# Multiple equilibria

I was reading about multiple equilibria reactions and can't quite comprehend some things.

Suppose we have two reactions: \begin{align} \ce{A + B &<=> C + D},& K_\mathrm{eq} &= 10^{20}\tag1\label1\\ \ce{A + E &<=> F + D},& K_\mathrm{eq} &= 10^{10}\tag2\label2 \end{align}

Both share the common components $$\ce{A}$$ and $$\ce{D}$$. Due to the large difference in equilibrium constants, is it fair to say that in a mixture that initially contains only $$\ce{A}$$, $$\ce{B}$$ and $$\ce{E}$$, reaction $$\eqref1$$ will basically take place to completion while reaction $$\eqref2$$ won’t really happen at all? Will the difference of concentrations of $$[\ce{C}]$$ to $$[\ce{F}]$$ be $$10^{10}$$?

Or will both reactions happen to a significant degree in equilibrium as both equilibrium constants are large?

And how would one go about calculating the exact concentrations for every component given starting values?

• K is practically more of a thermodynamic thing then kinetic. It doesn't really say about speed of reactions. Commented Jan 8 at 0:21
• The answer depends on initial conditions. E.g. if A is in excess and B is a limiting reagent then it does not matter much how big is the R1 eq. constant. // Mathematically, it is a set of 2 nonlinear equations of 2 variables. Well, in 4 variants, depending on reactant initial ratios. Commented Jan 8 at 7:26
• Let’s say A,B,E are used stoechiometrically. Like initial conditions for each of them is 1mol/l. Commented Jan 8 at 8:51
• Then reaction 1 should be favored heavily right? Commented Jan 8 at 8:52
• No ! The reaction (1) is not necessarily favored. It depends on the activation energy. If (1) has a rather high activation energy, much higher than (2), the yield of (1) would be negligible. Commented Jan 8 at 20:52

Ignoring kinetic considerations such as mentioned in comments, and limiting discussion to thermodynamic ones, you might consider what happens if you mix A, B and E in one pot. You can then combine the two reactions into one:

$$\ce{F + B <=> C + E},~~~~~~~~~~ K_\mathrm{eq} = 10^{10}\tag3\label3$$

Clearly product C and reactant E are strongly favored. If however the reactions you describe are not coupled (you have separate reaction vessels) this line of argument does not hold.

To put this on a quantitative footing you can solve the quadratic equation that follows when the initial conditions are that the reactants are present in equimolar amounts. This leads to the solution for uncoupled (1) or (2):

$$[\mathrm{product}] = A_0\frac{(K-K^{1/2})}{K-1}$$

When K>>1 this simplifies to

$$[\mathrm{product}] = A_0(1-K^{-1/2})$$

Therefore for $$K\geq 10^{10}$$, $$[\mathrm{product}] \approx A_0$$.

In addition for the two uncoupled reactions the ratio of products is

\begin{align}\frac{[C]}{[F]}&=\frac{(1-K_1^{-1/2})}{(1-K_2^{-1/2})}\\&\approx(1-K_1^{-1/2})(1+K_2^{-1/2})\\&\approx(1-K_1^{-1/2}+K_2^{-1/2})\\&\approx(1+K_2^{-1/2})\end{align}

In other words, the product concentrations for both uncoupled reactions differ by a tiny amount and are approximately $$A_0$$.

• I think that the equations are coupled because each is an equilibrium. So if we start with A+B and let these equilibrate then add E all the concentrations will adjust. Commented Jan 8 at 16:39
• @porphyrin the point in my answer is that F is hardly produced if you mix everything together (A,B and E). Instead if only A and E are mixed nearly all A is converted to F. Commented Jan 8 at 16:45

Yes, when the equilibrium is reached, there will be significantly more $$\ce{C}$$ than $$\ce{D}$$ present.

To calculate the equilibrium concentration, you just go through the standard method of using the ICE table. You obtain a system of two nonlinear equations for two variables, which you have to solve using a numerical solver.

Let's denote initial concentrations with $$[\ce{A}]_0$$, $$[\ce{B}]_0$$ and $$[\ce{E}]_0$$, and let $$x$$ and $$y$$ be the extents of the reaction (1) and (2), respectively. In the equilibrium, the following two equations hold:

$$\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]} = \frac{x(x+y)}{\left([\ce{A}]_0 - x - y\right)\left([\ce{B}]_0 - x\right)} = K_{\rm eq, 1}, \\ \frac{[\ce{F}][\ce{D}]}{[\ce{A}][\ce{E}]} =\frac{y(x+y)}{\left([\ce{A}]_0 - x - y\right)\left([\ce{E}]_0 - x\right)} = K_{\rm eq, 2}.$$

I will use Wolfram Mathematica* (but you can use any other software with a numerical solver) to get the solutions of these two equations, namely the $$x$$ and $$y$$, with $$[\ce{A}]_0 = \pu{1 M}$$, $$[\ce{B}]_0 = \pu{1 M}$$, $$[\ce{E}]_0 = \pu{1 M}$$, $$K_{\rm eq, 1}=10^{20}$$ and $$K_{\rm eq, 2}=10^{10}$$:

\begin{align}x &\approx \pu{1.0 M}, \\ y &\approx \pu{1.0e-5 M}.\end{align}

Therefore, the ratio of products is:

$$\frac{[\ce{C}]}{[\ce{F}]} = \frac{x}{y} \approx 10^5.$$

* This is the code I've used:

N[Solve[{
(x (x + y))/((A0 - x - y) (B0 - x)) == Keq1,
(y (x + y))/((A0 - x - y) (E0 - y)) == Keq2,
x < 1, y < 1
} /. {A0 -> 1, B0 -> 1, E0 -> 1, Keq1 -> 10^20,
Keq2 -> 10^10}, {x, y}, PositiveReals], 4]
(* {{x -> 1.000, y -> 0.00001000}} *)