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Unformatted text preview: 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 L1 s1 = k (0.20 mol L1 ) 2 (0.30 mol L1 ) k = 116.7 L 2 mol2 s1 = 120 L 2 mol2 s1 (d) Rate 4 = (116.7 L 2 mol2 s1 ) (0.17 mol L1 ) 2 (0.25 mol L1 ) Rate 4 = 0.84 mol L1 s1 156 2 4 4 4 2 4 1 1 MnO mol MnO 2 rate of formation of MnO . mol 2 0 L min mol MnO 3 . mol L min 1 33 =...
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This note was uploaded on 06/07/2009 for the course CHEM 6C taught by Professor Hoeger during the Spring '08 term at UCSD.
 Spring '08
 HOEGER
 Reaction

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