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6.2  A Decoupling Procedure for Removing the Coupling           127

                                                  2          2
                   ∂C AB    ∂C AB   ∂C AB        ∂ C AB     ∂ C AB
                 φ      + u      + ν      − D  ex     + D ey
                    ∂t       ∂x      ∂y           ∂x 2       ∂y 2
                                                                         (6.18)

                               k f  n f −1    n b −1
                        = φk b   r   C A C B − r  C AB .
                               k b
              Equations (6.16) and (6.17) are two conventional mass transport equations with-
            out any source/sink terms due to the redox chemical reaction so that they can be
            solved by the well-developed numerical methods available. Since the two new vari-
            ables, namely C I = C A + C AB and C II = C A + C AB , are independent of chemical
            reaction rates, they can be referred to as chemical reaction rate invariants, which are
            the analogues of the stress and strain invariants in the field of solid mechanics.
              If the redox chemical reaction is an equilibrium one, then both the forward and
            the backward reaction rates are theoretically infinite so that the chemical reaction
            becomes predominant in the reactive transport process. In this case, Eq. (6.18) can
            be written as

                                K e r n f −1 C A C B − r n b −1 C AB = 0,  (6.19)

            where K e = k f /k b is the chemical equilibrium constant.
              Inserting the two chemical reaction rate invariants, C I = C A + C AB and C II =
            C B + C AB , into Eq. (6.19) yields the following equation:

                             K e r(C I − C AB )(C II − C AB ) − C AB = 0.  (6.20)

              It is noted that n f = 2 and n b = 1 are substituted into Eq. (6.19) so as to
            obtain Eq. (6.20). Clearly, Eq. (6.20) has the following mathematical solution for
            the chemical product of the redox chemical reaction:


                                                               2 2
                                                         2
                     K e r(C I + C II ) + 1  [K e r(C I + C II ) + 1] − 4K r C I C II
                                                              e
              C AB =                −                  2 2            .  (6.21)
                          2K e r                    4K r
                                                       e
              This indicates that for an equilibrium chemical reaction, we only need to solve
            the mass transport equations of chemical reaction rate invariants (i.e. C I = C A +C AB
            and C II = C B + C AB in this particular example) using the conventional numerical
            methods. Once the distributions of the chemical reaction rate invariants are obtained
            in a computational domain, the chemical product distribution due to the chemical
            reaction can be calculated analytically. As a result, the distributions of the chemical
            reactants can be calculated using the distributions of both the chemical product and
            the chemical reaction rate invariants.
              However, for non-equilibrium chemical reactions, the chemical reaction rates
            are finite so that we need to solve at least one reactive transport equation with the
            source/sink term for each of the chemical reactions in the geochemical system. This
            means that for the general form of the redox chemical reaction considered in this
            study, we need to solve Eq. (6.18) numerically if the reaction rates of this redox
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