Rate equations reversible

Pulse radiolysis results (74) have led other workers to conclude that adsorbed OH radicals (surface trapped holes) are the principal oxidants, whereas free hydroxyl radicals probably play a minor role, if any. Because the OH radical reacts with HO2 at a diffusion controlled rate, the reverse reaction, that is desorption of OH to the solution, seems highly unlikely. The surface trapped hole, as defined by equation 18, accounts for most of the observations which had previously led to the suggestion of OH radical oxidation. The formation of H2O2 and the observations of hydroxylated intermediate products could all occur via... 

Models used to describe the growth of crystals by layers call for a two-step process (/) formation of a two-dimensional nucleus on the surface and (2) spreading of the solute from the two-dimensional nucleus across the surface. The relative rates at which these two steps occur give rise to the mononuclear two-dimensional nucleation theory and the polynuclear two-dimensional nucleation theory. In the mononuclear two-dimensional nucleation theory, the surface nucleation step occurs at a finite rate, whereas the spreading across the surface is assumed to occur at an infinite rate. The reverse is tme for the polynuclear two-dimensional nucleation theory. Erom the mononuclear two-dimensional nucleation theory, growth is related to supersaturation by the equation. 

The number of independent rate equations is the same as the number of independent stoichiometric relations. In the present example. Reactions (1) and (2) are reversible reactions and are not independent. Accordingly, C,. and C, for example, can be eliminated from the equations for and which then become an integrable system. Usually only systems of linear differential equations with constant coefficients are solvable analytically. 

Example 5 Percent Approach to Equilibrium For a reversible reaction with rate equation r = L[A — (1 — A)Vl6], the size function kV,./V of a plug flow reactor will be found in terms of percent approach to equilibrium ... 

A reversible reaction A B is conducted in a plug flow reactor. The rate equation is... 

FIG. 23-17 Multiple steady states of CSTRs, stable and unstable, adiabatic except the last item, (a) First-order reaction, A and C stable, B unstable, A is no good for a reactor, the dashed line is of a reversible reaction, (h) One, two, or three steady states depending on the combination Cj, Ty). (c) The reactions A B C, with five steady states, points 1, 3, and 5 stable, (d) Isothermal operation with the rate equation = 0 /(1 -I- C y = (C o Cy/t. 

Both the principles of chemical reaction kinetics and thermodynamic equilibrium are considered in choosing process conditions. Any complete rate equation for a reversible reaction involves the equilibrium constant, but quite often, complete rate equations are not readily available to the engineer. Thus, the engineer first must determine the temperature range in which the chemical reaction will proceed at a... 

All other reversible seeond-order rate equations have the same solution with the boundary eonditions assumed in Equation 3-176. Table 3-4 gives solutions for some reversible reaetions. 

Now we set up the solution of the rate equations that express the reversible chemical process ... 

The rate of reaction nearly always depends upon reactant concentrations and (for reversible reactions) product concentrations. The functional relationship between rate of reaction and system concentrations (usually at constant temperature, pressure, and other environmental conditions) is called the rate equation. 

The isolation experimental design can be illustrated with the rate equation v = kc%CB, for which we wish to determine the reaction orders a and b. We can set Cb >>> Ca, thus establishing pseudo-oth-order kinetics, and determine a, for example, by use of the integrated rate equations, experimentally following Ca as a function of time. By this technique we isolate reactant A for study. Having determined a, we may reverse the system and isolate B by setting Ca >>> Cb and thus determine b. 

To take the inverse Laplace transform means to reverse the process of taking the transform, and for this purpose a table of transforms is valuable. To illustrate, we consider a simple first-order reaction, whose differential rate equation is... 

In deciding whether to write an elementary reaction as either a reversible or an irreversible reaction, we take the practical view that if the reverse reaction is negligibly slow on the exp>erimental time scale, the reaction is essentially irreversible. Consider the alkaline hydrolysis of an ester, for which the rate equation is... 

The appearance of Cd in the denominator means that D is coupled to a reversible step prior to the rds. If k, and k- were so large that the fast preequilibrium assumption is valid, then the Cd term in the denominator would drop out, and we would have v = A 2 CaObCd, giving the composition of the rds transition state. If k2 is very much larger than it, and k i, Eq. (5-59) becomes v = A CaOb the first step is now the rds, and the rate equation gives the transition state composition. 

The reversible hydration of nitrogen-containing heterocyclic bases and their hydroxy and mercapto derivatives is acid-base catalyzed and, at constant pH, the reactions obey first-order rate equations. 2 26.27-33... 

The rate equation for reversible reactions with two substrates is defined. 

An interesting method, which also makes use of the concentration data of reaction components measured in the course of a complex reaction and which yields the values of relative rate constants, was worked out by Wei and Prater (28). It is an elegant procedure for solving the kinetics of systems with an arbitrary number of reversible first-order reactions the cases with some irreversible steps can be solved as well (28-30). Despite its sophisticated mathematical procedure, it does not require excessive experimental measurements. The use of this method in heterogeneous catalysis is restricted to the cases which can be transformed to a system of first-order reactions, e.g. when from the rate equations it is possible to factor out a function which is common to all the equations, so that first-order kinetics results. 

The rate equations determine the rate of change of the probability of a particular configuration, a, within an ensemble of growing crystals. They must include the rate constants for adding or subtracting units, which are assumed to obey microscopic reversibility. The net flux between configurations a and a which occur with probability P(a) and P(a ) respectively, and differ by one unit is ... 

Solution Example 4.5 was a reverse problem, where measured reactor performance was used to determine constants in the rate equation. We now treat the forward problem, where the kinetics are known and the reactor performance is desired. Obviously, the results of Run 1 should be closely duplicated. The solution uses the method of false transients for a variable-density system. The ideal gas law is used as the equation of state. The ODEs are... 

Example 7.11 showed how reaction rates can be adjusted to account for reversibility. The method uses a single constant, Kkinetic or Kthemo and is rigorous for both the forward and reverse rates when the reactions are elementary. For complex reactions with fitted rate equations, the method should produce good results provided the reaction always starts on the same side of equilibrium. 

Equilibrium Compositions for Single Reactions. We turn now to the problem of calculating the equilibrium composition for a single, homogeneous reaction. The most direct way of estimating equilibrium compositions is by simulating the reaction. Set the desired initial conditions and simulate an isothermal, constant-pressure, batch reaction. If the simulation is accurate, a real reaction could follow the same trajectory of composition versus time to approach equilibrium, but an accurate simulation is unnecessary. The solution can use the method of false transients. The rate equation must have a functional form consistent with the functional form of K,i,ermo> e.g., Equation (7.38). The time scale is unimportant and even the functional forms for the forward and reverse reactions have some latitude, as will be illustrated in the following example. 

In the other limiting case, when the denominator of Eq. (Ill) becomes large, the surface is heavily occupied because either the reaction from intermediate to product or the reverse reaction to reactant is slow. In this case the rate equation is approximately... 

Rapid exchange between Xi and Xi is reported in reference (3). This means that the forward and reverse reaction rates of this step are mnch faster than all others, and hence this particnlar step can be treated as a qnasi-eqnihbrium. The two intermediates in that step are present at all times in concentrations related to one another by a thermodynamic eqnihbrium constant and can be Inmped into one pseudo-intermediate [Xs]. This approach is very useM in reducing the number of terms in the denominator of the rate equation, which is equal to the square of the number of intermediates in the cycle (7). 

The rate equation for the reversible reaction of E and I must reflect both the forward (association) and reverse (dissociation) reactions ... 

If the forward and reverse reactions are nonelementary, perhaps involving the formation of chemical intermediates in multiple steps, then the form of the reaction rate equations can be more complex than Equations 5.33 to 5.36. 

A rate equation is required for this reaction taking place in dilute solution. It is expected that reaction will be pseudo first-order in the forward direction and second-order in reverse. The reaction is studied in a laboratory batch reactor starting with a solution of methyl acetate and with no products present. In one test, the initial concentration of methyl acetate was 0.05 kmol/m3 and the fraction hydrolysed at various times subsequently was ... 

If the forward reaction is pseudo first order and the reverse reaction second order, then, as discussed in Sections 1.4.4 and 1.4.5 in Volume 3, the rate equation may be written as ... 

The number of independent rate equations is the same as the number of independent stoichiometric relations. In the present example, reactions 2 and 3 are reversible and are not independent. 

The reversible rate equation, Equation 3.3, is easier to handle when the equilibrium constant is known. Then... 

The cis-trans isomerization of 1,2-dimethylcyclopropane at 453 C is a reversible first order reaction. The percentage of cis is shown as a function of t in sec in the table. Find the constants of the rate equation. 

The fraction, x, of -hydroxy crotonic acid that isomerizes into acetoacetic ester was measured at 25°C at various times. The data indicate reversibility. Find the rate equation. 

A reversible reaction is expected to have the rate equation,... 

The reversible gas phase reaction, A 2B, is conducted at 540 F and 3 atm in a PFR. Feed contains 30 mol% A and the balance inert materials, the total being 75 lbmols/hr. The rate equation is... 

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