

[TOP]
Elementary reaction kinetics In this section, we shall relate kinetic data to postulated reaction mechanisms (1) Definition of Elementary reactions
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(1) Definition of Elementary reactions
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(2) The molecularity of a reaction
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(3) Molecularity vs. order
It is most important to distinguish molecularity from reaction order:
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(4) The rate laws of elementary reactions
Rule: The rate law for elementary reactions may be deduced directly from their molecularity. In fact: (A) The rate law of a unimolecular elementary reaction is firstorder in the reactant: where P denotes products (several different species may be formed). A unimolecular reaction is firstorder because the number of A molecules that decay in a short interval is proportional to the number available to decay. Therefor the rate of decomposition of A is proportional to its molar concentration.(B) An elementary bimolecular reaction has a secondorder rate law: A bimolecular reaction is secondorder because its rate is proportional to the rate at which the reactant species meet, which in turn is proportional to their concentrations. Therefore, if we believe that a reaction is a singlestep, bimolecular process, we can write down the rate law (and then go on to test it). Bimolecular elementary reactions are believed to account for many homogeneous reactions.Note that the converse of this rule does not follow, that is, for example
secondorder
rate law does not imply that the reaction is bimolecular  the reaction
might be complex ! Also, care must be taken to ensure that the reaction
/step that we are considering is really elementary. For example, the reaction
H_{2}(g) + I_{2}(l) > 2HI(g) may look simple, but it is
not elementary, and in fact, it has a very reaction mechanism, and hence
a complex rate law.
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(5) Consecutive elementary reactions  Variations of Concentration with
time
(51) A full investigation Some reactions proceed through the formation of an intermediate (I), as in the consecutive unimolecular reactions: For such a process we have:(1) The concentration of A is (i) decreasing through unimolecular decomposition of A, and (ii) is not replenished. Thus we have: ………………….(eqn. 1)
(1) From eqn. 1 we obtain: Through these equations we may deduce that (see fig. 1):………………….(eqn. 4)
We may also deduce that the concentration of the intermediate I is at a maximum when: i.e.: i.e.:
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(52) Approximations It is important to note that there is considerable increase in mathematical complexity as soon as the reaction mechanism more than a couple of steps, and in fact, a reaction scheme which involving many steps is nearly always unsolvable analytically, and hence alternative methods of solution are necessary. One approach to such a problem is to integrate the rate laws numerically. However, other approximation methods are possible.
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(521) The steady state approximation (SSA)
An alternative and more common to numerical intregration approach is to invoke the so called steadystate approximation. The steadystate approximation (SSA) assumes that, after an initial induction period (i.e. an interval during which the concentrations of intermediates, I, rise from zero, see fig. 2) the rates of change of concentrations of all reach intermediates are negligibly small, i.e.: The SSA approximation significantly reduces the complexity. For example, in our case, where the meachinsm is given by: eqn. 2 would become: i.e. we may assume that: and hence: and so we see that P is formed by a first order decay of A, with rate constant k_{a} (= the rate constant of the slower, ratedetermining step). Hence from these deductions and from eqn. 4 we can write down: as obtained before (though through a much longer process).The SSA is however not always valid. To be valid, we require a scenario
where the concentrtaion of intermediates must remain small and
hardly change during most of the course of the reaction. In other
words, the concentration of I in the course of the reaction must be described
by Fig. 2 not by Fig. 1.
Mathematically, the required conditions for the SSA to be vaild are that [I] << [A] & [P]. Thus since we have: then under the SSA we must have: i.e. we requre that: In other words, the first reaction producing intermediate I must be slow relative to the rate at which intermediate I reacts and is depleated. This will not allow intermediate I to accumulate as happened in Fig. 1. In fact, the greater the difference between teh two rate constants, the more accurate the SSA will be.
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(522) The rate determining step Let us once agian consider the mechanism: Once agian we shall however assume that k_{b} >> k_{a} . In such a scenario, whenever a molecule I is formed, it decays rapidly to P. (Note that these are essentially the same conditions that were required for the SSA, and in fact, we will be getting the same result.)Mathematically speaking, referring to the full mathematyical derivatiuon in section (5.1) we have: i.e. eqn. 6 reduces to: which shows that the formation of the final product P depends on only the smaller of the two rate constants. In other words, the rate of formation of P depends on the rate at which I is formed, not on the rate at which I changes into P. For this reason, the step'A > I' is called the ratedetermining step of the reaction. Note: Similar remarks apply to more complicated reaction mechanisms,
and in general the ratedetermining step is the one with the smallest rate
constant.
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Let us now focus our attention to a slightly more complicated mechanism in which an intermediate I reaches an equilibrium with the reactants A and B, i.e.: where the rate constants are k_{a} and k'_{a} for the forward and reverse reactions of the equilibrium and k_{b} for the final step. This scheme involves a preequilibrium, in which an intermediate is in equilibrium with the reactants.A preequilibrium arises when the rates of formation of the intermediate and its decay back into reactants are much faster than its rate of formation of the products, that is, when k'_{a} >> k_{b} but not when k_{b} >> k'_{a}. Since we are assuming that A, B and I are in equilibrium we may write: From now onwards, we may take two approaches:(1) We assume that the rate of reaction of I to form P is too slow to affect the maintenance of the preequilibrium. In such cases we have: (i.e. the rate law has the form of a secondorder rate law).(2) We assume that reaction of I to form P may affect the maintenance of the preequilibrium. In such cases we have: where through the steady state approximation we obtain: i.e.: Note that if k_{a}`>>k_{b} then this last equation will reduce to the one in approach (1).Note: An application of all this theory may be found in the MichaelisMenten
mechanism which is discussed at a later stage.

