Reaction Mechanism

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					    Reaction Mechanisms
       Steps of a Reaction

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                            The Ozone Layer
             Ozone is most important in the stratosphere, at this level in
             the atmosphere, ozone absorbs UV radiation

    100 Km                                                       Mesosphere
                                   Mesosphere                    Troposphere

    50 Km

                  Ozone Layer
    10 Km       Mt.                    Troposphere

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              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
                                 O22                  s before it dissociates.
                                            320 nm
                                            or less   Ozone removal step:
    242 nm
    or less           2O
                      O                         h    O3 + O  2O2
               O2                      O3

                  Slow ozone
                  removal step         O

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        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   19 kJ
    the DHrxn as written is       O kJ
    -392 kJ.                              O3 + O
                                                              Eact (rev)
                                                              = 411 kJ
    A reaction energy
    diagram for this reaction                           2O2
    with the calculate Ea(rev) -392 kJ
    is shown.

                                         Reaction Progress
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            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
           Net:             O + O3        2O2

    3. CFCs                 CCl2F2        CClF2• + Cl•
    Chlorofluorocarbons     Cl• + O3      ClO• + O2
                            ClO• + O      Cl• + O2
           Net:             O + O3        2O2

         10,000 O3 will breakdown to O2 for every Cl•

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             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

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                  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.
         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.)
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                       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.
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           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.

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        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.

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              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.

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      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.
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            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.

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       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?

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          Modification of Rate Law

    RATE = k [int]
      Written in terms of reactants-

            Rate  k [int]
          [int ]
    keq          [int]  k eq [A]
     Rate  k • k eq [A]  K' [A]
    The rate law is now expressed in terms of the reactant.

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       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  k [N2O4 ] [O3]          Rate  k[NO 2 ] [O 3 ]
                 [N2O 4 ]                 bimolecular
       k eq 
              [NO2 ] [NO2 ]             Rate is based on
       [N 2O4 ]  keq [NO2 ]
                               2       slowes t elem. step

     Rate  k • k eq [NO2 ]2[O3 ]
       Rate  k[NO2 ]2 [O3 ]
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       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)
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                Enzyme Catalysis Reaction
    Consider oxidation of ethanol to aldehyde:
          CH3CH2OH (l)  ADH  CH3CHO + R-H2
     ADH - Alcohol dehydrogenase
                    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]

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    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.

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