Why doesn't chemical equilibrium lead to uncertainties?

Let's consider a reaction

$$\ce{a A + b B <=> c C + d D}$$

We will also assume that the concentrations are low and the activity coefficient is one. Now let's say initially one has $$p_0$$, $$q_0$$, $$r_0$$, $$s_0$$ as the concentrations of $$\ce{A}$$, $$\ce{B}$$, $$\ce{C}$$, $$\ce{D}$$, respectively. Let the equilibrium constant be $$K$$. Let's investigate about the equilibrium composition.

If $$x$$ equivalents of reactants reacted, then the equation for $$x$$ would be:

$$K = \frac{(r_0 + cx)(s_0 + dx)}{(p_0 - ax)(q_0 - bx)}.$$

Now, my question is how do we know all the time such equations have proper roots for $$x$$? I mean all the conditions $$x$$ needs to satisfy, like $$ax < p_0$$ and first of all, how do we know the equation has roots in first place? It's a quadratic and it need not have real roots necessarily. This is just a simple case of 2 reactants and 2 products.

I would like to see some rigorous proof that such equations always contain proper (roots which satisfy) all the conditions or a counter example of such a chemical equation. If such counter examples exist, how does the chemical reaction proceed?

• As long as K is between rs/pq and infinity, x is 0 or greater. Feb 1 '19 at 2:26
• @JonCuster Okay, now I see something. Thanks ! Feb 1 '19 at 2:48
• I think you are seeing this from the wrong end: we know that chemicals behave this way. There is always one equilibrium position, and there is a thermodynamic reason for that. And yes, formally we can most cases describe this equilibrium with simple equations like you just presented, but this equation is derived from several approximations, so if you find cases when it doesn't work, then it doesn't work. It is like proving that e.g. r0 is always positive: No, we know that r0 is positive, we don't prove that mathematically, and we care about only the examples which do not contradict this.
– Greg
Feb 1 '19 at 4:24
• Well, assuming an elementary reaction you have the wrong equation. The equilibrium is given by $$\ce{K = } \dfrac{\ce{[C]^c[D]^d}}{\ce{[A]^a[B]^b}}$$
– MaxW
Feb 1 '19 at 23:28
• @MaxW oops my bad. Feb 2 '19 at 2:23