How to determine steady state of multi-step reaction with both reversible and irreversible steps?

Question

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.

Answer 1

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).

Answer 2

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

https://www.youtube.com/watch?v=hCE_-GNnnMA

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:

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

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

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