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The Steady State Approximation

The preceding example points out one obvious problem, and implies another. The obvious problem is that rate laws for elementary processes are not immediately transferable to rate laws for the overall reaction. [Pg.724]

The implied problem is not so obvious. Typically, we determine rate laws in terms of amounts we can measure. For example, in the reaction between hydrogen and oxygen gases, we would want to express a rate law in terms of the amounts of H2 and O2. On the other hand, we certainly do not want to express a rate law [Pg.724]

Unless otherwise noted, all art on this page is Cengage Learning 2014. [Pg.724]

If the rate-determining step is the first step in the mechanism, then the rate law for the overall reaction is simply the rate law from the elementary process. (And because the first step does not have any intermediates as reactants, by definition the rate law can be expressed in terms of measurable quantities of chemical species.) Suppose, however, that the rate-determining step is the second step. Consider the following hypothetical two steps  [Pg.725]

B represents an intermediate product whose concentration is difficult to determine experimentally. A and C represent normal chemical reactants whose amounts can be measured. If the second step is the slow, rate-determining step, then it will force the first step to be backed up like cars on a one-lane road. [Pg.725]

In the simplest reaction mechanisms one particular step is usually rate determining. However, it is not unusual in complex, multistep reaction mechanisms for different steps to be rate determining under different sets of conditions. [Pg.732]

In cases where a specific rate-determining step cannot be chosen, an analysis called the steady-state approximation is often used. The central feature of this method is the assumption that the concentration of any intermediate remains constant as the reaction proceeds. An intermediate is neither a reactant nor a product but something that is formed and then consumed as the reaction proceeds. [Pg.732]

For example, the reaction between nitric oxide and hydrogen, [Pg.732]

In this mechanism the intermediate is N202. To apply the steady-state approximation to this mechanism, we assume the concentration of N202 remains constant. That is, [Pg.732]

we will identify the steps that produce N202 and those that consume N202 and write the rate law for each. Then we will apply the condition [Pg.732]

Solution Two questions must be answered. First, does the mechanism give the correct overall stoichiometry Adding the three steps does yield the correct balanced equation  [Pg.743]

Second, does the mechanism agree with the observed rate law Since the overall reaction rate is determined by the rate of the slowest step. [Pg.743]

Overall rate = rate of second step = 2[Cl][CHCl3] [Pg.743]

Since the chlorine atom is an intermediate, we must find a way to eliminate [Cl] in the rate law. This can be done by recognizing that since the first step is at equilibrium, its forward and reverse rates are equal  [Pg.743]

Taking the square root of both sides yields [Pg.743]

This set of two differential equations and one algebraic equation replaces the original set of three differential equations. In practice, kijk2 will not be infinitely small, but may nevertheless be small enough that dj/df 0, and [Pg.59]

The formal kinetics A B C occurs commonly in practice. Consider the reactivity of the free methyl radical, generated by azomethane pyrolysis, with an alkane, RH  [Pg.60]

If this reaction is carried out at a total pressure of 1 atm and at 700 K, [RH] 5i 10 molecules cm. Because of the high reactivity of CH3, one expects [RH]jj [CH3], and if the conversion of azomethane is small [RH] will be nearly constant over the course of the reaction. The bimolecular rate of the reaction of CH3 with RH, r = A 2[CH3][RH], can be expressed as /t2[CH3], where k 2 — A 2[RH]. The apparent kinetic order of this reaction is then 1, and with a constant concentration folded into the rate coefficient, is spoken of as pseudo-first order. This two-step reaction sequence is now formally equivalent to A- B- C. Using data from the literature, — 10 to 10 , and we expect the steady state ap- [Pg.60]

For example, if we apply the steady-state approximation to the catalytic intermediates of the 2-methylheptane/iso-octane isomerization from the previous section we obtain Eqs. (2.31)-(2.33). [Pg.53]

As Reiser describes it, once a radical chain is initiated, it propagates spontaneously until it is terminated by an encounter with another radical, by disproportionation, or in some other way. The average chain length, which ultimately determines the photosensitivity of the system, according to Reiser, is linked to the kinetic rate constants of the individual processes. Furthermore, the rate of initiation (i), propagation (p), and termination (t) can be fairly well described by the following equations  [Pg.270]

Reiser, Photoactive Polymers The Science and Technology of Resists, p. 128, John Wiley Sons, Hoboken, NJ (1989). [Pg.270]

Decker, A novel method for consuming oxygen instantaneously in photopolymerizable films, Makromol. Chem., 180, 2027 (1979). [Pg.270]

The ratio kp/ /ki. Reiser asserts, defines the characteristic ratio of the monomer system. It is not dependent on the model of initiation, and it measures the ability of the monomer to support a radical chain reaction. [Pg.271]

Traditionally, reaction mechanisms of the kind above have been analysed based on the steady-state approximation. The differential equations for this mechanism cannot be integrated analytically. Numerical integration was not readily available and thus approximations were the only options available to the researcher. The concentrations of the catalyst and of the intermediate, activated complex B are always only very low and even more so their derivatives [Cat] and [B]. In the steady-state approach these two derivatives are set to 0. [Pg.91]

Between about lO2 to 102 time units, the concentration [B] 10 4=[Cat]o, confirming the steady-state approximations. [Pg.91]

With the availability of numerical ODE solvers, exercises of the kind just presented are superfluous. While the results are the same for large parts of the data, numerical integration delivers a complete analysis that covers the whole reaction from time 0 to the end. [Pg.91]

Now suppose that the rate constants have values such that the rate of change of Cb is very small relative to the rates of change of other concentrations then in the conventional formulation it is stated that Cb is at steady state, and the assumption Cb = 0 is made. With this assumption and Eq. (3-140) we find [Pg.100]

The reaction displays simple first-order kinetics, with the observed first-order rate constant being equal to kik2l(k i + k.  [Pg.101]

This procedure constitutes an application of the steady-state approximation [also called the quasi-steady-state approximation, the Bodenstein approximation, or the stationary-state hypothesis]. It is a powerful method for the simplification of complicated rate equations, but because it is an approximation, it is not always valid. Sometimes the inapplicability of the steady-state approximation is easily detected for example, Eq. (3-143) predicts simple first-order behavior, and significant deviation from this behavior is evidence that the approximation cannot be applied. In more complex systems the validity of the steady-state approximation may be difficult to assess. Because it is an approximation in wide use, much critical attention has been directed to the steady-state hypothesis. [Pg.101]

The sufficient and necessary condition is therefore Cb iCa. As a consequence of imposing the more restrictive condition, which is obviously not correct throughout most of the reaction, it is possible for mathematical inconsistencies to arise in kinetic treatments based on the steady-state approximation. (The condition Cb = 0 is exact only at the moment when Cb passes through an extremum and at equilibrium.) [Pg.101]

A study of this system often is carried out with pseudo-order conditions relative to D. Then the apparent second-order rate constant is given by Eq. (3-146). [Pg.102]

we see that the rate of the reaction depends on the square root of the concentration of the initiator and linearly on the concentration of the monomer. The steady state approximation fails when the concentration of the monomer is so low that the initiation reaction cannot occur at the same rate as the termination reaction. Under these conditions, the termination reaction dominates the observed kinetics. [Pg.73]

One feature of the calculation so far has probably not gone unnoticed there is a considerable increase in mathematical complexity as soon as the reaction mechanism has more than a couple of steps. A reaction mechanism involving many steps is nearly always unsolvable analytically and alternative methods of solution are necessary. One approach is to integrate the rate laws numerically with a computer. An alternative approach, which continues to be widely used because it leads to convenient expressions and more readily digestible results, is to make an approximation. [Pg.252]

The steady-state approximation assumes that after an initial induction period, [Pg.252]

This approximation greatly simplifies the discussion of reaction mechanisms. For example, when we apply the approximation to the consecutive first-order mechanism, we set d[I]/df = 0 in eqn 7.11, which then becomes [Pg.253]

The product P is formed by unimolecular decay of I, so it follows that [Pg.253]

We see that P is formed by a first-order decay of A, with a rate constant the rate constant of the slower, rate-determining, step. We can write down the solution of this equation at once by substituting the solution for [A], eqn 7.10, and integrating  [Pg.253]

In industry, as well as in a test reactor in the laboratory, we are most often interested in the situation where a constant flow of reactants enters the reactor, leading to a constant output of products. In this case all transient behavior due to start up phenomena have died out and coverages and rates have reached a constant value. Hence, we can apply the steady state approximation, and set all differentials in Eqs. (142)-(145) equal to zero  [Pg.59]

The last equation is not independent of the others due to the site balance of Eq. (141) hence, in general, we have n-1 equations for a reaction containing n elementary steps. Note that steady state does not imply that surface concentrations are low. They just do not change with time. Hence, in the steady state approximation we can not describe time-dependent phenomena, but the approximation is sufficient to describe many important catalytic processes. [Pg.59]

Consider a closed system comprised of two, first-order, irreversible (one-way) elementary reactions with rate constants and ky. [Pg.105]

If C° denotes the concentration of A at time f = 0 and C% = Cc — 0, the material balance equations for this system are  [Pg.106]

Placing the functional form of x(t) into the equation for — gives  [Pg.106]

This first-order initial-value problem can easily be solved by the use of the integration factor method. That is. [Pg.106]

At tmax. the curve of Cc versus t shows an inflection point, that is, d w)/ dt ) = 0. [Pg.107]


The rate equation for the S z form is, from the steady-state approximation, as follows ... [Pg.107]

To be analytically useful equation 13.16 needs to be written in terms of the concentrations of enzyme and substrate. This is accomplished by applying the steady-state approximation, in which we assume that the concentration of ES is essentially constant. After an initial period in which the enzyme-substrate complex first forms, the rate of formation of ES... [Pg.636]

A second common approximation is the steady-state condition. That arises in the example if /fy is fast compared with kj in which case [i] remains very small at all times. If [i] is small then d[I] /dt is likely to be approximately zero at all times, and this condition is commonly invoked as a mnemonic in deriving the differential rate equations. The necessary condition is actually somewhat weaker (9). Eor equations 22a and b, the steady-state approximation leads, despite its different origin, to the same simplification in the differential equations as the pre-equihbrium condition, namely, equations 24a and b. [Pg.514]

Comparing this to the theoretical expression based on the steady state approximation suggests that the mechanism is not straightforward ... [Pg.524]

For a sequenee of reaetion steps two more eoneepts will be used in kinetics, besides the previous rules for single reaetions. One is the steady-state approximation and the seeond is the rate limiting step eoneept. These two are in strict sense incompatible, yet assumption of both causes little error. Both were explained on Figure 6.1.1 Boudart (1968) credits Kenzi Tamaru with the graphical representation of reaction sequences. Here this will be used quantitatively on a logarithmic scale. [Pg.123]

A useful approach that is often used in analysis and simplification of kinetic expressions is the steady-state approximation. It can be illustrated with a hypothetical reaction scheme ... [Pg.195]

Assume that the steady-state approximation can be applied to the intermediate TI. Derive the kinetic expression for hydrolysis of the imine. How many variables must be determined to construct the pH-rate profile What simplifying assumptions are justified at very high and very low pH values What are the kinetic expressions that result from these assumptions ... [Pg.506]

The overall rate of a chain process is determined by the rates of initiation, propagation, and termination reactions. Analysis of the kinetics of chain reactions normally depends on application of the steady-state approximation (see Section 4.2) to the radical intermediates. Such intermediates are highly reactive, and their concentrations are low and nearly constant throughout the course of the reaction ... [Pg.683]

The result of the steady-state condition is that the overall rate of initiation must equal the total rate of termination. The application of the steady-state approximation and the resulting equality of the initiation and termination rates permits formulation of a rate law for the reaction mechanism above. The overall stoichiometry of a free-radical chain reaction is independent of the initiating and termination steps because the reactants are consumed and products formed almost entirely in the propagation steps. [Pg.683]

Following the steady-state approximation, both propagation steps must proceed at the same rate otherwise, the concentration of A- or C- would build up. By substituting for the concentration of the intermediate C-, we obtain ... [Pg.684]

The concentration of the reaction intermediate AB may be determined by using the steady state approximation for intermediates,... [Pg.28]

Applying the rate expressions to Equations 1-222, 1-223, 1-224, 1-225 and 1-226, and using the steady state approximation for CH3, C2H5, and H, for a eonstant volume bateh reaetor yields ... [Pg.53]

This also accounts for the production of the small amount of butane. If the reaction mechanism were steps 1, 2, 3, 4, 5a, and 5b, then applying the steady state approximations would give the overall order of reaction as 1/2. [Pg.56]

Consider further Scheme XIV and rate equations (3-139) to (3-141). Evidently Cb will be small relative to (Ca + Cc) if ( -i + 2) i- Then B plays the role of a reactive intermediate in the overall reaction A— C. This is the usual condition that is taken as a warrant for the application of the steady-state approximation. If Cb is small, it is reasonable that Cb will be small throughout most the reaction, so it is set equal to zero. As Wong (53) has pointed out, however, the condition Cb = 0 is a sufficient but unnecessary condition for Eq. (3-142) to hold. Erom Eq. (3-140) we obtain... [Pg.101]

Proceeding in the usual way to apply the steady-state approximation to the complex ES ... [Pg.102]

Several features of this treatment are of interest. Compare the denominators of Eqs. (3-147) and (3-149) Miller has pointed out that the form of Eq. (3-147) is usually seen in chemical applications of the steady-state approximation, whereas the form of Eq. (3-149) appears in biochemical applications. The difference arises from the manner in which one uses the mass balance expressions, and this depends upon the type of system being studied and the information available. [Pg.103]

The quantitative description of enzyme kinetics has been developed in great detail by applying the steady-state approximation to all intermediate forms of the enzyme. Some of the kinetic schemes are extremely complex, and even with the aid of the steady-state treatment the algebraic manipulations are formidable. Kineticists have, therefore, developed ingenious schemes for writing down the steady-state rate equations directly from the kinetic scheme without carrying out the intermediate algebra." -" ... [Pg.104]

One way to examine the validity of the steady-state approximation is to compare concentration—time curves calculated with exact solutions and with steady-state solutions. Figure 3-10 shows such a comparison for Scheme XIV and the parameters, ki = 0.01 s , k i = 1 s , 2 = 2 s . The period during which the concentration of the intermediate builds up from its initial value of zero to the quasi-steady-state when dcfjdt is vei small is called the pre-steady-state or transient stage in Fig. 3-10 this lasts for about 2 s. For the remainder of the reaction (over 500 s) the steady-state and exact solutions are in excellent agreement. Because the concen-... [Pg.104]

In the preceding subsection we described the preequilibrium assumption. Let us now see how that assumption is related to the steady-state approximation. Scheme XIV will serve for the discussion. The equilibrium and steady-state expressions for the intermediate concentration are... [Pg.105]

It, therefore, appears that the equilibrium approximation is a special case of the steady-state approximation, namely, the case i > 2- This may be, but it is possible for the equilibrium approximation to be valid when the steady-state approximation is not. Consider the extreme but real example of an acid-base preequilibrium, which on the time scale of the following slow step is practically instantaneous. Suppose some kind of forcing function were to be applied to c, causing it to undergo large and sudden variations then Cb would follow Ca almost immediately, according to Eq. (3-153). The equilibrium description would be veiy accurate, but the wide variations in Cb would vitiate the steady-state description. There appear to be three classes of practical behavior, as defined by these conditions ... [Pg.105]

The relative fluctuations in Monte Carlo simulations are of the order of magnitude where N is the total number of molecules in the simulation. The observed error in kinetic simulations is about 1-2% when lO molecules are used. In the computer calculations described by Schaad, the grids of the technique shown here are replaced by computer memory, so the capacity of the memory is one limit on the maximum number of molecules. Other programs for stochastic simulation make use of different routes of calculation, and the number of molecules is not a limitation. Enzyme kinetics and very complex oscillatory reactions have been modeled. These simulations are valuable for establishing whether a postulated kinetic scheme is reasonable, for examining the appearance of extrema or induction periods, applicability of the steady-state approximation, and so on. Even the manual method is useful for such purposes. [Pg.114]

For Scheme XIV, and for each of the following sets of rate constants, calculate the exact relative concentration cb/ca as a function of time. Also, for each set, calculate the approximate values of cb/ca using both the equilibrium assumption and the steady-state approximation. [Pg.130]

Apply the steady-state approximation to Scheme XXII for ester hydrolysis to find how the experimental second-order rate constant qh is related to the elementary rate constants. [Pg.130]

If a reaction system consists of more than one elementary reversible reaction, there will be more than one relaxation time in general, the number of relaxation times is equal to the number of states of the system minus one. (However, even for multistep reactions, only a single relaxation time will be observed if all intermediates are present at vanishingly low concentrations, that is, if the steady-state approximation is valid.) The relaxation times are coupled, in that each relaxation time includes contributions from all of the system rate constants. A system of more than... [Pg.139]

Suppose in Scheme XVI that the steady-state approximation is applicable to the intermediate A. ... [Pg.183]

Either step could be rate determining. Study of many reactions has shown that most occur with a significant isotopic effect, but for some reactions the isotope effect is absent.If we apply the steady state approximation to the intermediate, this reaction scheme leads to... [Pg.298]

Because the cationic intermediate is unstable, it will be permissible to apply the steady-state approximation, leading to Eq. (8-65) for the reaction rate. [Pg.428]

Using the steady-state approximation, and taking into account the fact that the values of kQn are relatively insensitive to the ratio of radii rx/rB (Debye, 1942), so that one can set rx/rB l, one gets the expression in Scheme 3-33 as a good approximation for the encounter rate coefficient (R = gas constant). [Pg.56]

Applying the steady-state approximation to the concentration of the intermediate, C, one obtains the rate equation... [Pg.409]

Applying the steady-state approximation to both [II] and [IV] individually leads to... [Pg.418]

Applying the steady-state approximation to both [VI] and [VIII] one obtains (29) for the steady-state concentration of VI and (30) for the steady-state concentration of VIII, viz. [Pg.427]

Given the postulated reaction scheme, the net rate of reaction often takes a simple form when it is expressed in terms of the concentration of the intermediate. Such an expression is algebraically correct, and is the form one needs so as to propose and interpret the mechanism. This form is, however, usually not useful for the analysis of the concentration-time curves. In such an expression the reaction rate is given in terms of the concentration of the intermediate, which is generally unknown at the outset. To eliminate the concentration term for the intermediate, one may enlist certain approximations, such as the steady-state approximation. This particular method is applicable when the intermediate remains at trace levels. [Pg.70]


See other pages where The Steady State Approximation is mentioned: [Pg.22]    [Pg.57]    [Pg.100]    [Pg.102]    [Pg.109]    [Pg.109]    [Pg.219]    [Pg.352]    [Pg.367]    [Pg.245]    [Pg.259]    [Pg.412]    [Pg.416]    [Pg.419]   


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Steady-state approximation

The Approximations

The Steady State

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