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See. 5.5   Least-Squares Analysis                              251

                                We have three  linear equations  and  three  unknowns  which  we can  solve for:
                                a,,  a,, and a,.  A detailed example delineating the lunetics of the reaction


                                using  linear  least-squares  analysis  can  be  found in  Example  10-2. If  we  set
                                a,  = 0 and consider only two variables, Y and X, Equations  (5-31) and (5-32)
                                reduce to the familiar least-squares equations for two unknowns.
                                /   Example 5-5  Using Least-Squares Analysis to Determine Rate Luw Paramteters


                                   The  etching  of  semiconductors  in  the  manufacture  of  computer  chips  is  another
                                   important  solid-liquid  dissolution  reaction ,(see Problem  P5-12 and Section  12.10).
                                   The dissolution of the semiconductor MnO,  was studied using a number of diflerent
                                   acids  and salts. The rate  of  dissolution  was  found to  be  a function  of  the  reacting
                                   liquid  solution  redox  pptential  relative to  the  energy-level conduction  band  olf  the
                                   semiconductor I: was found that the reaction rate could be increased by  a factor of
                 A  lo5 fold increase   lo5 simply by changing the anion of the acid9!! From the data below, determine the
                  in reaction rate!  ! !   reaction  order  and specific reaction rate for the dissolution of  MnO,  in  HBr.
                                        C,  (mol HBrMm3)        0.1    0.5    1.0   2.0   4.0

                                        -ri,  (mol HBr/m2. h) X  lo2   0.073   0.70   f.84   4.86   12.84

                                   Solution
                                   We assume a rate law of  the form
                                                             -  rkB  =                       (E5 -5.1)
                                   Letting A  = HBr, taking the In  of  both  sides of  (E5-5.1), and using the initial rate
                                   and concentration  gives

                                                         In(-ri,)  = lnk+a lnCAo             (E5-5.2)
                                   Let  .Y  = In(-ri,),  a  = Ink, b  = a, and  X = lnCAo. Then
                                                               Y=a+bX                        (E5-5.3)
                                   The least-squares equations to be solved for the best values of a and b are for N runs




                                                         N        N       N
                                                         %X,Y,  =az~~+bx~,~                  (E5-5.5)
                                                         z=I      1=1    z=1
                                   where  i  = run number. Substituting  the  appropriate  values from Table E5-5.1  into
                                   Equations (E5-5.4) and (E5-5.5) gives






                                  S. E. Le Blanc and H. S. Fogler, AIChE J.,  32, 1702 (1986).
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