HW2_1 - [A] 2 x 0.080 [0.20] x Rate4 [B] 4 y 0.040 [0.40]...

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Homework due 2/1 Consider the reaction 3A + B + 2C ----> P for which the following data has been obtained. Trial [A] [B] [C] Initial Rate 1 0.10 0.20 0.10 0.020 2 0.20 0.20 0.10 0.080 3 0.20 0.20 0.20 0.160 4 0.10 0.40 0.10 0.040 Give the following: a) the rate law; b) the order with respect to each reactant and the overall order; c) the value of the rate constant k. Like the previous example the rate law is determined by by taking the ratio of two trials in which only the concentration of the desired reactant is changed. For reactant A use trials 1 & 2, for reactant B use trials 1 & 4, and For A: For B: Rate2
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Unformatted text preview: [A] 2 x 0.080 [0.20] x Rate4 [B] 4 y 0.040 [0.40] y-------- = ------ -------- = ---------4 = 2 x x = 2 --------- = ------ = ------- = ------------ 2 = 2 y y=1 Rate1 [A] 1 x 0.020 [0.10] x Rate1 [B] 1 y 0.020 [0.20] y For C: Rate 3 [C] 3 z 0.160 [0.20] z-------- = -------- ------- = ----------- 2 = 2 z z = 1 Rate Law: Rate = k[A] 2 [B][C] Rate2 [C] 2 z 0.080 [0.10] z This reaction is 2nd order in A, 1st order in B, 1st order in C, and 4th order overall k = Rate/[A] 2 [B][C] For trial 1: 0.020/(0.10) 2 (0.20)(0.10) = 100 The same value will be obtained for all of the other trials....
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This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.

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