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					 Chemical Kinetics
        Jens Poulsen
         5 x 2 hours



        AB
          Atkins
Quanta, matter, and Change.
     Chapters 19-21.
A reaction may be investigated
     on several levels…
              stochiometry:
  2 N 2O5 ( g )  4 NO2 ( g )  O2 ( g )
                rate law:
       v  k  [ N 2O5 ( g )]
               atomistic:
  N 2O5  NO2  NO3 k a
  NO2  NO3  N 2O5 k a
                            '


  NO2  NO3  NO2  NO  O2 kb
  NO  N 2O5  NO2  NO2  NO2 kc
   Chapter 22: Rates of chemical
       reactions. Concepts:
 Defining reaction   A B v 
                                   d [ A]
                                               k  [ A]
                                      dt
  rates.
 Integrated rate           [ A]  [ A]0  exp( kt)


  laws.                 A B C
 Elementary
  reactions.          A B C
 Consecutive
  reactions.
   Start with chemical kinetics
      on an empirical level:
2 N 2O5 ( g )  4 NO2 ( g )  O2 ( g )

  Rate equations: How much N2O5 (g) as a function of time etc?

           No information about reaction mechanism
                     (on an atomic scale).
  Experimental techniques
 concentrations can be followed by...

      Measuring pressure during
         chemical reaction:
    2 N 2O5 ( g )  4 NO2 ( g )  O2 ( g )

     Measuring conductivity during
         chemical reaction:
(CH 3 )3 CCl (aq)  H 2O(l )
                                        
 (CH 3 )3 COH (aq)  H (aq)  Cl (aq)
         Example 19.1
monitoring variation in pressure.
 2 N 2O5 ( g )  4 NO2 ( g )  O2 ( g )
       : fraction of N 2O5 that has decomposed.


  Amount :      N 2O5        NO2     O2          l
                                              tota
                                  1         3
              n(1   ) 2n         n n(1   )
                                  2         2
                                   3
                      p  p0 n(1   )
                                   2
  Def. of v - rate of reaction
2 HOIO (aq)  IO  3 (aq)  H  (aq)  HOI (aq)
                            
      1 d [ HOIO ]   d [ IO3 ]   d[H  ]   d [ HOI ]
  v                                  
      2      dt           dt       dt           dt

                  more generally
             aA  bB  cC  dD
        1 d [ A]    1 d [ B]   1 d [C ]   1 d [ D]
   v                                 
      ( a ) dt    (b) dt     c dt       d dt
     Reaction order
     A  B  ..  C  ...
Often has reaction rate of form

     v  k  [ A] [ B]  ...
                 m    n


 Order m with respect to A,..
   Overall order is m+n+..
                         Examples
                     first order reaction

           C2 H 6 ( g )  2CH 3 ( g ) v  k[C2 H 6 ( g )] T  700 0 C


                 second order reaction
2 NOBr ( g )  2 NO ( g )  Br2 ( g ) v  k[ NOBr ( g )]                2




                zero order reaction
        N 2  3H 2 on Iron  2 NH 3 ( g ) v  k
      Determination of rate law
   A  B  ..  C  ...    v  k  [ A]m [ B]n  ...

 isolation method: excess of B, C, …
  m can be determined etc.

 method of initial rates: vary the conc. of A,
B, C, .. In turn and check how rate changes.
                 Integrated rate laws
 First order rate

 v
       d [ A]
          dt
                  k  [ A]      [ A]  [ A]0  exp( kt)

 straight line: ln([ A] /[ A] )  kt
                                    0



 Second order rate
                                        1               1
                                  [ A]  [ A]0  kt
       d [ A]
v              k  [ A]   2

         dt
 straight line:                               1
                                   [ A]1  [ A]0  kt
           d [ A]
   v               k  [ A][ B]
             dt

                  leads to
    [ B] /[ B]0
ln(             )  kt  ([ B]0  [ A]0 )
    [ A] /[ A]0
Reactions approaching equil.
     A B         v  k[ A]
     BA          v  k '[ B]


     d[ A]
            k[ A]  k '[ B]
      dt
                  Solution:
                      k ' k exp((k  k ' )t )
       [ A]  [ A]0                                 for       [ B]0  0.
                               k ' k




                                                    k'
                 t   : [ A]eq  [ A]0
                                                  k ' k
                                                              k
                 [ B]eq  [ A]0  [ A]eq  [ A]0
                                                           k ' k
                         [ B]eq       k
                 K               
                         [ A]eq       k'




Relation between equilibrium constant
         and rate constants.
   Disturbing a system in eq:
                          k
                         
                        A B
                          k'

              Sudden change in k , k '
             e.g. by change in T.
[ A]  [ A]eq  x [ B]  [ B]eq  x
d [ A]
        k[ A]  k '[ B]  k ([ A]eq  x)  k ' ([B]eq  x)
 dt
d [ A]                     d [ A] dx
        kx  k ' x but         
 dt                         dt      dt
     dx
         (k  k ' ) x  x(t)  x(0)exp(-(  k ' )t )
                                          k
     dt




The rate to new equilibrium is
     given by the sum of
             k, k'

                  1
                       k  k'
                  
                 Example 19.4
      k'
                                                   2
H 2O(l ) H (aq)  OH (aq) KW  [ H ][OH ] / c
      
      k



                 37 s  37 10 s.
                                  6


              Calculate the rate constants      k, k'
             for water autoprotolysis, given
       Forward/backward reaction is first/second-
        order, respectively, and T=298 K, pH=7.

                    BLACK-BOARD!
Temperature dependence of
   chemical reactions.
A B C           v  k (T )[ A][B]

  What can be said about k(T) ?


   k (T )  A  exp(  Ea / RT )

       Arrhenius equation
Svante Arrhenius
Nobel prize 1903.
                 interpretation
k (T )  A  exp(  Ea / RT ) A  B  C     v  k (T )[ A][B]




       *Reaction coordinate: molecular distortion
      along which the reactants become products.
     *Transition state = activated complex = climax
                        of reaction.
         *Once the reactants have passed the
                transition state, products
                        are formed.
            interpretation
     k (T )  A  exp(  Ea / RT )




 Ea: activation energy (change in potential energy).
Only molecules having kinetic energy larger than Ea
                  get over barrier.
          exp(-Ea/RT): fraction of reactants
   having enough kinetic energy to pass barrier
  A: pre-exponential factor: measure of the rate of
                      collisions
k (T )  A  exp(  Ea / RT )   A B C   v  k (T )[ A][B]




 exp(-Ea/RT): fraction of reactants having
kinetic energy higher than barrier height Ea.

 A: proportional to collision frequency.

 K(T) = collision frequency times
  fraction of successful collisions
                 = A * exp(-Ea/RT)
   Accounting for the rate laws.
 Elementary reactions: The fundamental
  “building blocks” of chemical reactions.
 Describes what happens on an atomic
  scale.
 CH 3 I  CH 3CH 2O   CH 3OCH 2 CH 3  I 


 one ethanoate ion collides with one
 methyliodide molecule and forms one iodide
 ion plus one methylethylether….
        Elementary reactions
 The molecularity of a reaction: # of
  molecules participating.
 Unimolecular: one molecule.
 Bimolecular: two molecules.
 A unimolecular reaction is first-order.
 A bimolecular reaction is second-order.
            A P     d [ A] / dt   k[ A]
            A B  P      d [ A] / dt   k[ A][ B ]
        Elementary reactions
 If we know a reaction is single-step and
  bimolecular we can write down a rate
  equation directly:

             A B  P   d[ A] / dt  k[ A][B]




 Same applies to unimolecular reactions...
        Elementary reactions
 On the other hand, consider the reaction
                A( g )  B( g )  P( g )


 A rate law of form
                     d[ A] / dt  k[ A][B]


 does not imply the reaction to be simple bi-
 molecular:
                       A B  P
       Example.

v  k[ HBr ( g )][ O2 ( g )]
Consecutive elementary
      reactions
          ka   kb
         A I  P




 example: enzyme/substrate
  Solving time-dependence of
              [P].
                  ka   kb
                 A I  P




             BLACK-BOARD!

        ka exp(kbt )  kb exp(kat ) 
[ P]  1 
                                      [ A]0
                                       
                 kb  k a             
Consecutive reactions




Qualitative time-dependence of
         [A], [I] and [P].
         Steady State approx.
         VERY IMPORTANT!!
 The higher the molecularity, the more
  complex mathematics when solving rate
  equations. Need approximation.
 Introduce steady state approximation:
               ka              kb
           A I  P

                        d[ I ]
             [I ]  0          0   kb  ka
                         dt
 Steady state approx. continued.
  More generally: all intermediates I , I ,..., I        1   2   n




are assumed to have negligible concentration
  and rate of change of concentration:
             [ I1 ]  0, [ I 2 ]  0,... , [ I n ]  0



             d [ I1 ]      d[ I 2 ]           d[ I ]
                       0,           0,....., n  0
               dt           dt                 dt
    Apply steady state approx.
     to consecutive reaction.
                  ka       kb
              A I  P

d[ I ]
        k a [ A]  kb [ I ]  0  [ I ]  k a [ A] / kb
 dt
d [ P]
        kb [ I ]  k a [ A]  k a exp(k a t )[ A]0
 dt
d [ P]
        kb [ I ]  k a [ A]  k a exp(k a t )[ A]0 
 dt
[ P]             s

  
[ P ]0
         d [ P]   dtk a exp(k a t )[ A]0 
                 0

[ P]  [ P]0  (1  exp(k a t ))[A]0 
[ P]  (1  exp(k a t ))[A]0
           Steady state example.
         2 N 2O5 ( g )  4 NO2 ( g )  O2 ( g )
             N 2O5  NO2  NO3 k a
             NO2  NO3  N 2O5 k a
                                     '


             NO2  NO3  NO2  NO  O2 kb
             NO  N 2O5  NO2  NO2  NO2 kc

       Two intermediates: NO and NO3.
d [ NO3 ] / dt  0  0  k a [ N 2O5 ]  (k a  kb )[NO2 ][NO3 ]
                                            '


d [ NO] / dt  0  0  kb [ NO2 ][ NO3 ]  kc [ NO][N 2O5 ]
           d [ N 2O5 ] / dt  ka [ N 2O5 ]  ka [ NO2 ][NO3 ]
                                              '


            kc [ NO][N 2O5 ]




          Hence..[exercise]:

d [ N 2O5 ] / dt  2k a kb [ N 2O5 ] /(k  kb )                '
                                                                a
Rate determining step
 Example. when ka<kb then
       ka        kb
     A I  P
 may be treated as a simple
          reaction:
            ka
        A P
  ”the principle of the rate
     determining step”
                    Proof:
               When ka<kb then

        ka exp(kbt )  kb exp(kat ) 
[ P]  1 
                                      [ A]0
                                       
                 kb  k a             
                    becomes

         [ P]  1  exp(ka t ) [ A]0
          [ A]0  [ A]t

                  as expected
 Rate determining step – in general

 If one elementary step in a reaction is
  slower than others then this step controls
  the rate of the overall reaction.
 The slow step is rate determining if it can’t
  be sidestepped.
 Once the rate determining step is found, the
  rate expression can be written down
  immediately.
                      Example.
                         ka     kb
                      A I  P

                         ka '   kb '
                      A  I ' P

 Even if ka<< kb the upper reaction may be sidestepped if
  ka<ka’ and ka<kb’. Then
                                        ka
                                       A I

  is not rate determining.
        Preequilibria.
                  ka     kb
                  
              A B I P
                  
                  ka '




Assume A and B are in eq. then

      [I ]         ka
K               K '
   [ A][B]         ka
d [ P] / dt  kb [ I ]  kb K [ A][B]
    Example. Analysing pre-eq. by
           steady state.
 we do not assume eq.

                               ka     kb
                               
                     A B I P 
                               ka '



     d [ P ] / dt  kb [ I ]
     d [ I ] / dt   kb [ I ]  k a [ A][B ]  k a [ I ]  0 
                                                  '


            k a [ A][B ]                  k a kb [ A][B ]
     [I ]                d [ P ] / dt 
             kb  k a'
                                              kb  k a
                                                     '
When pre-eq exists kb<ka’
                    ka    kb
                   
          A B I P
                   ka '




                 ka kb [ A][B] ka kb [ A][B]
   d [ P] / dt               
                    kb  k a
                           '
                                     ka'


   the old pre - eq analysis result.
         Unimolecular reactions.
         Lindemann Hinshelwood
              mechanism.
     cyclo  C3 H 6  CH 3CH  CH 2   v  k[cyclo  C3 H 6 ]



 Found to involve bimolecular step, how can it be
  first order?
 The reaction is not given by an elementary first
  order mechanism.
 Can be described by the Lindemann Hinshelwood
  mechanism.
Lindemann Hinshelwood

             ka
                           d [ A ]
  A  A  A  A                     k a [ A]2
                             dt
        ka '
                     d [ A ]
 A  A  A  A                k a ' [ A][ A ]
                       dt

        kb             
                 d[ A ]
   A P                   kb [ A ]
                   dt
    Use steady state...
  d [ A ]
            kb [ A ]  k a ' [ A][ A ]  k a [ A]2  0 
    dt
            k a [ A]2
  [A ]                    
           kb  k a ' [ A]

 d [ P]                   d[ P] kb ka [ A]2
         kb [ A ]  0       
  dt                       dt    kb  ka ' [ A]

If kb is small: kb << ka’ [A] then first-
           order reaction
                    d [ P ] kb k a [ A]
                           
                     dt        ka'
On the other hand at low
       pressure:
            d [ P]
                    k a [ A]2
             dt
      Since the reaction
       ka
                      d [ A ]
  A  A  A  A                k a [ A]2
                        dt

   becomes ”bottleneck”.
Define effective rate constant:
     d [ P] kb k a [ A]2      k k [ A]
                            b a            [ A]  K [ A]
      dt     kb  k a ' [ A] kb  k a ' [ A]
               kb k a [ A]
     with K 
              kb  k a ' [ A]


      Lindemann Hinshelwood
     mechanism gives linear plot

             1            d [ P] 1    1  ka '
         K         [ A] /        [ A] 
                            dt    ka      k a kb
The kinetics of complex reactions
 Chain reactions
 Explosions
 Catalysis (enzymes)
         What is a chain reaction?
 Example, thermal
  decomposition of ethanal: CH 3CHO ( g )  CH 4 ( g )  CO( g )
 Elementary reactions
  (Rice-Herzfeld):
initiation : CH 3CHO  CH 3  CHO  v  ki [CH 3CHO]
propagation : CH 3  CH 3CHO  CH 4  CH 3CO  v  k p [CH 3 ][CH 3CHO]
propagation : CH 3CO  CO  CH 3  v  k ' p [CH 3CO]
termination : 2CH 3   CH 3CH 3   v  kt [CH 3 ]2

 Chain carriers are CH3
  and CH3CO radicals
        Derivation of rate law:
 Steady state:
     d [CH 3 ]
                 ki [CH 3CHO]  k p [CH 3 ][CH 3CHO]
          dt
      k p [CH 3CO]  kt [CH 3 ]2  0
         '


     d [CH 3CO]
                  k p [CH 3 ][CH 3CHO]  k p [CH 3CO]  0
                                             '

          dt

 Add these together:
                                               ki
   0  ki [CH 3CHO]  kt [CH 3 ]  [CH 3 ] 
                                2
                                                  [CH 3CHO]
                                               kt
                       Insert into
                                              ki
0  ki [CH 3CHO]  kt [CH 3 ]  [CH 3 ] 
                            2
                                                 [CH 3CHO]
                                              kt

              d [CH 4 ]
                         k p [CH 3 ][CH 3COH ]
                 dt
                    ki
               kp      [CH 3COH ]3 / 2
                    kt

            As observed experimentally.
  Key elements of chain reaction
 Initiation: involves formation of chain carriers
 propagation: done by chain carriers
 Termination: chain carriers are ”destroyed”.
New example:
                     Br2 ( g )  H 2 ( g )  2 HBr ( g )
Initiated by heat.
       New example:
  Br2 ( g )  H 2 ( g )  2 HBr ( g )

     Complicated rate law:
   d [ HBr]     k[ H 2 ][Br2 ]3 / 2
            
       dt     [ Br2 ]  k '[ HBr]

Can be explained by postulating
 a chain reaction mechanism.
  Mechanism: collision induced
Initiation : Br2  M  2 Br   M
Propagation :   Br   H 2  HBr  H  v  k p [ Br][H 2 ]
Propagation :   H   Br2  HBr  Br  v  k ' p [ H ][Br2 ]
Retardation :   H   HBr  H 2  Br  v  k r [ H ][HBr]
Termination :   Br   Br  Br2    v  kt [ Br]2


          Retardation step: A product is
                      removed.
        Chain carriers are H and Br radicals.
              Steady state analysis.
                         Rate of formation
  d [ HBr]
            k p [ H 2 ][Br]  k p '[ H ][Br2 ]  kr [ HBr][H ]
      dt

                            Steady-state:

d [ H ]
          k p [ H 2 ][Br]  k p '[ H ][Br2 ]  k r [ HBr][H ]  0
   dt
d [ Br]
          2ki [ Br2 ][M ]  k p [ H 2 ][Br]  k p '[ H ][Br2 ]  k r [ HBr][H ]
    dt
 2kt [ Br]2 [ M ]  0
       Adding these two equations give

                                                    ki
2ki [ Br2 ][M ]  2kt [ Br] [ M ]  0  [ Br] 
                          2
                                                       [ Br2 ]1/ 2
                                                    kt


        Inserting the expression for Br
           radical into first eq. gives
                              ki        1/ 2
                      kp         [ Br2 ] [ H 2 ]
                              kt
             [ H ] 
                      k ' p [ Br2 ]  k r [ HBr]
   Substituting into rate expression gives
d [ HBr]
            k p [ H 2 ][Br]  k p '[ H ][Br2 ]  k r [ HBr][H ]
    dt
           ki
    2k p       [ H 2 ][Br2 ]3 / 2
           kt

  [ Br2 ]  ( k r / k ' p )[HBr]


      Which has the same form as the
      experimental rate law provided
                         ki
         k  2k p               k '  (k r / k ' p )
                         kt
                   Explosions
 An explosion is a rapidly
  accellerating reaction arising
  from a rapid increase in reaction
  rate with increasing
  temperature


  fast reaction releasing heat (exothermic)  T rises 
  even faster reaction  T rises even more  ..
                     Example

          O2 ( g )  2 H 2 ( g )  2 H 2O( g )

 Has not been fully understod. # Chain carriers
             grow exponentially.

initiation : H 2  H   H  v  const  vinit
propagation : H 2  OH   H   H 2O v  k p [ H 2 ][OH ]
branching : O2  H   O   HO  v  kb [O2 ][H ]
             O   H 2  H   HO  v  k 'b [O][H 2 ]
                          1
termination : H   wall  H 2 v  kt [ H ]
                          2
              H  O2  M  HO2   M  v  kt '[ H ]
 chain carriers are     H , OH ,  O 
 Branching: one chain
  carrier becomes two or
  more.

 Occurence of explosion given by explosion region
                (regions of T,p).

      If T is low, rate constants are
       too small.
      If p is low, the reaction rates, v,
       are too low due to infrequent
       collisions.
                     Example
      Show that an explosion happens
       when rate of chain branching
     exceeds that of chain termination.
                   Answer:
      Focus only on the rate of prod of H 
     chain carrier as important. Rate of
          change of chain carrier:
d
    [ H ]  vini  k p [ H 2 ][ HO ]  kb [ H ][ O2 ] 
dt
kb ' [O ][ H 2 ]  kt [ H ]  k 't [ H ][ M ][ O2 ]
                             Example
             Steady state approximation:
    d
      [OH ]   k p [OH ][H 2 ]  kb [O2 ][H ]  k 'b [O][H 2 ]  0
   dt
    d
      [O]  kb [O2 ][H ]  k 'b [O][H 2 ]  0 
   dt
              k
   [OH ]  2 b [O2 ][H ] /[ H 2 ]
              kp
   [O]  kb [O2 ][H ] / k 'b [ H 2 ]

     Then chain carrier production rate is

d
   [ H ]  vini  (2kb [O2 ]  kt  k 't [ M ][O2 ])[H ]
dt
                           Example
                               write:
         kbranch  2kb [O2 ] measuring chain branching rate.

         kterm  kt  kt' [O2 ][ M ] measuring terminati on rate.

            Then chain carrier production rate is
          d
             [ H ]  vini  (kbranch  kterm )[H ]
          dt
                   vinit
a ) [ H ]                  (1  e ( kt e rm  kbranc h ) t ) kbranch  kterm
             kterm  kbranch
                 vinit
b) [ H ]                  (e ( kbranc h  kt e rm ) t  1) kbranch  kterm
            kbranch  kterm
      Homogeneous catalysis
 A catalyst lowers the activation energy for a
  reaction.
 It is not consumed in the reaction (reforms in
  the end).
 Examples are enzymes, metal catalysts etc.
 Homogeneous catalysis: the catalyst exists
  in same phase as reactants.
                   Example:
2 H 2O2 (aq)  2 H 2O(l )  O2 ( g )
              as catalyzed by bromide ions.
                     
H 3O  H 2O H 2O  H 3O2
            2
                                        
                              K  [ H 3O2 ] /[ H 3O  ][H 2O2 ]
                                     
H 3O2  Br   HOBr  H 2O v  k[ H 3O2 ][Br  ]
HOBr  H 2O2  H 3O   O2  Br     ( fast)

        Notice that bromide reappears in the end.
                   Enzymes   E  S  ES  P  E

 Enzymes (E) are
  biological catalysts.
 Typical proteins.
 Contains an active site
  that binds substrate
  (S):
 Induced fit model: S
  induces change in E
  and
  only then do they fit
  together.
  Michaelis Menten mechanism.
     Three exp. observations
 i) For a given amount of S, the rate of product
  formation is proportional to [E].
 ii) For a give amount of E and low values of [S],
  the rate of product formation is proportional to [S].
 iii) For a given amont of E and high values of [S],
  the rate of product formation reaches a max.
  value, v(max), and becomes independent of [S].
  i) and ii) - but not iii) - is consistent with
         E  S  P  E v  k[S ][E ]

   Michaelis-Menten explains i)-iii)
                   
             E  S ES ka , k 'a
                   

             ES  P  E kb

                    leads to
                   kb [ E ]0          [ E ][ S ]
v  kb [ ES ]                   KM 
                1  K M /[S ]0          [ ES ]
                                   Proof.
                                     d [ ES]
   Steady state approx.:                      ka [ E ][S ]  k 'a [ ES]  kb [ ES]  0
                                        dt
                               k a [ E ][ S ]
   then             [ ES ] 
                                k 'a  k b

                               k 'a  kb [ E ][ S ]
                     KM                
   Define                         ka       [ ES ]


Use [ E ]0  [ E ]  [ ES ] [ S ]  [ S ]0 substrate in excess

    [ E ]0  [ ES ](1  K M /[S ])  [ ES ](1  K M /[S ]0 )
                   then

     [ ES ]  [ E ]0 /(1  K M /[S ]0 )

v  kb [ ES ]  kb [ E ]0 /(1  K M /[S ]0 )

    Then i)-iii) is correct. show!
       Alternative formula:              vm ax  kb [E ]0

                         vm ax
    v  kb [ ES ]                   
                    1  K M /[S ]0
    1   1     1 KM
          
    v vm ax [ S ]0 vm ax

Plot of 1/v against 1/[S] gives straight
       line for MM mechanism.
              Plot is called
         Lineweaver-Burk plot
        Determine enzyme parameters.

 Lineweaver-Burke plot
  gives   vm ax  kb [E ]0          v  kb [ ES ] 
                                                           vm ax
                                                                       
                                                      1  K M /[S ]0
  and
          K m / vm ax               1
                                      
                                        1
                                           
                                              1 KM
                                    v vm ax [ S ]0 vm ax
 From knowing total
  enzyme concentration
  we calc. K m , kb

 Cannot determine k a ,     k 'a
             Catalytic constant.
 k b gives the number
  of ”turn overs” pr. unit        
  time. This number –
                             E  S ES k a , k 'a
  with units of pr time –    ES  P  E kb
  is called the catalytic
  constant, k
               cat
     Catalytic efficiency.
                 
           E  S  ES         k a , k 'a
           ES  P  E             kb
Catalytic efficiency: measure of overall
        enzym efficiency equals
  effective rate constant for overall
                 reaction
                            ka          kb
 v  [ES] kb  [ E ][S ]           kb     [ E ][S ]
                         k 'a  kb      KM
    kb
 
    KM
 If enzyme is very              
  efficient then kb is      E  S ES k a , k 'a
                                 
  large. Then               ES  P  E kb
         ka
              kb  k a
      k 'a  kb

 ka is determined by
  diffusion.                                1 1
                            k a  10  10 M s
                                     8   9
          Enzyme inhibition
 An inhibitor reduces rate of product
  formation by binding to E, ES or both,
  thereby decreasing [ES] and hence product
  formation.
                   
             E  S  ES k a , k 'a
             ES  P  E kb
               
             EI E  I , K I  [ E ][I ] /[ EI ]
                                 [ ES ][I ]
             ESI ES  I , K ' I 
                 
                                    [ ESI ]
  New rate of product formation.
                          vmax
     v  kb [ ES] 
                       ' K M /[S ]0
       1  [ I ] / K I  '  1  [ I ] / K 'I   vmax  kb [ E ]0

 Maximum velocity is no longer obtainable
  when competitive inhibition occurs.
                             Proof:
 Introduce conservation
  of enzyme molcs:
         [ E ]0  [ E ]  [ EI ]  [ ES ]  [ ESI ]
            then
           [ E ]0  [ E ]  [ ES ] '
            since
            K M  [ E ][S ] /[ ES ]  [ E ][S ]0 /[ ES ]
           [ E ]0  K M [ ES ] /[S ]0  [ ES ] ' 
           [ ES ]  [ E ]0 /( ' K M  /[ S ]0 )
[ ES ]  [ E ]0 /( ' K M /[ S ]0  )
v  kb [ ES ] 
v  kb [ E ]0 /( ' K M  /[ S ]0 )
v  vm ax /( ' K M  /[ S ]0 )
    Different types of inhibition
 Case 1:   1  '  1 meaning only EI not ESI
  is formed. Once EI is formed, S cannot bind
  to E. Termed competitive inhibition.
  Case 2:   1  '  1 meaning that only ESI
  not EI is formed. I can only bind to E if S is
  already present i.e. ES. ESI does not lead to
  product. Termed uncompetitive inhibition.
    Different types of inhibition
 Case 3:   1  '  1 meaning that both EI
  and ESI are formed. Inhibitor binds to a site
  different from the active site in both cases.
  EI cannot bind S. Termed non-competitive
  inhibition.
          How to determine  ,  '
 Do a unhibited exp. with S & E and find
  vmax and Km.
 Add inhibitor with known conc. and plot
  Lineweaver-Burk from which  ,  ' and
  thereby K I , K 'I can be determined.
     Example of use of competitive
              inhibition.
 To kill bacteria (B). B
  has enzyme dihydro-
  pteroate synthease
  producing folate which
  is crucial for survival of
  B.
 Active substrate p-
  aminobenzoic acid.
 Inhibitor Sulfanilamide.

				
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