Page 574 - Bird R.B. Transport phenomena
        P. 574
     554   Chapter 18  Concentration Distributions in Solids and in Laminar Flow
                                                                Fig. 18.4-1.  Absorption  of Л by  В with a
                                                                homogeneous reaction in the liquid  phase.
     §18.4   DIFFUSION WITH A      HOMOGENEOUS
             CHEMICAL REACTION
                           As  the  next  illustration  of  setting  up  a mass  balance, we  consider  the system  shown  in
                           Fig. 18.4-1. Here gas  A  dissolves in liquid  В in a beaker and diffuses  isothermally  into the
                           liquid  phase. As  it diffuses,  A  also undergoes  an irreversible  first-order  homogeneous re-
                           action: A  + В  —> AB.  An  example  of  such  a system  is the absorption  of CO  by  a concen-
                                                                                          2
                           trated aqueous solution  of  NaOH.
                               We  treat this as a binary  solution  of A  and  B, ignoring  the small amount of AB  that is
                           present  (the pseudobinary  assumption).  Then the mass  balance  on  species  A  over  a  thick-
                           ness  Az  of  the liquid  phase becomes
                                                  N \ S  -        -  k'"c SAz  =  0            (18.4-1)
                                                   Az z                A
                           in which k'" is a first-order  rate constant for  the chemical decomposition  of A, and  S is the
                           cross-sectional  area  of  the liquid.  The product k'"c  represents  the moles  of  A  consumed
                                                                     A
                           by  the reaction per  unit volume  per unit time. Division  of  Eq.  18.4-1  by  SAz  and  taking
                           the limit as  Az  —» 0 gives
                                                              +  k'l'c  =  0                   (18.4-2)
                                                          dz       A
                           If the concentration of A  is small, then we may to a good  approximation write  Eq. 18.0-1 as
                                                                    dc
                                                         дг  = -qt  — A                        (18.4-3)
                                                           Л2     AB
                           since the total molar concentration с is virtually  uniform  throughout the liquid. Combin-
                           ing the last two equations gives
                                                             2
                                                        „   d c  ,„,
                                                              A
                                                                                               (18.4-4)
                           This is to be solved  with  the following  boundary conditions:
                           B.C.I:                atz  = 0,   c  =  c                           (18.4-5)
                                                                  AO
                           B.C. 2:               at z  = L,  N  A  = 0 (or dcjdz  = 0)         (18.4-6)
                                                             Az
                           The first  boundary  condition asserts  that the concentration of  A  at the surface  in the liq-
                           uid  remains at a fixed  value  c .  The second states  that no A  diffuses  through the bottom
                                                    A0
                           of the container at z  = L.
                                                        2
                               If  Eq. 18.4-4  is multiplied  by  L /C Q% ,  then it can be written  in dimensionless  vari-
                                                          A
                                                             AB
                           ables  in the form  of  Eq. C.l-4
                                                                 2
                                                          ^   -  ф Т  = О                      (18.4-7)





