The Steady State and Equilibrium Assumptions in chemical Kinetics Dwight C. Tardy and E. David Cater University of lowa, lowa City, IA 52242
In introductory chemistry courses (particularly physical chemistry), the technique of obtaining or deriving a rate law from a mechanism comprised of elementary steps is usually discussed. For simplification purposes two approximation methods are often used in obtaining the desired rate law: a) steady state approximation b) equilibrium approximation The former method has recently been discussed by Porter and and is appropriate for a low Skinner ( I ) in THIS JOURNAL concentration intermediate that is relatively constant with time. The latter is a specific case of the steady state approximation, namely for species which have equilibrium concentrations. The validity of these approximations has also been discussed (2). In many cases both approximations lead to the same overall rate law. In other cases, such as the thermal decomposition of ozone, they do not. An elusive factor of two in the rate law
k~ = 2.Yfi X
lop2enp
[
"I
(
)
'C-1.98T mole-sec
and M is any collision partner. The net rate for the disappearance of ozone a t time t , Rolnet(t),is
For 0 2 and 0 we may write .. . .
a t the wrong point in the rate law de;iuation. At least one popular physical chemistry text (3) includes the incorrect factor of two in deriving the ozone rate law, for just this reason. Examination of current textbooks of physical chemistry show that most gloss over how one might judge whether (or not) the steady state or equilibrium approximation should he used. In this paper we will delineate between the two approximations and demonstrate the importance of making the equilibrium assumption a t the appropriate step in the calculation. The ozone decomposition was chosen as the example because of the current interest in the ozone depletion prohlem, and because the elementary rate constants are known. We first discuss the principles, then show numerically how closely both approximations approach the exact case. The problem of the factor of two arises when the final product of the overall reaction appears also in one of the steps in the mechanism. Thus, for the overall reaction A B E the first mechanism gives no prohlem, but the second mechanism may.
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The exact time dependence of the 0 , 0 z , and On concentration can he calculated by solving eqns. (31, (41, and (5) simultaneously, a procedure requiring the use of a computer. However, simplification in ohtaining the rate law from eqns. (3)-(5) is found by assuming a steady state (or more precisely a quasi-steady state) for the concentration of oxygen atoms. This requires that [0](or more precisely [0]/[03]) to he constant with time, i.e., d[O]/dt = 0. Thus, from eqn. ( 5 )
so that substituting [Om]into (3) gives
A+BeC+D no problem
A+C-E
AfC-D
steady state and equilibrium approximations give different rate laws if not done correctly
For the overall reaction
The factor of two which appears in eqn. (6) has conceptual importance. In terms of Roal(t),Ro,-l(t), and Roa2(t),eqn. (5) along with the steady state assumption gives Ro3W = Ro3-'(t) +Ro?(t)
the proposed mechanism is Physically, under the steady state assumption the rate of 0 atom production is equal to the net rate of consumption so that the net rate of producing 0 atoms from reaction (1)must be balanced by the rate of depletion of 0 atoms by reaction (21, i.e., where kl, k-I and kz (4) are
Roal(t)R o 3 - ' ( t ) Volume 60 Number 2
= Ro?(t)
February 1983
109
Hence the net depletion of O3 must he given by the sum of depletions caused by reactions (1) and (2) which gives:
This accounts for the factor of two. Under certain conditions this solution can he further simplified. Namely, if the reverse step in reaction (1) is much faster than (2) then it can be assumed that the concentration of O atoms is given by the equilibrium constant for the reaction
sumption [0] starts off high, decreases, and approaches the steady state and real concentration for large percent conversion. The term k2[O3]becomes much less than h-1[02] for this reaction only as O3 disappears toward high extent of reaction. Thus, the equilihrium assumption is a poor one in this example for practical calculations. The above example is not one in which application of the equilibrium assumption is appropriate as shown by the figure. The uniqueness of the factor of two is due to the fact that the same reactant is consumed in two separate steps. This factor does not appear, for example, in the following mechanism for the reaction: A C D.
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Then Here reaction (2) is really assumed to be a small (negligible) werturbation on the O atom concentration and the dvnamic kquilihrium concept is valid. Then the net rate of 0;disappearance is Using the steady state approximate for B which is equivalent to the expression that arises when i t is assumed that h-1[02] >> kz[Oa] and this limit is used in eqn. (7). This limitation puts restrictions both on h-1 and hz and the concentrations of 0 2 and 03. However, the above procedure is not normally used in arriving at the rate law with the equilibrium assumption. Instead the rate expression for 0 3 (eqn. (3)) is used with the equilibrium oxygen atom concentration.
In this expression the factor of two is missing. The reason is obvious. By assuming equilibrium we tacitly assumed that so that eqn. (1) gives In other words, we have neglected the net disappearance of O3 via reaction (1).Ro32(t)is a small number compared to the absolute magnitude of Ro,'(t) or Ro,-'(t), but it is just this difference which drives the reaction. Clearly substituting into the proper equation is necessary; the proper equation can he deduced from the nature of the problem. T o clarify the implications of using the steady state andlor equilihrium approximations, three plots of [O] versus time and extent of reaction are shown in the figure. The plot laheled "exact" plot was ohtained by solving eqns. (3)-(5) simultaneously to obtain exact concentrations. The second and third plots were obtained after simplifying the calculation by applying the steady state and equilihrium approximations, eqns. (7) and (81, respectively. The temperature chosen was 10O0C,and the initial ozone concentration was 100 tom. These are values at which actual experiments might be performed. The calculations were performed on an IBM 360165 digital comwuter nsine a Gear Inteeration Routine ( 5 ) .Note that it
At the beginning of the reaction the equilibrium condition is not fulfilled (the [Oz] is extremely low), however, as the reaction proceeds, [Oz] increases. For the equilihrium as110
Journal of Chemical Education
we get
As expected from the stoichiometry of the bulk reaction the rates of disappearance of A and C are equal. If equilihrium is assumed for the first reaction, then k-I >> kz[C] and the rates are h2 h ~ l h - 1[C][A].If the equilihrium expression for B is initially assumed, then by necessity
As before, incorrect use of the equilibrium assumption leads to an erroneous rate if A and C are identical. A direct comparison can be seen by adding R ~ " ~ l ( t )R p t ( t ) = Riotaln"(t)
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Plots of 101 (atomslcc) versus time tor the decomposition of Os at 10O0C calculated using the (11 ateady state approximation (0).(2) equilibrium approxiand (31exact solution (.I The extent of reaction is provided on the mation (A), upper scale.
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where A is replaced by B, so that for the general case (from the balanced equation: 2A D) And then assuming equilibrium, k-l[Dl
>> k z [ C ]
and then Rt,,wD'Yt)
=
2 R ~ = ~=~ 2( R t )p t ( t )
While if equilibrium is initially assumed, Rstp(t)
= RA"*'(t) = Ra"*Yt)
In summary, the equilibrium assumption, taken literally, does not give a net contribution to the rate for a reactant involved in an equilibrium step. The ozone reaction also is degenerate in this sense, namely that 02 (product) is produced in both elementary steps. This contrasts with the earlier discussion in which the reactant, 03, was followed.This would affect the expression for the rate of oxygen production since step one produces one molecule of Oz and step two produces two molecules of 0 2 . Thus, the false equilibrium assumption would give 213 of the current equilibrium assumption. This can be analyzed by considering the sequence A$B+D
and calculating the rate of production for D and E. Assuming a steady state, we get RnnSt(t)= k,[A] - k-I[B][D]
if equilibrium is literally assumed, then the appearance of a product from that step will be zero. In conclusion, as shown for the ozone system, the concentrations predicted by steady state and equilibrium assumptions are different and, depending on the progress of reaction, can be far from the exact concentration. This fact has always been known. We have also shown, that under the equilibrium assumption, the rate of product production or consumption of reactant in the eauilibrium sten is not zero. and careful manipulation must be made to arrive a t the correct rate constant in the rate expression. The derived orders will be correct, however, the combined rate constant will be off by a ratio of small whole numbers. The eauilibrium assumntion must be made after the steady state a k m p t i o n . Acknowledgment Thr aurl~t.rignarly apl~rc.< inrr rwti\.in:: c w n p u ~ funds ~r i r m 1I1e1nirersit?.
