Steady states and stability analysis of a bimolecular non-equilibrium reaction scheme with general hinshelwood-langmuir saturation-inhibition law

The stability analysis of a simple two-component bimolecular reaction scheme involving a Hinshelwood-Langmuir law of nth-order [X/(1 + qX)n, n ~ 2], is presented. LE JOURNAL DE PHYSIQUE LETTRES TOME 38, 15 OCTOBRE 1977, 1 Classification Physics Abstracts 82.60 87.15 Recently [1, 2, 3] the authors have discussed the stability analysis of a simple bimolecular reaction scheme involving a first-order Hinshelwood-Langmuir saturation law. Such a saturation law is of relevance in combustion kinetics [4] and also corresponds to the Michaelis-Menten enzyme kinetics in biochemistry and to the Holling law in Ecology (see Fig. 1). The model reaction scheme is FIG. 1. First-order Hinshelwood-Langmuir (Michaelis-MentenHolling) saturation law. in which X and Y are variable intermediate reactants, and A and P are products whose concentration is kept constant. All steps in eq. (1) are taken to be irreversible. In eq. (Ie) we denote by S/1 a saturation/ inhibition law of the following type in which n denotes the order of the HinshelwoodLangmuir law. The general case n > 2 is of relevance in reactor kinetics [4] (see Fig. 2). In contradistinction to figure 1 the case n > 1 in eq. (2) describes a saturation plus inhibitory step in (1). q accounts for the strength of this step. FIG. 2. N-th-order Hinshelwood-Langmuir inhition law (n > 2). The maximum of the reaction rate is at Xc. Article published online by EDP Sciences and available at http://dx.doi.org/10.1051/jphyslet:019770038020040500 L-406 JOURNAL DE PHYSIQUE LETTRES We associate the following differential equations with the scheme (1), (2) for isothermal processes, which we take in dimensionless form. The variables X, Y, and A represent dimensionless concentrations. The stationary solutions of (3) are given by the relations There would be as many solutions as there are real and positive roots of the polynomial (4b), The number of such roots is given by the HarriotDescartes theorem. We have : (i) If Aq > l/n, there is no real and positive solution of (6) and so no steady states of (3). (ii) If Aq l/n either we have two steady states or there is none. In the latter case the alternative is solved by constructing the intersects of ( 1 + qX )n with X/A. Therefore two steady solutions of (3) exist if the straight line X/A has a greater slope than the tangent to the curve (1 + qX)n at the origin. Otherwise there is no steady solution of (3). Thus we have the alternative