Summary of IUPAC definition

$\def\d{\mathrm{d}}$In IUPAC recommendations from 1994[1, 1166–1167], the authors discuss the process

$$\ce{A <-->[$k_1$][$k_{-1}$] X\\ X + C ->[$k_2$] D}$$

A kinetic stationary state or steady state is defined very reasonably:

$$\frac{\d\ce{[X]}}{\d t}=0.\tag2$$

They go on to emphasise that condition $(2)$ is not equivalent to $\ce{[X]}$ being constant, 'not even approximately'; simply that the change in $\ce{[X]}$ is small compared to changes in concentrations of $\ce{[A]}$ and $\ce{[D]}$ because $\ce{[X]}$ itself is small. If $\ce{[X]}$ were constant in a stationary state, they claim it would lead to the following contradiction.

  • Say reactant $\ce{C}$ is in excess. Assume that $\ce{[X]}$ is constant. Then the law of mass action $v = k_2\ce{[X][C]}$ would reduce to some $v = k'\ce{[C]} \overset{\mathrm{excess}}{\approx} k\in\Bbb{R}$. In other words, production of $\ce{D}$ could continue at a constant rate even after $\ce{A}$ has run out.

The document concludes that $\ce{[X]}$ cannot possibly be constant in steady-state conditions.

How I would resolve the apparent contradiction

I do not see a way around $\ce{[X]}$ being constant. Indeed, for any reasonable function[a] $\ce{[X]}:t'\to \Bbb{R^{0+}}$ equation $(2)$ implies

$$\int_0^t \frac{\d\ce{[X]}(t')}{\d (t')}\d(t')= \ce{[X]}(t) = r \in \Bbb{R^{0+}}\tag3$$

where $t'$ was simply a dummy variable.

  • Resolution to contradiction: If $\ce{C}$ is in a large excess, $\ce{[X]}$ is small, and $\ce{A}$ has run out, it is simply not possible for steady-state conditions to hold. In other words, it is our stationary state approximation $(2)$ which fails to hold for $\ce{X}$. So we can easily keep condition (3) for a steady state.

    More accurately, if $\ce{A}$ has run out, the only possible stationary state would have been $\ce{[X]} = 0$. This is because, at a stationary state, $$\ce{[X]} = \frac{k_1\ce{[A]}}{k_{-1} + k_2\ce{[C]}} \overset{\ce{[A]} = 0}{=} 0.$$ If $\ce{[X]} = 0$, it is unfeasible for the reaction to continue at non-zero constant velocity since $v = k'\ce{[C]} = 0$ (where $k' = k_2\ce{[X]}$).


  • Is / is not the concentration of $\ce{X}$ constant, approximately or otherwise, in a kinetic stationary state (steady state)?

IUPAC states $\ce{[X](steady\ state)}$ is not even approximately constant because it would lead to contradictions.

I claim that $\ce{[X](steady\ state)}$ is constant, and contradictions surface when a steady state itself is impossible given certain constraints (or the constraints themselves are contradictory).

[a] Among other things that it is diffentiable everywhere in its domain (i.e., reaction rate is defined for every time $t'$).

[1] Muller, P. Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994). Pure and Applied Chemistry 2009, 66 (5), 1077–1184. DOI: 10.1351/pac199466051077. pages 1166–1167

  • $\begingroup$ The trick to understanding this mathematically is through singular perturbation. At short times the PSSH is invalid; at longer times it is. $\endgroup$
    – Curt F.
    Nov 26, 2017 at 17:29
  • $\begingroup$ Also exactly what is meant by "approximately" is very important. \$100 is approximately \$0 to Bill Gates, but not to me. $\endgroup$
    – Curt F.
    Nov 26, 2017 at 17:31
  • $\begingroup$ @CurtF. Are you busy at this time? Would you consider expanding your comments into an answer (with mathematical argumentation)? Or perhaps join me in chat? $\endgroup$ Nov 26, 2017 at 17:59
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    $\begingroup$ I deleted my answer for now. I shot from the hip and was talking at you not understanding what you wrote. After really reading the IUPAC definition I'm having trouble understanding how their #1 definition applies to a steady state. For #1 I think that the use of steady state for [X] is flawed. The #2 definition is ok. // I'm chewing on this for the moment. I'll repost with a new answer and a very different analysis. $\endgroup$
    – MaxW
    Nov 29, 2017 at 5:25
  • $\begingroup$ @MaxW Thank you for the update. It could also have been that my question was badly worded or a bit misleading. I look forward to your answer. Take your time, there is no rush. $\endgroup$ Nov 29, 2017 at 17:04

2 Answers 2


IUPAC should say they are defining the "steady state approximation" rather than the steady state.

There are two approximations in this technique.

First, that final reaction step is not an equilibrium (reverse reaction rate is zero).

Second, that the intermediate is so unstable that its concentration (and therefore the rate of change of the concentration) is approximately zero.

Compare the IUPAC definition to the following 1954 U.S. National Research Council explanation:


The most generally applicable of these simple methods is the so-called "steady-state approximation." A steady state may be defined as a condition in which the rates of change of the concentrations of the several intermediates are very small compared to the rates of change of the concentrations of the reactants and products. This condition is realizable whenever the ratio of the concentrations of the intermediates to the concentrations of the reactants is very much less than unity. When this condition is not attained, the method is not applicable; however, it should not then be necessary since the (larger) concentrations of the intermediates could be measured by experimental means. ... The steady-state approximation consists in setting the rates of change of each of the intermediates equal to zero and in solving simultaneously the resulting algebraic equations.

Overall, if the situation was truly, not just approximately, steady state, yes the concentration of the intermediate would be constant, but the point is that the "stead state approximation" is useful even when the concentration of the intermediate, though always near zero in absolute terms, decreases to say half of its original concentration over the course of an experiment (as the starting material, A, decreases to half its original concentration).

Additionally, a very small $\ce{[X]}$ does not purely-mathematically imply a small rate of change of $\ce{[X]}$, because $\ce{[X]}$ could rapidly oscillate within a small, near-zero, range, resulting in brief instances of large rate of change. Instead, $\ce{[X]}$ being small over a given period of time really just places a limit on how long of a time the rate of change can exceed a certain value. For example, if $\ce{[X]} < 0.001\ \mathrm{M}$,
$|\mathrm{d}[\ce{X}]/\mathrm{d}t|$ must not exceed $1\ \mathrm{M/s}$ for more than $0.001\ \mathrm{s}$.

  • 2
    $\begingroup$ I agree that IUPAC should say they are defining the "steady state approximation" rather than the steady state. // I disagree that Second, that the intermediate is so unstable that its concentration (an therefore the rate of change of the concentration) is approximately zero. It really should be that the rate of change of the intermediate is insignificant to the rates of change of the concentrations of the reactants and products. It is a significant figures argument, not that the intermediate isn't changing concentration. $\endgroup$
    – MaxW
    Nov 29, 2017 at 17:16
  • $\begingroup$ @MaxW I agree with your phrasing "the rate of change of the intermediate is insignificant to the rates of change of the concentrations of the reactants and products", but I also think my phrasing "is approximately zero" is equivalent to your phrasing "is insignificant". $\endgroup$
    – DavePhD
    Nov 29, 2017 at 17:41
  • $\begingroup$ I up voted your answer, but I'll make a different point. There are two definitions in the IUPAC text. You're talking about the first definition which is what the OP was questioning. The second part of the IUPAC definition for a flow reaction would have a true steady state. $\endgroup$
    – MaxW
    Dec 1, 2017 at 17:53
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    $\begingroup$ @LinearChristmas yes, I agree, [X] could rapidly oscillate, even though it is very small, resulting in a large rate some of the time. There needs to be an additional, so far unstated, assumption to rule that out. $\endgroup$
    – DavePhD
    Dec 4, 2017 at 19:19
  • $\begingroup$ Sorry, I managed to delete my comment before you replied. My own example was a poor (since the constant will not change sign), but yes the main point stands. If you add this point to your answer, I will accept it. $\endgroup$ Dec 4, 2017 at 19:31

Ok, this isn't an answer, but below is the text in question from Muller's Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994). The definition contains two parts.

I agree with DavePHD that the first definition, which is for a batch reaction, would be better called what is known as the STEADY STATE APPROXIMATION

The second definition is for a flow reactor which is at a true STEADY STATE. In this case all the reactants, intermediates, and products would have constant concentrations. That is to say the relative fluctuations (%) in those concentrations would be on the order of the precision with which the flows could be controlled.

Text is below

steady state (or stationary state)

(1) In a kinetic analysis of a complex reaction involving unstable intermediates in low concentration, the rate of change of each such intermediate is set equal to zero, so that the rate equation can be expressed as a function of the concentrations of chemical species present in macroscopic amounts. For example, assume that $\ce{X}$ is an unstable intermediate in the reaction sequence:

$$\ce{A <-->[$k_1$][k_{-1}] X}$$

$$\ce{X + C ->[$k_2$] D}$$

Conservation of mass requires that:

$$\ce{[A] + [X] + [D] = [A]_0}$$

which, since $\ce{[A]_0}$ is constant, implies:

$$\dfrac{-d\ce{[X]}}{dt} = \dfrac{d\ce{[A]}}{dt} + \dfrac{d\ce{[D]}}{dt}$$

Since $\ce{[X]}$ is negligibly small, the rate of formation of $\ce{D}$ is essentially equal to the rate of disappearance of $\ce{A}$, and the rate of change of $\ce{[X]}$ can be set equal to zero. Applying the steady state approximation ($d\ce{[X]}/dt = 0$) allows the elimination of $\ce{[X]}$ from the kinetic equations, whereupon the rate of reaction is expressed

$$ \dfrac{d\ce{[D]}}{dt} = -\dfrac{d\ce{[A]}}{dt} = \dfrac{k_1k_2\ce{[A][C]}}{k_{-1} + k_2\ce{[C]}}$$

Note: The steady-state approximation does not imply that $\ce{[X]}$ is even approximately constant, only that its absolute rate of change is very much smaller than that of $\ce{[A]}$ and $\ce{[D]}$. Since according to the reaction scheme $\ce{d[D]/dt = k_2[X][C]}$, the assumption that $\ce{[X]}$ is constant would lead, for the case in which $\ce{C}$ is in large excess, to the absurd conclusion that formation of the product $\ce{D}$ will continue at a constant rate even after the reactant $\ce{A}$ has been consumed.

(2) In a stirred flow reactor a steady state implies a regime so that all concentrations are independent of time.

I think definitions in

A GLOSSARY OF TERMS USED IN CHEMICAL KINETICS, INCLUDING REACTION DYNAMICS (IUPAC Recommendations 1996) edited by KEITH J. LAIDLER, Pure & Appl. Chem., Vol. 68, No. 1, pp. 149-192, 1996.

are much better.

Pre-Equilibrium or Prior Equilibrium

The mechanism of a reaction may involve two or more consecutive reactions: If any step except the first is rate-controlling,those steps that precede it are essentially at

$\ce{A <=> B <=> C <=> ... <=> X -> Y ->Z}$

equilibrium, and there is said to be a pre-equilibrium, or a prior equilibrium; for example

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{rate-controlling step}$
$\ce{A <=> B <=> C <=> ... <=> X -> Y ->Z}$

$\text{|}\blacktriangleleft\text{---- pre-equilibrium ------} \blacktriangleright\text{|}$

Steady State (or Stationary State)

If during the course of a chemical reaction the concentration of an intermediate remains constant, the intermediate is said to be in a steady state.

In a static system a reaction intermediate reaches a steady state if the processes leading to its formation and those removing it are approximately in balance. The steady-state hypothesis leads to a great simplification in reaching an expression for the overall rate of a composite reaction in terms of the rate constants for the individual elementary steps. Care must be taken to apply the steady-state hypothesis only to appropriate reaction intermediates. An intermediate such as an atom or a free radical, present at low concentrations, can usually be taken to obey the hypothesis during the main course of the reaction.

In a flow system a steady state may be established even for intermediates present at relatively high concentrations.

  • $\begingroup$ If possible, please add quotes around what is from another source, mark what is edited (e.g., the word 'approximation'), and what is your own comment on the matter. (+1) $\endgroup$ Dec 4, 2017 at 18:58
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    $\begingroup$ @LinearChristmas - OK I tried quoting with > character which indents and shades. That what you had in mind? $\endgroup$
    – MaxW
    Dec 4, 2017 at 20:48
  • $\begingroup$ Yes, that's exactly what I meant. If you still bear with me, wouldn't it be more correct to also have full-headed arrows in the first steady state definition? I believe the harpoons explicitly signify an equilibrium which does not have to be met. (If you disagree, feel free to argue.) Though I admit, mhchem currently doesn't render the proper symbol too well (Win8.1/Firefox Quantum). $\endgroup$ Dec 4, 2017 at 21:07
  • $\begingroup$ @LinearChristmas - I think this is the reaction to which you are referring. $\ce{A <=>[$k_1$][k_{-1}] X}$ // You want this instead $\ce{A <-->[$k_1$][k_{-1}] X}$? $\endgroup$
    – MaxW
    Dec 4, 2017 at 21:18
  • 1
    $\begingroup$ @LinearChristmas - Thanks for pointing out problem. I absolutely spaced on that. It has been 40 years since I took kinetics in grad school. $\endgroup$
    – MaxW
    Dec 4, 2017 at 21:27

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