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 &quot;pure&quot; 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:
$A_0 B_0 - [C]_{eq}(A_0 +  B_0)  + [C]_{eq}^2 = K_{eq} {[C]_{eq}}$
Sort the terms and you have your quadratic equation:
$A_0 B_0 - [C]_{eq}(A_0 +  B_0 - K_{eq})  + [C]_{eq}^2 =  0$