Lecture+3+-+Chemical+Kinetics

Lecture+3+-+Chemical+Kinetics - Problem Set#1 due at end of...

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Problem Set #1: due at end of class on Tue Aug 9 th .
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( ) ( ) g 2NO g O N 2 2 [] 2 2 2 2 2 1 O N k dt NO d dt O N d rate = + = = = t kt NO dt e O N k NO d 0 0 2 2 0 2 [ ] t kt k e O N k NO 0 0 2 2 ( 2 = [ ] () ( ) kt kt e O N e k O N k NO = = 1 2 1 1 2 0 2 2 0 2 2 [][ ] ( ) kt e O N NO = 1 2 0 2 2 0 1 02 03 04 05 0 0.0 0.4 0.8 1.2 1.6 2.0 Concentration Time [N 2 O 2 ] = [N 2 O 2 ] 0 exp(-kt) [NO] = 2[N 2 O 2 ] 0 (1- exp(-kt)) [ ] [ ] kt e O N O N = 0 2 2 2 2 NO(t) N 2 O 2 (t) kt e O N k O N k dt NO d rate = = + = 0 2 2 2 2 2 1 First-order Rise and Decay
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Second-Order Kinetics Products 2A Products B A + [] 0 1 1 A kt A + = [] [] [ ][ ] kt B A A B A B = 0 0 0 0 ln 1 2 d d A k t A rate = = [ ] [ ] B A k t B t A rate = = = d d d d
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Second-order Half-life (t 1/2 ) = kt + 1 [A] 0 [A] t 1 2A products = kt 1/2 + 2 [A] 0 [A] 0 1 = k[A] 0 1 t 1/2 t 1/2 = 0.5[A] 0 /k (zero-order) t 1/2 = 0.693 k (1 st -order) (2 nd -order)
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Second-order Kinetics Sample Problem The second order reaction (2A product) has the following kinetic data: Time (min) [A] (M) 0 0.0169 12.18 0.0144 24.55 0.0124 42.50 0.0103 68.05 0.00845 What is the half-life (t 1/2 )? What is the value of the rate constant, k? At what time (t) is [A] t = 0.005M? = kt + 1 [A] 0 [A] t 1 t = 1 [A] 0 [A] t 1 - 1 k = 1 ( 0.0169 ) ( 0.005 ) 1 - 1 0.87 = 2.0 x 10 2 min = k[A] 0 1 t 1/2 By inspection, t 1/2 is 68.05 min: 1 1 0 2 / 1 87 . 0 ) 0169 . 0 )( 05 . 68 ( 1 ] [ 1 = = = s M A t k
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Pseudo First-Order Kinetics CH 3 COOC 2 H 5 + H 2 O l CH 3 COOH + C 2 H 5 OH [ ] [ ] O H H COOC CH k Rate 2 5 2 3 = [H 2 O] = 55.5 M and remains nearly constant during rxn. [] 5 2 3 H COOC CH k Rate = O H k k 2 = (k´ = pseudo 1 st order rate constant) (k = 2 nd order rate constant) ethyl acetate acetic acid ethanol = k[H 2 O] time ln[A] t -k’ [ ] 5 2 3 ) 5 . 55 ( Rate H COOC CH M k = k’
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2 nd Order Kinetics if [A] = [B] products B A ⎯→ + [] B A k dt A d r = ] [ = ate kdt - = B A ] A [ d [] [] [ ][ ] 0 0 0 0 ln 1 A B B A B A kt = [ ] () [ ] 0 0 0 0 ln ln A B B A kt B A = [ ] B A ln time [ ] 0 0 ln A B [ ] [ ] ( ) 0 0 B A k Slope = Can simplify kinetic expression if set initial concentrations equal ( [ ] [ ] 0 0 B A = ). Since 1 = = B A ν , then [ ] B A = throughout the course of the reaction: [ ] 2 A k B A k dt A d = = , which we solved already. kt A A + = 0 1 1 or kt B B + = 0 1 1 [ ] [ ] 0 0 0 0 A B e B A B A kt =
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Second-Order Renaturation of DNA A G G C A C T C G T A G G C A C T C G T [ ][ ][ ] 2 A k B A k rate = = [] 0 1 1 A kt A + = [ ] kt A f A A 0 0 1 1 + = = [ ] kt A A A 0 0 1 + = or [A] = [B] because AB B A + k 5’ 3’ 5’ 3’
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