Page 166 - Modeling of Chemical Kinetics and Reactor Design
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136    Modeling of Chemical Kinetics and Reactor Design

                                Generally, these occur simultaneously.  The relative amount of
                              ethylene and acetaldehyde obtained therefore depends on the relative
                              speed of the two reactions competing with each other for the available
                              alcohol. These speeds, in turn, are determined by the choice of catalyst
                              and temperature.
                                Parallel reactions often have mechanisms of different reaction
                              order.  An example is the solvolysis reactions in the presence of
                              added nucleophiles:

                                                                          +
                                                     
                                 C H SO Cl +  H O  →     C H SO OH +    H +  Cl −
                                                      k 1
                                  6
                                                              5
                                     5
                                        2
                                                2
                                                                  2
                                                            6
                                 C H SO Cl +  F −  k 2 2  →  C H SO F +  Cl −           (3-106)
                                        2
                                     5
                                  6
                                                          6
                                                                2
                                                             5
                                                  HO
                                In the solvolysis reaction, H O is the solvent as well as a reactant,
                                                           2
                              and it is present in large excess. It does not appear in the rate equation,
                              which is expressed by
                                 A →    C                                              (3-107)
                                    
                                     k 1
                                 AB+ →     D                                           (3-108)
                                        
                                        k 2
                                The rate equation is:
                                 − ( r  )  =−  dC A  = kC  + k C C                      (3-109)
                                   A net           1  A  2  A  B
                                            dt
                                An excess of component B is used, and its concentration remains
                              constant, so that the rate equation reduces to a first order. That is,
                                 −  dC A  =KC A                                         (3-110)
                                          e
                                   dt
                              where

                                K  = k  + k C                                           (3-111)
                                  e    1    2 B
                                Measurements of K  at different concentrations of B will enable the
                                                  e
                              rate constants for each of the reactions to be determined. Parallel
                              reactions give a reduction in the yield of the product.  This can be
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