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I read that every chemical reaction is theoretically in equilibrium in an old textbook. If this is true how can a reaction be one way?

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    $\begingroup$ Can you explain the contradiction? $\endgroup$ – SplitInfinity Mar 2 '16 at 5:35
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    $\begingroup$ Wait, wait, what do you mean by in equilibrium? Possibility of achieving equilibrium? $\endgroup$ – Mithoron Mar 3 '16 at 12:25
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    $\begingroup$ Irreversible reactions by definition only go in one direction and hence there is no reverse reaction. Without a forward and reverse reaction (a reversible reaction) there can be no equilibrium between the products and the reactants. $\endgroup$ – MaxW Mar 3 '16 at 18:19
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  1. Yes, every chemical reaction can theoretically be in equilibrium. Every reaction is reversible. See my answer to chem.SE question 43258 for more details.

  2. This includes even precipitation reactions and reactions that release gases. Equilibrium isn't just for liquids! Multiphase equilibria exist.

  3. The only thing that stops chemical reactions from being "in equilibrium" is the lack of the proper number of molecules. For a reaction to be in equilibrium, the concentrations of reactants and products must be related by the equilibrium constant.
    $$ \ce{ A <=> B} $$ $$ K = \frac{[B]}{[A]} $$

    When equilibrium constants are extremely large or small, then extremely large numbers of molecules are required to satisfy this equation. If $K = 10^{30}$, then at equilibrium there will be $10^{30}$ molecules of B for every molecule of A. Another way to look at this is that for equilibrium to happen, there need to be at least $10^{30}$ molecules of B, i.e. more than one million moles of B, in the system for there to be "enough" B to guarantee an equilibrium, i.e. to guarantee that there will be a well-defined "equilibrium" concentration of A.

    When this many molecules are not present, then there is no meaningful equilibrium. For very large (or very small) equilibrium constants, it will be very difficult to obtain an equilibrium. In addition to needing a megamole-sized system (or bigger), the system will have to be well-mixed, isothermal, and isobaric. That's not easy to achieve on such large scales!

Update Commenters suggest that "irreversible" reactions do not have an equilibrium. This is true, but tautological. In the real world, all reactions are reversible, at least to a (perhaps vanishingly small) degree. To say otherwise would violate microscopic reversibility. A reaction that was 100% irreverible would have an equilibrium constant of infinity. But if $K= \infty$, then $\Delta G^{\circ} = -RT \ln{K}$ would turn into $\Delta G^{\circ} = -\infty$. So to get infinite energy we would just have to use 100% irreversible reactions! Hopefully the problems with the idea of "irreversible" reactions are becoming apparent.

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    $\begingroup$ @user109987 - Unfortunately this answer is just wrong. Irreversible reactions by definition only go in one direction and hence there is no reverse reaction. Without a forward and reverse reaction (a reversible reaction) there can be no equilibrium between the products and the reactants. $\endgroup$ – MaxW Mar 3 '16 at 18:17
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    $\begingroup$ I'm confused why you commented on my answer but flagged @user109987. Also, the answer is not wrong because no reaction is 100% irreversible. Saying that there is no reverse reaction violates the principle of microscopic reversibility. $\endgroup$ – Curt F. Mar 3 '16 at 18:40
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    $\begingroup$ I flagged user109987 because that user accepted your answer which is wrong by definition of an irreversible reactions. // There is a fine difference here. There is not an equilibrium constant K between reactants and products for the reaction of barium chloride and sulfuric acid to give a barium sulfate precipitate and hydrogen chloride. If there is in fact such an equilibrium constant, then derive it. Show how the equilibrium shifts in the solution when solid barium sulfate is added after the reaction. $\endgroup$ – MaxW Mar 3 '16 at 19:47
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    $\begingroup$ Yes of course there is an equilibrium! If you want to know how to calculate the equilibrium constant and want to know how the composition of the system shifts in response to adding more barium sulfate, then I suggest you ask a new question right here at chem.SE. $\endgroup$ – Curt F. Mar 3 '16 at 20:49
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    $\begingroup$ Ok, I'll pick up the gauntlet. Question posted here chemistry.stackexchange.com/questions/47384/… $\endgroup$ – MaxW Mar 4 '16 at 5:09
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Equilibrium can only apply to a closed system.

Reactions which form insoluble precipitates or gases which escape do not exhibit the behavior of a closed system. Therefore, these reactions may not be in equilibrium. However, these claims are pragmatic rather than real.

As it turns out, in the above answer, barium sulphate has a $\ce{K_{sp}}$ of $\ce{1.1 x 10^{-10}}$, so formally there is some small equilibrium related to the amount of barium sulphate in solution $\ce{1.05 x 10^{-5}}$

As gasses are escaping in solution, they may be readsorbed, an thus there would be some small equilibrium for processes like that.

But pragmatically, these reactions are not at equilibrium.

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    $\begingroup$ If there were an "equilibrium" for the reaction then the equation would be something like: $$\ce{K_{eq}} = \dfrac{\ce{[BaSO4][HCl]^2}}{ \ce{[BaCl2][H2SO4]}}$$ and such an equilibrium just doesn't exist. $\endgroup$ – MaxW Mar 2 '16 at 17:29
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    $\begingroup$ I cannot rationally interpret the idea of a chemical equilibrium not existing. I can rationally interpret is it pragmatically zero. $\endgroup$ – Lighthart Mar 2 '16 at 19:18
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    $\begingroup$ Look at it this way - adding solid BaSO4 to the solution won't shift the "equilibrium" to the left . So how could there be an equilibrium?!? $\endgroup$ – MaxW Mar 3 '16 at 2:20
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    $\begingroup$ (1) Equilibrium can be applied to open systems in some cases. (2) insoluble precipitates and gases can and do participate in equilibrium reactions, whether or not they are "reabsorbed". An example is rising bicarbonate levels in the ocean due to higher CO2 levels in the atmosphere. There is an equilibrium. (3) It is unclear what definition of "pragmatic" you are using. $\endgroup$ – Curt F. Mar 3 '16 at 12:15
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    $\begingroup$ @MaxW if the amount of BaSO4 was under the incredibly small solubility limit, adding more of it would shift the equilibrium. The effect would be so small that you could pragmatically say there is no change in the equilibrium, but that is not quite true. It is however, a useful imprecision. $\endgroup$ – Lighthart Mar 3 '16 at 19:16
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Yes every reaction is an equilibrium. A complete reaction is a equilibrium with high equlibrium constant. If you write the expression for equilibrium constant, you will find that high equilibrium constant implies that the conc. of the products is very high, i.e. the reaction has reached completion.

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$$ \ce{BaCl2 (aq) + H2SO4 (aq) -> BaSO4 (s) + 2 HCl (aq)} $$ Let's just think about what (aq) means; it means you have ions floating about in there which are in equilibrium with their solids.

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

Note if I had $\ce{BaSO4}$ in water, which would be in equilibrium (so tiny dissolved/tiny bit of $\ce{Ba^{2+}}$ ions and $\ce{SO4^{2-}}$ ions), and I added $\ce{Cl-}$ ions, a negligible amount more $\ce{BaSO4}$ will dissolve, as $\ce{BaCl2}$ would go to equilibrium, reducing $\ce{Ba^{2+}}$ ions, leading to negligible amounts of $\ce{BaSO4}$ dissolving to remain at $K_{\mathrm{s}}$. This also shows Le Chatelier's principle.

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No, every reaction isn't in equilibrium with its products. Consider the following irreversible reaction: $$\ce{BaCl2(aq) + H2SO4(aq) -> BaSO4(ppt) + 2HCl(aq)}$$.

By definition if the reaction is irreversible then there is no equilibrium for that reaction.

  • If there were an "equilibrium" for the reaction then the equation would be something like: $$\ce{K_{eq}} = \dfrac{\ce{[BaSO4][HCl]^2}}{ \ce{[BaCl2][H2SO4]}}$$ and such an equilibrium just doesn't exist since when the barium sulfate precipitates there could be a microgram or a kilogram as the product.
  • Think of it another way - adding $\ce{HCl}$ (in dilute solution) or $\ce{BaSO4}$ won't shift the reaction to the left. (Adding more HCl would shift $\ce{HSO4^{-} <-> H+ + SO4^{2-}}$ in concentrated solutions, which is besides the point I'm trying to make.)
  • There is a solubility product for barium sulfate, but the solubility product doesn't depend on the amount of barium sulfate precipitate, nor the concentration of HCl. So the solubility product isn't for the overall reaction but rather for part of the system:

$$\ce{[Ba][SO4^{2-}] = K_{sp}}$$

(Full disclosure - Theoretically the barium sulfate solubility product wouldn't depend on the HCl concentration, but really that isn't quite true. The barium sulfate solubility product really depends on the activity of the barium and sulfate ions, so the ionic strength of the solution matters.)

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    $\begingroup$ Isn't this theoretically in equilibrium, just that for all intents and purposes the equilibrium always lies far to the product side? $\endgroup$ – bon Mar 2 '16 at 10:19
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    $\begingroup$ There is no equilibrium for the BaSO4 product. Since the ppt is another phase there could be 1 microgram or 1 kilogram. $\endgroup$ – MaxW Mar 2 '16 at 14:35
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    $\begingroup$ This seems wrong. Multiphase equilibria exist! Liquid water at 100 °C is actually in equilibrium with gaseous water (steam) at 100 °C even though there could be 10 microgram or 100 exagrams of steam. $\endgroup$ – Curt F. Mar 3 '16 at 12:18
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    $\begingroup$ The OP's question was if "all reactions" are in equilibrium, not if at one equilibrium exists for every chemical system. For example in aqueous solution $\ce{[H^+][OH^-]} = 1 \times 10^{-14}$. But by definition if the reaction is irreversible then there is no equilibrium for the reaction. $\endgroup$ – MaxW Mar 3 '16 at 17:53

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