Talk:Steady state (chemistry)

Latest comment: 6 years ago by Dirac66 in topic Merger proposal

Untitled edit

Does chemical equilibrium imply that you have a steady state? Richard Giuly 08:32, 4 March 2007 (UTC)Reply

Steady state and equilibrium are not the same thing, but the two are related. Steady state (in any context, not just kinetics) means that the rate of increase is equal to the rate of decrease. The steady-state approximation, for example, assumes intermediates are so reactive that they do not accumulate to any measurable extent, but rather, react almost immediately. Hence, they will exist at some very low, negligible concentration. Clearly, such a system is not at equilibrium, since reactants are being consumed and products are forming, but the concentration of the intermediate is, nonetheless, at steady state, since it is being consumed at the same rate it is being formed. When the system does reach equilibrium, the reactants and products will be at steady state as well, since their concentrations are not changing with time. So in summary, the fact that one component of a chemical system is at steady state does not imply that the entire system is in equilibrium; however, when the system is in equilibrium, all components are at steady state. -71.195.18.33 (talk) 08:48, 20 September 2009 (UTC)Reply

This is a useful subject, could use some more elaboration. For example, what is the meaning of the equation for [B] when k1 = k2 (the solution has a singularity due to k2-k1 in denominator)? Sb4 (talk) 01:12, 5 November 2008 (UTC)Reply

Practically speaking, no two reactions can possibly have the exact same microscopic rate constants, so the question is irrelevant. Furthermore, nature does not obey equations; equations are our attempts to describe nature. The singularity is a mathematical artifact with no physical meaning. If you consider the physical situation the equations are trying to describe, you see that when k1 = k2, it means that B is consumed as quickly as it forms, as if a single reaction A -> C were occurring. In this case, the kinetics become first-order and the singularity disappears. I would add that the principle of microscopic reversibility requires that all reactions be reversible, even if the kinetics of the reverse reaction are incomprehensibly slow. When reversible reactions occur in series (A <-> B <-> C), the system of differential equations is no longer analytically solvable, so mathematical concerns like a singularity are even less of a concern. -71.195.18.33 (talk) 08:48, 20 September 2009 (UTC)Reply


There are several approaches to solving a system of first-order linear differential equations. Two of the most popular approaches are the Laplace transform method and the matrix algebra approach. To address your specific question, when the rate constants are equal, the solution to the system of differential equations breaks down into cases. In the case of a first-order consecutive reaction, there will be two solutions to the rate law for B = f(t) and C = f(t). Having the two rate constants equal does not imply that the intermediate product is consumed as fast as it is formed, however. While the rate of formation/consumption depends on the magnitude of the rate constants, the actual rate also depends on the instantaneous concentration of the particular species in solution. If the two rate constants are equal, then application of the steady-state approximation no longer applies. The potential for reversibility is correct. However, the system of equations still has an analytical solution. All systems of ordinary first-order linear differential equations have analytical solutions including the system A <--> B <--> C. -ZPOT (talk) 19:03, 23 December 2009 (UTC)Reply

The content of this article diverges. edit

The introduction to this article explains the concept of a steady state. However, the section "Steady state approximation in chemical kinetics" diverges from the concept of a steady state! The reason is that when using steady state approximation one imposes a steady state on a subset of the reactants which is not of immediate interest. One is rather interested in the slow dynamics of the remaining system which is not at steady state. I propose to transfer this section to a separate article like "Steady state approximation" or the more common term "Quasi-steady state approximation" which is also dealt with in the context of Michaelis–Menten_kinetics. — Preceding unsigned comment added by Initialfluctuation (talkcontribs) 16:25, 8 August 2012 (UTC)Reply

Assessment comment edit

The comment(s) below were originally left at Talk:Steady state (chemistry)/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

This is a useful subject, could use some more elaboration. For example, what is the meaning of the equation for [B] when k1 = k2 (the solution has a singularity due to k2-k1 in denominator)? Sb4 (talk) 01:02, 5 November 2008 (UTC)Reply

Last edited at 01:02, 5 November 2008 (UTC). Substituted at 06:58, 30 April 2016 (UTC)

Merger proposal edit

I propose that Steady state (chemistry) be merged into Dynamic equilibrium. I think that the content in the Steady state (chemistry) article can easily be explained in the context of Dynamic equilibrium. One can see that the two definitions overlap and that they are, in fact, synonymous (I'd like to underline that we're talking about DYNAMIC and not thermodynamic equilibrium, there is a difference). I noticed the overlapping of the definitions very well when I tried to connect the article Dynamic equilibrium in English with its German counterpart called Fließgleichgewicht. In this German Wikipedia article it is explained that the term Fließgleichgewicht (literally translated as "flow equilibrium") is synonymous with Dynamisches Gleichgewicht (lit. "dynamic equilibrium"). There are even two references to back that fact up in the article itself. However, I couldn't connect the English article Dynamic equilibrium with this German article about dynamic equilibrium (called Fließgleichgewicht), because Wikipedia told me that the German article is already connected to another English article, in this case Steady state (chemistry). This makes no sense, as the definitions overlap and as is stated and backed with references in the German article Fließgleichgewicht (dynamic equilibrium). It only makes sense to merge these two English articles as well (Steady state (chemistry) and Dynamic equilibrium).

Above edit by Petrumila 5 June 2017.
Oppose. These two concepts are not the same. In dynamic equilibrium, the concentrations of all reactants and products are constant, so that there is no net reaction. In a steady state as defined in chemical kinetics, the concentration of one or more intermediate species is (approximately) constant, but the concentrations of reactants and products do vary so that there is a net reaction. See the concentration versus time graphs in this article. So it is better to have two articles to help the reader distinguish the two concepts. The second paragraph of the intro also explains the difference between the two concepts.
And perhaps there should be two distinct German articles as well, although it is difficult for me to judge since my German is too poor to read the German article properly. Dirac66 (talk) 00:33, 6 June 2017 (UTC)Reply