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Given a reversible chemical reaction, e.g. $$\ce{CuCl4^2-(aq) <=> Cu^2+(aq) + 4Cl-(aq) + \text{heat}}$$ occurring in a medium, the equilibrium constant is defined as $$K = \frac{[\ce{Cu^2+}][\ce{Cl-}][\ce{Cl-}][\ce{Cl-}][\ce{Cl-}]}{[\ce{CuCl4^2-}]}$$ where [X] is the molarity or molar concentration of X = amount$_X$ / volume.

Then, as you can read anywhere (e.g. google 'equilibrium constant equals 1' to see (1st hit) - 'The Equilibrium Constant, K - Chemistry LibreTexts' -

If K is about equal to 1, the reaction will reach equilibrium as an intermediate mixture, meaning the amounts of products and reactants will be about the same.

Yet, if the 'amounts of products and reactants' are the same, that means to me the sum of the molarities of the products would equal the sum of the molarities of the reactants, but K = 1 means that the products of molarities are the same.

I can't attach any intuitive meaning to the product of the molarities.

Note: the first answer is exactly what I needed. This question was a pedagogical question - as noted above the first google hit gives a textbook with incorrect info on the equilibrium constant. That had me confused. Also, I note that my kid's 10 grade honors chemistry text defines the equilibrium constant as the ratio the product of the molarities of the products divided by the product of molarities of the reactants when the system is in equilibrium, and not as in answer 1 below as the ratio of forward and reverse reaction coefficients, and the latter is a much better definition.

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closed as unclear what you're asking by Todd Minehardt, Mithoron, A.K., Karsten Theis, Jon Custer Mar 31 at 17:46

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    $\begingroup$ Equilibrium is not when the concentrations of products and reactants are about the same. It is when they don't change anymore even though none of them are zero (i.e. reaction does not go to completion). K can be a million or a millionth, and the reaction will still be able to reach equilibrium. $\endgroup$ – Karsten Theis Mar 30 at 23:54
  • $\begingroup$ Yes, so the Chemistry LibreTexts quoted in the question is ... wrong. $\endgroup$ – DrWill Mar 31 at 14:02
  • $\begingroup$ There is some wiggle room. K about 1 might mean between a millionth and a million. Amounts of products and reactants will be about the same might mean their concentrations are within the same couple of orders of magnitude. $\endgroup$ – Karsten Theis Mar 31 at 15:26
  • $\begingroup$ Here are two correct statements. (1) If we start with all concentrations at standard state (1 M for solutes), the reaction will go forward if K > 1 and reverse if K < 1. (2) Everything else being the same, the higher the equilibrium concentration the more product at equilibrium. Both statements are based on Q vs K. For some more information why the textbook statement is a bit misleading, have a look at my answer to a previous question. $\endgroup$ – Karsten Theis Mar 31 at 15:29
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I think the most intuitive way to think about this is to view it through a kinetic lens. That is, recall that at equilibrium, the rate of the forward reaction and the rate of the reverse reaction must be equal. When we write expressions for the rates of reactions, we multiply the reactants because the probability of collision is a function of the product of the two concentrations, not the sum. This only really makes sense for an elementary step, but the result applies to the whole reaction.

Let's consider the simpler reaction $\ce{A + B <=> C}$.

If there's is only one step in the reaction, the rate of the forward reaction will be given by $k_\mathrm{fwd}[A][B]$ and the rate of the reverse reaction by $k_\mathrm{rev}[C]$.

At equilibrium, these are equal, so

$$k_\mathrm{fwd}[\ce{A}][\ce{B}] = k_\mathrm{rev}[\ce{C}].$$

Rearranging,

$$\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]},$$

and we see the product of the concentrations in the equilibrium constant.

You can also go the route of deriving the expression from the free energies and Boltzmann distribution, but I don't think that's nearly as intuitive as thinking in kinetic terms.

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  • $\begingroup$ Thanks. Is there anything that can be concluded if the equilibrium constant equals 1, other than that the products(*) of the concentrations of the products and reactants are equal? $\endgroup$ – DrWill Mar 30 at 23:41
  • $\begingroup$ Yes, K=1 means that the Gibbs free energy of the product state is exactly equal to that of the reactant state when all concentrations are 1 M. (And if there are gases, they are at 1 bar. Or more technically, the activities of all species are 1.) $\endgroup$ – Andrew Mar 30 at 23:53
  • $\begingroup$ K = 1 is nothing fundamentally special because it depends how we choose the standard state. Alien chemists not aware of IUPAC conventions might be using c = 3.1415 mol / L as their standard state for solutes (because that is equal to 1 in the units they happen to be using), but after converting from one to the other standard state, the measurements on earth would be entirely consistent with the alien measurements. $\endgroup$ – Karsten Theis Mar 31 at 17:30
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I can't attach any intuitive meaning to the product of the molarities.

The answer by Andrew gives a good rationale for the product of concentrations in the equilibrium constant expression. In my answer, I will try to give an intuitive argument why the sum of concentrations would not work out.

When a reaction is at equilibrium, forward and reverse rates are non-zero, i.e. the reaction is still going on at the particular level. That means all reactants and products have to be present when the reaction reaches equilibrium. This is in contrast to a reaction that goes to completion, where the limiting reactants are depleted, and the reaction stops.

So let's say you measure the reactant and product concentrations of the reaction $$\ce{A + B <=> C}$$ at equilibrium. Then, using your "sum equilibrium expression", you take the sum of the product concentrations and divide those by the sum of the reactant concentrations to arrive at the "equilibrium sum-constant".

I could now go ahead and make a mixture containing A and C only, with the concentration of C equal to that you measured at equilibrium, and the concentration of A equal to the sum of concentrations of reactants you measured at equilibrium. This reaction would have the same ratio of concentration sums, so it should be at equilibrium. Yet, the concentration of B is zero, so there can't be any forward reaction, and it can't be at equilibrium, leading to a contradiction.

Another contradiction is that we expect if we start at equilibrium and dilute the reaction mix, C should dissociate a bit (Ostwald's dilution law, see this question). With your "sum equilibrium expression", dilution would affect the numerator and denominator equally, so the equilibrium would not be disturbed.

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