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Chapter 130 - 13.6(a MnO 2 2 mol MnO 4 4 rate of formation...

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13.6 (a) (b) (c) 13.16 rate = k [NO] a [O 2 ] b , where the orders a and b are to be determined. When the concentration of NO is doubled the rate increases by a factor of 4, that is 2 a = 4, so a = 2. When both the concentration of NO and O 2 are doubled, the rate increases by a factor of 8, that is 2 2 × 2 b = 8, so b = 1. Therefore, the rate law is, rate = k [NO] 2 [O 2 ]. 13.18 (a) ( 29 ( 29 1 12.6 0.6 2 1.4 0.2 9 3 2 1 12.6 0.3 3 4.2 0.1 3 3 1 a a b b rate rate a rate rate b = = = = = = = = Therefore, 2 nd order with respect to A and 1 st order with respect to B. The overall order of the reaction is 3. (b) Rate = k [A] 2 [B] 1 (c) Using data from experiment 2, 1.4 mol L -1 s -1 = k (0.20 mol L -1 ) 2 (0.30 mol L -1 ) k = 116.7 L 2 mol -2 s -1 = 120 L 2 mol -2 s -1 (d) Rate 4 = (116.7 L 2 mol -2 s -1 ) (0.17 mol L -1 ) 2 (0.25 mol L -1 ) Rate 4 = 0.84 mol L -1 s -1 156 2 4 4 4 2 4 1 1 MnO 2 mol MnO rate of formation of MnO 2.0 mol L min 3 mol MnO 1.33 mol L min - - - - - - =  = 2 + + 4 2 4 + 1 1 MnO 4 mol H rate of reaction of H (aq) 2.0 mol L min 3 mol MnO = 2.7 mol H L min - - - - 
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