Page 233 - Modeling of Chemical Kinetics and Reactor Design
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Reaction Rate Expression  203

                                                                     A                    B
                              Amount at t = 0                       C                    C
                                                                      AO                  BO
                              Amount at t = t                        C                   C
                                                                      A                    B
                              Amounts that have reacted          C   – C  =            C   – C
                                                                   AO   A               BO   B
                              In terms of the fractional conversion of B, X B


                                 X =  C BO  − C B
                                  B                                                     (3-299)
                                        C BO

                              Equation 3-299 is further expressed as:

                                 C =  C (1 − X )                                        (3-300)
                                       BO
                                  B
                                              B
                              Since the concentration of A and B is the same, then

                                 C = C =  C (1  − X )                                   (3-301)
                                       B
                                           BO
                                  A
                                                   B
                              and
                                 −  dC B  = C BO  dX B                                  (3-302)
                                    dt         dt

                                Substituting Equations 3-300 and 3-302 into Equation 3-298 gives

                                            dX B     2        2
                                 − ( r  ) =C    = kC     − (1  X  )                     (3-303)
                                   B     BO          BO     B
                                             dt
                                Rearranging Equation 3-303 and integrating between the limits t =
                              0, X  = 0 and t = t, X  = X  gives
                                                    B
                                                         B
                                  B
                                 X B  dX           t
                                 ∫     X ) 2  =  kC BO∫  dt                             (3-304)
                                        B
                                 0  ( 1−  B        0

                                      1  
                                 X B
                                           =  kC BO t                                 (3-305)
                                  0  1 X−  B 
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