ws13s08key - CH 302 Spring 2008 Worksheet 13 [A] [B] [C]...

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CH 302 Spring 200 8 Worksheet 13 [A] [B] [C] rate 0.10 M 0.10 M 0.10 M 1.4 x 10 -4 M/s 0.20 M 0.10 M 0.10 M 2.8 x 10 -4 M/s 0.37 M 0.25 M 0.10 M 8.09 x 10 -3 M/s 0.37 M 0.25 M 0.05 M 3.24 x 10 -2 M/s For the data given above, find the order of the reaction with respect to the indicated species. 1. A Answer: When you double the amount of A, the reaction rate doubles, so the reaction is first-order in A . 2. B Answer: We know the reaction is first-order in A. So multiplying A by 3.7 (from the first one) would yield a rate of 3.7(1.4 x 10 -4 M/s) = 5.2 x 10 -4 if B and C were unchanged. Thus, multiplying B by 2.5 yielded a net change of (8.09/0.52) = 15.6 = 2.5 3 . So the reaction is third-order in B . 3. C Answer: Halving C quadruples the reaction order. So the reaction is of order -2 in C . 4. Assume the reaction does not depend on any other species besides A, B, and C. Write the expression for the rate of the reaction in terms of the rate constant k. Answer: rate = k[A][B] 3 /[C] 2 5. Calculate the rate constant k. k = (rate)[C]/[A][B] 3 = (1.4 x 10 -4 M/s)(0.1 M) 2 /((0.1 M)(0.1 M) 3 ) = 0.014 M -1 s -1 6. If A = 10 8 M -1 s -1 and T = 298 K, what is E a for this reaction? k = Aexp(-E
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This note was uploaded on 04/12/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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ws13s08key - CH 302 Spring 2008 Worksheet 13 [A] [B] [C]...

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