equilibrium - Which rate, the forward or reverse rate of acid dissociation, is more strongly affected when diluting acetic acid in aqueous solution?

Acetic acid is a weak acid. It is in equilibrium with acetate and hydronium ions in aqueous solution:

$$\ce{CH3COOH(aq) + H2O(l) <=> CH3COO-(aq) + H3O+(aq)}$$

Ostwald's law states that the degree of ionisation of a weak acid increases upon dilution (in quantitative terms: as the square root of the volume).

Consider two scenarios:

1. We start out with 1.001 moles of water and 1.000 moles of acetic acid (so water is the solvent and acetic acid is the solute), and dilute it with water to twice the volume.


2. We start out with 100 mM acetic acid, and dilute it with water to twice the volume.

How do the rates of the forward and reverse reaction change in these two scenarios, and how do these changes help to explain Ostwald's dilution law? (I know that once equilibrium has been reached after dilution, the forward and reverse rates will be equal again, so I am asking about the rates after dilution, but before changes to concentrations by the reaction going towards equilibrium. If mixing is slower than the acid-base reactions and you can't do the actual experiment, treat it as a thought experiment.)

This question was inspired by Why degree of dissociation/ionisation affected by dilution? . I am specifically interested in addressing the following ideas from the OP of that question:

as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

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since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

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since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

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What motivates or pushes [the acid] to dissociate more?

If it helps, it would be fine to discuss what happens with a single acetic acid molecule and a single acetate anion upon dilution.

Answer

Let's consider case 2. To make the math easier, I'm going to assume that acetic acid and any mixture of acetic acid/water has the same density as water. Thus, the molarity of pure acetic acid is approximated as 16.7 mol/L. That isn't exactly right, but conceptually that doesn't change anything. Also, since the mole fraction of water is a bit less than 1, let's use its actual molarity of approx 55.6 M.

So, to start, we have 0.1 M of acetic acid in water, which means only ~6 mL/L water, and we can assume the water is still 55.6 M. And assume a total volume of 1 L. At equilibrium, we know that the forward and reverse rates of ionization are the same, so

$k_f[\ce{H2O}][\ce{AcOH}]=k_r[\ce{AcO-}][\ce{H3O+}]$

Based on the published $K_a$, we can calculate that [AcOH]~0.1 M and [AcO-]=[H3O+]=0.0013 M.

If we instantly add an additional 1 L water to double the volume, the AcOH, AcO- and H3O+ concentrations all drop by half, but the water concentration stays at about 55.6 M. Intuitively, it should be clear that the result is a much greater drop in the rate of reverse reaction than the rate of the forward reaction. More rigorously, we relative rate of the forward reaction is (55.6x0.05)/(55.6x0.1)=0.5, while that of the reverse reaction is (0.00065)^2/(0.0013)^2=0.25. Since the rates were equal before dilution, the forward rate is now approx twice that of the reverse, and in order to return to equilibrium, the AcOH concentration must get even lower and the ion concentrations must come back up.

Conceptually, this comes about because the chance of an acetate ion and a hydronium ion encountering each other diminishes as the solution gets more dilute. The dissociation of the acid in bulk water, on the other hand, is just a function of how many undissociated molecules are present.

With regard to the quotes you included, some brief responses are

as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

The concentration of water is not the same as H3O+, which is decreased by dliution. Water self-ionization will partially compensate, but not very much.

since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

The dielectric constant of water does have an influence on the $K_a$, but it doesn't change when you add more water.

since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

If we start with dilute acetic acid in water, we assume that the acetate ions are far enough apart from each other that they do not interfere with each other's solvation, so additional dilution doesn't factor in.

What motivates or pushes [the acid] to dissociate more?

I think it's more appropriate to say that it reassociates less rather than dissociates more, but that's mostly semantics.