Page 85 - Modeling of Chemical Kinetics and Reactor Design
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Reaction Mechanisms and Rate Expressions  55

                              If the chain length is large, it follows that

                                 kC    *  >>  k C                                       (1-239)
                                  3
                                             1
                                                 2
                                     2
                                    CH 5        CH 6
                              and Equation 1-236 becomes
                                 kk C 2    −  k k C 2  =  0                             (1-240)
                                  3 5    *    14  CH 6
                                                    2
                                       2
                                      CH 5
                              Thus,
                                         kk  05 .
                                 C   *  =   14    C                                   (1-241)
                                                    2
                                   2
                                  CH 5   kk      CH 6
                                           3 5
                                Substituting Equation 1-241 into Equation 1-237 gives


                                                                .
                                           dC            kk   05
                                               2
                                 + ( r  ) =   CH 4  = k    14    C
                                    2
                                   CH 4       dt       3  kk     CH 6
                                                                    2
                                                           3 5
                                   kk k   05 .
                                 =   13 4   C                                         (1-242)
                                    k 5      CH 6
                                                2
                              The rate of decrease in ethane C H  is:
                                                             2  6
                                             dC
                                                 2
                                 − ( r  ) =−    CH 6
                                     2
                                    CH 6        dt
                                  dC
                                 −   CH 6  = kC     + k C   * C    + k C  * C    − k C  * C  *
                                      2
                                                                                    5
                                                                      4
                                                       2
                                             1
                                                2
                                                                             2
                                                               2
                                     dt        CH 6      CH 3  CH 6     H   CH 6      H   C H 5
                                                                                           2
                              which together with Equation 1-232 yields
                                              dC
                                                  2
                                 − ( r  ) =−    CH 6  = kC      + k C    *              (1-243)
                                                            2
                                    2
                                   CH 6         dt       1  CH 6   3  CH 5
                                                                       2
                                From Equations 1-237, 1-239, and 1-242, it follows that
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