# how to derive equilibrium concentrations from initial conditions for mass action reactions?

I am reading a tutorial on biochemical reactions and mass action kinetics (https://www.math.utah.edu/~keener/books/control.pdf, pp. 1-2) and would like to derive an analytic solution to confirm simulation results. This is NOT a homework question, I am simply trying to understand things.

The tutorial considers a simple reaction where A and B combine, reversibly, to produce C:

$$A + B \rightarrow^{f} B$$

$$C \rightarrow^{r} A + B$$

where $$f, r$$ are the forward/reverse rate constants. It is clear that the change in [C] over time is:

(Eq. 1) $$\displaystyle\frac{d[C]}{dt} = f[A][B] - r[C]$$

It's simple to then show that the equilibrium constant $$K_{eq}$$ is:

(Eq. 2) $$K_{eq} = \displaystyle\frac{r}{f} = \frac{[A]_{eq}[B]_{eq}}{[C]_{eq}}$$.

The question: how can we use this to derive the equilibrium concentration of one of the species (like $$[C]_{eq}$$) as a function of the initial concentrations of A and B, $$A_0, B_0$$ and rate constants $$f, r$$? Simulation of Eq. 1 for $$A_0 = 200, B_0 = 100, f = 0.0001, r = 0.001$$ shows that $$[C]_{eq}$$ is around $$90$$. How can this be confirmed analytically?

Attempt at solution: try to rewrite Eq. 2 in terms of initial concentrations. We can use the fact that $$[A]_{eq}$$ and $$[B]_{eq}$$ can each be rewritten in terms of $$A_0, B_0, [C]_{eq}$$:

$$[A]_{eq} = A_0 - [C]_{eq}$$

$$[B]_{eq} = B_0 - [C]_{eq}$$

this is because the equilibrium concentration of "pure" A (A that wasn't used with B to make C) has to the total amount of A we started with, minus the amount of A that went into producing C. Same argument for B.

But it's not clear if this helps get a solution? Plugging these quantities into Eq. 2 we get:

$$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} = K_{eq}$$

$$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} - K_{eq} = 0$$

which seems too messy to be correct. I was expecting a simpler quadratic equation for such a simple problem. Guidance on solution or references to derivations will be great.

Here is the expression from the question:

$$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} = K_{eq}$$

Get rid of the fraction:

$$(A_0 - [C]_{eq})(B_0 - [C]_{eq}) = K_{eq} {[C]_{eq}}$$

Distribute the sums in the product:

$$A_0 B_0 - [C]_{eq}(A_0 + B_0) + [C]_{eq}^2 = K_{eq} {[C]_{eq}}$$

$$A_0 B_0 - [C]_{eq}(A_0 + B_0 - K_{eq}) + [C]_{eq}^2 = 0$$