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ch12 - 12 12.1(b Chemical Kinetics 3 I(aq H3AsO4(aq 2 H(aq...

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12 Chemical Kinetics 12.1 3 I(aq) + H 3 AsO 4 (aq) + 2 H + (aq) I 3 (aq) + H 3 AsO 3 (aq) + H 2 O(l) (a) t ] I [ _ = 4.8 x 10 4 M/s M/s) 10 x (4.8 3 1 = t ] I [ _ 3 1 = t ] I [ 4 _ _ _ 3 = 1.6 x 10 4 M/s (b) t ] I [ 2 = t ] H [ _ 3 + = (2)(1.6 x 10 4 M/s) = 3.2 x 10 4 M/s 12.2 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) time [N 2 O 5 ] [O 2 ] 200 s 0.0142 M 0.0029 M 300 s 0.0120 M 0.0040 M Rate of decomposition of N 2 O 5 = s 200 _ s 300 M 0.0142 _ M 0.0120 _ = t ] O N [ _ 5 2 = 2.2 x 10 5 M/s Rate of formation of O 2 = s 200 _ s 300 M 0.0029 _ M 0.0040 = t ] O [ 2 = 1.1 x 10 5 M/s 12.3 Rate = k[BrO 3 ][Br][H + ] 2 1st order in BrO 3 , 1st order in Br, 2nd order in H + , 4th order overall Rate = k[H 2 ][I 2 ], 1st order in H 2 , 1st order in I 2 , 2nd order overall Rate = k[CH 3 CHO] 3/2 , 3/2 order in CH 3 CHO, 3/2 order overall 12.4 H 2 O 2 (aq) + 3 I(aq) + 2 H + (aq) I 3 (aq) + 2 H 2 O(l) Rate = t ] I [ _ 3 = k[H 2 O 2 ] m [I] n (a) M/s 10 x 1.15 M/s 10 x 2.30 = Rate Rate 4 _ 4 _ 1 3 = 2 M 0.100 M 0.200 = ] O H [ ] O H [ 1 2 2 3 2 2 = 2 Because both ratios are the same, m = 1. M/s 10 x 1.15 M/s 10 x 2.30 = Rate Rate 4 _ 4 _ 1 2 = 2 M 0.100 M 0.200 = ] I [ ] I [ 1 _ 2 _ = 2 Because both ratios are the same, n = 1. The rate law is: Rate = k[H 2 O 2 ][I] (b) k = ] I ][ O H [ Rate _ 2 2 35
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Using data from Experiment 1: k = M) M)(0.100 (0.100 M/s 10 x 1.15 4 _ = 1.15 x 10 2 /(M s) (c) Rate = k[H 2 O 2 ][I] = [1.15 x 10 2 /(M s)](0.300 M)(0.400 M) = 1.38 x 10 3 M/s 12.5 Rate Law Units of k Rate = k[(CH 3 ) 3 CBr] 1/s Rate = k[Br 2 ] 1/s Rate = k[BrO 3 ][Br][H + ] 2 1/(M 3 s) Rate = k[H 2 ][I 2 ] 1/(M s) Rate = [CH 3 CHO] 3/2 1/(M 1/2 s) 12.6 (a) The reactions in vessels (a) and (b) have the same rate, the same number of B molecules, but different numbers of A molecules. Therefore, the rate does not depend on A and its reaction order is zero. The same conclusion can be drawn from the reactions in vessels (c) and (d). The rate for the reaction in vessel (c) is four times the rate for the reaction in vessel (a). Vessel (c) has twice as many B molecules than does vessel (a). Because the rate quadruples when the concentration of B doubles, the reaction order for B is two. (b) rate = k[B] 2 12.7 (a) ln kt _ = ] Br ) NH [Co( ] Br ) NH [Co( o + 2 5 3 t + 2 5 3 k = 6.3 x 10 6 /s; t = 10.0 h x h 1 s 3600 = 36,000 s ln[Co(NH 3 ) 5 Br 2+ ] t = kt _ + ln[Co(NH 3 ) 5 Br 2+ ] o ln[Co(NH 3 ) 5 Br 2+ ] t = s) /s)(36,000 10 x (6.3 _ 6 _ + ln(0.100) ln[Co(NH 3 ) 5 Br 2+ ] t = 2.5294; After 10.0 h, [Co(NH 3 ) 5 Br 2+ ] = e 2.5294 = 0.080 M (b) [Co(NH 3 ) 5 Br 2+ ] o = 0.100 M If 75% of the Co(NH 3 ) 5 Br 2+ reacts then 25% remains. [Co(NH 3 ) 5 Br 2+ ] t = (0.25)(0.100 M) = 0.025 M ln kt _ = ] Br ) NH [Co( ] Br ) NH [Co( o + 2 5 3 t + 2 5 3 ; t = k _ ] Br ) NH [Co( ] Br ) NH [Co( ln o + 2 5 3 t + 2 5 3 t = /s) 10 x (6.3 _ 0.100 0.025 ln 6 _ = 2.2 x 10 5 s; t = 2.2 x 10 5 s x s 3600 h 1 = 61 h 12.8 Slope = 0.03989/min = 6.6 x 10 4 /s and k = slope 36
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A plot of ln[cyclopropane] versus time is linear, indicating that the data fit the equation for a first-order reaction. k = 6.6 x 10 4 /s (0.040/min) 12.9 (a) k = 1.8 x 10 5 /s t 1/2 = /s 10 x 1.8 0.693 = k 0.693 5 _ = 38,500 s; t 1/2 = 38,500 s x s 3600 h 1 = 11 h t 1/2 t 1/2 t 1/2 t 1/2 (b) 0.30 M 0.15 M 0.075 M 0.0375 M 0.019 M (c) Because 25% of the initial concentration corresponds to 1/4 or (1/2) 2 of the initial concentration, the time required is two half-lives: t = 2t 1/2 = 2(11 h) = 22 h 12.10 After one half-life, there would be four A molecules remaining. After two half-lives, there would be two A molecules remaining. This is represented by the drawing at t = 10 min.
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