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© University of South Carolina Board of Trustees Trial[NO] M [H 2 ] M Initial Rate, M/s 10.0570.1304.50x10 -3 20.0570.2609.00x10 -3 30.110.1301.80x10 -2 Student Problem Write the rate law for the reaction given the following data 2NO + 2H 2 N 2 + 2H 2 O

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© University of South Carolina Board of Trustees Chapt. 13 Sec. 13.3 Properties of Common Rate Laws

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© University of South Carolina Board of Trustees Common Rate Laws A products a) First-order Rate = k [A] b) Second-order Rate = k [A] 2 c) Zero-order Rate = k [A] 0 = k

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© University of South Carolina Board of Trustees Common Rate Laws A products a) First-order Rate = k [A] b) Second-order Rate = k [A] 2 c) Zero-order Rate = k [A] 0 = k

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© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t )

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© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt

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© University of South Carolina Board of Trustees [A] = [A] 0 exp(- k t)

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© University of South Carolina Board of Trustees Concentration at a Later Time C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6 sucrose glucose fructose This reaction is 1 st order with a rate constant of 6.2 x10 -5 s -1. If the initial sucrose concentration is 0.40 M, what is the concentration after 2 hrs (7200 s)?

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© University of South Carolina Board of Trustees Time to Reach a Concentration C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6 sucrose glucose fructose This reaction is 1 st order with a rate constant of 6.2 x10 -5 s -1. If the initial sucrose concentration is 0.40 M, at what time does the concentration fall to 0.30 M ?

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© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt ln [A]( t ) = ln [A] 0 - k t y = b +mx y = b +mx

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© University of South Carolina Board of Trustees Graphing a 1 st -Order Reaction

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© University of South Carolina Board of Trustees Graphing a 1 st -Order Reaction

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© University of South Carolina Board of Trustees Graphing a 1 st -Order Reaction straight line this is a first-order reaction.

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© University of South Carolina Board of Trustees [A] vs t Data Rate Law Method of Initial Rates (Sec. 13.2) Trial and Error with Common Laws (Sec. 13.3)

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© University of South Carolina Board of Trustees ln [A] = ln [A] 0 - k t intercept = ln [A] 0

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© University of South Carolina Board of Trustees ln [A] = ln [A] 0 - k t

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© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt ln [A]( t ) = ln [A] 0 - k t Half-life

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© University of South Carolina Board of Trustees 1 st -Order Half-Life [A] [A]/2 t 1/2

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© University of South Carolina Board of Trustees 1 st -Order Half-Life [A] [A]/2 t 1/2 5 2.51.75 s t 1/2

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© University of South Carolina Board of Trustees 1 st -Order Half-Life [A] [A]/2 t 1/2 5 2.51.75 s 3 1.51.72 s t 1/2

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© University of South Carolina Board of Trustees Half-life [A] [A]/2 t 1/2 5 2.51.75 s 3 1.51.72 s 1 0.51.73 s t 1/2

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© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt ln [A]( t ) = ln [A] 0 - k t Half-life always the same

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