CHEM-6 Midterm: CHEM 6 Dartmouth kinsolns Exam

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31 Jan 2019
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This problem illustrates how the differential rate law can be simplified when the concentration of one or more of the participating species remains effectively constant during the course of the reaction. (i) we will assume 1 liter of solution. Thus, we have 0. 1 mols of ch3cocl in 1000 ml of h2o. Since the density of h2o (at room temperature) is ~1. 0 g/ml, the weight of. No. of mols h2o present at time t 0 = (1000 g)/(18 g mol 1) = 55. 6 (ii) from the reaction stoichiometry, when the reaction is complete 0. 1 mols ch3cocl will have reacted with 0. 1 mols h2o. Since h2o is present in great excess, [h2o] remains effectively constant during the course of the reaction. We will take the effective constant concentration of h2o to be the average of the initial and final values; [h2o]avg = 55. 55 m. therefore, may be simplified to: where k" = k [h2o] = k (55. 55 m)

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