230_Sp10_L33_post - CHEM 230 Sp10 Lecture 33 Chapter 13...

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CHEM 230 Sp10 Lecture 33 Chapter 13: Chemical Kinetics 1) Summary 2) Temperature effect on rates 3) Catalysts Session ID: 230
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Review Rate: stoichiometric relationships a A + b B c C + d D Reaction rate = - 1 a " [ A ] " t = # 1 b " [ B ] " t = 1 c " [ C ] " t = 1 d " [ D ] " t Rate Law: order in each reactant -determine experimentally by a) comparing intial concentrations of reactants and initial rates OR b) by measuring [reactant] over time and trying out integrated rate law plots
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Experimental Determination of Order and k Initial [A] Initial [B] Measured Rate 1 M 1 M 10 M/s 2 M 1 M 40 M/s 1 M 2 M 10 M/s a) A + B ˠ C b) [ A ] t = " kt + [ A ] 0 ln[ A ] t = " kt + ln[ A ] 0 1 [A] t = 1 [A] 0 + kt Zero order First order Second order
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The hydrolysis of sucrose (C 12 H 22 O 11 ) produces fructose and glucose (each C 6 H 12 O 6 ). Two mechanisms are proposed: I) Step 1: C 12 H 22 O 11 C 6 H 12 O 6 + C 6 H 10 O 5 (slow) Step 2: C 6 H 10 O 5 + H 2 O C 6 H 12 O 6 (fast) II) C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6 Would these mechanisms be distinguishable in a dilute solution? A) Yes B) No Concept Question
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12 H 22 O 11 ) produces fructose and glucose (each C 6 H 12 O 6 ). Two mechanisms are proposed: I) Step 1: C 12 H 22 O 11 C 6 H 12 O 6 + C 6 H 10 O 5 (slow) rate=k[C 12 H 22 O 11 ] Step 2: C 6 H 10 O 5 + H 2 O C 6 H 12 O 6 (fast) II) C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6 rate=k[C 12 H 22 O 11 ][H 2 O] pseudo first order
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This note was uploaded on 01/24/2012 for the course PCHEM 230 taught by Professor Gottfried during the Fall '11 term at University of Michigan.

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230_Sp10_L33_post - CHEM 230 Sp10 Lecture 33 Chapter 13...

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