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Unformatted text preview: Version 074 Final Mccord (52450) 1 This printout should have 52 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. McCord CH302 1pm This exam is only for McCords MWF 1pm CH302 class. R = 8 . 314 J/mol K R = 0 . 08206 L atm/mol K R = 62 . 36 L torr/mol K 1 L atm = 101.325 J G = H T S T f = k f m T b = k b m P A = x A P A , pure = cRT ln parenleftbigg P 2 P 1 parenrightbigg = H vap R parenleftbigg 1 T 1 1 T 2 parenrightbigg ln parenleftbigg K 2 K 1 parenrightbigg = H rxn R parenleftbigg 1 T 1 1 T 2 parenrightbigg ln parenleftbigg k 2 k 1 parenrightbigg = E a R parenleftbigg 1 T 1 1 T 2 parenrightbigg pH = log[H + ] K w =[H + ][OH ] K w = 1 . 10 14 at 25 C pH = p K a + log [ A ] [ HA ] G = nFE G = RT ln K anode  solution  solution  cathode E = E  RT nF ln Q E = E  . 0257 n ln Q E = E  . 05916 n log Q I t n F = moles ln parenleftbigg [A] [A] parenrightbigg = kt t 1 / 2 = ln2 ak 1 [A] 1 [A] = kt t 1 / 2 = 1 ak [A] [A] [A] = kt t 1 / 2 = [A] 2 ak 001 10.0 points 2 . 37 moles of a weak electrolyte is dissolved into 588 grams of water. The freezing point of the solution is 9 . 52111 C. What is the percent ionization for this substance in this solution? K f = 1 . 86 C /m for water. Assume that the electrolyte is a simple 1:1 ratio. 1. 50.0 2. 34.0 3. 25.0 4. 29.0 5. 42.0 6. 24.0 7. 43.0 8. 41.0 9. 33.0 10. 27.0 Correct answer: 27%. Explanation: T f = 9 . 52111 C T f = 0 C m H 2 O = 588 g n = 2 . 37 mol K f = 1 . 86 C /m T f = T f T f = 0 . C ( 9 . 52111 C) = 9 . 52111 C T f = K f m eff m eff = T f K f = 9 . 52111 C 1 . 86 C /m = 5 . 11888 m Version 074 Final Mccord (52450) 2 initial molality of solution = 2 . 37mol MX . 588kg water = 4 . 03061 m MX M + + X ini, m 4 . 03061 , m x x x fin, m 4 . 03061 x x x Total molality = 4 . 03061 m x + x + x = 4 . 03061 m + x m eff = 4 . 03061 m + x = 5 . 11888 m x = 1 . 08827 m % ionization = x m ini 100% = 1 . 08827 m 100% 4 . 03061 m = 27% 002 10.0 points The graph describes the energy profile of a reaction. A B Time Energy(kJ) 50 300 400 What are the values for H and E a , respec tively, for the reaction in the direction writ ten? 1. 250 kJ, 100 kJ 2. 250 kJ, 100 kJ 3. 250 kJ, 350 kJ correct 4. 250 kJ, 100 kJ 5. 250 kJ, 350 kJ Explanation: E a H A B Time Energy(kJ) 50 300 400 H = 300 kJ 50 kJ = 250 kJ E a = 400 kJ 50 kJ = 350 kJ 003 10.0 points The following figure represents the progress of a given reaction at 298 K. b b b b b G A B C D E rxn progress At point B on this figure, what is the relation ship of Q to K ?...
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 Spring '07
 Holcombe

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