Chapter_16_Feb_8_201 - I mpor t ant New office hour s M on and Fr i 11 12 Wed 4 5 Chemist r y Tut or Room Fleming 236 It is critically important

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It is critically important that you completely follow the instructions regarding the Blackboard homework to get full credit, e.g., significant figures. Important…. New office hours: Mon and Fri 11 – 12; Wed. 4 – 5 Chemistry Tutor Room – Fleming 236
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Integrated Rate Laws rate = - [A] t = k [A] rate = - [A] t = k [A] 0 rate = - [A] t = k [A] 2 first order rate equation second order rate equation zero order rate equation ln [A] 0 [A] t = - kt ln [A] 0 = - kt + ln [A] t 1 [A] t 1 [A] 0 - = kt 1 [A] t 1 [A] 0 + = kt [A] t - [A] 0 = - kt
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Sample Problem 16.5 PLAN: SOLUTIO N: Determining the Reactant Concentration at a Given Time PROBLEM: At 1000 o C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87 s -1 , to two molecules of ethylene (C 2 H 4 ). (a) If the initial C 4 H 8 concentration is 2.00 M, what is the concentration after 0.010 s? (b)
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This note was uploaded on 03/09/2010 for the course CHEM 1332 taught by Professor Halasyamani during the Spring '10 term at University of Houston-Victoria.

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Chapter_16_Feb_8_201 - I mpor t ant New office hour s M on and Fr i 11 12 Wed 4 5 Chemist r y Tut or Room Fleming 236 It is critically important

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