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Else_AIEC-INGLE_cH003.qxd  7/13/2006  1:45 PM  Page 90
                    90                               3. Heterogeneous Processes and Reactor  Analysis


                    to derive the definition of the expansion factor (eq.(3.88)).

                    Solution
                    Under constant   P and   T ,

                                             V  (  x  1)    V  (  x 0)    N  (  x  1)    N  (  x 0)

                                          R
                                                 V  (  x  0)  N  (  x  0)
                    Since

                                                   N     N
                                                     x (   0)  tot,i
                                                N  x (   1)  N    tot,i  N
                                                          A,i

                    v and this gies the definition equation (3.88)

                                                        N  A,i

                                                    R
                                                        N
                                                         tot,i
                    Example 2
                    Suppose a gas mixture consisting of 18% O  2  , 3% SO  2  and 79% nitrogen is fed into a reac-
                    tor so that the following reaction takes place:

                                                SO    O    1  SO
                                                   2   2  2   3

                    Express the concentration of SO  2  and O  2  as a function of con v ersion.
                    Solution
                    The limiting reactant is SO  2  and thus, by using the stoichiometry of the reaction, we ha e v

                                                   18379
                                              N  tot,i     100
                                                      1
                                                N  2 SO  3 (  
  x )
                                                          2
                                                          SO
                                                    18 1.5
                                                N  O2      SO2  x
                                                   N  3 SO  3   x
                                                         2
                                                         SO
                    Then

                                                 1 0 1 0.5
                                                                0.5
                                                      1
                                                   3
                                                     (    0.5)  0.015
                                              R
                                                  100
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