Lecture 32: Complex Reactions and Mechanisms (Cont.)

Lecture #32 page 1 Complex Reactions and Mechanisms (continued) III) Reversible Reactions k1 A B k-1 If 1st order, Keq = [B]eq [A]eq Rforward = Rf = k1[A] Rbackward = Rb = k-1[B] Rf = Rb At Equilibrium, Keq = k1[A]eq = k-1[B]eq ⇒ k1 k−1 a) 1st order reversible reactions A k1 k-1 B − d[A] = k1 [A] − k −1 [B] dt [B] = [B]o + ([A]o – [A]) So… − d[A] = k1 [A] − k−1 ([B]o + [A]o − [A]) dt Lecture #32 At Equilibrium, ⇒ − ( ) dt ⇒ d[A] =0 dt [A]eq = d [A] − [A]eq page 2 =− k−1 ([B]o + [A]o ) k1 + k−1 d([A]) = (k1 + k−1 ) [A] − [A]eq dt ( ) [A] − [A]eq = ([A]o − [A]eq )e −(k1 +k−1 )t ln (|[A] - [A]eq|) Conc [A]0 > [A]eq Slope = -(k1+k-1) [A]eq [A]0 < [A]eq time time Can measure: Keq = k1 k −1 And extract and k1 + k −1 ≡ kobs k1 and k-1 Lecture #32 page 3 b) Higher order reactions e.g. − A+B k2 2nd order forward, 1st order backward C k-1 d[A] = k2 [A][B] − k −1 [C] , dt K= After much calculation, get… [C ]eq [A]eq [B]eq , K= k2 k −1 A mess! We must begin simplifying from the beginning! Use Flooding in this case: [B]o >> [A]o,[C]o k1 ≡ k2[B]o ≈ k2[B] Then − d[A] = k1 [A] − k −1 [C] dt This is now pseudo 1st order in A ⇒ Looks the same as in part a) Measure: K = k2 k −1 , kobs ≡ k1 + k-1 = k2[B]o + k-1 By changing [B]o over a few experiments, can extract k2 and k-1 Lecture #32 IV) A − page 4 Series Reversible Reactions (1st order) k1 k-1 k2 B C d[A] = k1 [A] − k −1 [B] dt − d [ B] = k1[ A] − k−1[ B] − k2 [ B] dt d[C] = k2 [B] dt Can solve this, but it is an even bigger mess than in part IIIb)!! And here Flooding, as an approximation, is not going to do much for us. We need to find new approximations for more complicated mechanisms! Lecture #32 IV) Steady State and Equilibrium Approximations a) Steady State Approximation A k1 k-1 B k2 C Assume that [B] is small and slowly varying d[B] ≈ 0 and (k2 + k-1) >> k1 dt e.g. [B] reaches a steady state concentration [B]SS and remains there d[B] = k1 [A] − k−1 [B]SS − k2 [B]SS ≈ 0 dt Steady State approximation Solving… So − [B]SS = k1 [A] k−1 + k2 d[A] = k1 [A] − k−1 [B]SS dt − d[A] k1k2 [A] = dt k−1 + k2 k k [A] d[C] d[A] = k2 [B]SS = 1 2 =− dt k −1 + k2 dt page 5 Lecture #32 Looks like k' A (first order) with k'= C page 6 k1k2 k −1 + k2 **Necessary Condition for use of Steady State Approximation** Data must be taken after B has built up to a steady i) state value. (k2 + k-1) >> k1 ⇒ [B]SS is small ii) ______________________________________ b) Equilibrium Approximation A k1 k-1 Assume That is… B B k2 C k2 << k-1 and k1 k2 C is the rate limiting step. Then… A and B quickly come into equilibrium, while C slowly builds up. Keq = k1 [B] ≈ k −1 [A] [B] = k1 [A] = Keq [A] k −1 Equilibrium approximation kk d[C] = k2 [B] = k2Keq [A] = 1 2 [A] So… dt k−1 Lecture #32 page 7 k k d[C ] = 1 2 [A ] dt k −1 Or, Looks like A k' C (first order) with k'= k1k2 k−1 In general, for a mechanism with multiple pre-equilibria… e.g. A A A A ↔ ↔ ↔ ↔ I1 I2 I3 I4 In ↔ B K1 K2 K3 K4 (Rate Determining) ⎡n ⎤ d[B] =k n [In ] = kn ⎢∏ Ki ⎥[A] dt ⎣ i=1 ⎦ Examples: A) Apparent Termolecular Reactions (Reaction Chaperones) Lecture #32 k I+I+M page 8 I2 + M M is a rare gas molecule or the wall of the reaction vessel Mechanism: k1 I+I I2* k-1 I2* + M (M* k2 I2 + M* M) where (k2 + k-1) >> k1 , So d[I2* ] dt that is the Steady State approximation! = k1 [I] 2 − k −1 [I2* ]SS − k2 [I2* ]SS [M] ≈ 0 Steady State approximation Solving… And … [I2* ]SS k1 [I] 2 = k −1 + k2 [M] d[I2 ] k1 [I] 2 * = k2 [I2 ]SS [M] = k2 [M] dt k−1 + k2 [M] Limiting Cases Lecture #32 i) k2[M] >> k-1 page 9 d[I2 ] =k 1 [I] 2 dt then (high pressure) ii) k2[M] << k-1 second order d[I2 ] k 1 k2 = [M][I] 2 dt k −1 then (low pressure) B) third order Gas decomposition (Lindemann Mechanism) A(g) → products Mechanism: k1 A+M A* k-1 k2 A* + M products (B + …) M is a rare gas molecule and/or A, k1 is fast, k-1 So… (k2 + k-1) >> k1 , is very fast, k2 is slow Steady State approximation again. d[A * ] = k1 [A][M] − k −1 [A * ]SS [M] − k2 [A * ]SS ≈ 0 dt Lecture #32 page10 Steady State approximation [A * ]SS = − k1 [A][M] k −1 [M] + k2 k k [A][M] d[A] d[B] = = k2 [A * ]SS = 1 2 dt dt k−1 [M] + k2 Limiting Cases i) High pressure (1 bar) − ii) k-1[M] >> k2 d[A] k1k2 = [A] = k∞ [A] dt k −1 Low pressure (~10-4 bar) − d[A] = k1 [A][M] dt (1st order) k-1[M] << k2 (if M≡A, then 2nd order in A)