# How to estimate reaction products of competitive binding?

I have two reactions,

$$\ce{A + C \rightleftharpoons AC}$$

$$\ce{B + C \rightleftharpoons BC}$$

I'd like to estimate the relative steady state proportions

$$\frac{[\ce{BC}]_{SS}}{[\ce{AC}]_{SS}+[\ce{BC}]_{SS}}$$

and

$$\frac{[\ce{AC}]_{SS}}{[\ce{AC}]_{SS}+[\ce{BC}]_{SS}}$$

without numerically solving any differential equations, given $[\ce{A}]_{init} = [\ce{B}]_{init}$ and thermodynamic properties: $\Delta G, \Delta H,$ and $\Delta S$.

I'm hoping that there is a way to generalize this for more than two reactions in a linear way. It is for a DNA microarray optimization problem in which computational speed is important. Any thoughts?

• Are the A,B, and C in the same vat? Also, what is the limiting reagent? What are the rates? Lastly, sorry for being nit-picky, but the double headed arrow is for resonance structures, not equilibrium reactions. – CoffeeIsLife Jan 7 '14 at 1:04
• A, B, and C are in the same solution. We can assume that the forward rates are equal for both reactions and the reverse rates are determined by $K_{eq}$ from the Van't Hoff equation. We can also assume excess $C$. If the formula turns out to be pretty ugly, I was hoping for an order of magnitude estimate (some function of both $K$'s). – bravetang8 Jan 7 '14 at 1:23

Why would you have to solve differential equations for this problem? What algebra have you tried so far?

If C is in excess, then there is no competition! You should be able to model the reactions as just $\ce{A <=> AC}$ and $\ce{B <=> BC}$. The equilibrium equation would be (for A) $K_a=\frac{ac}{A_{init}-ac}$, and you can apparently already calculate $K_a$ from the thermodynamic parameters, so that is an equation where the only unknown is $ac$, the amount of AC formed. The equation is linear!

$ac=A_{init}\frac{K_a}{1+K_a}$

A similar result holds for B of course.

Once you can calculate $ac$ and $bc$, you should be able to calculate the parameters you want to know very easily.

If C is not in excess, the equilibrium equations will not be linear and the balance on C will couple the two equilibria (competition happens!).