9 Thermodynamics & Kinetics-Complex Reactions and Mechanisms 3, Lecture notes of Chemistry

Download 9 Thermodynamics & Kinetics-Complex Reactions and Mechanisms 3 and more Lecture notes Chemistry in PDF only on Docsity! A B Ck-1 k2 5.60 Spring 2007 Lecture #32 page 1 Complex Reactions and Mechanisms (continued) IV) Steady State and Equilibrium Approximations a) Steady State Approximation k1 Assume that [B] is small and slowly varying e.g. d[B] ≈ 0 and (k2 + k-1) >> k1dt [B] reaches a steady state concentration [B]SS and remains there d[B] dt = k1[A] − k−1[B]SS − k2[B]SS ≈ 0 Steady State approximation Solving… [B]SS = k1[A] k−1 + k2 d[A]So − dt = k1[A] − k−1[B]SS d[A] k1k2[A] − = dt k−1 + k2 ______________________________________ 5.60 Spring 2007 Lecture #32 page 2 d[C] = k2[B]SS = k1k2[A] = − d[A] dt k−1 + k2 dt k' k k Looks like A C (first order) with k'= 1 2 k−1 + k2 **Necessary Condition for use of Steady State Approximation** i) Data must be taken after B has built up to a steady state value. ii) (k2 + k-1) >> k1 ⇒ [B]SS is small A B C k-1 k2 b) Equilibrium Approximation k1 Assume k2 << k-1 and k1 k2That is… B C is the rate limiting step. Then… A and B quickly come into equilibrium, while C slowly builds up. k [ ]B [ ] k A eq [ ]Keq = 1 ≈ [ ] B = k 1 [ ] = K A k−1 A −1 Equilibrium approximation 5.60 Spring 2007 Lecture #32 page 5 Limiting Cases i) k2[M] >> k-1 then d[I2] =k1 [I]2 dt (high pressure) second order ii) k2[M] << k-1 then d[I2] = k1k2 [M][I]2 dt k−1 (low pressure) third order B) Gas decomposition (Lindemann Mechanism) A(g) → products Mechanism: k1 A + M A* + M k-1 A* k2 products (B + …) M is a rare gas molecule and/or A, k1 is fast, is very fast, k2 is slow k-1 So… (k2 + k-1) >> k1 , Steady State approximation again. 5.60 Spring 2007 Lecture #32 page 6 d[A*] * * dt = k1[A][M] − k−1[A ]SS[M] − k2[A ]SS ≈ 0 Steady State approximation k [A][M] [A*]SS = 1 k−1[M] + k2 − d[A] = d[B] = k2[A*]SS = k1k2[A][M] dt dt k−1[M] + k2 Limiting Cases i) High pressure (1 bar) k-1[M] >> k2 − d[A] = k1k2 [A] = k∞ [A] (1st order) dt k−1 ii) Low pressure (~10-4 bar) k-1[M] << k2 d[A] − dt = k1[A][M] (if M≡A, then 2nd order in A)