I n Equations 3 and 4 the difference between the reverse and forward rates of reaction have been canceled because of the assumption of instantaneous equilibrium. The resulting Equations 3a and 4a are both incorrect. Equations 1 and 2 correctly predict that the CSTR has no dynamics while Equations 3a and 4a predict an exponential decay. I n physical systems it is impossible to achieve instantaneous equilibrium.If the actual rates of reaction, R f and R , were known and used, then there would be a short period of time where R -R , was nonzero. This in turn would couple Equations 3 and 4 so that they would be correct.Suppose, however, that for a system under study, the assumption of instantaneous equilibrium is an excellent assumption, and that the exact R f and R, are unknown. All that is known is that R and R , are very fast. The question then arises as to how correct mass balances can be formulated assuming instantaneous equilibrium. The answer can be stated in the following principle. First, mass balances are written down for all components which are not involved in any equilibrium reactions. Next, all equilibrium equations are written down. N o attempt is made to balance any component in equilibrium. If more equations are required, then they can be obtained by balancing groupings of atoms for which there is no net loss or gain via reactions. These last equations will be of the type: Accumulation = flow inflow out. A good choice for these last equations would be elemental balances. I n the previous study (McAvoy et al., 1972), the balance on the variable 6, which was the sum of acetate ion and acetic acid concentration, was of this last category. Effectively the 6 balance was an elemental carbon balance.To illustrate the above principle, two examples will be considered.Example 1, NaOH Neutralizing HzC03 in a CSTR.The different chemical species whose concentrations are to be found are: Na+, OH-, H+, H2C03, HC03-, and C032-.The equilibrium reactions are HZCO, iS HCO3-+ H + HC03-FI COS2-+ H S €I20 8 H f + OH-The following approach will yield a well-defined set of equations. Balance NB+, write down three equilibrium equations and an electroneutrality equation. Last, balance carbon by balancing the sum of the concentrations of H2COa, HCOsand c03'-, to give six equations in six unknowns. Example 2, a Tubular Reactor in Which the Following Reactions Occur :2 NzOs + 2 NzO4 + 02Nz04 F! 2 NO2The concentration of the following species must be found as functions of distance and time: NzOs, Nz04, Oz, and NO*.Direct balances can be made on NzOs and 02. (The balance would consider only 02 and not the oxygen in the other species. It would not be an elemental balance in the sense previously discussed.) To finish the formulation, an equilibrium equation and a balance on nitrogen can be made.A direct balance on Nz04 should not be made. Dynamic energy balances on fast equilibrium systems would have to be achieved by the fundamental approach of writing internal energy as a function of such items as composition and teniperature. It ...