:19 PM 1 Reaction Mechanisms Steps of a Reaction

:19 PM 2 The Ozone Layer Ozone is most important in the stratosphere, at this level in the atmosphere, ozone absorbs UV radiation 10 Km 50 Km 100 Km Ozone Layer Stratosphere Mesosphere Troposphere Mt. Everest Troposphere Stratosphere Mesosphere

:19 PM 3 Mechanism of Ozone; Chapman Cycle Chapman cycle shows that O 3 exist at steady state. It is constant in the stratosphere. O2O2O2O2 O2O2O2O2 O3O3O3O3 2 O O 320 nm or less 242 nm or less Slow ozone removal step O 3 lives for ~ s before it dissociates. Ozone removal step: O 3 + O  2O 2

:19 PM 4 Reaction Coordinate for ozone destruction A Key reaction in the upper atmosphere is  O 3 (g) + O (g)  2O 2 (g) The E a (fwd) is 19 kJ, and the  H rxn as written is -392 kJ. A reaction energy diagram for this reaction with the calculate E a(rev) is shown. O 3 + O 2O 2 O kJ 19 kJ -392 kJ E act (rev) = 411 kJ Reaction Progress

:19 PM 5 Path to Destruction: Ozone  1. Water Vapors: H 2 O   OH + H  H + O 3   OH + O 2  OH + O   H + O 2  Net:O + O 3   2O 2  2.. N 2, Dinitrogen : N 2 + O 2   2NO  NO + O 3   NO 2 + O 2  NO 2 + O   NO + O 2  Net:O + O 3   2O 2  3. CFCs CCl 2 F 2   CClF 2 + Cl  ChlorofluorocarbonsCl + O 3   ClO + O 2  ClO + O   Cl + O 2  Net:O + O 3   2O 2  10,000 O 3 will breakdown to O 2 for every Cl

:19 PM 6 Influence by CFC: Ozone Comparison of activation energies in the uncatalyzed decompositions of ozone. The destruction of ozone can be catalyzed by Cl atoms which leads to an alternative pathway with lower activation energy, and therefore a faster reaction. Progress of reaction Energy (kJ)

:19 PM 7 Reaction Mechanism The mechanism of a reaction is the sequence of steps (at a molecular level) that leads from reactant to products. Elementary Steps Sequence of steps which describes an actual molecular event. Stoichiometry The overall stoichiometric reaction is the sum of the elementary steps. Scientist want to learn about mechanism because an understanding of the mechanism (how bonds break and form) may lead to conditions to improve reaction product yield, (or prevent side products formation. i.e, depletion of ozone.)

:19 PM 8 Ozone: Revisited Chapman’s Cycle  O 3  O 2 + O  O + O 3  2O 2  2O 3  3O 2 Elementary Steps give rise to Rate Law Since elementary steps describes a molecular collision, the rate law for an elementary step (unlike the overall reaction) can be written from the Stoichiometry. Consider an elementary step  iA + jB  Product (slow step) rate = k [A] i [B] j The rate of the reaction is directly proportional to concentrations of the colliding species.

:19 PM 9 Elementary Step: Rate Law Consider the following proposed mechanism for the conversion of NO 2 to N 2 O 5. What is the rate law. Step1 NO 2 + O 3  NO 3 + O 2 (slow) Step2 NO 3 + NO 2  N 2 O 5 (fast) rate = k 1 [NO 2 ] 1 [O 3 ] 1 In a series of steps, the slowest step determines the overall rate. In the mechanism for a chemical reaction, the slowest step is the rate-determining step.

:19 PM 1010 Elementary Steps: Order of reaction Elem. StepRate LawOrderMolecularity 1A  ProductRate = k[A]1st orderunimolecular 22A  ProductRate = k[A] 2 2nd orderbimolecular A + B  Prod.Rate = k[A][B] 33A  ProductRate = k[A] 3 3rd orderTermolecular 2A + B  ProductRate = k[A] 2 [B] A + B + C  ProdRate = k[A][B][C] * Termolecular mechanism (elementary step) is very rare. Scientist who propose such a mechanism must make careful measurements.

:19 PM 1 Multiple Elementary Steps Most reactions involve more than one elementary step. Rate-Limiting - When one step is much slower than any other, the overall rate is determined by the slowest “Rate- determining” step. Reaction is only as fast as the slowest elementary step Analogy: Leaving class after an exam. On way skiing, speed only as fast as creepy crawler 12-cars ahead.

:19 PM 1212 Rate Determining Step from Rate Law Consider: NO 2 + CO  NO + CO 2 Mechanism: (1) NO 2 (g) + NO 2 (g)  NO 3 g) + NO (g) (2) NO 3 (g) + CO (g)  NO2 + CO 2 (g) Net: NO 2 (g) + CO (g)   CO 2 (g ) + NO (g) Rate = k[NO 2 ] 2 Which is Rate determining step (1) or (2) ? RDS is the step that determines the rate law. When scientists propose a mechanism, they can only say that it is consistent with the experimental data. There may be other mechanism that is consistent with experimental data. If experiments are done in the future to disprove the mechanism, then his proposed mechanism must be revised.

:19 PM 1313 RDS and Rate Law: Example Consider the reaction : NO (g) + O 3  NO 2 + O 2 Two mechanisms (elementary steps) are proposed: Mechanism 1NO + O 3  NO 2 + O 2 Rate = k [NO] [O 3 ] Proposed rate law:Rate = k [NO] [O 3 ] Mechanism 2O 3  O 2 + O (slow) NO + O  NO 2 (fast) Rate = k [O 3 ] Proposed rate law:Rate = k [O 3 ] What are the Rate Laws ? When a potential mechanism is proposed, 2 factors must be considered - Rate determining step must be consistent with observed rate law. Sum of all the steps must yield the observed stoichiometry.

:19 PM 1414 Complicated Reaction Mechanism Reaction mechanism in which slow step (rate determining step) involves an intermediate. Consider: A  B Mechanism: A  int (fast) int  B (slow) NET: A  B RATE = k[int] -but the rate law cannot be written in terms of an intermediate (catalyst okay, but not intermediate). -It must be expressed in terms of stable species How is the Rate Law modified?

:19 PM 1515 Modification of Rate Law RATE = k [int ] Written in terms of reactants- The rate law is now expressed in terms of the reactant.

:19 PM 1616 Rate Laws from Mult. Steps Mechanism. Consider the reaction: what would be the rate law based on the two proposed mechanism: 2 NO 2 (g) + O 3  N 2 O 5 + O 2 Mechanism (1) Mechanism (2) NO 2 + NO 2  N 2 O 4 (fast) NO 2 + O 3  NO 3 + O 2 (slow) N 2 O 4 + O 3  N 2 O 5 + O 2 (slow) NO 3 + NO 2  N 2 O 5 (fast)

:19 PM 1717 Rate Laws from Mult. Steps Mechanism. The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed via a two-step mechanism: H 2 O 2 (aq) + I - (aq)  H 2 O (l) + IO - (aq)(slow) H 2 O 2 (aq) + IO - (aq)  H 2 O (l) + I - (aq) + O 2 (g)(fast) a) Rate Law: Rate Law = k [ H 2 O 2 ] [I - ] b) Overall reaction: 2 H 2 O 2 (aq)  H 2 O (l) + O 2 (g) c) Intermediate: IO - (aq) Catalyst: I - (aq)

:19 PM 1818 Enzyme Catalysis Reaction Consider oxidation of ethanol to aldehyde: CH 3 CH 2 OH (l)  ADH  CH 3 CHO + R-H 2 ADH - Alcohol dehydrogenase Mechanism E + S  ES ES  E + P (slow) E + S  E + P Rate = k [ES] K eq = [ES]  K eq [E] [S] = [ES] [E] [S] Rate = K K eq [E] [S] = K’ [E] [S]

:19 PM 1919 Summary The dynamics of the series of steps of a chemical change is what kinetics tries to explain. Variation in reaction rate are observed through concentration and temperature changes, which operate on the molecular level through the energy of particle collision. Kinetics allows us to speculate about the molecular pathway of a reaction. Modern industry and biochemistry depend on its principles. However, speed and yield are very different aspects of a reaction. Speed is in the kinetic domain, likelihood (spontaneity) is in the thermodynamic domain.