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 NO 2 and F 2 indicate a second-order rate Overall reaction: 2 NO 2 (g) + F 2 (g)  2FNO 2 (g) Rate = k [NO 2 ] [F 2 ] 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 NO 2 and F 2 in 1:1 ratio Possible mechanism Step 1 NO 2 (g) + F 2 (g)  FNO 2 (g) +F(g)slow Step 2 NO 2 (g) + F(g) -> FNO 2 (g)fast Overall: 2 NO 2 (g) + F 2 (g)  2FNO 2 (g) Rate for step 1 = k 1 [NO 2 ] [F 2 ] rate determining step Rate for step 2 = k 2 [NO 2 ] [F]

3. For the reaction:2 H 2 (g) + 2NO(g)  N 2 (g) + H 2 O(g) The observed rate expression is: rate = k[NO] 2 [H 2 ] The following mechanisms have been proposed. Based on the rate law can any mechanism be ruled out? Mechanism I 2 H 2 (g) + 2NO(g)  N 2 (g) + H 2 O(g) k1k1 H 2 (g) + N 2 O(g)  N 2 (g) + H 2 O(g) fast k6k6 Mechanism III H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) slow k5k5 Mechanism II H 2 (g) + NO(g)  N(g) + H 2 O(g)slow k2k2 NO(g) + N(g)  N 2 (g) + O(g)fast k3k3 O(g) + H 2 (g)  H 2 O(g)fast k4k4

4. Mechanism Irate = k 1 [H 2 ] 2 [NO] 2 not possible Mechanism IIrate = k 2 [H 2 ] [NO] not possible Mechanism IIIrate = k 5 [H 2 ] [NO] 2 possible If mechanism III is a possible mechanism, try to detect N 2 O experimentally to confirm mechanism.

5. 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) + O 2 (g)  2 NO 2 (g) rate = k [NO] 2 [O 2 ] Experiments indicate that an intermediate is involved in this reaction Step 1 Step 2 OONO(g) + NO(g)  k2k2 2 NO 2 (g) slow NO(g) + O 2 (g)   k1k1 k -1 OONO(g)fast, equilibrium

6. Rate law for second step = k 2 [NO] [OONO] Cannot be compared with experimental rate law since OONO is an intermediate Rate of production of OONO = k 1 [NO] [O 2 ] Rate of consumption of OONO to NO and O 2 = 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 NO 2 At equilibrium k 1 [NO] [O 2 ] = k -1 [OONO]

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

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

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 + O 3  NO 2 + O 2 NO 2 + O.  NO + O 2 Overall: O. + O 3  2O 2 Contributed to reaction (4) in Chapman mechanism 1995 Nobel Prize in Chemistry

10. Kinetics and Equilibrium For a reaction which occurs in a single elementary step NO(g) + O 3 (g) NO 2 (g) + O 2 (g) k1k1 k -1 Rate of forward reaction = k 1 [NO] [O 3 ] Rate of reverse reaction = k -1 [NO 2 ] [O 2 ] At equilibrium: rate of forward reaction = rate of reverse reaction k 1 [NO] eq [O 3 ] eq = k -1 [NO 2 ] eq [O 2 ] eq where the eq denotes equilibrium concentrations

11. where K is the equilibrium constant

12. For the reaction:2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g) The reaction occurs through three elementary reactions NO(g) + NO(g) N 2 O 2 (g) k1k1 k -1 N 2 O 2 (g) + H 2 (g) N 2 O(g) + H 2 O(g) k2k2 k -2 N 2 O(g) + H 2 (g) N 2 (g) + H 2 O(g) k3k3 k -3 K = ([N 2 ] [ H 2 O] 2 ) /([ NO] 2 [H 2 ] 2 )

13. At equilibrium k 1 [NO] 2 eq = k -1 [N 2 O 2 ] eq k 2 [N 2 O 2 ] eq [H 2 ] eq = k -2 [N 2 O] eq [H 2 O] eq k 3 [N 2 O] eq [H 2 ] eq = k -3 [N 2 ] eq [H 2 O] 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: H 2 (g) + Cl 2 (g)  2 HCl(g) Cl 2 (g) + light  2  Cl(g) initiation  Cl(g) + H 2 (g)  HCl(g) +  H(g) propagation  H(g) + Cl 2 (g)  HCl(g) +  Cl(g) propagation  Cl(g) +  Cl(g)  Cl 2 (g)termination  H(g) +  H(g)  H 2 (g)termination  H(g) +  Cl(g)  HCl(g)termination

17. CCl 2 F 2 + h  CF 2 Cl + Cl  CCl 2 F 2 + O   CF 2 Cl + ClO   Cl + O 3   ClO + O 2  ClO +  O   Cl + O 2 Overall Reaction  O + O 3 -> 2O 2 “Ozone Depletion” 1995 Nobel Prize in Chemistry reactions which generate free radicals Stratospheric Ozone depletion by chlorofluorocarbons (CFC’s)

18. initiation propagation termination Polymerization