A general way to combine equilibrium constants in reaction networks?

Question

I have a network of states, each linked with neighboring states by unique forward and reverse transition rates ($k_{f}$ and $k_{r}$) - let's just say these are chemical species with multiple intermediate reactants and reversible intermediate reactions. At long timescales, this system will reach a stationary distribution or thermodynamic equilibrium where forward and reverse transition rates are balanced by corresponding occupancy of each state, and it is typical to characterize the system with an equilibrium constant $K_{eq}$ for each reversible transition e.g. $K_{1} = \frac{k_{-1}}{k_1}$. My question is about how one goes about legally aggregating or combining these states to deal with hidden variables and apparent rates. In the example below, let us say that states B,C,D,and E are all within a black box and only flux from A into this collection of states can be measured (i.e. an apparent equilibrium constant). What exactly are the rates comprising this apparent constant? Or how could one form an expression for that apparent equilibrium constant by combining the constituent equilibrium constants?

I am familiar with a related example of how this is done in a simpler case of $X \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} Y \underset{k_{-2}}{\stackrel{k_2}{\rightleftharpoons}} Z$ where an overall equilibrium constant from X to Z is simply the sum of the constituent constants for X to/from Y and Y to/from Z. However I do not know how to generalize to more complex topologies such as the one below (which contains cycles for example). I vaguely suspect that there is an analogy to Ohm's law somewhere here and the procedure for adding currents in parallel circuits.

$A \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} B$;

$A \underset{k_{-2}}{\stackrel{k_2}{\rightleftharpoons}} C$;

$B \underset{k_{-3}}{\stackrel{k_3}{\rightleftharpoons}} D$;

$C \underset{k_{-4}}{\stackrel{k_4}{\rightleftharpoons}} D$;

$B \underset{k_{-5}}{\stackrel{k_5}{\rightleftharpoons}} E$;

$C \underset{k_{-6}}{\stackrel{k_6}{\rightleftharpoons}} E$

Answer 1

Not sure if this answers the question, but I will give some thoughts.

In order to obtain an apparent equilibrium constant, I would assume you need to specify two reactants - it doesn't make much sense to me to talk about the apparent equilibrium constant from $[A]$ to a bunch of variables, maybe you could explain a bit more? Your case seems to be however fairly easy to do, because you don't have any bimolecular interactions and you can therefore solve it with linear algebra. In this case you can set up a system of differential equations and set them to zero to obtain stationary quantities:

$$ 0 = -(k_1 + k_2)[A]_{eq} + k_{-2}[B]_{eq} + k_{-1}[C]_{eq}\\ 0 = -(k_{-1} + k_3 + k_5)[B]_{eq} + k_1[A]_{eq} + k_{-3}[D]_{eq} + k_{-5}[E]_{eq}\\ \vdots\\ n = [A]_{eq} + [B]_{eq} + [C]_{eq} + [D]_{eq} + [E]_{eq} $$

Then you can have a lot of fun with the resulting matrix if you are looking for an analytical solution using Gaussian elimination (but it's still doable) or you can just solve numerically. You will have more equations than variables and some of these will be linearly dependent but should be just enough to give you a unique value. Bear in mind that depending on how physical your rate constants are you may get an equilibrium distribution or a steady state. The way to check this is to see whether there is a net flow of material around any closed cycle (see here). Finally you can define any apparent equilibrium constant you want using any two different reactants. You can also readily check that this method works for the simple example you gave.

In the special case of all your rate constants obeying detailed balance, you can obtain any equilibrium concentrations by multiplying any relevant equilibrium constants along any path and you don't need to solve the above matrix equation.