Quasi-equilibrated step

This result indicates that the sensitivity of the overall rate to the reverse rate constant for a step depends on the reversibility of the step. For example, the sensitivity of the overall rate to the reverse rate constant is equal to zero for an irreversible step (z = 0) and to the negative value of the sensitivity of the overall rate to the forward rate constant for a quasi-equilibrated step (zt = 1). 

Therefore, the sum of the sensitivities for the forward and reverse rate constants of a particular step i is equal to zero for a quasi-equilibrated step (.Zi = 1). Since the sum of sensitivities for all steps is equal to unity, the sensitivity of a quasi-equilibrated step is passed on to the subsequent steps in the reaction sequence. Similarly, the sum of the sensitivities for the forward and reverse rate constants of a particular step i is equal to the sensitivity for the forward rate constant for an irreversible step (z = 0). Since the sum of sensitivities for all steps is equal to unity, the sensitivity of an irreversible step is not passed on to the subsequent steps in the reaction sequence. Accordingly, the sensitivities of the rate constants for a step approach zero as the preceding steps that produce the reaction intermediates become irreversible. 

It must be emphasized that step 2 is not an elementary step, but a sum of all of the quasi-equilibrated steps that must occur after dinitrogen adsorption. According to this abbreviated sequence, the only species on the surface of the catalyst of any kinetic relevance is N. Even though the other species (H, NH, etc.) may also be present, according to the assumptions in this example only N contributes to the site balance ... 

Derive a rate expression for the hydrogenation of ethylene on Pt assuming steps 1, 2, and 3 are quasi-equilibrated, step 4 is virtually irreversible, and C2H5 is the most abundant reaction intermediate covering almost the entire surface ([ ]o [ C2H5]). Discuss why the rate expression cannot properly account for the experimentally observed half order dependence in H2 and zero-order dependence in ethylene. Could the observed reaction orders be explained if adsorbed ethylene ( C2H4 ) were the most abundant reaction intermediate Explain your answer. 

Finally, this step may be essentially at equilibrium if T and T are both very large compared to the slow step(s) and if T = T, thus this quasi-equilibrated step is denoted by [2] ... 

Boudart has discussed in detail the fact that the rate law derived from a complex catalytic cycle comprised of a number of elementary steps can frequently be represented by only two kinetically significant steps if the assumptions of a RDS and a MARI are invoked however, ambiguities can develop which prevent one from distinguishing among different reaction models [11,26]. In similar fashion, but with perhaps less dramatic results, a L-H-type or H-W-type model [27] invoking more than one elementary surface reaction step can be greatly simplified by the presence of quasi-equilibrated steps which precede the RDS or, if a RDS does not exist, the series of slow steps on the surface. The SSA may also be required in the latter case to eliminate all unknown surface reaction intermediates from the rate law. Significant simplification is achieved with the assumption of a RDS. 

It is worth noting at this point that a H-W model invoking product desorption as the RDS could also give equation 7.30 for values of n = 1 or 2, provided the fraction of vacant sites is very low and the fraction of sites covered by the two reactants is very high. It may also be possible to propose a more realistic sequence of steps comprised of reversible and irreversible elementary steps (perhaps even including some quasi-equilibrated steps) that could result in a rate expression of the Mars-van Krevelen form. Consequently, the Mars-van Krevelen rate expression should be considered to be only a mathematical fitting equation with no theoretical basis. 

Note that four steps are quasi-equilibrated, step 2 is reversible and step 6 is irreversible. Derive the rate expression for NH3 disappearance assuming that adsorbed nitrogen (N ) is the MARI. Note that K5 = I/Khj. 

Equation (1.58) can be solved for any reaction condition if the equilibrium constants, Ki, for steps (1) and (2), and the forward and reverse rate constants, kf and kj, for steps (3) and (4) are known. Note that we do not need to know the rate constants of the quasi-equilibrated steps (1) and (2) and the equilibrium constants are sufficient. We will use this analytical solution later and compare it to a full numerical solution of the CO oxidation model. 

The value of Zi approaches zero as step i becomes irreversible, and it approaches unity as step i becomes quasi-equilibrated. Therefore, the value of Zi may be termed the reversibility of step i. The value of z, remains bounded between 0 and 1, provided that step i proceeds in the forward direction. 

Therefore, Campbell s degree of rate control, XRCl, provides an excellent measure of the sensitivity of the overall reaction rate to the kinetic parameters for each step. The value of A"rc>( approaches zero as step i becomes quasi-equilibrated, and the value of 3lrC, becomes small as the preceding steps that produce the reaction intermediates for step i become irreversible. 

Finally, we note that some confusion exists regarding the concept of a rate-determining step. For the three-step reaction scheme of this example, consider the case for which step 1 is quasi-equilibrated and steps 2 and 3 are irreversible ... 

The fraction of the platinum surface that is free of adsorbed species, 0+) is equal to 0.084 under these conditions. The most abundant surface intermediates are adsorbed II and C4II indicating that the only significant kinetic parameters in the expression for a, are Kjeq and q. It is clear from these values of zt that steps 2-4 are quasi-equilibrated (z,- 1 or At 0) and step 1 is rate determining (zi ztot or A 4totai) under these reaction conditions. 

Since steps 2-4 are quasi-equilibrated under the conditions of the kinetic studies, the following simplified rate expression may be derived to described the hydrogenation/dehydrogenation reactions ... 

The previous discussion involved a two-step sequence for which the adsorption of reactant is nearly equilibrated (quasi-equilibrated). The free energy change associated with the quasi-equilibrated adsorption step is negligible compared to... 

If desorption of B from the surface is rate-determining, then all elementary steps prior to desorption are assumed to be quasi-equilibrated ... 

Since the binding of olefin is not quasi-equilibrated at high pressure, the subsequent hydrogenation step cannot be considered as rate-determining. The relative concentrations of reactive intermediates and products are now given as ... 

A composite event can be substituted for any reaction sequence that has a single rate-controlling step, provided that each intermediate species is quasi-equilibrated with its particular set of terminal species. To obtain the composite event, start with Eq. (2.4-1) for the controlling step, including... 

In this case the remaining elementary reactions may be considered at quasiequilibrium and may be lumped into a single, intermediate, quasi-equilibrated reaction [8]. This is similar to the case of the two-step mechanism of Boudart [28]. 

The value of XRCi approaches zero as step i becomes quasi-equilibrated, while this value is close to unity for the step with the maximum sensitivity. 

It can be seen that, if there is a rate-determining step, only the rate constants of that step enter into the rate equation. The rate constants of the other steps in quasi-equilibrium appear only as ratios which, as is already known, are equal to the equilibrium constants of these equilibrated steps. This is a considerable simplification of the kinetic problem, since, at least in principle, equilibrium constants are more easily arrived at than rate constants. Even when this is not so, the number of arbitrary constants in the rate equation is reduced considerably. 

Changes in observed activation energy as a function of the temperature can, however, also be the result of a more complex reaction mechanism, for example, when a surface reaction is preceded by a quasi-equilibrated adsorption step. In the case of an irreversible reaction with immediate product desorption, the reaction rate can be written as (see also Section 3.1.2) ... 

If there exists a rate determining step (rds), all other steps being equilibrated or quasi-equilibrated, i.e., Ai =0, Vi = v-i except for i = d, the (2.88) becomes ... 

One can view the formation of the activated complex, X, as being quasi-equilibrated while its decomposition into the products is an irreversible rate determining step consequently, for the reaction A -h B C, this process can be represented as ... 

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