Is it valid to calculate concentrations given equilibrium constant and initial concentration?

Question

Suppose we have the following reaction:

$$\ce{2A <=> B + C}\tag{1}$$

which can be thought the sum of the following reactions:

$$\ce{A <=> B}\tag{2}$$ $$\ce{A <=> C}\tag{3}$$

with equilibrium constants $K_1$ and $K_2$ respectively. The $K$ for the first reaction will be the product of $K_1$ and $K_2$ i.e. $K=K_1K_2 $. Now if we are given that initial we have $x$ mol of $\ce{A}$ and no amount of $\ce{B}$ and $\ce{C}$, one can calculate the amount of B and C present at equilibrium using the equilibrium constant $K$:

$$K=\frac{y\cdot y}{(x-2y)^2}\tag{4}$$

If we solve (4) for a given $x$ (initial amount) then concentrations of $\ce{B}$ and $\ce{C}$ must be equal. But how can this be possible if $K_1$ and $K_2$ have different values? I mean the following equality must holds:

$$[\ce{C}]= [\ce{B}] \Longleftrightarrow K_1{[\ce{A}]} = K_2\ce[{A}] \tag{5}$$

Edit I am asking if it is valid to use the equilibrium constant $K$ to find the concentrations of $\ce{B}$ and $\ce{C}$ because every reaction can be thought of as sum of other reactions. It is a common exercise in many general Chemistry textbooks where you are given a reaction with an equilibrium constant and you must find the concentrations at equilibrium. Well the substitutions into equilibrium constant that they make is the one that I have also did (assuming 0 initial concentration for both $\ce{B}$ and $\ce{C}$. So what is going wrong?

Answer 1

The equation

$$\ce{2A <=> B + C}$$

implies a fixed stoichiometry between B and C. If the two separate reactions (2) and (3) were happening, the correct net equation would be

$$\ce{(x + y) A <=> x B + y C}$$

Here is an example that would be well-described by reaction (1):

$$\ce{2 H2O <=> H3O+ + OH-}$$

One water molecule cannot make a hydronium ion (goes against the conservation of mass and charge). The same goes for a water reacting to form a hydroxide. Instead, for every hydronium that forms, exactly one hydroxide is formed. So the separate reactions do not exist, and the separate equilibrium constants do not either.

Of course, there are other reactions where two independent products exist, such as linear glucose forming the alpha or beta anomer of the pyranose ring. In this case, there are two equilibrium constants, but there is no combined equilibrium constant because the two reactions are not the steps of an overall reaction, but rather two distinct reaction paths.

Answer 2

You have stated that the reaction:

$$\ce{2A <=> B + C}\tag{1}$$ [...] can be thought the sum of the following reactions: $$\ce{A <=> B}\tag{2}$$ $$\ce{A <=> C}\tag{3}$$

However, this is not always the case. In particular, your assertion that the overall reaction (1) is a linear combination of reactions (2) and (3), is based upon the assumption that the rate of forward reaction ($r_f$) and rate of reverse reaction ($r_r$) are equivalent.

I have re-written your reactions (2) and (3) in more detail below (including the forward and reverse reaction rates):

$$\ce{A <=>[\ce{r_1}][\ce{r_2}]}B\tag{2*}$$ $$\ce{A <=>[\ce{r_3}][\ce{r_4}]}C\tag{3*}$$

Chemical systems are dynamic. At equilibrium, even though the concentrations specified by the reaction equation may not be changing, it does not imply that reaction is NOT occurring. In actuality, at equilibrium, all reactions are occurring, just at equal and opposite rates. If one looks at the equation for $K$, the equilibrium constant, one would see that this can also be expressed as the ratio of the forward and reverse reactions for the reaction system, such that:

$$K = \frac{r_f}{r_r} $$

In your case, you are assuming that in equations (2*) and (3*), that $r_1=r_3$ and $r_2=r_4$. Assuming this allows equations (2*) and (3*) to be "added" together. As such, these equations revert to the forms you've outlined in (2) and (3) and when combined they constitute equation (1).

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Now, regarding the portion of your question:

The $K$ for the first reaction will be the product of $K_1$ and $K_2$ i.e. $K=K_1K_2 $. Now if we are given that initial we have $x$ mol of $\ce{A}$ and no amount of $\ce{B}$ and $\ce{C}$, one can calculate the amount of B and C present at equilibrium using the equilibrium constant $K$: $$K=\frac{y\cdot y}{(x-2y)^2}\tag{4}$$

The equilibrium constant, $K$, is specified in terms of the equilibrium concentrations of each reaction constituent. For the reaction system below, we wouls specify $K$ as being,

$$\ce{2A <=> B + C}\tag{4*}$$ $$K = \frac{[B][C]}{[A]^2}$$

In particular, $[A]$, is the equilibrium concentration of A. Modifying $[A]$ to be dependent on the initial amount of A is contradictory, as $[A]$ should not be dependent on the initial amount of itself.