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Reaction mechanism and reaction order (PowerPoint)

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					Reaction mechanism and reaction order
• Chemical reaction: single or sum of many steps =>reaction mechanism
• A step: (elementary reaction)
    –   one, two or three reacting molecules => molecularity
    –   unimolecular: isomerization, decomposition
    –   bimolecular: energy transfer, combination
    –   termolecular: energy transfer + combination
• Reaction order vs. molecularity
    – Reaction takes place in an elementary reaction
         • unimolecular: first order
         • bimolecular: second order
         • termolecular: third order


    – Is a first order reaction is always an unimolecular reaction?
            Consecutive reaction processes
• RX + H2O = ROH + H+ + X-
    – Unaffected by the PH of solution, SN1 reaction
    – RX  R+ + X-
    – R+ + H2O  ROH + H+


• RX + OH- = ROH + X-
    – Alkaline solution
    – RX  R+ + X-
    – R+ + OH-  ROH


• What happen for 2nd order reaction?
          Consecutive reaction processes

• A  B <k1>,                 B  C <k2>
  – -d[A]/dt = k1 [A]; -d(a-x)/dt = k1 (a-x)

  – d[B]/dt = k1(a-x) - k2[B] = k1 a e-k1t - k2[B]
  – d[B]/dt + k2[B] = k1 a e-k1t



     [ B] 
                 ak1
              k 2  k1
                                     
                       e  k1t  e k2t , k1  k 2 , [ B]  0..t  0
  Plot of concentrations of A, B, C
        in consecutive process
                Consecuti

       1

      0.8

      0.6                               [A]
1/a




                                        [B]
      0.4                               [C]
      0.2

       0
            0    20           40   60
                       time
                         Maximum [B]


d [ B]
 dt
       
            ak1
         k 2  k1
                                      
                  k 2 e  k2t  k1e  k1t  0


        1        k1 
tm          ln  
     k1  k 2  k 2 
                 
     Formation of an intermediate complex

• A + B  X (1) X  A + B(2) X  C + D(3)
• Principles of Stationary state
   – 1913 Chapman: the net rate of formation of a reaction intermediate
     may be put equal to zero.
   – X so reactive, [X] is low


      d[ X ]
                k1[ A][ B ]  k 1[ X ]  k 2 [ X ]  0
        dt
                                k k [ A][ B ]
      d [C ] / dt  k 2 [ X ]  1 2
                                  k 1  k 2
    Formation of an intermediate complex

• A + B  X .. k1; X  A + B … k-1
• X + C  P + Q … k2
   d[ X ]
           k1[ A][ B ]  k 1[ X ]  k 2 [ X ][C ]
     dt
   d [ P]                  k k [ A][ B ][C ]
           k 2 [ X ][C ]  1 2
    dt                        k 1  k 2 [C ]

   Two limiting cases: k-1 > k2[C], k-1 < k2[C],
     Third body effect in atomic recombination

• Two atoms molecule + lot of vibrational energy
  – short life time production: needs a collision with other
    molecular species to shed some excessive energy
• X + X + M  X2 + M
  – before first vibration (10-13 s)  effective collision
  – effective molecule=highly negative activation energy
     • higher temperature: lower reaction rate
  – efficiency is different depend on molecule
     • large molecule
            Porter’s interpretation(1961)
• I + M = I.M (K1); I.M + I  I2 + M(k2)
   – I.M: spectroscopically identified for benzene
   – [I.M]=K1[I][M]
   – d[I2]/dt = k2[I][I.M] = K1k2[I]2[M]
       • RT ln K = -G = -(H-TS)
       • k = A exp[-Ea/RT]

              d ln K1k 2       d ln K1     d ln k 2 
     Ea   R                R            R
               d (1 / T )        d (1 / T )    d (1 / T )
            
      H1  E2
                     Parallel reaction

• AB (1)                         AC (2)
   – -d[A]/dt = (k1 + k2)[A]
   – (a-x)=a exp[-(k1+k2)t]
   – k1/k2 = Yield of B/Yield of C              B
• AC (1), A+BD (2)
   – -d[A]/dt = (k1 + k2[B])[A]             A
• AB (1), A+AC (2)                            C
       Reactant participating in equilibria


• H2 + I2 = 2HI
   – Bodenstein 1899
   – first order with respect to each reactants
   – Bimolecular process, elementary reaction, four center
     transition state

   – Sullivan 1967: 0-200C ; non elementary reaction
   – I2 + h  I2*, I2*  2 I       2 I + H2  2 HI
        A probable mechanism for H2 + I2

• I2 = I + I                              …        (1)
• I + I+ H2  HI + HI                     …        (2)


• K1 = [I]2/[I2]             [I]=K11/2[I2]1/2
• d[HI]/dt = 2k2[H2][I]2 = 2k2K1[H2][I2]

   – Ea(k2) = 21 kJ/mole
   – Ea(k2) + Bond dissociation energy of I2 == Overall activation
     energy of Bodenstein thermal experiment.
        A probable mechanism for H2 + I2

• I2 = I + I                          …         (1)
• I + H2 = IH2                        …         (2)
• I + IH2  HI + HI           …       (3)

• K1 = [I]2/[I2]             [I]=K11/2[I2]1/2
• K2 = [IH2]/[I][H2] [IH2]=K2 [I][H2]
• d[HI]/dt = 2k3[IH2][I] = 2k3K1K2[H2][I2]
             Hydrogen-halogen kinetics

• H2-F2, H2-Br2, H2-Cl2
   – complicated experimental rate law
   – interpreted in terms of free radical chain reaction
• H2-I2
   – simple kinetic behavior
   – the least understood system
         Reactant participating in equilibria

• nA=An
  – [A]=Kn-1/n [An]1/n = Kn-1/n [c/n]1/n

  –   [A]: a part of aggregate,
  –   [aggregate]  [normal concentration of A]/n
  –   Butyl lithium initiator of anionic polymerization, rate  initiator 1/6
  –   butyl lithium associate to hexamer
      Reactant participating in equilibria

• A = B- + H+ (Ka)
  – [B-]=Ka[A]/[H+]; [A] = [H+] [B-]/ Ka.;

  – [A] + [B-] = [A]{([H+] + Ka)/[H+]}
  –            = [B]{(Ka + [H+])/Ka} => a
  – [B-] = a·Ka/(Ka + [H+])
  – [A] = a·[H+]/(Ka + [H+])


  – We can formulate rate equation by [A] & [B-]
  – PH dependent
               Opposing reaction

• AB        (k1)            B A               (k-1)
 – dx/dt = k1(a-x) -k-1 x = k1a - k1x - k-1 x

 – final equilibrium.
 – k1(a-xe) = k-1 xe         k1a = (k1 + k-1)xe

 – dx/dt = (k1 + k-1)(xe-x)
 – ln {(xe)/(xe-x)} = (k1+k2)t
            Isotope exchange reaction
• AX + BX* = AX* + BX
  –   Assume      K=1  G = -RT ln K = 0
  –   [AX]t=0 = a; [BX*]t=0 = b;    [AX*]t=0 = [BX]t=0 = 0
  –   At equilibrium
  –   x2/(a-x)(b-x) = 1  x = ab/(a+b)
  –   [AX]= a-x = a2/{a+b}, [BX*]= b-x = b2/{a+b}
  –   [AX*]= ab/{a+b}, [BX]= ab/{a+b}
  –   [AX]t+ [AX*]t = a, [BX]t+ [BX*]t = b


      dx      a  x  b  x  x 2 
                                    
                             
      dt      a  b  ab 
                                    
       dx      a  x  b  x  x 2 
                           
       dt      a  b  ab 
– = {(a+b)/ab}(xe-x)
– Integration with x=0 at t=0
– ln{xe/(xe-x)} = {(a+b)/ab}t
– no information on reaction order for 
    • Determined by initial rate with different a and b
    •  = kn apbq
– ln{xe/(xe-x)} = kn ap-1bq-1 (a+b)t

				
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posted:3/10/2012
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