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Question on Keq and how water affects certain reactions

  1. Oct 2, 2014 #1
    Say I have a reaction like the one described below
    [(NH3)5CoCl]^(2+) + [OH]^(1-) <-----> [(NH3)4Co(NH2)Cl]^(1+) + H2O

    The Keq will be equal to [products(without H2O)] / [reactants]. Now when I add water to this reaction mixture, the reaction will move to the left(reactants side). This is not because H2O is being added which one might think would push the reactants to the left but rather because of the concentration effects that water has correct? The Keq constant has two values in the denominator while only one in the numerator, so adding water will decrease the denominator more relative to the numerator. This produces an increase in the reaction quotient, Q, which results in a shift in the reaction back to the reactants side in order to decrease Q back to the value of Keq. Is my thinking correct?

    Thanks
     
  2. jcsd
  3. Oct 2, 2014 #2

    TeethWhitener

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    Yes, this is an illustration of Le Chatelier's principle. Adding species to one side of an equilibrium forces the reaction in the other direction to re-establish equilibrium.

    Strictly speaking though, leaving [H2O] out of [itex]K_{eq}[/itex] is just an approximation. Since many reactions are done in dilute aqueous solution, the concentration of water doesn't change all that much over the course of the reaction. This can be seen intuitively if you imagine a solution that's 98wt.% water and 2wt.% salt. If you add enough water to double the mass of the solution, then the salt concentration changes to 1wt% (a decrease by half), whereas the water concentration changes from 98wt.% to 99wt.%, barely a change at all. So [H2O] is actually a part of [itex]K_{eq}[/itex], but its concentration usually changes so little compared to the rest of the species in the reaction that it can be ignored.
     
  4. Oct 2, 2014 #3
    Hmmm icic, thanks for your reply! However, just to reaffirm my understanding with this other scenario I just came up with. What if you had a hypothetical reaction such as

    A + B <----> C + D + H2O

    A,B,C and D are aqueous. Since H20 is eliminated via approximation, its only effect is through the changing concentration of A,B,C and D. Unlike the other example that I posted up there, this reaction will not experience any significant change in reaction quotient, correct? Since the volume values will properly cancel out this time in the numerator and denominator.

    Thanks again!
     
  5. Oct 2, 2014 #4

    TeethWhitener

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    Kind of. If H2O is just a solvent, then technically it would be in both the numerator and the denominator and they would cancel. But if H2O is consumed or produced during the course of a reaction, it can potentially affect the equilibrium constant. For example, [tex] H_{2}CO_{3(aq)} \rightleftharpoons H_2O_{(aq)} + CO_{2(aq)}[/tex] has an equilibrium constant [tex]K_{eq}=\frac{[H_2O][CO_2]}{[H_2CO_3]}[/tex] Since H2O is produced by the dissociation of carbonic acid in this case, the equilibrium will be affected by the water. How much it's affected, and whether water can be neglected, is a matter of how dilute or concentrated the solution is.
     
  6. Oct 3, 2014 #5

    TeethWhitener

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    Just to clarify, because this is kind of tricky to get at first, if you took our hypothetical reaction: [tex]H_2CO_{3(aq)} \rightleftharpoons H_2O_{(aq)}+CO_{2(aq)}[/tex] and wrote the equilibrium constant with every single species, it would look like this: [tex]K_{eq}=\frac{[H_2O]_{prod}[CO_2]}{[H_2O]_{reac}[H_2CO_3]}[/tex] But in a dilute solution (where you have a large amount of water as a solvent relative to dissolved CO2), [itex][H_2O]_{prod}[/itex] won't be that much more than [itex][H_2O]_{reac}[/itex], meaning [tex]\frac{[H_2O]_{prod}}{[H_2O]_{reac}}\approx 1[/tex] and therefore [tex]K_{eq}\approx \frac{[CO_2]}{[H_2CO_3]}[/tex] Hope that makes things a little clearer.
     
  7. Oct 3, 2014 #6
    hmm, I understand. Luckily we can just ignore water for the most part. Thanks for sticking to this thread. Made things much clearer for me.
     
  8. Dec 17, 2014 #7
    Sorry to resurrect, but I don't get this. You only wrote H2O (aq) on one side of the reaction, so how does it appear on both sides of the K equation?
     
  9. Dec 17, 2014 #8

    TeethWhitener

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    The decomposition of [itex]H_2 CO_3[/itex] produces [itex]H_2 O[/itex] as one of its products. However, in the example above, the reaction is being run as a dilute solution in water. Therefore, if you were to include all of the species in the reaction, you'd have to include the solvent water too:
    [tex]H_2 CO_{3(aq)} + H_2 O (solvent) \rightleftharpoons H_2 O (product) + CO_{2(aq)} + H_2 O (solvent)[/tex]
    But in a dilute solution, [itex]H_2 O (solvent) \approx H_2 O (solvent) + H_2 O (product)[/itex], so they basically cancel out in the final equilibrium constant. Caveat: if the solution isn't dilute (that is, if the approximation above doesn't hold), then you have to include the [itex]H_2 O (product)[/itex] in the equilibrium constant explicitly.
     
  10. Dec 18, 2014 #9
    How is this calculation done then if the solution is not dilute and you must use full K? let's say you know initial concentration of water and final concentration of water; are you suggesting putting the initial in the denominator of K, even though it is an initial (not an equilibrium) concentration of its species?
     
  11. Dec 18, 2014 #10

    TeethWhitener

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    No, in that case you'd have to use the [itex]H_2 O (product)[/itex], because it's the actual product of the reaction. Leaving [itex]H_2 O (product)[/itex] out of the equilibrium constant was just an approximation in the first place. The point that I was trying to make by including the solvents explicitly is that in a dilute solution, the "concentration" of the solvent (water) is so overwhelmingly large, that even if you produce the solvent as a product in the reaction, you don't change the overall concentration of the solvent all that much, so it doesn't affect the equilibrium. In the case of a non-dilute solution where the concentration of solvent does change appreciably, then you absolutely must take that into account.

    However, keep in mind that equilibrium constant is just that: constant. In the case for the carbonic acid reaction, the equilibrium constant is [itex]K=588[/itex] in pure water. This means that if you attempt to have a very concentrated solution of carbonic acid (low water concentration), you'll be fighting Le Chatelier's principle. The reaction will swing strongly back toward equilibrium by turning [itex]H_2 CO_3[/itex] into [itex]H_2 O[/itex] and [itex]CO_2[/itex].
     
  12. Dec 18, 2014 #11

    Borek

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    I was observing the discussion out of curiosity where it will land. Interesting approach, but I think I will stay with a more classic one. If

    [tex]K' = \frac{[H_2CO_3]}{[H_2O][CO_2]}[/tex]

    and concentration of water can be treated as constant, we can redefine the constant as

    [tex]K = K'[H_2O] = \frac{[H_2CO_3]}{[CO_2]}[/tex]

    I feel like it is less confusing.
     
  13. Dec 18, 2014 #12

    TeethWhitener

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    Thanks Borek. You're right; your approach is much less confusing. The important point is that ignoring the water is only appropriate if the concentration of water doesn't change much.
     
  14. Dec 19, 2014 #13
    Sure, makes sense. Can we estimate K = K'[H2O] using the initial (55.5 M) value of [H2O] and measuring K' in reasonably dilute solution? Since there is no way to measure [H2O] exactly directly after reaction

    The point being that [H2O] = 55.5 M + [H2O]prod?

    I still think it is wrong to write H2O on both sides of Keq, it does not fit any of borek's equations either.
     
    Last edited: Dec 19, 2014
  15. Dec 19, 2014 #14

    Borek

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    Sure. But it is all a matter of what we measure and how we define the constant - i.e. we can measure (I am ignoring technicalities here) concentrations of the reactants and products and plug them into equation not taking water into account at all - that will yield exactly the same result. Say, measure concentrations of H+ and OH- and multiply them and you get Kw.

    You can calculate it from the mass balance.
     
  16. Dec 19, 2014 #15

    TeethWhitener

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    I'm writing this from my phone so I'm not quite as adept with all the latex commands and various other functionalities and whatnot, so I apologize in advance.

    Astudious said "The point being that [H2O] = 55.5 M + [H2O]prod?"

    [H2O] is not additive; furthermore, 55.555M is the molarity for pure water. In solution, therefore, [H2O] will be less than 55.555M to begin with (call it 55.555-x). As the reaction approaches equilibrium, [H2O] will approach 55.555-x+y, but still be less than 55.555. The approximation of [H2O] as constant is technically the limit of an infinitely dilute solution.

    Astudious said "I still think it is wrong to write H2O on both sides of Keq, it does not fit any of borek's equations either."

    You're right. H2O only goes on the product side. This is always true. But since H2O is a product in addition to being the solvent, the fact that you start the reaction with x amount of water affects where the equilibrium lies. A comparable situation is the common ion effect, where the presence of two salts with an ion in common in a solution affects the solubility equilibrium for the overall mixture.

    I worry that I'm just making this more confusing. :/
     
  17. Dec 19, 2014 #16
    No, this makes sense, so long as in the Keq expression for dissociation, [H2O] should only be in the numerator (not in the denominator). Sure, there was some H2O to begin with, but this just has to be included in the [H2O] term in the numerator. In the same sense, with the common ion effect we include contribution from the common ion into the term in the Ksp expression for tat species; we don't write any terms in the numerator and denominator.
     
    Last edited: Dec 19, 2014
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