Lecture8_S08

Lecture8_S08 - Lecture 8 Last Time First Order reactions: 2...

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Lecture 8 Last Time First Order reactions: [H 2 O 2 ] t Time, t 2 H 2 O 2 (l) O 2 (g) + 2 H 2 O(l) t 1/2 = 0.693 k This looks like some Mathematical decay function [A] t = [A] 0 e -kt ln[A] t = -kt + ln[A] 0 ln[A] 0 ln[A] t Slope = -k Time, t 2 NOBr(g) Br 2 (g) + 2 NO(g) Try: [NOBr] t Time, t ln[NOBr] t Time, t ln[A] 0 RATS!
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We should have done experiments to determine exponential in rate law: Rate (M/s) [NOBr] 50 0.15 200 0.30 Exp# 1 2 Rate 2 Rate 1 = 2 x Doubling concentration 200/50 = 4 = 2 x So, x = 2 Reaction is second order: rate = k[NOBr] 2 Calculator party 1 [NOBr] t 1 [NOBr] 0 = kt + y = mx + b 1/[NOBr] t Time, t 1/[NOBr] 0 Slope = k
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Half-life of a second-order reaction: 1 1/2[A] 0 1 [A] 0 = kt 1/2 + @ t 1/2 [A] t = ½[A] o 2
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This note was uploaded on 04/17/2008 for the course CHEM 2 taught by Professor Ryan during the Spring '08 term at Marquette.

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Lecture8_S08 - Lecture 8 Last Time First Order reactions: 2...

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