Page 202 - Modeling of Chemical Kinetics and Reactor Design
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172    Modeling of Chemical Kinetics and Reactor Design

                                Consider a mole balance on a constant volume batch reactor repre-
                              sented by


                                 −  dC A  = kC C b B                                    (3-227)
                                            a
                                            A
                                   dt
                              Using the method of initial rates gives

                                  dC A               a   b
                                  −
                                                 ) = kC
                                   dt   O  =− ( r A O  AO C BO                        (3-228)

                              Taking the logarithms of both sides of Equation 3-228 gives

                                   
                                                  +
                                 ln −  dC A   =  ln ka  ln C  + b  lnC                 (3-229)
                                      dt               AO        BO
                              This can be represented in the form


                                Y = C  + C X  + C X                                     (3-230)
                                      0     1  1   2  2
                              where Y = ln (–dC /dt), C  = ln k, X  = ln C  , X  = ln C  , C  = a,
                                               A      O         1       AO   2      BO   1
                              and C  = b. If N experimental runs are performed, then for the ith
                                    2
                              run, Equation 3-230 becomes

                                Y  = C  + C X  + C X   2,i                              (3-231)
                                               1,i
                                  i
                                            1
                                       O
                                                     2
                              where X  = ln C  AOi , with C AOi  being the initial concentration of A
                                      1i
                              for the ith run. Solving for the unknowns C , C , and C  for N experi-
                                                                          1
                                                                                 2
                                                                      O
                              mental runs, i = 1, 2, 3 . . . N, gives
                                 N                N          N
                                 ∑ Y =  NC +   C 1 ∑ X +  C 2 ∑ X  i 2                  (3-232)
                                     i
                                                       i 1
                                           O
                                 i=1              i=1        i=1
                                 N            N          N          N
                                 ∑ XY =   C O ∑ X +   C 1∑ X +   C 2 ∑  X X  i 2        (3-233)
                                                             2
                                                             i 1
                                                                         i 1
                                                   i 1
                                        i
                                      i 1
                                                         =
                                 =
                                                                    =
                                              =
                                 i 1          i 1        i 1        i 1
                                 N             N          N              N
                                 ∑ XY =    C O ∑ X  i 2  +  C 1∑  X X  i 2  +  C 2 ∑ X 2 i 2  (3-234)
                                        i
                                                              i 1
                                      i 2
                                                          =
                                               =
                                 =
                                 i 1          i 1        i 1            i 1=
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