermic/exothermic)  Usually a spontaneous reaction is exothermic and gives off heat, but some spontaneous solutions are endothermic  Types of Interactions: o Solvent-solvent o Solute-solute o Solute-solvent  The Solution Process: o First need to undo solvent and solute (requires energy) o Separation of solute and of solvent is endothermic o Bringing the two together is exothermic o See “Enthalpy of Solution” figure in lab manual ΔH solutionH 1st stepH 2nd stepΔH 3rd step st o 1 step:  Absorbs energy to separate solute o 2 step: rd Absorbs energy to separate solvent o 3 step:  Releases energy to combine both  Spontaneity: o Solid < Liquid < Gas (least spontaneity to most spontaneity)  Spontaneity and Solubility: o What causes a solute to be dissolved?  Like dissolves like  Intermolecular Forces: o Dipole-dipole  Polar molecules o London Dispersion Forces  Everything has it, but it’s the only force nonpolar molecules have, so it’s typically associated with nonpolar molecules o Hydrogen Bonding  Molecule must be polar and have H-F, H-O, or H-N o Strengths:  Strongest = covalent bonds  All intermolecular forces are significantly weaker than covalent bonds  Weak = hydrogen bonds  Very weak = dipole-dipole  VERY very weak = London dispersion forces  Relative Solubility: o Ex: hexane and water C C C H H C C C :O: (Hexane) (Water)  Won’t dissolve—not alike  London dispersion forces holding hexane won’t break for water  Water has hydrogen bonds  Water is polar, and hexane is nonpolar o Will iodine dissolve better in hexane or water? I I  Iodine has London dispersion forces, but no hydrogen bonds  Answer: Hexane o What about methanol (CH OH)2  Answer: H 2 o NOTE: If it’s not an ionic salt or doesn’t have hydrogen bonds, it won’t dissolve in water  Why is water a universal solvent? o For an ionic compound to dissolve, ions have to separate and solvent (water) has to separate o Next step: hydration  Water molecules separate ions Ionic Compound Water Ions Separate Water Separates Increase disorder Increase disorder Hydration  Gas can be a solute o Pressure will have a big impact in this case o Henry’s Law:  C = KD (directly proportional)  C—concentration (could be molarity-M, molality-m, or partial pressure- PA)  K-value gets smaller as temperature increases  As temperature increases, solubility increases (trend) o Example of Henry’s Law worked out:  Given:  H2O at 20°C  Partial pressure = 8.00 torr  K = 3.91 x 10 molal/atm  Conversions:  1atm = 760torr 8.0???????????????? ×1???????????? o ???????? = 760???????????????? = 0.01052????????????  Work: −2 ????????????????????  ???? = 3.91 × 10 ???????????? × 0.01053???????????? −4  ???? = 4.12 × 10 ???????????????????? o A + B C + heat (exothermic—releases heat) o A + B + heat C (endothermic—absorbs heat  If you increase the temperature of an exothermic reaction (-ΔH), you’re adding more product, so it will go in the other direction to reach equilibrium o ↑T of exothermic reaction (-ΔH)—equilibrium shifts to reactants (decreases solubility because you are shoving solute out of the solution) o ↑T of endothermic reaction (ΔH)—equilibrium shifts to products (increase solubility because you are helping the solute dissolve) Solubility and Temperature:  Solubility of gas—think in terms of pressure o Solubility of gas A = partial pressure of gas A AP ) o ↑P ,A↑solubilityA o ↓P ,A↓solubilityA  Biggest factor in solubility = ΔH (change in enthalpy) o If we have +ΔH (endothermic), heat = reactant (A + B + heat → C)  Increase heat, increase solubility (shoving heat towards product) (↑heat, ↑solubility) o If we have –ΔH(exothermic), heat = product (A + B → C + heat)  Increase heat, decrease solubility (↑heat, ↓solubility)  As we increase temperature, solubility increases (in general)—exothermic is exception Colligative Properties:  Rely/depend on the concentration of the solute  Add solute—lowers vapor pressure (changing the equilibrium of solution)  If we have a pure solvent, we have equilibrium o # molecules evaporating = # molecules condensing Pure solvent Add solute Molecules evaporating Molecules evaporating = = Molecules condensing Molecules condensing  Rauoult’s Law: o Vapor pressure solvent propotional tχ solvent o To find new vapor pressure: PsolventP°solvent χsolvent o To find change in vapor pressure: ΔPsolute P°solute χsolute o Example problem:  Given:  T = 27°C  P° = 104 torr  0.100 mol naphtha  9.90 mol benzene  Find:  Partial pressure of benzene (benzene  Work using formula P benz= (P°benz(χbenz: χ = 9.90= 0.99 benzene 10.0 Pbenz= (104 torr)(0.99) = 102.96 torr  Work using formula ΔP benz= (P°benz(χnapth: ΔP benz (104)(0.01) = 1.04 torr P° - ΔP = 104 – 1 = 103 torr  Boiling Point Elevation: o Boiling Point: when vapor pressure equals 1atm o Add solute, decrease vapor pressure  This means we need to heat to a higher temperature to get it to boil o Boiling point elevation is different for different solvents  Boiling point elevation is proportional to the molality (m) of the solute  ΔT B (K )Bm) o Add solute, decrease freezing point  Ex: salt on ice  Salt lowers freezing point of the ice  Salt does NOT melt ice. It just causes the freezing point to be lowered once the ice is melted and it is able to mix with the water o Add solute:  Decrease vapor pressure  Decrease freezing point  Increase boiling point o Ex: Calculate K - f  T° = 48.1°C  1.05 g urea  Work: 1.05???? ????????????????  = 0.0175???????????? ???????????????? 60.06????  30.0 ???? ???????????????? = 0.0300???????? ????????????????  Tf= new temperature = 42.4°C  ΔTf= (K f(m)  ΔTf= (T° - T )/m = (K f)/m = 48.1°????−42.4°= 9.8°????/???? 0.0300???????? ????????????????????  Osmosis: o Diffusion of fluid through semipermeable membrane  When water diffuses, it creates pressure (osmotic pressure)  Colligative property  ???? = ????????????  ???? = osmotic pressure  M = molar concentration of solute  R = ideal gas law constant (×???????????? ) ????????????×????  T = temperature in kelvins ????  Similar to ???? = ???????? (difference is the units) ????  Δ[solute] = Δπ  Change in concentration of solute = change in osmotic pressure o Ex: Osmotic Pressure-  Given:  5.70mg protein sample  1.00mL solution  Calculate molar mass:  Need to find ???????????????????? ???????? ???????????????????????????? ???????????????????? ???????? ????????????????????????????  π = 6.52torr = 0.00858atm  T = 20°C = 293K  π = MRT ???? 0.00858???????????? ????????????  ???? = = = 3.57 × 10 −3 ????−4 ???? × 293????−3 ????  3.57×10 ????????????× 1.0×10 = 3.57 × 10 ???????????? ???????????????????????????? ???? 1  5.70???????? ???????????????????????????? = 5.7 × 10 ???? ???????????????????????????? 5.70×10−3????  3.75


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