16.3 Reaction Mechanisms Steps of a Reaction

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

Mechanism of Ozone; Chapman Cycle Chapman cycle shows that O3 exist at steady state. It is constant in the stratosphere. O3 lives for ~200 - 300 s before it dissociates. Ozone removal step: O3 + O  2O2 O2 320 nm or less 242 nm or less 2 O O2 O3 Slow ozone removal step O

Reaction Coordinate for ozone destruction A Key reaction in the upper atmosphere is O3 (g) + O (g)  2O2 (g) The Ea (fwd) is 19 kJ, and the DHrxn as written is -392 kJ. A reaction energy diagram for this reaction with the calculate Ea(rev) is shown. O3 + O 2O2 O kJ 19 kJ -392 kJ Eact (rev) = 411 kJ Reaction Progress

Path to Destruction: Ozone 1. Water Vapors: H2O  OH• + H• H• + O3  OH• + O2 OH• + O  H• + O2 Net: O + O3  2O2 2.. N2 , Dinitrogen : N2 + O2  2NO NO + O3  NO2 + O2 NO2 + O  NO + O2 3. CFCs CCl2F2  CClF2• + Cl• Chlorofluorocarbons Cl• + O3  ClO• + O2 ClO• + O  Cl• + O2 10,000 O3 will breakdown to O2 for every Cl•

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. Energy (kJ) Progress of reaction

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.)

Ozone: Revisited Chapman’s Cycle O3  O2 + O O + O3  2O2 2O3  3O2 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.

Elementary Step: Rate Law Consider the following proposed mechanism for the conversion of NO2 to N2O5. What is the rate law. Step1 NO2 + O3  NO3 + O2 (slow) Step2 NO3 + NO2  N2O5 (fast) rate = k1[NO2]1 [O3]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.

Elementary Steps: Order of reaction Elem. Step Rate Law Order Molecularity 1 A  Product Rate = k[A] 1st order unimolecular 2 2A  Product Rate = k[A]2 2nd order bimolecular A + B  Prod. Rate = k[A][B] 3 3A  Product Rate = k[A]3 3rd order Termolecular 2A + B  Product Rate = k[A]2[B] A + B + C  Prod Rate = k[A][B][C] * Termolecular mechanism (elementary step) is very rare. Scientist who propose such a mechanism must make careful measurements.

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.

Rate Determining Step from Rate Law Consider: NO2 + CO  NO + CO2 Mechanism: (1) NO2 (g) + NO2 (g)  NO3 g) + NO (g) (2) NO3 (g) + CO (g)  NO2 + CO2 (g) Net: NO2 (g) + CO (g)  CO2 (g ) + NO (g) Rate = k[NO2] 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.

RDS and Rate Law: Example Consider the reaction : NO (g) + O3  NO2 + O2 Two mechanisms (elementary steps) are proposed: Mechanism 1 NO + O3  NO2 + O2 Proposed rate law: Rate = k [NO] [O3] Mechanism 2 O3  O2 + O (slow) NO + O  NO2 (fast) Proposed rate law: Rate = k [O3 ] 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.

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?

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

Rate Laws from Mult. Steps Mechanism. Consider the reaction: what would be the rate law based on the two proposed mechanism: 2 NO2 (g) + O3  N2O5 + O2 Mechanism (1) Mechanism (2) NO2 + NO2  N2O4 (fast) NO2 + O3  NO3 + O2 (slow) N2O4 + O3  N2O5 + O2 (slow) NO3 + NO2  N2O5 (fast)

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: H2O2 (aq) + I- (aq)  H2O(l) + IO- (aq) (slow) H2O2 (aq) + IO- (aq)  H2O(l) + I- (aq) + O2 (g) (fast) a) Rate Law: Rate Law = k [H2O2] [I- ] b) Overall reaction: 2 H2O2 (aq)  H2O(l) + O2 (g) c) Intermediate: IO- (aq) Catalyst: I- (aq)

Enzyme Catalysis Reaction Consider oxidation of ethanol to aldehyde: CH3CH2OH (l)  ADH  CH3CHO + R-H2 ADH - Alcohol dehydrogenase Mechanism E + S  ES ES  E + P (slow) E + S  E + P Rate = k [ES] Keq = [ES]  Keq [E] [S] = [ES] [E] [S] Rate = K Keq [E] [S] = K’ [E] [S]

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.