a1s - Chemistry 2000B Spring 2002 Assignment 1 Solutions...

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Unformatted text preview: Chemistry 2000B Spring 2002 Assignment 1 Solutions (a) 1 L of solution contains 1.875 mol of sodium sulfate (Na2 SO4 ) and weighs 1.2106 kg. The mass of sodium sulfate is 266 3 g £ ¡ ¡ 1 875 mol 142 043 g mol ¢ ¤ ¢ £ ¢ mNa2 SO4 ¢ 1. The mass of water is therefore 0 2663 kg 0 9443 kg ¢ ¢ ¥ 1 2106 ¢ mH2 O ¢ The molality is thus 1 986 mol kg ¤ 1 875 mol 0 9443 kg ¢ ¢ ¢ c ˜ ¢ (b) The solution contains sodium and sulfate ions and water. Taking, as before, 1 L of solution, we have ¢ 3 750 mol 3 750 1 875 mol © ¢ § ¢ © ¤ ¢ ¦ ©§ 4 0 06461 ¢ ¢ ¢ § § © ¨ XNa ¦ 944 3 g 52 42 mol 18 0152 g mol nNa nH2 O nNa nSO2 52 42 ¢ nH2 O ¢ 3 750 mol ¢ 4 ¢ 2nSO2 ¦ nNa ¢ 1 875 mol ¢ 4 ¢ nSO2 2. 1 mol L 1 s 1 ¢   ¤ ¢     mol L 1 s 5   5 10  7 89    ¢ ¢  ¢    rate 1 42 10 6 mol L 18 0 10 3 s 1∆ B 3 94 10 2 ∆t ¢ ∆B ∆t ¨ 3. Comparing experiments 1 and 2, we see that cutting the concentration of ammonium ion in half reduces the rate by the same factor. In experiments 2 and 3, the ratio of the NO2 concentrations is 1.5 and the ratio of the rates is 5 4 3 6 1 5. Thus, the reaction is first order in each of the two reactants:  ¢ ¢ ¤ ¢ k NH4 NO2 ¢       v The rate constant can be recovered from any one of the experiments. For instance, using data from experiment 1: 10   ¢  £ ¤  ¢   ¡ £ 1 30 4 L mol 1s  ¤ ¢ ¢ ¡ NH4 NO2 7 2 10 6 mol L 1 s 1 0 24 mol L 0 10 mol L  v 1 ¢      k (a) Br2 g NO g NO g Br2 g # # © © © ## NOBr2 g 2NO g  © ## ©  # !  # # Br2 g NOBr2 g 2NOBr g "  !  # $©   ¨ NO g NOBr2 g 2NOBr g 2NOBr g  NO g NOBr2 g   4. ©  (b)  ¥ %    k2 NO NOBr2        ¥ 2k2 NO NOBr2    ¥ k1 NO Br2      k1 NO Br2 k2 NO NOBr2   k1 NO Br2  ¥ &       d NO dt d Br2 dt d NOBr2 dt d NOBr dt   (c) If the first step was rate determining, the rate law would be v k1 NO Br2 . Since this doesn’t agree with the experimental rate law, the first step is not rate determining. In fact, using techniques you will learn in Chemistry 2710 (if you opt to take this course), we can show that this rate law is consistent with the second step being rate determining.     (d) At the very least, we need to do a set of experiments in which we measure the rate at fixed [NO] varying Br2 , and another set at fixed Br2 varying [NO]. For instance, we might perform the following series of experiments:  Br2 1 1 1 2 4      Experiment [NO] 1 1 2 2 3 4 4 1 5 1 Note that the units of concentration don’t matter so I haven’t written any in. I’m essentially proposing to look at two doublings of each of the concentrations. It’s important to study at least three different concentrations for each reactant because it is possible to be fooled if we use the minimum amount of data. If the experimental rate law is correct, then the measured rate in experiment 2 should be four times the rate in experiment 1, and the rate in experiment 3 should be four times the rate in experiment 2. Also, the rate in experiment 4 should be twice the rate in experiment 1 and the rate in experiment 5 should be twice the rate in experiment 4. (a) If it’s a first-order reaction, we should get a straight line when we plot ln NOBr vs t :  2  5. -3 -3.2 ln [NOBr] -3.4 -3.6 -3.8 -4 -4.2 0 10 20 t (s) 30 40 We do not get a straight line, therefore it is not a first-order reaction. If it’s a second-order reaction, a plot of 1 NOBr vs t should be linear:   '¤ 65 60 1/[NOBr] (L/mol) 55 50 45 40 35 30 25 20 0 10 20 t (s) 30 40 This is a very good fit. This is therefore a second-order reaction. The rate constant is the slope of this relationship. Because it was slightly easier than some of the alternatives, I decided to get the slope by measuring the points at which the line enters and leaves 3 ¢   '¤ ¤ 0 806 L mol 1s   ¢ 61 5 21 2 L mol 45 5s 1 ¢ ¢ ¤ ¢ £ ¥ ¥ ¡ ¢ ¥   '¤ k 21 2 L mol. At ¤ the graph. At t 5 s, the line passes through the point 1 NOBr t 45 s, 1 NOBr 61 5 L mol. The slope is therefore ¥ For comparison, the value obtained by linear regression is k 0 803 L mol 1s 1 . Obviously, there are greater uncertainties in this determination than in most other calculations you might carry out in this class, so your answer might be a little different from mine. I would think that, from a carefully drawn graph with a similarly carefully drawn line through the data, you should be able to get the slope to within 0 02 L mol 1s 1 or so.   ¢  ¢  ( (b) The equation relating concentration to time is  4 00 10 4 mol  ¢    0 0400 mol L and we want NOBr ¤    ¤ ¢   L. Thus 3082 s 0 1 0 0400 mol L ¢ ¤ ¢ ¥ 1 10 4 mol L ¤  1 1 NOBr NOBr 0 1 0 803 L mol 1s 1 4 00 0   ¢  )  ¥    1 k ) t kt © We have NOBr 0 1 NOBr 0 ¢ 1 NOBr ¢ I used the value of k obtained by linear regression, but I could just as easily have used the other value. Obviously you would use whatever you calculated in the previous part of this question. Note also that there’s no point worrying too much about significant figures in this problem since there are significant uncertainties in the rate constant.  © 1  4 NO g were the rate-limiting step. NOBr2 g  2NOBr g k NOBr 2 . This  (c) The rate law, as determined in the previous part of this question, is v would be the rate law for the reaction if ...
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This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.

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