6
$\begingroup$

Just like decreasing volume in a gaseous reversible reaction shifts the position of equilibrium towards the side that produces more moles (increasing pressure), shouldn't adding water shift equilibrium to side that produces more moles in a liquid equilibrium mixture (increasing concentration)?

On research, I've seen an explanation that it does not because the concentration of both reactants and products decrease evenly, but I don't understand this. The side with more moles should have a greater decrease in concentration because the concentration is proportional to the number of moles. Also, by the same argument, decreasing volume in a gaseous reaction should not shift equilibrium.

My guess is that adding water does affect equilibrium for a reaction with different number of moles on each side. Because the equilibrium constant is affected. Is this right?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Apparently it is right (I asked at work, to be sure). If you consider a reversible reaction $\ce{A + B = C}$ as the effect of two opposite reactions with rates $\ce{r_f = k_f \cdot [A] \cdot [B]}$ and $\ce{r_b = k_b \cdot [C]}$, equilibrium is reached when the forward and backward rates are the same: $\ce{r_f = r_b}$, which also gives you the definition of the equilibrium constant $\ce{K=\frac{k_f}{ k_b}}$. If you suddenly double the volume, the two rates change, so all concentrations have to readapt, for the equilibrium to be re-established (as both rate constants stay the same). $\endgroup$
    – user6376297
    Commented May 3, 2017 at 17:01
  • 1
    $\begingroup$ The equilibrium constant is not affected. But the number of moles of products and educts may change to hold the equilibrium when the amount of solvent changes. $\endgroup$
    – aventurin
    Commented Nov 2, 2017 at 22:24

2 Answers 2

Reset to default
6
$\begingroup$

Yes, the equilibrium does indeed readjust after a change in concentration. This can be seen by careful examination of the formula of the equilibrium constant $K$. Consider an arbitrary reaction that produces two products from a single reactant and is in equilibrium:

$$\begin{align}\ce{A &<=> B + C}\tag{1}\\[0.5em] K &= \frac{[\ce{B}][\ce{C}]}{[\ce{A}]}\tag{2}\end{align}$$

Let’s assume initial concentrations of $\pu{2M}$ for each species; that gives us $K = 2$ in dimensionless units (you can also use M as a unit here). Remember that $K$ is a constant so we can examine whether equilibrium has been reached by examining $K$.

Now let’s double the volume of solvent. The new concentration of each reactant is $\pu{1M}$. Therefore:

$$K' = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]} = \frac{1\times 1}{1} = 1\tag{2'}$$

Since $1\ne2$, we are no longer at equilibrium. Thus the concentrations must readjust. How? Obviously, the value of the fraction is too low. To increase its value, we can increase the numerator or decrease the denominator — luckily for us, both of these mean that we must perform the reaction in a forward direction to get think right again. Since $[\ce{B}]=[\ce{C}]$, we can actually calculate the ratio of concentrations and from $2[\ce{A}]+[\ce{B}]+[\ce{C}]=\text{const.}=4$, we can calculate exact values.

$$\begin{align}&&K &= \frac{[\ce{B}]^2}{[\ce{A}]}\\[0.5em] \Longrightarrow&&[\ce{A}] &= \frac{[\ce{B}]^2}2\\[1.3em] &&2\times\frac{[\ce{B}]^2}{2} + [\ce{B}] + [\ce{B}]&=4\\ &&[\ce{B}]^2 + 2[\ce{B}]-4 &= 0\\ \Longrightarrow&&[\ce{B}]_{1/2} &= \frac{-2\pm\sqrt{2^2+4\times4}}{2}\\[0.5em] &&[\ce{B}]_{1/2} &= -1\pm\sqrt{5}\\[0.5em] &&[\ce{B}]_{1/2} &\approx 1.24\end{align}$$

Thus, the product side has increased.

A more real-world example is maybe the dimerisation of hydrogenchromate $\ce{HCrO4-}$ according to $(3)$:

$$\ce{2 HCrO4- <=> Cr2O7^2- + H2O}\tag{3}$$

This is also pH-dependent, since hydrogenchromate is a weak acid and can deprotonate to give chromate $\ce{CrO4^2-}$. However, it is fairly easy to try this experiment yourself: take an orange coloured dichromate solution and add water. A colour change to yellow should be observed indicating that the reaction has equilibrated to the reactants’ side.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So just to confirm, for the final example, this occurs because the reactant side produces a greater amount of moles, not because the amount of water has increased thus increasing the reaction rate? I take that it is the former since you're not actually increasing the concentration of water however I just wanted to make sure. $\endgroup$
    – bio
    Commented Sep 10, 2022 at 2:30
  • $\begingroup$ @bio The real underlying reason for the hydrogenchromate/dichromate example is a change in pH which corresponds to more or less water being present in the system. In fact, it is often written with $\ce{CrO4^2- + H+}$ instead of $\ce{HCrO4-}$. $\endgroup$
    – Jan
    Commented Oct 1, 2022 at 15:02
0
$\begingroup$

Dilution of an aqueous ionic equilibrium favors the reaction that produces more ions. This changes the position of equilibrium. When in doubt, consider the expression of K and know that the equilibrium shifts to satisfy the constant K at a given temperature.

Improve this answer
$\endgroup$