Big Chemical Encyclopedia

Chemical substances, components, reactions, process design ...

Articles Figures Tables About

Equality of the rates

H. R. Gersmann (Koninklyke Shell Laboratories, Amsterdam, Netherlands) The results obtained by Russell correlate with those obtained by Gersmann and Niewenhuis (Organic Reaction Symposium, Cork, 1964) in the study of autoxidation of esters and ketones. Here weakly acidic esters also showed rates of ionization equal to the rate of oxidation as shown by the equality of the rate of racemization of an optically active ester to the rate of oxidation. [Pg.212]

From the equality of the rates, the concentration of reactant at the outer surface of the catalyst particle can be expressed in terms of bulk concentration ... [Pg.67]

This equation can be interpreted as the equality of the rate of diffusion to the depth r0D(C0/r0) and the reaction rate in a layer whose depth is KSCqTq (both values are with respect to a unit external surface of the layer). [Pg.75]

The equalization of the rates in both directions at high H results from the splitting of the triplet that disables the 7+) and T ) states when the spin-conversion is carried out by a weak hyperfine coupling. A similar effect was predicted and backed experimentally in the classical works with Py [217-219]. [Pg.314]

Comparison of the kinetic curves in Figs. 12a and b shows that there is reasonable agreement between the experimental and calculated data for the case under discussion. Also the data in Fig. 12 provide support for the theoretically based conclusion with respect to the equality of the rates of the forward and reverse exchange reactions (Ifig. 12) when the isotherm is close to linear. An interesting feature of the kinetic curves (solid and dashed lines in Figs. 12a and b) is that they cross at F = 0.85 for computerized dependencies (Fig. 12a) and at F = 0.55 for the experimental ones (Fig. 12b). The reason for the different response to time of the forward (solid lines) and reverse (dashed lines) IE can be established through analysis of the Vg and concentration distribution curves. [Pg.183]

A quasi-stationary treatment of the intermediates gives equality of the rates of all five steps and of the overall reaction... [Pg.22]

When a reaction proceeds through several successive elementary steps, and one of these reactions is very much slower than any of the others, then the rate will depend on the rate of this single slowest step. The slow step is the rate-determining step. In the above example, reaction (9a) constitutes such a rate-determining step. In reactions (9a) and (9b), a steady state for [HBr02] is rapidly reached (so-called pseudosteady state). HBrOz formed in reaction (9a) is almost immediately consumed in reaction (9b). Hence, the formation of HOBr in (9b) occurs at the same rate as in (9a). From the condition of equality of the rates of the two reactions in a pseudosteady state, ra = rb, we obtain... [Pg.130]

The kinetic method, where the equilibrium conditions are determined by equality of the rates of adsorbtion and desorption. [Pg.279]

The Langmuir isotherm is the most widely used in the derivation of kinetic models for gas-solid catalytic reactions. The derivation of a Langmuir isotherm depends on the kinetic method where the equilibrium conditions are determined by the equality of the rates of adsorption and desorption. At equilibrium the following relation holds,... [Pg.279]

The equality of the rates of the second and the third reaction is a necessary consequence of the condition of a steady-state. The equality of the rate of the first reaction with the rate of the second and the third reaction shows that the first reaction proceeds predominantly via reactions (VI.5) and (VI.6), whereas the contribution of a direct bimole-cular reaction between COS and Ha is insignificant. [Pg.359]

Figure 6.2. Illustration of the steady-state concept in enzymatic catalysis From equality of the rates k] [E] [S] = k[ES] + k 2 [ES]... Figure 6.2. Illustration of the steady-state concept in enzymatic catalysis From equality of the rates k] [E] [S] = k[ES] + k 2 [ES]...
Also the rate of initiation r< must be specified, although the nature of the initiation step remains immaterial. The condition of steady state will be expressed, as in the case of copolymerization, by the equality of the rates of the two chain transfer steps, as shown on the scheme on the left and right ... [Pg.116]

The kinetics of methane combustion over a perovskite catalyst (Lao.9Ceo.iCo03) has been studied in Micro-Berty and fixed bed reactors. Discrimination among twenty-three rival kinetic models from Eley-Rideal, LHHW and Mars-van Krevelen (MVK) types has been achieved by means of (a) the initial rate method as well as by (b) integral kinetic data analysis. Two MVK type models could be retained as a result of the two studies, with a steady-state assumption implying the equality of the rate of three elementary steps. [Pg.599]

The experimental initial rate curves in the Berty reactor (Fig. 2.b) shows a trend that can represent either type b ox e among the theoretical graphs in the Figure 1. Since the model Ml 7 belongs to type c it can be removed in the final selection from likely mechanisms. The two retained models. Ml 8 and M20 both are of Mars-van Krevelen types where the steady-state assumption involves the equality of the rate of three elementary steps as it has been shown in the Table 1. [Pg.603]

The oxidation of thiophene (Th) by peroxynitrous acid (HOONO), generated in the H2O2-HNO2 system, proceeds simultaneously in both the gaseous and the liquid phases of the reactor. The conclusion that the active oxidant in both cases is the HO radical, formed by HO-ONO bond homolysis, is based on the equality of the rate constants of the reaction in the two phases. ... [Pg.143]

Trend of kinetic curves for each participant of the reaction is modeled in Fig. 1.7. It is seen, that in some time the concentrations of the substances become time-undependable, which corresponds to the equilibrium condition. Time-dependences of the rate of each stage are also shown here, and that illustrates the dynamic character of the chemical equilibrium reaching equilibrium condition does not mean the end of the reaction. In equilibrium condition the total rate of the reaction is equal to zero because of the equality of the rates of direct and reverse reaction, which have quiet definite values. [Pg.14]

Equation [50] assumes that the system is in a quasi-equilibrium state, where an approximate equality of the rates of addition and fragmentation holds, namely... [Pg.142]

The initial autoacceleration of the reaction is completed in a very short time (Fig. 5.1), at a negligible degree of conversion of the reactants, and is not observed experimentally. The induction period in the oxidation of methane at high pressures is, in fact, a manifestation of a branched-chain process occurring in the steady-state mode, which is characterized by the approximate equality of the rates of chain-branching steps (II)—(IX) and quadratic chain-termination steps (XIII)—(XIV). [Pg.79]

The equilibrium constant for a reaction arises from the equality of the rates of all the elementary processes. However, it is not possible starling from the rate law for the forward process and the equilibrium constant to obtain the rate law for the reverse process. We will illustrate this with the same reaction. Writing the above reactions in the reverse direction... [Pg.22]

But according to the properties of the multiplying coefficients, we have And taking into account the theorem of the equality of the rates, we get ... [Pg.218]


See other pages where Equality of the rates is mentioned: [Pg.258]    [Pg.588]    [Pg.29]    [Pg.190]    [Pg.351]    [Pg.107]    [Pg.153]    [Pg.642]    [Pg.148]    [Pg.151]    [Pg.109]    [Pg.196]    [Pg.214]    [Pg.156]    [Pg.8]    [Pg.192]    [Pg.80]    [Pg.389]    [Pg.282]   


SEARCH



Equal

Equaling

Equality

Equalization

© 2024 chempedia.info