#Chem 162-2008 Exam II review session draft-2

#Chem 162-2008 Exam II review session draft-2 - CHEM...

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CHEM 162-2008 HOURLY II REVIEW SESSION DR. ED TAVSS Sunday, March 23 rd , 8 – 10 PM, Hickman Hall 138 Chemical Equilibria (chapter 14) Solubility Product Equilibria (chapter 16A) Monday, March 24 th , 8 – 10 PM, Food Sci. Aud. Acid and Base Equilibria (chapter 15) Chem 162-2007 exam II review session 1
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CHEMICAL EQUILIBRIUM CHAPTER 14 Chem 162-2007 exam II review session 2
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SKIP Relationship of Kinetics and Equilibria Begin with 20 coins and 2 circles, A & B. Put all coins in pile A. Count how many coins are in eachpile and move 40% (rounding up to the nearest coin) of the coins that were in pile A to pile B, and then 25% of the coins that were in pile B to pile A. Record the number of coins in each pile. Repeat the cycles until equilibrium is reached. ET: Show (if true) that starting with 20 coins in B (with a 25% loss) and 0 coins in A (with a 40% loss) will result in the same equilibrium values. A B 1. 0 Rate f = 0.40 x 20 = 8 2. 12 8 -25% Rate r = 0.25 x 8 = 2 3. 14 -40% 6 Rate f = 0.40 x 14 = 6 4. 8 12 -25% Rate r = 0.25 x 12 = 3 5. 11 -40% 9 Rate f = 0.40 x 11 = 5 6. 6 14 -25% Rate r = 0.25 x 14 = 4 7. 10 -40% 10 Rate f = 0.40 x 10 = 4 8. 6 14 -25% Rate r = 0.25 x 14 = 4 9. 10 -40% 10 Rate f = 0.40 x 10 = 4 10. 6 14 -25% Rate r = 0.25 x 14 = 4 Chem 162-2007 exam II review session 3 -4 -4 -4 -4 -4 -5 -3 -6 -2 -8
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11. 10 -40% 10 Chem 162-2007 exam II review session 4
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CHAPTER 14 - CHEMICAL EQUILIBRIUM A + B C + D (gases or solutes, only) Law of mass action (or molar interaction ): K c = ([C][D])/([A][B]) eA + fB gC + hD K c = ([C] g [D] h )/([A] e [B] f ) e.g., 2 NO + O 2 2 NO 2 NO + NO + O 2 NO 2 + NO 2 K c = ([NO 2 ] 2 )/([NO] 2 [O 2 ]) K A B [B]/[A] = K 1 C D [D]/[C] = K 2 A + C B + D ([B][D])/([A][C]) = K 1 x K 2 * * Addition of ([B]/[A]) + ([D]/[C]) = (([B][C] + [A][D]))/([A][C]) Equations K(s) Add Multiply Reverse Invert Double Square Halve Square root K p inatm = K c inM (RT) ∆ngas Chem 162-2007 exam II review session 5
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CHEM162-2006 4th WEEK RECITATION CHAPTER 13 - NON ACID-BASE CHEMICAL EQUILIBRIA NON-ACID-BASE EQUILIBRIUM CONCEPTS ET: Discuss K for gases or solutes, only. 31 (mod.) Write expressions for K for the following reactions. a. P 4 (s) + 5O 2 (g) P 4 O 10 (s) K = 1/[O 2 ] 5 = [O 2 ] -5 MWP4 = 123.89 g/mol; D = 2.70 g/mL; therefore, concentration = 0.0022 mol/mL = 2.2 mol/L b. NH 4 NO 3 (s) N 2 O(g) + 2H 2 O(g) K = ([N 2 O][H 2 O] 2 ) c. CO 2 (g) + NaOH(s) NaHCO 3 (s) K = 1/([CO 2 ] d. S 8 (s) + 8O 2 (g) 8SO 2 (g) K = [SO 2 ] 8 /[O 2 ] 8 e. 5Fe 2+ (aq)+MnO 4 - (aq) +8H + (aq) 5Fe 3+ (aq) +Mn 2+ (aq)+ 4H 2 O(l) K = ([Fe 3+ ] 5 [Mn 2+ ])/([Fe 2+ ] 5 [MnO 4 - ][H + ] 8 ) Chem 162-2007 exam II review session 6
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CHEM162-2005 4th WEEK RECITATION CHAPTER 13 - NON ACID-BASE CHEMICAL EQUILIBRIA NON-ACID-BASE EQUILIBRIUM CONCEPTS ET: Discuss K conversions. Refer to formula sheet, including addition of equilibrium equations (corresponding to multiplying the K’s). 13.21 At a given temperature, K = 1.3 x 10 -2 for the reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) Calculate values of K for the following reactions at this temperature. d. 2N 2 (g) + 6H 2 (g) 4NH 3 (g) Double the equation = square the K = (1.3 x 10 -2 ) 2 = 1.69 x 10 -4 a. 1/2N 2 (g) + 3/2H 2 (g) NH 3 (g) Half of the equation = square root of K = √(1.3 x 10 -2 ) = 0.114 b. 2NH 3 (g) N 2 (g) + 3H 2 (g) Reverse the equation = inverse of K = 1/(1.3 x 10 -2 ) = 76.9 c. NH 3 (g) ½ N 2 (g) + 3/2H 2 (g) Reverse and ½ the equation = the inverse square root of K = √(1/(1.3 x 10 -2 )) = 8.77 Chem 162-2007 exam II review session 7
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Chemical Equilibria – Types of Review Problems Given equilibrium concentrations, find K
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This note was uploaded on 04/04/2008 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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#Chem 162-2008 Exam II review session draft-2 - CHEM...

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