Page 66 - The engineering of chemical reactions
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50   Reaction Rates, the Batch Reactor, and the Real World
                            For the reversible reaction
                                            A+BZCC,      Y =  k&ACB -k&c

                       among ideal gases, if CAM =  CB~  and Cco = 0, the substitution of  CjS  in terms of X yields




                       which again can be integrated for X(t)  and substituted back to find CA,  Cs,  and Cc.
                        Example 2-5  Find the conversion versus time for the reaction

                                                 A  -+  3B,  r=kCi
                        n a constant-pressure batch reactor assuming  A  and  B  are ideal gases starting with pure A
                       it CAo.
                           The number of moles of  A  and  B  are found by setting up the mole table

                       Species          Moles initially    Moles
                       1                NAO                NAO(~ --xl
                       3                0                  ~NAoX
                        otal moles      NAO                NAO(~ +2x)
                       volume           VO                 Vdl   +2x)

                           Therefore, the concentrations are



                           and
                                                                3 x
                                                      NB
                                                 CB  =  - =  cAo------
                                                       V       1 + 2 x
                           In a constant-pressure batch reactor the mass balance on  A  becomes
                                           dNA                   = -kCiV
                                           ~  _  -hTAog   =  -rV
                                            dt
                                                                  V,(l   +2X)

                           so that
                                                  dX
                                                  -  =kCA,(;;ZX:’
                                                   dt
                           This equation can be separated to yield
                                                         X
                                                      1     1 + 2 x
                                                    -      ___
                                                t  = kc,.,,, s (1 - X)2 dX
                                                         0
                           which must be  integrated by partial fractions to give
                                                     3 x
                                                    -     + 2 ln(1 -  X)
                                                 t=  1-x
                           This equation cannot be inverted to find  X(t)  explicitly, and so we can only find
                           CA   (t  ) numerically.
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