The Lindemann Mechanism
(This mechanism is sometimes called the Lindemann-Hinshelwood mechanism. Lindemann found it and Hinshelwood polished it.)
There is a class of gas-phase reactions of which one example is:
N2O5 → NO2 + NO3. (1)In chemical kinetics experiments sometimes this reaction looks like a first order reaction and sometimes it looks like a second order reaction, depending on the reaction conditions. For example, you might change the apparent order of the reaction by changing the pressure.
The postulated mechanism is:
(2)The N2O5* is said to be "activated" N2O5. The rate fo the reaction is given by the rate at which the activated N2O5 falls apart into products. We can see from Equation 3 that the formation of products is a unimolecular process which has a rate law of
. (4)The activated N2O5 (that is, the N2O5*) is a transient species because it is neither a reactant nor a product. It is produced during the course of the reaction and it is gone at the end of the reaction.
We can also write a rate law for the change in concentration of this transient species. We do this by looking at all of the elementary steps in the proposed mechanism which produce or destroy this transient species. Notice that from Equation 2 the transient species is produced by the forward process with rate constant, k2, and it is destroyed by the reverse process with rate constant, k-2. The transient species is also destroyed by the process in Equation 3, where it "falls apart" to give the products at a rate, k1. So the elementary reaction steps that produce the transient species must appear in the rate equation for the transient with a positive sign and the steps that destroy the transient appear with a negative sign. Thus the complete description for the rate of change of the concentration of the transient is,
. (5)Equations 4 and 5 form a set of coupled first order differential equations in time which can be solved exactly, but it is easier to make an approximation which will allow us to obtain a simple, but approximate, solution.
This approximation is called the steady state approximation and it will allow us to obtain, relatively easily, a simple solution which is good enough for our purposes. It turns out that the concentrations of transient species usually build up quickly and decay away very slowly during the course of the reaction, and the concentrations of transient species usually do not ever get very large. If this is the case (and we will assume that it is the case in this course) we can say that the rate of change of the concentration of transient species is essentially zero over most of the course of the reaction.
Our procedure is then to look for transient species and approximate their rate of change as small (say zero) over most of the course of the reaction. That is, we set
. (6)Then Equation 5 becomes,
. (7)Solve this for [N2O5*] to get,
. (8)Then we can insert this transient species concentration back into our original rate equation for the formation of product (Equation 4) to obtain an overall rate equation for the reaction,
. (9)We now must ask whether or not this rate law fits the experimental fact that sometimes the reaction appears to be first order in N2O5 and sometimes it appears to be second order in N2O5.
To check this we look at Equation 9 in two limiting cases:
If k1 >> k-2 [N2O5],
that is, when [N2O5] is small,
Rate ≈ k2[N2O5]2, (10)and the reaction looks like it is second order.
If k1 << k-2 [N2O5],
that is, when [N2O5] is large,
, (11)and the reaction looks like it is first order.
There are lots of variations to the Lindemann mechanism. (From now on we will use the letters A, B, C, ..., etc to designate molecules.)
Let's look at a variation where there is an inert gas, M, present. The proposed mechanism is,
(12)The rate is still given by an equation similar to Equation 4,
. (15)But for the transient species we get
. (16)In Equation 16 we have already applied the steady state approximation by setting this rate as approximately equal to zero. Using the steady state approximation we solve for [A*],
, (17)and plug this into Equation 15 to get the reaction rate as,
. (18)If [M] = 0 then we are back to the Lindemann mechanism. However, there are new possibilities. We can change the appearance of the kinetics by manipulating [M]. For example, if we make [M] very large, large enough to overwhelm all the other terms in the rate law, the effective rate law becomes,
. (19)That is, the reaction becomes effectively first order in A independent of the values of either [A] or [M] (as long as [M] is large).
Sometimes the conditions are such that there is virtually no "self activation" of the A molecules and all of the activation comes from collisions of A molecules with the inert gas molecules, M. Our mechanism covers this case. We can obtain the rate law in this case by setting k2 and k-2 equal to zero. This gives
. (20)In this situation the reaction is first order in A regardless of the M concentration, but the actual reaction rate depends on [M]. One can define an effective first order rate constant by rewriting the rate equation as
, (21)so that the effective rate constant is
. (22)Clearly you can change the effective first order rate constant, and hence the half-life of the reaction, by adjusting [M]. These examples illustrate a statement that was made earlier in our study of chemical kinetics, namely that the rate law may contain species that are not part of the balanced chemical reaction. Other examples of species not in the balanced reaction occuring in the rate law would include catalysis, where a catalyst does not normally appear in the balanced reaction but does appear in the rate law.
Copyright 2004, W. R. Salzman
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