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From rate law to reaction mechanism

From rate law to reaction mechanism Products of a reaction can never be produced faster than the rate of the slowest elementary reaction - rate determining step Experimental data for the reaction between NO 2 and F 2 indicate a second-order rate

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From rate law to reaction mechanism

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  1. From rate law to reaction mechanism Products of a reaction can never be produced faster than the rate of the slowest elementary reaction - rate determining step Experimental data for the reaction between NO2 and F2 indicate a second-order rate Overall reaction: 2 NO2(g) + F2(g)  2FNO2(g) Rate = k [NO2] [F2] From the rate law : reaction does not take place in one step How can a mechanism be deduced from the rate law?

  2. Rate determining step must involve NO2 and F2 in 1:1 ratio Possible mechanism Step 1 NO2(g) + F2(g)  FNO2(g) +F(g) slow Step 2 NO2(g) + F(g) -> FNO2(g) fast Overall: 2 NO2(g) + F2(g)  2FNO2(g) Rate for step 1 = k1 [NO2] [F2] rate determining step Rate for step 2 = k2 [NO2] [F]

  3. For the reaction: 2 H2(g) + 2NO(g)  N2(g) + H2O(g) The observed rate expression is: rate = k[NO]2 [H2] The following mechanisms have been proposed. Based on the rate law can any mechanism be ruled out? k1 Mechanism I 2 H2(g) + 2NO(g)  N2(g) + H2O(g) k2 Mechanism II H2(g) + NO(g)  N(g) + H2O(g) slow k3 NO(g) + N(g)  N2(g) + O(g) fast k4 O(g) + H2 (g)  H2O(g) fast k5 Mechanism III H2(g) + 2NO(g)  N2O(g) + H2O(g) slow k6 H2(g) + N2O(g)  N2(g) + H2O(g) fast

  4. Mechanism I rate = k1[H2]2 [NO]2 not possible Mechanism II rate = k2[H2] [NO] not possible Mechanism III rate = k5[H2] [NO]2 possible If mechanism III is a possible mechanism, try to detect N2O experimentally to confirm mechanism.

  5. fast, equilibrium k2 k1 NO(g) + O2 (g)  OONO(g)  Step 2 OONO(g) + NO(g) 2 NO2(g)  k-1 slow Mechanism involving an initial fast reaction that produces an intermediate, followed by a slower second step. Rate law of an elementary step must be written with respect to the reactants only; an intermediate cannot appear in the rate law 2 NO(g) + O2(g)  2 NO2(g) rate = k [NO]2 [O2] Experiments indicate that an intermediate is involved in this reaction Step 1

  6. Rate law for second step = k2 [NO] [OONO] Cannot be compared with experimental rate law since OONO is an intermediate Rate of production of OONO = k1 [NO] [O2] Rate of consumption of OONO to NO and O2 = k-1 [OONO] Rate of forward and reverse of 1 > rate of 2; equilibrium is established in 1 before significant amount of OONO consumed to form NO2 At equilibrium k1 [NO] [O2] = k-1 [OONO]

  7. Rate = k2 [NO] [OONO] = k2 [NO] (K [NO] [O2]) = k2 K [NO]2 [O2] = k’ [NO]2 [O2]

  8. Stratospheric Ozone Chapman mechanism (1930’s) (1) O2 + light  2O. (2) O. + O2 + M  O3 + M (3) O3 + light  O. + O2 (4) O. + O3 2 O2 For O3 concentration to be in balance: Rate of production of O3 by reactions (1) and (2) must equal rate at which is O3 consumed in (3) and (4) In the 1960’s data published showed that reaction (4) was much too slow to balance levels of O3

  9. Paul Crutzen realized that rate constants could not explain measured distribution of ozone in the stratosphere. Using experimental data, Crutzen presented a model: NO + O3 NO2 + O2 NO2 + O. NO + O2 Overall: O. + O3 2O2 Contributed to reaction (4) in Chapman mechanism 1995 Nobel Prize in Chemistry

  10. k1 k-1 Kinetics and Equilibrium NO(g) + O3 (g) NO2 (g) + O2 (g) For a reaction which occurs in a single elementary step Rate of forward reaction = k1 [NO] [O3] Rate of reverse reaction = k-1 [NO2] [O2] At equilibrium: rate of forward reaction = rate of reverse reaction k1 [NO]eq [O3]eq = k-1 [NO2]eq [O2]eq where the eq denotes equilibrium concentrations

  11. where K is the equilibrium constant

  12. 2NO(g) + 2H2(g) N2(g) + 2H2O(g) k1 k2 k3 k-1 k-2 k-3 For the reaction: K = ([N2] [ H2O]2) /([ NO]2 [H2]2) The reaction occurs through three elementary reactions NO(g) + NO(g) N2O2(g) N2O2(g) + H2(g) N2O(g)+ H2O(g) N2O(g) + H2(g) N2(g)+ H2O(g)

  13. At equilibrium k1 [NO]2eq = k-1 [N2O2]eq k2 [N2O2]eq [H2]eq = k-2 [N2O]eq [H2O]eq k3 [N2O]eq [H2]eq = k-3 [N2]eq [H2O]eq

  14. Chain reactions Reaction which proceeds through a series of elementary steps, some of which are repeated many times. A highly reactive intermediate reacts to produce another highly reactive intermediate

  15. Chain reactions include reactions in: Atmosphere - ozone depletion Explosions Polymerization Nuclear Fission Anti-oxidants Many chain reactions involve free-radicals - atoms or molecules with one or more unpaired electron; formed by homolytic cleavage of a covalent bond

  16. Overall:H2(g) + Cl2(g)  2 HCl(g) Cl2(g) + light  2Cl(g) initiation Cl(g) + H2 (g)  HCl(g) + H(g) propagation H(g) + Cl2(g) HCl(g) + Cl(g) propagation Cl(g) + Cl(g) Cl2(g) termination H(g) + H(g) H2(g) termination H(g) + Cl(g) HCl(g) termination

  17. Stratospheric Ozone depletion by chlorofluorocarbons (CFC’s) CCl2F2 + hn CF2Cl + Cl CCl2F2 + O CF2Cl + ClO reactions which generate free radicals Cl + O3ClO + O2 ClO + O Cl + O2 Overall Reaction O + O3 -> 2O2 “Ozone Depletion” 1995 Nobel Prize in Chemistry

  18. Polymerization initiation propagation termination

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