Page 160 - Modeling of Chemical Kinetics and Reactor Design
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130    Modeling of Chemical Kinetics and Reactor Design

                                  (1− X  ) −+ 1  X A
                                          n
                                                       −
                                      A         = kC n AO 1 t                          (3-80)
                                    1− n     0 
                                ( 1− X A ) 1−n  −= kC n AO 1 ( 1− ) n t                  (3-81)
                                                    −
                                             1

                                Table 3-2 gives solutions for some chemical reactions using the
                              integration method.



                                             METHOD OF HALF-LIFE t           1/2

                                As previously discussed, the half-life of a reaction is defined as the
                              time it takes for the concentration of the reactant to fall to half of its
                              initial value. Determining the half-life of a reaction as a function of
                              the initial concentration makes it possible to calculate the reaction
                              order and its specific reaction rate.
                                Consider the reaction A → products. The rate equation in a constant
                              volume batch reaction system gives


                                 − ( r  ) =  −dC A  = kC n                               (3-82)
                                   A               A
                                          dt
                                Rearranging and integrating Equation 3-82 with the boundary
                              conditions t = 0, C  = C   and t = t, C  = C  gives
                                                A    AO             A    A

                                  C A        t
                                             ∫
                                 −  ∫  dC A  = dt
                                      C n                                                (3-83)
                                  C AO  A    0

                                      n
                                  C −+1  C A
                                 −   A      = kt                                       (3-84)
                                   −+1   C AO
                                     n
                                   1    C {  A } C A
                                          −
                                         1 n
                                 ( n − 1)    C AO  =  kt                                 (3-85)
                                 C {  1− n  −  C } = (  −
                                                kt n 1)
                                          1−
                                            n
                                   A      AO                                             (3-86)
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