midterm_soln - Chemistry 2000 Spring 2002 Midterm...

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Unformatted text preview: Chemistry 2000 Spring 2002 Midterm Examination Solutions (a) The concentrations in experiments 1 and 2 are in a 3:1 ratio, as are the rates. This is therefore a first-order reaction with rate law ¢ £¡ 2 ¤ k CH3 N v ¦ §¥ 1. The rate constant can be determined from either experiment: 25 4 10 ©  ¦    © ©  ¦ ¦ ¥ 2 1 s  6 0 10 6 mol L 1s 2 4 10 2 mol L 1  ¤ v CH3 N ¦ k ¢ ¨¡ (b) ∆ C2 H6 ∆t 10 4 mol L 6 0 10 6 mol L 1s ¡ ¦ 17 s ¦  1  ¡ ¥    ©  ¡ ¥ ¦ ∆t d C2 H6 dt ∆ C2 H6 v ¥ v  ¢ ¤ t (min) ln %bp  2. If it’s a first-order reaction, a graph of ln percent vs t should be a straight line of slope In this case, we have k. 2.21 5.16 13.26 22.84 40.52 60.41 4.55 4.42 4.17 3.78 3.14 2.30 ¢ ¤ My graph is shown in Fig. 1. The data fit a straight line reasonably well, which confirms that the decomposition of bp is a first-order process. The rate constant is found by taking the slope of the line. The line passes through the points 0 4 65 and 70 1 98 (estimated from the graph). The slope is therefore ¢ ¢ ¤ 1  0 0381 min ¦  ¦  ¦  ¦ ¦  0 0381 min 1 .  ¦ k, this means that k   (a) 17 5 103 J mol 8 314 472 J K 1mol 1 243 15 K ¤   ¤ ¦ ¦ 1 ¦ ¢ 1 ! "¤  exp ¦    106 L mol 1s 1 ©  © © ¢ ¦ 2 09 1010 L mol 1 s  12 ¢ Ea RT  k∞ e   k  ¦ 3. 4 65 1 98 0 70 min ¦ Since the slope is ∆y ∆x ¤ slope 5 4.5 ln(percent) 4 3.5 # 3 2.5 2 1.5 0 10 20 30 40 50 60 70 t (min) ¢ Figure 1: Plot of ln %bp vs t from the data of question 2. ¤ ¡ d OH k OH CF3 CH2 OCHF2 dt (c) If the initial concentrations are equal, then by stoichiometry the two concentrations d OH remain equal for all time. Thus k OH 2 , which is a simple second-order dt reaction. ¡ ¥ ¡  ¥ ¥ (b) ¡ ¡  ¥ ¥ (d) According to the second-order integrated rate law,  ¦ 1 0 0405 mol L !  ¦    ¦ ¢ $ 0 140 K ¦ ¦ ¤ ¦ ¦  ¤ ¤ ¥ ¡  ¦ © ¢ ¢ ¦ © ¦ ¦ ¡ ¥  0 223 0 111 0 177 ¤ aPCl3 aCl2 aPCl5 ¥ ¦ ¡  ¦ ¢ Q ¡ 4. ¥ t ! R 1 R0 0 004 05 mol L 1 1 6 L mol 1 s 1 0 004 05 mol L 2 09 10 1 06 10 4 s ¦ 11 kR 0 10 R 0 t This means that the amounts of products will have to increase to reach equilibrium, which in turn means that the reaction is spontaneous as written. 2  10 4 mol L. Since Kw  © 13 ¦ ¡  ¥ 5. Potassium hydroxide is a strong base. Accordingly, OH aH aOH , ¢ ¢ ¤& 10 © 22 11   ¦ © ¦ ©  ¦ ¦ ¦ & % % pH Kw 2 9 10 15 aOH 1 3 10 4 log10 aH 10 65 ¦  ¤ % aH  ¢ a2 PBr P 2 Br . We need to know the equilibrium pressures of bromine atoms and aBr2 PBr2 P molecules. The number of moles of Br2 at equilibrium is 100 1 20% 98 8% of the initial number, i.e. 0 988 1 05 mol 1 0374 mol. The number of molecules dissociated is 0 0120 1 05 mol 0 0126 mol. The number of bromine atoms produced is twice as large (by stoichiometry), so nBr 0 0252 mol. The volume of the flask is 0 980 L 1000 L m3 9 80 10 4 m3 . The temperature is 1600 273 15 K 1873 K. The pressures of the reactant and product are therefore ¤ ('   )¤ ¦ ¤ ¢ ¢   © © ¦ ¦ ¦ 1  © © ¢ ¦ ¦ ¦ 1 ¦ ¦ ¦ ¤ ¦  ¦ ¢  ¢ ¤ ¦ ¢ ¤ ¦ ¤ ¢ ¤  ¦  ¦ ¦ 0 ¦ ¢ ¤ 1 ¤ ¦ 1873 K  ¢ ¤ ¦ ¢ 1873 K 1 ¤ ¤ nBr2 RT V 1 0374 mol 8 314 472 J K 1mol 9 80 10 4 m3 1 65 107 Pa 163 atm nBr RT V 0 0252 mol 8 314 472 J K 1mol 9 80 10 4 m3 4 00 105 Pa 3 95 atm 3 952 9 60 10 2 163 ¢ ¢ ¦  © © ¦ ¦ ¦ K ¦ PBr  ' PBr2 ¦  6. K  7. In these problems, we only ever need to worry about the first proton. By Le Chatelier’s principle, we know that the acidic environment created by the rst proton will inhibit the dissociation of the second proton. The relevant equilibrium is thus 4 0    2 3 ¢ ¦ % aC2 O4 H aH aC2 O4 H2 ¤ ¤&  ¢ Ka H aq  for which we have C2 O4 H aq  C2 O4 H2 aq The Ka is quite large. It seems extremely likely that we will be able to ignore the autoionization of water. However, we probably should not treat this equilibrium as if little dissociation occurred because of the size of the equilibrium constant. Suppose that the amount of oxalic acid which dissociates, expressed in terms of activities, is x. Then 3 aH 0 x x % & aC2 O4 H2 aC2 O4 H Initial 0.043 0 Change x x Final 0 043 x x   ¦ The Ka expression becomes  10 3 ! ¦ ¢  ¦ ¢  ¤ ©  ¦   6 75 ©  © ¦ ¦ ¤  ©  ¦ ©  ©  %  ¦ ¦ ¢ 0 ¦ ¦ 4 1 53 © 2 795  4 ¤ ©  ¦ ¦  ©  x ¦  ¦ 0 aH pH  10 x2 0 043 x x2 6 5 10 2x 2 795 10 3 1 6 5 10 2 6 5 10 2 2 2 2 96 10 2 2 96 10 2 log10 aH log10 2 96 10 2 ¦  %  65 2 ¦ ...
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