Page 207 - Modeling of Chemical Kinetics and Reactor Design
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Reaction Rate Expression  177

                              reactor design. This is because knowledge of the mechanism will make
                              if possible to fit the experimental data to a theoretical rate expression,
                              which will be more reliable than an empirical fit. Also, the mechanism
                              may require some modifications and optimization for the final design.


                                                        Example 3-1
                                                                     −
                                                   (
                                                               (
                                                        −
                                                                                            2−
                                                       4
                                                                    3
                                The oxidation of  Fe CN)  to  Fe CN)  by peroxidisulfate,  SO ,
                                                       6            6                     2  8
                              can be monitored spectrophotometrically by observing the increase in
                              absorbance at 420 nm, D 420  in a well-mixed batch system. Assume
                              that the kinetic scheme is:
                                 Fe CN) 4 −  +  1  S O 2 −   → (  3 −  +  SO 2 −
                                   (
                                                           Fe CN)
                                                      
                                                       k 2
                                       6       2  8               6      4
                                            2
                                                    [
                                 −  d [ Fe (CN ) 4 − ]  = k Fe (CN ) 4 − ][ S O 2 8 − ]
                                                                2
                                                   2
                                  dt        6              6
                                                                                   ×
                                Using pseudo-first order conditions with  [SO 2− ] =  . 18 10 − 2  M and
                                                                           8
                                                                        2
                                      4−
                                              ×
                              [Fe (CN ) ] =  . 65 10 − 4 M, the following absorbances were recorded
                                      6
                              at 25°C:
                              t/s       0       900    1,800    2,700   3,600    4,500      ∞
                              D       0.120    0.290   0.420    0.510   0.581    0.632    0.781
                               420
                                                                                       2−
                                Calculate the pseudo-first order rate constant  k = k S O ] and,
                                                                                  [
                                                                                       8
                                                                                    2
                                                                                  2
                                                                              1
                              hence, k .
                                      2
                              Solution
                                Table 3-6 gives D  – D with time t. For a first order rate law, the
                                                 ∞
                              rate equation is expressed by
                                   ( D −  D)
                                 ln  ∞       =−  kt
                                   ( D − D )     1                                      (3-246)
                                     ∞
                                          O
                                Equation 3-246 is further expressed by
                                ln (D  – D) = ln (D  – D ) – k t                        (3-247)
                                     ∞
                                                    ∞
                                                         O
                                                               1
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