Ch13_L4-6_115b07

Ch13_L4-6_115b07 - First exam: Chemical Kinetics (Chapter...

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First exam: Chemical Kinetics (Chapter 13 only!) Material covered until the end of Monday lecture (in book and in class) - rates of chemical rxns - integrated rate laws - zero-, first, and second order rxns; - pseudo first or second order kinetics - mechanisms - rate expression <-> mechanism - rate limiting rxn, fast equilibrium, steady state apprx. - Michaelis-Menten kinetics and catalysis - T-dependence of rxn rates, Arrhenius law Not included: activated complex and activation energy Kinetics and Chemical Equilibrium Reaction Dynamics (section 13.6) You need to know all definitions, in particular rate (R). Equations and constants will be given to you.
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kC dt dC = kt C C = 0 ln ln kt e C C = 0 k k t 693 . 0 2 ln 2 / 1 = = Integrated rate law: 1st order rxns L4: Review
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2 2 1 kC dt dC = kt C C 2 1 1 0 + = Integrated rate law: 2nd order rxns Half-life: when C = C 0 /2 2 / 1 0 0 2 1 2 kt C C + = k C t 0 2 / 1 2 1 = products A = 2 2 kC dt dC = kt C C + = 0 1 1 products A = Also for A + B when [A] = [B]
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k kC dt dC = = 0 Integrated rate law: zero order rxns kdt dC = = t C C dt k dC 0 0 kt C C = 0 kt C C = 0 When would a zero order rxn be observed?
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The Decomposition Reaction 2N 2 O(g) --> 2N 2 (g) + O 2 (g) takes Place on a Platinum Surface Only molecules on surface can react. No dependence on N 2 O concentration. This is a zero order reaction.
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Pseudo first order reaction: When one reactant, C, is in great excess over another, its concentration does not change during the reaction, even when the second reactant is totally depleted. In this case [C] t = [C] 0 = constant at ANY time Example: C + D A + B; [C] >> [D] [D] is limiting reagent It is found that R = k[C][D] At the end of the reaction [C] ~ [C] 0 So R ~ k[C] 0 [D] = k'[D] where k' = k[C] 0
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First lab: Measuring kinetics of Methyl Orange/Sn(II) reaction Sn 2+ + 4H + + MeO - > Sn 4+ + other products orange colorless Trick: reduce the rxn order by using large excess of H + and Sn 2+ -> determine m: l n m Sn H MeO k dt MeO d R ] [ ] [ ] [ ] [ 2 + + = = m eff MeO k dt MeO d R ] [ ] [ = = n l eff H Sn k k ] [ ] [ 2 + + =
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The rate for the reaction O+NO 2 -> NO + O 2 was studied under conditions of a large excess of NO 2 : [NO 2 ] 0 =1.0x10 13 molec/cm 3 . The overall rate law is given by: R = k [NO 2 ] [O] n . A plot of ln[O] as a function of time gave a straight line with -slope= 1.0x10 2 s -1 . - What is the order of reaction (n)? - What is k for this reaction? 1. [NO 2 ]>>[O] so [NO 2 ] ~ const: where k' = k[NO 2 ] 0 . 2. Straight ln[O] vs. t plot means that this is a first order reaction in [O], i.e., n=1. This is a pseudo first order rxn. 3. -slope=k' -> k'=1.0x10 2 s -1 . 4. Conclusion: if one reactant is in a large excess, it reduces the effective observed order of the rxn. But still need to get k from k obs ... 1 1 3 11 3 13 1 2 0 2 10 0 . 1 10 0 . 1 10 0 . 1 ] [ ' = = = s molec cm cm molec s NO k k 22 0 [] ' nn n dO RO O O d kN t Ok N O k =− = =
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2 2 2
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Reaction mechanisms: How does the reaction happen? Why rate laws are what they are?
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Ch13_L4-6_115b07 - First exam: Chemical Kinetics (Chapter...

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