Ii there are really too many reactants for it to be

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ii. There are really too many reactants for it to be likely that this reaction is elementary. Chapter 15 #25 and 43 2. (a) k = Ae - E a / ( RT ) = ( 5 . 4 × 10 16 s - 1 ) exp - 114700J / mol ( 8 . 314472JK - 1 mol - 1 )( 293 . 15K ) = 2 . 0 × 10 - 4 s - 1 . t 1 / 2 = ln2 k = ln2 2 . 0 × 10 - 4 s - 1 = 3 . 5 × 10 3 s . 1
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(b) The reaction is Additional prob- lems on the relationship between kinetics and equilibrium PAN k 1 k - 1 - * ) - radical + NO 2 . At equilibrium, the rates of the forward and reverse reactions are equal: k 1 [ PAN ] = k - 1 [ radical ][ NO 2 ] . K = [ radical ][ NO 2 ] [ PAN ] = k 1 k - 1 = 2 . 0 × 10 - 4 s - 1 5 . 74 × 10 9 Lmol - 1 s - 1 = 3 . 4 × 10 - 14 mol / L . Chapter 15 #19 (c) ln x = ln x 0 - kt . ln ( 0 . 05bar ) = ln ( 0 . 45bar ) - ( 2 . 0 × 10 - 4 s - 1 ) t . t = ln ( 0 . 45bar ) - ln ( 0 . 05bar ) 2 . 0 × 10 - 4 s - 1 = 1 . 1 × 10 4 s 1 . 1 × 10 4 s 3600s / h = 3 . 1h . 3. If the reaction is elementary, then according to the law of mass action it Chapter 15 #47 and a second- order plot should follow the second-order rate law v = k [ C 2 F 4 ] 2 . A plot of 1 / [ C 2 F 4 ] vs t should be linear. Here is my graph: 2
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0 50 100 150 200 250 300 350 0 500 1000 1500 2000 2500 3000 3500 4000 1/[C 2 F 4 ] (L/mol) t (min) The data fit the line nicely which supports the hypothesis that this is an elementary reaction. The rate constant is equal to the slope of the graph. Using a couple of points on the graph, you should get (roughly) k = 8 . 0 × 10 - 2 Lmol - 1 min - 1 . 4. (a) Hg ( aq ) Chapter 15 #49 (b) The equilibrium approximation can be applied to the first step: k 1 [ Hg 2 + 2 ] k - 1 [ Hg 2 + ][ Hg ] . [ Hg ] k 1 [ Hg 2 + 2 ] k - 1 [ Hg 2 + ] . The rate of reaction is v = k 2 [ Hg ][ Tl 3 + ] k 1 k 2 [ Hg 2 + 2 ][ Tl 3 + ] k - 1 [ Hg 2 + ] . (c) According to our rate law, doubling the mercury (II) ion concentration cuts the rate in half. 3
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