#Chem 162-2008 final exam review session

#Chem 162-2008 final exam review session - CHEM 162-2008...

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CHEM 162-2008 FINAL EXAM REVIEW SESSION DR. ED TAVSS Final Exam : Monday, 5/12, noon – 3 PM Mon, May 5 th , 8-10 PM, CDL-110 Chapter 12 – Solutions (barely covered) Chapter 13 – Kinetics (covered) Chapter 14 – Chemical equilibria (covered) Chapter 15 – Acid-base equilibria (v strongly covered) Tue, May 6 th , 8-10 PM, Hck 138 Chapter 16A–Slightly soluble salts equilibria (covered) Chapter 16B – Complex ion equilibria (covered) Chapter 22 – Coordination chemistry (barely covered) Chapter 17 – Thermodynamics (barely covered) Chapter 18 – Electrochemistry (Only electrolysis strongly covered) Chapter 19 – Nuclear chemistry (strongly covered) Attendance Chem 162-2008 final exam review session 1
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Chem 162-2008 final exam review session 2
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~42 HOURS OF LECTURE 4 HOURS OF REVIEW THEREFORE, SCRATCHING THE SURFACE REMEDY: o STUDY, STUDY, STUDY o E-MAIL ME IF YOU NEED HELP Chem 162-2008 final exam review session 3
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CHAPTER 12 – SOLUTIONS FORMULAS * Conversion strategy: Begin with given concentration. End with desired concentration units. Convert numerator and denominator separately to achieve desired concentration units. g solute + g solvent = g solution ; mol solute + mol solvent = mol solution Mass percent = grams of solute/100 g solution Mole percent = moles of solute/100 moles solution Molarity = moles of solute/L of solution ET: Molarity is temperature dependent, but molality is not . Difficult but important conversion Molality = moles of solute/kilogram of solvent PPM = grams of solute/1,000,000 grams solution Volume percent = volume of solute/100 mL of solution Proof = 2 x Vol. %; e.g., 2 x 40 mL/100 mL solution = 80 proof Mole fraction: X A = n A /(n A + n B ) X A + X B = 1 Particle fraction: i X A = in A /(in A + in B ) i X A + i X B = 1 Raoult’s law: Don’t use: P soln = X solvent P o solvent Use: P soln = i X solvent P o solvent For two volatile components: P soln = i X solventA P o solventA + i X solventB P o solventB i X solvent = in solvent /(in solvent + in solute ) i X A + i X B = 1 van’t Hoff factor, “i”; i = moles of particles in solution/moles of solute dissolved Boiling-point elevation: ΔT = T f - T i = K b im solute Freezing-point depression: ΔT = T f - T i = -K f im solute (+K f gives absolute change in temp; -K f gives actual T f or T i ) Osmotic pressure: πV = inRT π = (n/V)iRT = iM solute RT ΔH soln = ΔH solute-solutebondbreaking + ΔH solvent-solventbondbreaking + ΔH solute-solventbondforming Henry’s Law: S A =kP A ; M A = Solubility of dissolved gas in solution “S” may be any unit of concentration (depending on the units of “k”), e.g., X, M, %. Chem 162-2008 final exam review session 4
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CHAPTER 13 – KINETICS: F ORMULAS C 12 H 22 O 11 + H 2 O ----> C 6 H 12 O 6 + C 6 H 12 O 6 Rate of sucrose disappearance = -(∆ [sucrose])/(∆ time) = -([sucrose f ] - [sucrose i ])/(t f - t i ) aA + bB ----> cC + dD General rate of reaction = -(1/a)(∆[A]/∆t) = -(1/b)(∆[B]/∆t) = (1/c)(∆[C]/∆t) = (1/d)(∆[D]/∆t) Determination of order of reaction through method of initial rates: For a single reactant: Rate 1 = k[A 1 ] m Rate 2 = k[A 2 ] m (Rate 1/Rate 2) = (k[A 1 ] m )/(k[A 2 ] m ) = ([A 1 ]/[A 2 ]) m For two reactants: Rate1 = k[A 1 ] m [B 1 ] n Rate2 = k[A 2 ] m [B 1 ] n (Rate1/Rate2) = (k[A 1 ] m [B 1 ] n )/(k[A 2 ] m [B 1 ] n ) = (A 1 /A 2 ) m Integrated Reaction
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This note was uploaded on 09/23/2009 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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#Chem 162-2008 final exam review session - CHEM 162-2008...

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