Sp 08 FINAL EXAM SOLUTIONS

Sp 08 FINAL EXAM SOLUTIONS - CHM2045 Final Exam 1 Consider...

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CHM2045 Final Exam Spring 2008 Form Code A 1. Consider the reaction and rate data: 2NO(g) + Cl 2 (g) 2NOCl(g) Exp [NO] [Cl 2 ] Rate (x 10 6 mol/L . s) The experimentally determined rate equation (Rate = ) for this reaction is: (a) k [NO][Cl 2 ] (b) k [NO] (c) k (d) k [NO][Cl 2 ] 2 (e) k [NO] 2 [Cl 2 ] 1 0.250 0.250 1.43 2 0.500 0.250 5.72 3 0.250 0.500 2.86 4 0.500 0.500 11.4 If hold [Cl 2 ] constant and double [NO], as in Exp 1 and 2, the Rate quadruples. If hold [NO] constant and double [Cl 2 ], as in Exp 2 and 4, the Rate doubles. Therefore, Rate = k [NO] 2 [Cl 2 ] 2. If the reaction, 2A + B C , proceeds by the mechanism: A + B D (fast) A rate law (Rate = ) consistent with this mechanism is: (a) k [A][B] (b) k [A][B] 2 (c) k [B] (d) k [A] 2 [B] (e) k [A] D + B E (slow) E + A C + B (fast) In the case of the above fast-slow-fast mechanism with one reactant D in the initial fast step that is merely a reaction intermediate and not part of the overall reaction, the rate law for the overall reaction should include the reactants up to and including the slow rate-determining step, excluding the reaction intermediate. Therefore Rate = k [A][B][B] = k [A][B] 2 3. The dimerization of tetrafluoroethylene (C 2 F 4 ) to octafluorocyclobutane (C 4 F 8 ) is second order in reactant, C 2 F 4, and second order overall. At 450 K, the rate constant k = 0.0448 L mol 1 s 1 . If the initial concentration of C 2 F 4 is 0.100 M, the initial half-life is closest to: (a) 1.1 s (b) 15.5 s (c) 22.3 s (d) 223 s (e) 930 s Rate = k [C 2 F 4 ] 2 and the half-life for a 2nd-order reaction = 1/ k [C 2 F 4 ] 0 = (1 / [(0.0448)(0.100)]) = 223 s 4. A solution prepared from 360. mg of a sugar (ideal nonelectrolyte) in 1.00 g of water freezes at – 3.72 o C. (For water, K fp = 1.86 K/m, and the normal freezing point is 0.00 o C) What is the molar mass of this sugar? (a) 90 g/mol (b) 125 g/mol (c) 180 g/mol (d) 336 g/mol (e) 397 g/mol ΔT fp = 0.00 ºC – 3.72 ºC = – 3.72 ºC (or 273.15 K – 269.43 K = – 3.72 K) ΔT fp = K fp • molality = 1.86 K/m • m So, molality m = (3.72 K) / (1.86 K/m) = 2 m 2 moles sugar / 1000g water = 0.002 moles / 1.00 g water 0.360 g sugar / 0.002 moles = 180 g/mol 5. The data shown are for the reaction: 2H 2(g) + 2NO (g) → 2H 2 O (g) + N 2(g) . Experiment [H 2 ] [NO] Rate (M/s) What is the closest value to the missing rate (in M/s) for Experiment 5? (a) 0.0397 (b) 0.146 (c) 0.336 (d) 0.397 (e) 6.36 1 0.212 0.136 0.0248 2 0.212 0.272 0.0991 3 0.424 0.544 0.793 4 0.848 0.544 1.59 5 0.848 0.272 ?????????? If hold [H 2 ] constant and double [NO], as in Exp 1 and 2, the Rate quadruples. If hold [NO] constant and double [H 2 ], as in Exp 3 and 4, the Rate doubles. Based on the resulting rate law of Rate =
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This note was uploaded on 12/28/2011 for the course CHEM 2045 taught by Professor Gower during the Spring '11 term at University of Florida.

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Sp 08 FINAL EXAM SOLUTIONS - CHM2045 Final Exam 1 Consider...

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