Chapter 15 Even Problem Solutions

Chapter 15 Even Problem Solutions - CHAPTER FIFTEEN...

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Unformatted text preview: CHAPTER FIFTEEN CHEMICAL KINETICS Reaction Rates 10. ∆ [P 4 ] / ∆t = 6.0 × 10-4 mol L-1 s-1 ∆ [H 2 ] / ∆t = 3.6 × 10-3 mol L-1 s-1 12. a. The units for rate are always mol L-1 s-1 . b. Rate = k; k has units of mol L-1 s-1 . c. Rate = k[A], mol / L s = k (mol / L), k must have the unit of s-1 d. Rate = k[A] 2 , mol / L s = k (mol / L) 2 , k must have the units L mol-1 s-1 e. L 2 mol-2 s-1 14. 7.47 × 10 8 L mol-1 s-1 Rate Laws from Experimental Data: Initial Rates Method 16. a. Rate = k[I — ] [S 2 O 8 2- ] b. For all the experiments, k = 3.9 × 10-3 L mol-1 s-1 Therefore, k mean = 3.9 × 10-3 L mol-1 s-1 18. a. Rate = k[Hb] x [CO] y Comparing the first two experiments, [CO] is unchanged, [Hb] doubles, and the rate doubles. Therefore, x = 1 and the reaction is first order in Hb. Comparing the second and third experiments, [Hb] is unchanged, [CO] triples and the rate triples. Therefore, y = 1 and the reaction is first order in CO. b. Rate = k[Hb][CO] c. From the first experiment, k = 0.280 L μmol-1 s-1 The second and third experiments give similar k values, so k mean = 0.280 L µmol-1 s-1 d. Rate = 2.26 µmol L-1 s-1 20. Rate = 5.68 × 10 18 molecules cm-3 s-1 Integrated Rate Laws 22. a. Therefore, the rate law, the integrated rate law and the rate constant value are: Rate = k[A] 2 ; 1 / [A] = kt + 1/ [A o ] ; k = 3.60 × 10-2 L mol-1 s-1 b. t 1/2 = 9.92 × 10 3 s Note: We could have used the integrated rate law to solve for t 1/2 where [A] = (2.80 × 10-3 /2) mol/L. c. t = 2.98 × 10 4 24. a. The rate law, the integrated rate law and the rate constant value are: Rate = k[C 2 H 5 OH] o =k; [C 2 H 5 OH] = -kt + [C 2 H 5 OH] o ; k = 4.00 × 10-5 mol L-1 s-1 b. t 1/2 = 156 s Note: we could have used the integrated rate law to solve for t 1/2 where [C 2 H 5 OH] = (1.25 × 10-2 /2) mol/L. c. t = 313 s 26. The natural log plot is not linear, so the reaction is not first order. Since the second order plot of 1/[C 4 H 6 ] vs. t is linear, then we can conclude that the reaction is second order in butadiene. The differential rate law is: Rate = k[C 4 H 6 ] 2 For a second order reaction, the integrated rate law is: k = slope = 1.4 × 10-2 L mol-1 s-1 28.a. First, assume the reaction to be first order with respect to O. Hence, a graph of ln [O] vs. t should be linear if the reaction is first order. Since the graph is linear, we can conclude the reaction is first order with respect to O. b. The overall rate law is: Rate = k[NO 2 ][O] Since NO 2 was in excess, its concentration is constant. So for this experiment, the rate law is: Rate: k ’ [O] where k’= k[NO 2 ]. In a typical first order plot, the slope equals -k. For this experiment, the slope equals –k ’ = -k[NO...
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This note was uploaded on 04/07/2008 for the course CHEM 027 taught by Professor Mccallum during the Spring '08 term at Pacific.

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Chapter 15 Even Problem Solutions - CHAPTER FIFTEEN...

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