I am trying to determine the relative concentrations of $\ce{U1}$, $\ce{U2}$, and $\ce{E}$ when the following reaction is at steady state. Note that $k_t$ is the zero order rate constant of new $\ce{U1}$ entering the system and $k_d$ is the first order rate constant of substances leaving the system.

$k_1$ and $k_2$ are the first order rate constant for the forward and reverse reaction, respectively, of $\ce{U1}$ turning into $\ce{U2}$, and $k_e$ is the first order rate constant for $\ce{U2}$ reacting to $\ce{E}$. First order reactions are all first order in the respective reactant.

All the individual rate constants are known but the concentrations are not.

Can anyone point me in the right direction for how to determine the relative concentrations at steady state? I'm imagining an equation or set of equations such that when values for each rate constant are plugged in, the concentrations of the products are determined.


  • 3
    $\begingroup$ I find this rather confusing. I'm a chemist not a chemical engineer. It seems however that "k" is being used for rate constants and flow rates. So kt and kd seem to be flow rates and k1, k2, and ke seem to be rate constants. // I'm imaging some kind of reaction chamber with an input and output assuming complete mixing at any instant. // I don't get how kd relates to U1, U2, and E. $\endgroup$ – MaxW Mar 8 '19 at 22:16
  • 1
    $\begingroup$ In other words I'm thinking that kt is a flow rate like 1 liter per minute where U1 has a concentration of 0.100 mole per liter. Hence the flow rate of kd must be 1 liter per minute also, and for the output [U1] + [U2] + [E] = 0.100 moles per liter since all of the reactions are 1:1. $\endgroup$ – MaxW Mar 8 '19 at 22:28

Can anyone point me in the right direction for how to determine the relative concentrations at equilibrium?

This system will not reach equilibrium as long as there are irreversible reactions removing U2. If someone claims to have found an equilibrium state, the rates associated with k1 and k2 would be equal, but the [U2] would decrease because of those irreversible reactions.

I know that kt=3kd must be true for the whole system to be in balance, but I have been struggling to figure out how to relate the other rate constants.

Usually, lowercase k is used for rate constants, with the rate of reaction given by a rate law that involves k, but also concentrations of the reactants and perhaps other species. The way you are talking about it, all reactions would be zero order. For zero order reactions, there is no equilibrium (a melting process has no equilibrium - if the temperature is above the melting point, everything melts, and if it is below the melting point, everything freezes).

I have read that the Keq for multi-step reactions is the product of each single equilibrium constant, but I'm not sure how to apply this with the irreversible reactions in the system.

You can't. There is the concept of steady state, though, where concentrations of intermediates remain constant while the forward and reverse rate of a given reaction are different. In this case, the rates increasing the concentration of an intermediate are equal to the rates decreasing the concentration of that same intermediate, but there is a net reaction. This is typical for living organisms (they can, however, reach equilibrium after they die).

  • 1
    $\begingroup$ RE: This system will not reach equilibrium as long as there are irreversible reactions removing U2. ... // Not really true. An equilibrium can be established between k1 and k2 assuming that kd and ke are negligible. // Also in reaction chambers it is typical to have a quasi-stable steady state even though there is no true equilibrium. $\endgroup$ – MaxW Mar 8 '19 at 22:11
  • $\begingroup$ Sure, if kt, kd, and ke are zero, U1 and U2 will be at equilibrium. $\endgroup$ – Karsten Theis Mar 8 '19 at 22:13
  • 1
    $\begingroup$ Re: if kt, kd, and ke are zero, U1 and U2 will be at equilibrium. Not necessarily. It seems to me that this problem is about a reaction chamber that reaches some sort of quasi-steady state, not a true equilibrium. $\endgroup$ – MaxW Mar 8 '19 at 22:23
  • $\begingroup$ Karsten, thanks for the thoughtful reply. You've helped me realize that I'm muddling the concepts of equilibrium and steady state in my mind. I now am clear on the differences and am looking for concentrations at steady state. $\endgroup$ – CephBirk Mar 10 '19 at 16:53

Thanks to Karsten's comments clarifying the differences between equilibrium and steady state, I found this great resource:


Although the math is slightly more complex than in the example video, the logic is the same:

at steady state, $dX/dt = 0$, so each concentration $U_1, U_2$, and $E$ can be calculated.

$\frac{dU_1}{dt}=k_t-U_1 k_1+U_2 k_2-U_1 k_d = 0$

$\frac{dU_2}{dt}=U_1 k_1-U_2 k_2- U_2 k_e-U_2 k_d = 0$

$\frac{dE}{dt}=U_2 k_e- E k_d = 0$

Solving this system of three linear equations for the three unknowns gives:

$U_1=(k_t+\frac{k_1 k_2 k_t}{k_1 k_e+k1 k_d+k2 k_d+k_e k_d+k_d^2})\times\frac{1}{k_1+k_d}$

$U_2=\frac{k_1 k_t}{k_1 k_e+k_1 k_d+k_2 k_d+k_e k_d+k_d^2}$

$E=\frac{k_1 k_t k_e}{k_1 k_e k_d+k_1 k_d^2+k_2 k_d^2+k_e k_d^2+k_d^3}$

  • $\begingroup$ Now that you got your answer, could you edit the question to tell us the context of the problem? The question was closed, and if it remains closed, it will be deleted and others won't be able to learn from your thought process. $\endgroup$ – Karsten Theis Mar 12 '19 at 16:07
  • $\begingroup$ Karsten, it looks like my original wording has been pretty extensively rehashed and generally made more clear. What would make this more clear? $\endgroup$ – CephBirk Mar 12 '19 at 16:49
  • 1
    $\begingroup$ I think the problem is clear, and your answer is clear. What is not clear is what the context is. Is this a transport problem or a series of chemical reactions? What do U1, U2 and E represent - conformations of the same molecule, different molecules, different compartments? Why is the rate for the decay of U1 identical to that of U2 and E? If you add the specifics why your are interested in this model, it would be more relevant and easier to see what the assumptions are that led you to the model. It's great that it also contains the abstract description to use it on other systems. $\endgroup$ – Karsten Theis Mar 12 '19 at 17:18

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.