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# lec 4 - Chemical Kinetics(Chapter 9 Ch 9 Homework...

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Sucrose + Air Metabolic Energy Chemical Kinetics (Chapter 9) O 11H 12CO 12O O H C 2 2 2 11 22 12 + + Δ = -5693 kJ/mol Ch. 9 Homework : 2,4,6,8,10,12,14,16,20,22,23,26,27,32,34,38,56,57,64 Δ Sucrose CO 2 + H 2 O E a Rate (k) exp(-E a /RT) K eq exp(- Δ Gº/RT) Enzyme Catalyzed (fast) No enzyme (slow) K eq ~ 10 100

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Instantaneous Rate of Reaction [] [ ] dt R d dt P d = = rate Reaction rate at specific time point (t 1 ) is called instantaneous rate. [R](t) [P](t) Time Concentration t 1 P R Instantaneous rate equals the slope of the tangent at a specific time (t 1 ). 0 ] [ < dt R d Slope of tangent expressed by a derivative from calculus (e.g. d[P]/dt). Minus sign because Units of rate: M s -1 or M min -1 (i.e. [R] decrease vs t) dP dt
Reaction Rate & Rate Law General rate expression determine from balanced eqn: dD cC bB aA + + [ ] [ ] [ ] [ ] dt D d 1 dt C d 1 dt B d 1 dt A d 1 d c b a rate = = = = Divide by stoichiometric coefficient. Add minus sign for each of reactants. Rate law determined empirically: [] [ ] y x B A k rate = [A] vs time Rate vs [A] Reaction order given by x and y

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Zero-Order Kinetics A products rate = - Δ [A] = k [A] 0 rate = k [units of k] = mol L -1 s -1 Rate is constant at all concentrations rate [A] k Δ t Slope = -k Examples are enzyme reactions when the enzyme is saturated with substrate. [A] t = - kt + [A] 0
Zero-Order Half Life (t 1/2 ) [A] = [A] 0 -kt Time [A](t) [A] 0 Slope = -k Half-life (t 1/2 ) is time for [A] to decrease in half: [ ][ ] [] k A t kt A A kt A A 0 2 / 1 2 / 1 0 0 0 5 . 0 5 . 0 = = = t 1/2 Zero-Order Half Life: [ ] k A t 0 2 / 1 5 . 0 = P A

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Kinetic Data [A] Time (s) rate [A] t 1/2 = 0.5[A] 0 = 0.357 M = 72 s k 0.005 M/s Zero order kinetics Zero-order half-life What is the half-life for the reaction: A P? Time(s) [A] (M) Δ [A] Δ tR a t e ( M / s ) 0 0.715 22 0.605 74 0.345 132 0.055 -0.11 22 0.005 -0.26 52 0.005 -0.29 58 0.005 rate = k [A] t = - kt + [A] 0 Zero-order Kinetics Sample Problem t 1/2 = 72 s Δ t Δ [A] -
First-Order Kinetics A products = - k dt [A] d[A ] [A] 0 [A] t 0 t = -kt ln [A] t [A] 0 ln[A] t = -kt + ln[A] 0 [units of k] = s -1 1 x dx = ln x Recall calculus: ln x 2 ln x 1 = ln x 2 x 1 First order reaction: rate = k[A] 1 = k [A] d[A ] dt rate = - [A] t = [A] 0 exp(-kt) ln[A] time [A] Slope = -k time What is [A] vs time?

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