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Chem 300 - Ch 28/#2 Todays To Do List l 1 st -Order Reaction Kinetics l ½-life & Reaction Order l 2 nd -Order Reactions l Reversible Reactions

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Analysis of the Rate Law l In general: For a reaction: A + B products Rate law: v(t) = -d[A]/dt = k [A] m(A) [B] m(B) l Find time-dependence of [A]. l Integrate the rate-law. l Consider 1 st & 2 nd –order rate-laws.

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1 st -Order Reactions l v(t) = - d[A]/dt = k [A] (d [A]) / [A] = - k dt d [A] /[A] = - k dt From [A] 0 [A] & t = 0 t = t ln [A] – ln [A] 0 = -kt [A] = [A] 0 e -kt

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1 st Order Reaction k = (dot)0.0125 s -1, 0.025, 0.050, 0.10

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N 2 O 5 2 NO 2 + ½ O 2 @ 318K slope = -k = 3.04 x 10 -2 min -1

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Reaction ½-Life (t 1/2 ) l t 1/2 = Time required for ½ of reactant to disappear (when [A] = ½ [A] 0 ) l For 1 st -order reaction: ln [A] – ln [A] 0 = -kt ln (½ [A] 0 ) – ln [A] 0 = -k t 1/2 ln (½) + ln [A] 0 ) – ln [A] 0 = -k t 1/2 ln ½ = - ln2 = -k t 1/2 t 1/2 = 0.693/ k Independent of [A] 0

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N 2 O 5 2 NO 2 + ½ O 2 @ 318K

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Wide range of Reaction Speeds (k-values) l cyclopropene propyne k = 5.7 x 10 -4 /s at 500 K l cyclobutane 2-ethene k = 1.8 x 10 -12 /s at 500 K

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2 nd -Order Reactions l - d[A]/dt = k [A] 2 l Separate variables: (d [A]) / [A] 2 = - k dt l Integrate: 1/[A] = 1/[A] 0 + kt Plot 1/[A] vs t gives a straight line. Slope = - k intercept = 1/[A] 0

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NOBr NO + ½ Br 2 2 nd -Order Reaction

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2 nd -Order ½-Life l 1/[A] = 1/[A] 0 + kt l Substitute [A] = ½ [A] 0 at t = t 1/2 2/[A] 0 = 1/[A] 0 + kt 1/2 l Rearrange: t 1/2 = 1/k[A] 0 l Depends upon [A] 0 (initial conc.)

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2 nd -Order Rate Law l A + B products l -d[A]/dt = -d[B]/dt = k[A][B] l The integrated form: kt = {1/([A] 0 – [B] 0 )} ln[A][B] 0 /[B][A] 0

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Reversible Reactions l A = B k 1 = forward reaction k -1 = reverse reaction l At equilibrium: -d[A]/dt = d[B]/dt = 0 Rate forward = k 1 [A] Rate reverse = k -1 [B] k 1 [A] eq = k -1 [B] eq k 1 /k -1 = [B] eq /[A] eq = K eq

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Next Time Relaxation Methods & Fast Reactions Temperature Dependence Transition-State Theory

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