13
$\begingroup$

The term equilibrium is used in the context of reversible reactions that reach a point where concentrations no longer change. The term steady-state is used in enzyme kinetics when the concentration of the enzyme-substrate complex no longer changes (or hardly changes, in case of a quasi steady state). It is also used to describe multi-step biochemical pathways. Is there a difference between the two, given that both concern a situation where concentrations don't change over time?

$\endgroup$
  • 2
    $\begingroup$ From a (simplified) biochemistry perspective when an organism reaches equilibrium that means it died. Organisms strive for steady state consumption of energy to stay alive. $\endgroup$ – A.K. Mar 11 at 15:57
20
$\begingroup$

Yes, equilibrium and steady-state are distinct concepts.

A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.

Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.

In both cases, there are rates ($\mathrm{rate}_1$ and $\mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.

$$\ce{A <=>[rate_1][rate_2] B}\ \ \ \ \ vs \ \ \ \ \ \ce{source->[rate_1]C->[rate_2]sink} $$

For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.

$\endgroup$
  • 1
    $\begingroup$ If I may attempt a paraphrase: equilibrium is when they system is not changing over time in isolation (no inputs/output from the system), while steady state is when the system is not changing over time because it is being forced to remain that way (such as adding inputs and removing outputs at the rate they form). Thus if we take a steady state system, and grow the system to include the systems which added/removed products, that larger system might be at equilibrium. $\endgroup$ – Cort Ammon - Reinstate Monica Mar 11 at 16:18
  • $\begingroup$ @Cort_Ammon - I don't agree with your last statement. A steady state implies that one reaction makes C and another removes C, looking at the net reactions. Those two reactions are not at equilibrium. If they were, the first reaction would not contribute to increasing the concentration of C, and the second reaction would not contribute to decreasing the concentration of C. $\endgroup$ – Karsten Theis Apr 20 at 18:22
1
$\begingroup$

In short, equilibrium is a static process characterized by an equilibrium constant and does not have a time component. Steady state is a kinetic process characterized by rate constants and determined by the half-life of the substance with respect to the process. Time is an important dimension here. An example of a steady state process is seen in the formation of an enzyme-substrate complex as described so nicely by Karsten Theis in the post above. Another example is the administration of a drug to a person. The drug is continuously absorbed into the systemic circulation, continuously be distributed to the target site(s), continuously be metabolized and continuously be removed by excretion out of the body. These processes are collectively known as the 4 pharmacokinetic parameters.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.