Page 214 - Modeling of Chemical Kinetics and Reactor Design
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184    Modeling of Chemical Kinetics and Reactor Design

                                Substituting Equations 3-258 and 3-261 into Equation 3-256 gives


                                 −  dC A  = kC AO (1  − X A  )(C BO  − C AO  X A )      (3-262)
                                                             3
                                   dt
                                Substituting Equation 3-259 into Equation 3-262 and rearrang-
                              ing gives

                                          C AO  dX A         =
                                 C (  1− X )( C BO  −  3 C AO X )  kdt                  (3-263)
                                          A
                                                          A
                                  AO
                                Integrating Equation 3-263 between limits t = 0, X  = 0 and t = t,
                                                                                A
                              X  = X  gives
                               A
                                     A
                                 X A         dX                t
                                                               ∫
                                 ∫  (  X )(     A     X )  =  kdt                       (3-264)
                                 0  1−   A  C BO −  3 C AO  A  0

                                Equation 3-264 can be further expressed by


                                 X A          dX                  t
                                                                 ∫
                                 ∫               A            =  kdt                    (3-265)
                                 0  C ( 1−  X )    C BO  −  3 X A   0
                                             A
                                     AO
                                                 C AO      
                              where θ  = C   /C   . Equation 3-265 becomes
                                     B     BO  AO

                                 X A         dX                t
                                                               ∫
                                 ∫  C (        A      X )  =  kdt                       (3-266)
                                             A
                                 0   AO  1− X )(θ B  −  3  A   0
                                Converting Equation 3-266 into partial fraction gives

                                         1         ≡    A    +    B
                                 ( 1− X A  )( B  − 3X A )  1− X A  θ B  − 3X A          (3-267)
                                         θ


                                    A θ  −    )+         )
                                             A         A
                                1= ( B    3X       1− ( B  X
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