; Chapter 25 The rates of chemical reactions
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Chapter 25 The rates of chemical reactions

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  • pg 1
									       Tutorial on Monday and
             Wednesday

• No. 2 (Chapter 22: The rates of chemical
  reactions) by Bryan Lucier

• March 7 & 9 (from 2:30 to 3:50pm, Essex
  Hall room 182)
             Kinetic isotope effect
• Kinetic isotope effect: the decrease in the rate of a chemical
  reaction upon replacement of one atom in a reactant by a heavier
  isotope.

• Primary kinetic isotope effect: the kinetic isotope effect observed
  when the rate-determining step requires the scission of a bond
  involving the isotope:
                                            ~
                 k (C  D)     with    hc v (C  H )        CH 1 / 2 
                           e                        1  (     ) 
                 k (C  H )                              2kT            CD   

• Secondary kinetic isotope effect: the variation in reaction rate
  even though the bond involving the isotope is not broken to form
  product:
                  k (C  D)                          ~              ~
                              e              hc{v  (C  H )  v (C  H )}       CH 1 / 2 
                  k (C  H )                                                1  (     ) 
                                      with                  2kT                     CD       
Kinetic isotope effect
      (primary)
Kinetics isotope effect
     (secondary)
      22.8 Unimolecular reactions
•   The Lindemann-Hinshelwood mechanism
         A reactant molecule A becomes energetically excited by collision with
    another A molecule:
                  A + A → A* + A
                               d[ A*]
                                       k1[ A]2
                                 dt
         The energized molecule may lose its excess energy by collision with
    another molecule:
                 A + A* → A + A
                              d [ A*]
                                        k1 [ A][ A*]
                                           '
                                 dt
         The excited molecule might shake itself apart to form products P
                          A* → P
                                  d[ A*]
                                           k 2 [ A*]
                                    dt
         The net rate of the formation of A* is
                       d [ A*]
                                k1 [ A]2  k1 [ A][ A*]  k 2 [ A*]
                                             '
                          dt
• If the reaction step, A + A → A* + A, is slow enough to be
  the rate determining step, one can apply the steady-state
  approximation to A*, so [A*] can be calculated as
                   d [ A*]
                            k1[ A]2  k1[ A][ A*]  k2 [ A*]  0
                                        '
                      dt


  Then                                k1[ A]2
                          [ A*] 
                                    k1[ A]  k2
                                     '



  The rate law for the formation of P could be reformulated as
                         d [ P]              k1k2 [ A]2
                                 k2 [ A*]  '
                          dt                k1[ A]  k2


  Further simplification could be obtained if the deactivation of A* is
  much faster than A*  P, i.e., k1' [ A ][A*]  k2[ A*] then k1[ A]  k2
                                                                '


                                    d [ P ] k1 k 2
                                            ' [ A]
                                      dt     k1
  in case k1[ A]  k2
           '

                                 d[ P ]
                                         k1 [ A]2
                                  dt
                     d [ P]              k1k2 [ A]2
• The equation        dt
                             k2 [ A*]  '
                                        k1[ A]  k2
                                                      can be
  reorganized into
                     d [ P]    k1k 2 [ A]
                            ( '           )[ A]
                      dt      k1[ A]  k 2

• Using the effective rate constant k to represent
                            k1k 2 [ A]
                     k
                          k1' [ A]  k 2

• Then one has
                               '
                     1    k1              1
                                     
                     k   k1k2           k1[ A]
    The Rice-Ramsperger-Kassel
           (RRK) model
• Reactions will occur only when enough of required energy
  has migrated into a particular location in the molecule.

                          s 1
                   E 
                     *
           P  1 
                     
                   E 
                      
                           s 1
                     *
                       E 
      k b ( E )  1 
                         kb
                      E 
                         

• s is the number of modes of motion over which the energy
  may be dissipated, kb corresponds to k2
 The activation energy of combined reactions
• Consider that each of the rate constants of the following reactions
               A + A → A* + A                   (E1)
               A + A* → A + A                   (E1’)
               A* → P                           (E2)
  has an Arrhenius-like temperature dependence, one gets
                       k1 k 2 A1e  E1/ RT A2 e  E2 / RT
                             
                        k1'            '  E1 / RT
                                            '
                                    A1e
                              A1 A2
                                      e ( E1  E2  E1 ) / RT
                                                       '
                               '
                               A1
   Thus the composite rate constant also has an Arrhenius-like form
   with activation energy,
        E = E1 + E2 – E1’

   Whether or not the composite rate constant will increase with
   temperature depends on the value of E,

   if E > 0, k will increase with the increase of temperature
Combined activation energy
• Theoretical problem 22.20
     The reaction mechanism
                 A2 ↔ A + A (fast)
                 A + B → P (slow)
     Involves an intermediate A. Deduce the
  rate law for the reaction.

• Solution:
                        Chain reactions
• Chain reactions: a reaction intermediate produced in one
  step generates an intermediate in a subsequent step, then that
  intermediate generates another intermediate, and so on.


• Chain carriers: the intermediates in a chain reaction. It could be
  radicals (species with unpaired electrons), ions, etc.


• Initiation step:
• Propagation steps:
• Termination steps:
         23.1 The rate laws of chain
                 reactions
• Consider the thermal decomposition of acetaldehyde
                  CH3CHO(g)       →     CH4(g) + CO(g)
                        v = k[CH3CHO]3/2
  it indeed goes through the following steps:
      1. Initiation:  CH3CHO → . CH3 + .CHO            ki
                         v = ki [CH3CHO]
     2. Propagation: CH3CHO + . CH3 → CH4 + CH3CO.                                 kp
        Propagation: CH3CO. → .CH3 + CO                                           k’p
     3. Termination:         .CH3 + .CH3 → CH CH                                  kt
                                             3   3


• The net rates of change of the intermediates are:
   d [ .CH 3 ]
                ki [CH 3CHO ]  k p [ .CH 3 ][CH 3CHO ]  k ,p [CH 3CO . ]  2kt [ .CH 3 ]2
        dt
                d [CH 3CO. ]
                              k p [ .CH 3 ][CH 3CHO ]  k ,p [CH 3CO. ]
                    dt
• Applying the steady state approximation:
      0  k i [CH 3CHO]  k p [ .CH 3 ][CH 3CHO]  k ,p [CH 3CO. ]  2k t [ .CH 3 ]2

         0  k p [ .CH 3 ][CH 3CHO]  k ,p [CH 3CO. ]

• Sum of the above two equations equals:
                ki [CH 3CHO]  2kt [.CH 3 ]2  0

• thus the steady state concentration of [.CH3] is:
                                             1/ 2
                                      k    
                          [.CH 3 ]   i
                                      2k   
                                                   [CH 3CHO ]1 / 2
                                      t    

•    The rate of formation of CH4 can now be expressed as
                                                                  1/ 2
              d [CH 4 ]                                 k    
                         k p [ .CH 3 ][CH 3CHO]  k p  i
                                                        2k   
                                                                        [CH 3CHO]3 / 2
                 dt                                     t    
    the above result is in agreement with the three-halves order
     observed experimentally.
• Example: The hydrogen-bromine reaction has a complicated rate
   law rather than the second order reaction as anticipated.
                 H2(g) + Br2(g) → 2HBr(g)
 Yield
                       k[ H 2 ][Br2 ]3 / 2
                   v
                      [ Br2 ]  k '[ HBr ]

The following mechanism has been proposed to account for the above
  rate law.
     1. Initiation:     Br2 + M → Br. + Br. + M              ki
     2. Propagation:    Br. + H2 → HBr + H.                  kp1
                        H. + Br2 → HBr + Br.                 kp2
     3. Retardation:    H. + HBr → H2 + Br.                  kr
     4. Termination:    Br. + Br. + M → Br2 + M*             kt
derive the rate law based on the above mechanism.
• The net rates of formation of the two intermediates are
                d[ H . ]
                          k p1 [ Br . ][ H 2]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr ]
                  dt
         d [ Br . ]
                     2k i [ Br2 ][ M ]  k p1[ Br . ][ H 2 ]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr ]  2k t [ Br . ]2 [ M ]
            dt
• The steady-state concentrations of the above two intermediates can
  be obtained by solving the following two equations:
                       k p1 [ Br . ][ H 2]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr ]  0

                  2k i [ Br2 ][ M ]  k p1 [ Br . ][ H 2 ]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr ]  2k t [ Br . ]2 [ M ]  0
                                                                1/ 2
                                                       k     
                                            [ Br . ]   i
                                                       k     
                                                                      [ Br2 ]1 / 2
                                                        t    

                                                   k p1 (k i / k t )1 / 2 [ H 2 ][ Br2 ]1 / 2
                                        [H ] 
                                             .
                                                        k p 2 [ Br2 ]  k r [ HBr ]

•   substitute the above results to the rate law of [HBr]
                       d[ HBr ]
                                 k p1[ Br . ][ H 2 ]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr ]
                          dt
                                                            1/ 2            3/ 2
                                d [ HBr ] 2k p1 (k i / k t ) [ H 2 ][ Br2 ]
                                         
                                    dt      [ Br2 ]  (k r / k p 2 )[ HBr ]
continued




• The above results has the same form as the empirical rate law, and
  the two empirical rate constants can be identified as
                                   1/ 2
                             k                               kr
                    k  2k p  i                      k, 
                             k                              k p2
                              t   

• Effects of HBr, H2, and Br2 on the reaction rate based on the
  equation
                                             1/ 2            3/ 2
                 d [ HBr ] 2k p1 (k i / k t ) [ H 2 ][ Br2 ]
                          
                     dt      [ Br2 ]  (k r / k p 2 )[ HBr ]
• Self-test 23.1 Deduce the rate law for the
  production of HBr when the initiation step
  is the photolysis, or light-induced
  decomposition, of Br2 into two bromine
  atoms, Br.. Let the photolysis rate be v =
  Iabs, where Iabs is the intensity of absorbed
  radiation.
• Hint: the initiation rate of Br. ?
Exercises 23.1b: On the basis of the
   following proposed mechanism, account
   for the experimental fact that the rate law
   for the decomposition
         2N2O5(g) → 4NO2(g) + O2(g)
               is v = k[N2O5].

(1) N2O5 ↔ NO2 + NO3          k 1, k 1’
(2) NO2 + NO3 → NO2 + O2 + NO k2
(3) NO + N2O5 → NO2 + NO2 + NO2 k3

								
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