I would like to understand the classic kinetic model of association / dissociation that tries to describe the concentration of a compound $[\ce{AB}]$. Let's say we have a model:

$\ce{A + B} \xrightleftharpoons[k_{\text{off}}]{k_{\text{on}}} \ce{AB}$

So $k_{\text{on}}$ is the rate of binding and $k_{\text{off}}$ is the rate of unbinding and the rate equation is:


I don't understand why texts say that if $k_{\text{on}}$ and $k_{\text{off}}$ are both constant then we reach a steady state, i.e., $k_{\text{on}}\ce{AB}=k_{\text{off}}[\ce{AB}]$ I mean: if, for example, the rate of binding, is greater then the rate of unbinding, shouldn't the concentration of the final compound increase up to infinity? What I have in mind is: we produce 4 proteins per second and degrade 2 proteins per second, then the net production is 2 proteins per second so, as time goes by, the concentration should increase and doesn't reach a plateau.

Why do we instead reach a plateau in the final concentration? I can understand the plateau in the plot of the reaction velocity but still, if the velocity is constant, still the concentration of the final compound should increase.

  • 2
    $\begingroup$ As @David said, this reaction has nothing to do with Michaelin-Menten model of enzyme catalysis. It looks like one reversible reaction (which can be physical or chemical). $\endgroup$
    Commented May 29 at 8:34
  • $\begingroup$ I edited to remove the references to Michaelis-Menten, as we seem to all be in agreement that this is not that. $\endgroup$ Commented May 29 at 16:01
  • $\begingroup$ @MaximilianPress — That’s a lot better. However, unless either A or B are a protein — not stated — we are left with a question for SE Chemistry. $\endgroup$
    – David
    Commented May 29 at 17:31

2 Answers 2


There is a misunderstanding here that is very common when first learning about chemical kinetics. First you say:

if $k_{\rm on}$ and $k_{\rm off}$ are both constant

but then:

if the velocity is constant

This is not one and the same!

$k$ is the rate constant of reaction.* The actual reaction rate (i.e., the change in concentration over time) is given in your example by:

$v_{\rm on} = k_{\rm on}[A][B]$ for the forward (binding) reaction, and

$v_{\rm off} = k_{\rm off}[AB]$ for the reverse (unbinding) reaction.

Suppose that $k_{\rm on} > k_{\rm off}$. Then at first $v_{\rm on} > v_{\rm off}$ and the product AB is produced faster than it degrades. Therefore the concentration of AB increases. On the other hand, A and B are consumed in the reaction, so their concentrations decrease.

Now look at the equations for reaction rates above! The rate of the forward reaction, $v_{\rm on}$, depends on concentrations of starting materials, $[A][B]$; and the rate of the reverse reaction, $v_{\rm off}$, depends on the concentration of product, $[AB]$. Therefore, as the reaction proceeds, $v_{\rm on}$ must decrease and $v_{\rm off}$ must increase until they reach equilibrium where both rates are equal.

* Note: As a convention, reaction rate constants are usually denoted by lowercase $k$. Uppercase $K$ is reserved for equilibrium constants, or the Michaelis constant $K_M$, or a number of other constants.

  • 2
    $\begingroup$ "Suppose that 𝑘on>𝑘off." But kon is a second-order rate constant whereas koff is a first-order rate constant, and have very different units (M(-1).s(-1) versus s(-1), for example) and as such is it surely meaningless to compare them? $\endgroup$
    – user338907
    Commented May 28 at 14:09
  • $\begingroup$ @user338907 Not necessarily. For example, the equilibrium constant is equal to the ratio of forward and reverse reaction rate constants, so here K = k_on/k_off. Then k_on > k_off means that K > 1, i.e., equilibrium favours the products. $\endgroup$
    – knomologs
    Commented May 29 at 9:52
  • 1
    $\begingroup$ You cannot directly compare the magnitude of a first order rate-constant with a second-order rate constant. This IS a fundamental mistake in dealing with enzyme kinetics. How can something with dimensions of M(-1).s(-1) be greater or less than (or equal to) something with dimensions of s(-1)? To take a more obvious example, how can pH 7 be considered greater than or less than 100 degrees celsius? (Going back to MM, the steady-state dfn of Km is the sum of two first-order rate constants, divided by a second-order rate content, a good eg of the value of dimen. analysis) $\endgroup$
    – user338907
    Commented May 29 at 10:29
  • 1
    $\begingroup$ This is probably a more relevant answer than mine. $\endgroup$ Commented May 29 at 16:01

Update: As noted by commenter, the example stated in the question is not actually Michaelis-Menten but is rather an example of binding kinetics. See also the other answer, which is likely more relevant.

I would suggest, if you would like to find the steady-state concentration of $[AB]$, that you can solve this yourself mathematically. Try differentiating and setting the $\Delta [AB]$ side of the equation to zero, and solving. You will find:

$ \Delta [AB] = (k_{on}[A][B] - k_{off}[AB])\Delta t$

$ 0 = (k_{on}[A][B] - k_{off}[AB])\Delta t$


$k_{on}[A][B] = k_{off}[AB] $

And finally:

$\frac{[AB]}{[A][B]} = \frac{k_{on}}{k_{off}}$

Note that if the $k$ are constant, this amounts to the law of mass action.

You can also do a forward simulation (see R code):

# association/dissociation kinetics
kinetic_sim = function(k_on, k_off, a, b) {
  ab = 0
  ab_ts = c()
  a_ts = c()
  b_ts = c()
  for (i in 1:1000) {
    ab_old = ab
    ab = ab * (1 - k_off) + k_on * (a * b)
    a = a - a*b*k_on + ab_old * k_off
    b = b - a*b*k_on + ab_old * k_off
    #cat(i, a, b, ab, k_on*a*b, k_off*ab, "\n", sep=" ")
    ab_ts = append(ab_ts, ab)
    a_ts = append(a_ts, a)
    b_ts = append(b_ts, b)
  return(list(ab_ts, a_ts, b_ts))
out = kinetic_sim(.1, .2, 1, 1)
plot(out[[1]] / (out[[2]] * out[[3]]), ylab="[AB] / ([A] * [B])", xlab="time step")
plot(out[[1]], ylab="[AB]", xlab="time step")

concentration of ratio [AB]:[A]*[B] over time concentration of ratio [AB] over time

You can see that both the ratio and the final concentration equilibrate.

  • 2
    $\begingroup$ A few suggestions. 1. The reaction example shown by the OP does not depict an enzyme catalysis. So using the term "Michaelis-Menten" would be misleading. 2. The system has 3 variables but there are only two independent differential equations. The third equation should be the conservation of mass (a + b + ab = constant). Alternatively (deviating from OP's example), you can model the system as an open system with rates of synthesis and degradation of the different components. $\endgroup$
    Commented May 29 at 8:43
  • 1
    $\begingroup$ @WYSIWYG thanks for the thoughtful comment. 1) You are absolutely correct that this is not michaelis-menten, my brain went immediately to the kinetic example shown without thinking about its name. 2) you are correct, the constant mass is implicitly modeled as shown in the simulation. $\endgroup$ Commented May 29 at 15:51

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