Page 180 - Modeling of Chemical Kinetics and Reactor Design
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150    Modeling of Chemical Kinetics and Reactor Design

                                Examples of first-order reversible reactions are gas phase cis-trans
                              isomerization, isomerizations in various types of hydrocarbon systems,
                              and the racemization of α and β glucoses. An example of a catalytic
                              reaction is the ortho-para hydrogen conversion on a nickel catalyst.

                                    SECOND ORDER REVERSIBLE REACTIONS

                                Bimolecular-type second order reactions are:


                                 A + [    C +  D                                        (3-170)
                                       k 1
                                     B
                                       k 2
                                     1 k
                                 2A[   C +  D                                           (3-171)
                                    k  2

                                     1 k
                                 2A [  2C                                               (3-172)
                                    k 2


                                       k 1
                                     B
                                 A + [ 2   C                                            (3-173)
                                       k 2
                                With the restrictions that C AO  = C  and C  = C DO  = 0 in Equation
                                                                       CO
                                                               BO
                              3-170, the rate equation for a constant volume batch system becomes

                                 − ( r A ) =−  dC A  =C AO  dX A  = kC 2 A  − k C 2 C   (3-174)
                                                            1
                                                                  2
                                           dt        dt
                                                     2
                                      = kC 2  (1 − X  ) −  k 2  X 2
                                         1  AO     A         A                          (3-175)
                                                        k 1
                              where the fractional conversion X  = 0 at t = 0.  At equilibrium,
                                                                A
                              dX /dt = 0 and the final solution is:
                                 A
                                     X  −(2 X  −  X ) 1         1     
                                 ln   Ae    Ae      A   =2 kC AO   −1   t           (3-176)
                                                            1
                                        X Ae  −  X A             X Ae  
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