Overall equilibrium expression for competitive equilibria I have a case where I have two reactions that depend on a common reactant. The reactions can be written as:

I) $\ce{A + B &lt;=&gt; AB}$

II) $\ce{A + C &lt;=&gt; AC}$

We can thus write equilibrium expressions for both reactions:

I) $K_\mathrm{I} = \frac{[\ce{AB}]}{[\ce{A}][\ce{B}]}$

II) $K_\mathrm{II} = \frac{[\ce{AC}]}{[\ce{A}][\ce{C}]}$

How can I combine these equations to get an equation that relates the equilibrium concentration of B to that of C?

Secondarily, can this be extended to give an equation that relates the equilibrium concentrations of B and C to the initial concentrations of A, B and C?

I already know the equilibrium and rate constants for both reactions in isolation. If it is relevant, A is a solid with a fixed number of surface sites, to which B and  are adsorbed. B and C are in aqueous solution.
OK, so essentially we have $[A],[B],[C],[AB],\text{ and }[AC]$, and a few equations: three conservation laws and two equilibrium constants.
$$\left\{
\begin{align}
[A]+[AB]+[AC] &amp; = A_0\\
[B]+[AB] &amp; = B_0\\
[C]+[AC] &amp; = C_0\\
{[AB]\over[A][B]}&amp; =K_1 \\
{[AC]\over[A][C]}&amp; =K_2
\end{align}
\right.$$
With 5 unknowns and 5 equations, this should be solvable (because the nature manages to solve it somehow, if not for other reason). There is no guarantee, however, that the solution will be nice. Indeed, if we deal with the system exactly as it stands now, it is equivalent to a certain 4th degree algebraic equation which is better solved numerically.

Things get somewhat more manageable if we may disregard something. For example, if A is a relatively minor component, so that $[A]&lt;&lt;[B]$, then we may assume $[B]\approx B_0$ (and the same for C). All of a sudden, the problem becomes linear:
$$[AB]=K_1[A]B_0 \\
[AC]=K_2[A]C_0 \\
[A]+K_1[A]B_0+K_2[A]C_0=A_0
$$
which yields
$$
[AB]={K_1B_0\over1+K_1B_0+K_2C_0} \\
[AC]={K_2C_0\over1+K_1B_0+K_2C_0}
$$

Whether or not this approximation is realistic in your particular case is up to you.