# Chemistry - Does an irreversible reaction have an equilbrium between reactants and products?

## Solution 1:

The reaction you are interested in is the precipitation of barium sulfate, a relatively insoluble salt.

The reverse of this reaction is the dissolution of barium sulfate crystals. The dissolution kinetics of barium sulfate have been studied in a variety of systems over the decades. Here are a few links:

• Kornicker et al. 1991 reported that barium sulfate dissolution into water reached equilibrium in less than 30 minutes at 25 °C and took about 5 minutes at 60 °C. They also presented a rate law for $$\ce{BaSO4}$$ dissolution of $$r=k A (C_{eq} - C)^2$$, where $$A$$ is the specific area of the barium sulfate crystals, $$C_{eq}$$ is the equilibrium concentration of barium sulfate, and $$C$$ the instantaneous concentration of dissolved barium sulfate. This rate law had been used historically (see refs.) but these authors advocate instead for a first-order rate law.

• Dove and Czank 1995 also disagreed with second-order rate law of Kornicker et al. and proposed a first-order reaction for $$\ce{BaSO4}$$, i.e. $$r = k A (C_{eq} - C)$$.

I couldn't find any references for barite solubility in sodium chloride solution, but I think it is safe to say that the solubility is finite. Probably the kinetics of dissolution would not be more than an order of magnitude different than in pure water.

Since this dissolution is the reverse of the precipitation reaction you are interested in, and since the literature is clear that dissolution happens at non-zero rates, then both "reverse" reaction and forward reaction must be happening in your system. That sounds like the definition of a dynamic equilibrium to me.

Note the appearance of $$A$$ in the equations. The surface area of barium sulfate solid involved in the equilibrium determines the kinetics. Usually we assume that the affect of $$A$$ is the same for dissolution and precipitation kinetics, so that the effects cancel out. This is why we say that the activity of a solid phase is 1 and does not vary. However, the effects of $$A$$ do not always cancel out. The solubility of nanoparticles of $$\ce{BaSO4}$$ is probably higher than the solubility of bulk $$\ce{SO4}$$ for this reason.

Experimentally, one could probe the dynamics of the equilibrium by mixing $$\ce{^131BaSO4(s)}$$, i.e. radiolabeled barium sulfate, with a saturated solution of $$\ce{^138BaSO4(aq)}$$. This could be done even in a sodium chloride solution. At regular time intervals, samples of the solution would be taken, which could be filtered to remove any traces of solid particulates, and the levels of the radiotracer that had entered solution could be measured.

I don't know, quantitatively, what the outcome of the experiment would be, especially in sodium chloride solution. But I am 100% confident that at some time scale, probably within tens of minutes, radioactivity would appear in the liquid solution. That indicates that there is an equilibrium.

The relevant equilibrium constant in this case is the $$K_{sp}$$ for barium sulfate, as was alluded to in both your question and in the comments.

## Solution 2:

To continue with this question but in a more general way after a comment on the particular reaction in question.

The case of $\ce{BaSO4}$ has been chosen carefully to argue a particular point, but it seems to me that there are two equilibria involved, the first is chemical $\ce{Ba^{2+} + SO_4^{2-} <=> BaSO4_{aq}}$ the second involves no chemical reaction but is between equilibrium of $\ce{BaSO4_{aq}}$ and its solid. Because of the huge insolubility the reaction appears to be irreversible as the product is effectively removed from solution which drives the reaction very much towards products. If the reaction were to be studied in another polar solvent, one might try acetonitrile or dichloromethane for example, where insolubility is probably far less, then perhaps the true equilibrium $\ce{Ba^{2+} + SO_4^{2-} <=> BaSO4}$ would be measured.

Technically, all reactions are reversible since thermodynamics does not impose an energy difference or time-scale limit on an equilibrium, and assumes that reactants & products are not artificially separated from one another.
One might consider that an irreversible reaction is one in which after some time no appreciable concentration of reactants are left, but this assumes that the forward reaction is rapid enough to occur on what ever time-scale we set ourselves. Thus a better way is to know the reaction’s free energy and then, as this becomes the minimum size of the reverse reaction’s activation barrier, subjectively estimate whether the reaction is going to be sufficiently exothermic to make it ‘irreversible’. As only order of magnitudes of rate constants are required, this can be reasonably be done from a simple Arrhenius type equation. A reaction with a first order rate constant of 1/sec might be considered to be irreversible in some cases, but 1/year in others. Geochemists who study rock formation may perhaps consider these reactions as being very fast.

At the molecular level a reaction occurs because there is always a chance that, by random collisions with the surrounding solvent molecules (or other molecules in a gas or vapour), enough energy is imparted to the reactant molecules to overtop the activation barrier and transform them to products, and similarly for products returning to reactants. As the probability of reacting decreases exponentially with energy (Boltzmann distribution) this probability soon becomes vanishingly small with an increase in energy and hence for practical purposes a reaction may, in practice, be considered to be ‘irreversible’ or similarly products ‘un-reactive’ even though there is still technically equilibrium between reactants and products.