Retrospective analysis 2/13/2017 -- The barium sulfate example is a poor choice. Equilibrium equations should really be defined using activities, and the activity of solid barium sulfate is by definition 1.

A previous question, Is every chemical reaction in equilibrium?, started much discussion. I objected to the answer that Curt F. gave and challenged him to derive the equilibrium reaction for a particular irreversible reaction. He countered with a reply indicating that I should make the specific case a separate question - so here it is. I'm going to change the reaction slightly. Given the following reaction between barium chloride and sodium sulfate, does "the" chemical equilibrium exist?

$$\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4 v + 2NaCl(aq)}$$

I contend that "the" equilibrum between reactants and products such as $$K_{\text{eq}} = \frac{\ce{[BaSO4][NaCl]^2}}{\ce{[BaCl2][Na2SO4]}}$$ doesn't exist since when the barium sulfate precipitates there could be a microgram or a kilogram as the product. Furthermore adding solid barium sulfate to the product will not shift the reaction to the left.

I'd agree that calling the reaction irreversible and saying that $K_{\text{eq}}$ doesn't exist is a tautology. So given reaction between $\ce{aA + bB}$ to form products $\ce{cC + dD}$ then the reaction is a reversible reaction if the equilibrium such as $$K_{\text{eq}} = \frac{\ce{[C]^{c}[D]^{d}}}{\ce{[A]^{a}[B]^{b}}}$$ exists and if such an equilibrium doesn't exist then it is an irreversible reaction.

There are obviously "some" equilibrium in the reaction, but not "the" equilibrium between products and reactants.

  • Water has an autoionization equilibrium and $\ce{H2SO4}$ has two $\mathrm{p}K_\mathrm{a}$'s.
  • Barium sulfate has a $K_{\text{sp}}$.
  • Also the barium sulfate precipitate isn't static once formed. It is dissolving and reprecipitating at the same rate, but the rates depend on the surface area of the precipitate not the "concentration" (or mass) of the precipitate.

So, if I'm wrong, how do you calculate $K_{\text{eq}}$ for the overall reaction given? Assume you mix $500\mathrm{ml}$ of $0.1$ molar barium chloride with $500\mathrm{ml}$ of $0.1$ molar sodium sulfate. What is $K_{\text{eq}}$?!

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    $\begingroup$ I like the question, but I think that it comes down to a purely philosophical question. What do you consider to be the reactants. If you consider $\ce{BaCl2(aq)}$ the reactant, then you need to include, that these are in solution, hence it is $\ce{Ba^2+ + 2Cl- }$. Same for sodium sulfate and sodium chloride. Therefore as reactants only barium ions and sulfate ions are left, and yes, they are in equilibrium. $\endgroup$ Commented Mar 4, 2016 at 5:23
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    $\begingroup$ Why not? I would use $$K^\circ = \exp\{−\Delta_\mathrm{r}G^\circ/ R T\}$$ and then you can formulate your equilibrium constant in terms of activity $$K=\frac{a(\ce{Ba^2+})a(\ce{SO4^2-})}{a(\ce{BaSO4})}.$$ It obviously boils down to $K_\mathrm{sp}$ with all approximations included, but it is still a reaction which is in equilibrium. $\endgroup$ Commented Mar 4, 2016 at 5:53
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    $\begingroup$ @ ChesterMiller - The notion was that K isn't a "true" equilibrium constant between reactants and products since adding more solid barium sulfate to the products wouldn't shift the equilibrium to the left. // @Curt_F. also has made a detailed analysis of why an equilibrium exists using a thermodynamic argument. chemistry.stackexchange.com/a/43262/22102 My thermodynamics is very rusty and I forgot that the activity of a solid is 1 until Martin posted his comment. So the chemistry works, but sidesteps the amount of barium sulfate ppt produced in the "equilibrium." $\endgroup$
    – MaxW
    Commented Mar 4, 2016 at 18:04
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    $\begingroup$ @Shadock - the evaporation of water isn't really a chemical reaction of reactants to yield a product. $\endgroup$
    – MaxW
    Commented Mar 4, 2016 at 18:30
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    $\begingroup$ @Shadock - Considering thermodynamics you end up with two phases like the barium sulfate. It doesn't matter how much liquid water is left, just that there is some. The activity of liquid water will be 1. $\endgroup$
    – MaxW
    Commented Mar 4, 2016 at 19:41

3 Answers 3


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.


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.


Also realised it change from H to Na can't be bothered changing it as it's 11:33pm and I have school tomorrow BaCl2(aq)+H2SO4(aq)⟶BaSO4(ppt)+2HCl(aq) Let's just think what (aq) means; it means you have ions floating about in there which have Ks with there solids

(If you start from thinking no ppt just ions dissolved) we have Ba^2+ , H^1+ , Cl^1- , and SO4^2- Then you consider the Ks of the different salts which are BaCl2 HCl BaSO4 and H2SO4. They will all go to Ks so all the salts would be forming and being dissolved (unless blocked eg things can become supersaturated). BaSO4 has extremely low Ks so most will ppt at the same time BaCl2 will be go to Ks with the ions Ba^2+ and Cl^1- which are still in solution and H2SO4 would also go to Ks, which means BaCl2 is being formed and therefore there is a reverse reaction.

(((note if I had BaSO4 with water which would be in equilibrium (so tiny dissolved/tiny bit of Ba^2+ ions and SO4^2- ions) and I added Cl^-1 ions a negligible amount more BaSO4 will dissolve as BaCl2 would go to equilibrium reducing Ba^2+ ions leading to negligible amount of BaSO4 dissolving to remain at Ks)))-this also show " Le Chatelier's principle" Yea I was trying to post this on original thred but this one everybody seems to be saying this


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