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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?

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    $\begingroup$ As long as K is between rs/pq and infinity, x is 0 or greater. $\endgroup$ – Jon Custer Feb 1 at 2:26
  • $\begingroup$ @JonCuster Okay, now I see something. Thanks ! $\endgroup$ – Esha Manideep Feb 1 at 2:48
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    $\begingroup$ 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. $\endgroup$ – Greg Feb 1 at 4:24
  • $\begingroup$ Well, assuming an elementary reaction you have the wrong equation. The equilibrium is given by $$\ce{K = } \dfrac{\ce{[C]^c[D]^d}}{\ce{[A]^a[B]^b}}$$ $\endgroup$ – MaxW Feb 1 at 23:28
  • $\begingroup$ @MaxW oops my bad. $\endgroup$ – Esha Manideep Feb 2 at 2:23
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The reaction quotient Q approaches zero when at least one of the product concentration approaches zero (as the reaction goes to completion in the reverse direction and the extent of reaction x is at the lowest possible value), and it approaches infinity when at least one of the reactant concentrations approaches zero (as the reaction goes to completion in the forward direction and the extent of reaction x is at the highest possible value). Also, the reaction quotient increases monotonically with x.

No matter what K is, there is always a Q that matches K because Q assumes all values between zero and infinity.

The only real case where this argument does not work (real as in concentrations have to be non-negative real numbers) is when at least one reactant and one product concentration is zero. In that case, there will be no reaction at all, and Q is not defined.

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  • $\begingroup$ @Esha_Manideep Congrats regarding the Maths olympiad! $\endgroup$ – Karsten Theis Feb 1 at 4:54
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    $\begingroup$ Thanks ! JohnCusters comment does it all $\endgroup$ – Esha Manideep Feb 1 at 5:23
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    $\begingroup$ Sometimes, if you allow initial product concentrations to be non-zero, the reaction has to go backwards to reach equilibrium, and x will be negative. (Non of the concentrations will become negative though because the reaction has reached its limit when one species is depleted.) With negative values of x, the fraction assumes values in the range of zero to r s / p q. Bottom line: for some values of r, s, p, q, and K, a negative value of x is a physically (or perhaps chemically) meaningful result. $\endgroup$ – Karsten Theis Feb 1 at 12:24

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