According to my book,

In the real world [...] it is unlikely that any reaction would have exactly equal concentrations of products and reactants at equilibrium.

This was mentioned in context to the fact that it was an ideal situation for $K_\mathrm{eq}$, wherein $K_\mathrm{eq} = 1$

What confuses me is the term "unlikely", which I take to mean that it is uncommon yet possible. Can you explain why having equal concentrations of products and reactants at equilibrium such that $K_\mathrm{eq} = 1$ is both possible yet uncommon?

  • $\begingroup$ Can you give some more context? In what context did your book mention that statement? And equilibrium does not mean there is equal concentration of product and reactant, there are many reactions like $\ce{2A +C ->B} where that isn't true. (If someone is very pedantic then yes, technically perfect equilibrium is impossible to achieve) $\endgroup$
    – S R Maiti
    May 31, 2021 at 18:45
  • $\begingroup$ The book said it was an ideal situation for Keq, wherein Keq = 1 $\endgroup$
    – Ibby
    May 31, 2021 at 18:52
  • $\begingroup$ @ShoubhikRMaiti I assumed this meant ideal equilibrium $\endgroup$
    – Ibby
    May 31, 2021 at 18:53
  • 1
    $\begingroup$ Related: chemistry.stackexchange.com/questions/57075/… $\endgroup$ May 31, 2021 at 18:59
  • 1
    $\begingroup$ When you specifically want a value, everything is unlikely. While the thread contains right comments and answers, it is hard to understand the meaning of ideal in the context "to the fact that it was an ideal situation....". One can place 2 instead of 1. $\endgroup$
    – Alchimista
    Jun 1, 2021 at 11:13

1 Answer 1


For a chemical reaction,

$$\Delta G = -R T \ln K_{eq}$$ $$\implies K_{eq} = e^{\frac{-\Delta G}{RT}} $$

Thus, in order for a chemical reaction to have $K_{eq}=1$*, it is necessary to have $\Delta G = 0$

For some chemical reactions**, we can achieve this by adjusting the temperature and/or pressure. Thus it is possible. However, it's unlikely that the temperature and pressure at which a chemical reaction is taking place is coincidentally just that needed to achieve $\Delta G = 0$.

On the other hand, in finding phase transition temperatures (melting and boiling points), we are sweeping through a range of temperatures specifically to find the points at which the phases coexist, i.e., where their free energies are the same and thus where $\Delta G = 0$. Thus these are equilibria for which $K_{eq}=1$.

*We can never exactly get $\Delta G = 0$, because we can never exactly adjust T and p; consequently, we can never exactly have $K_{eq}=1$. Thus what I mean (and presumably what your textbook means) when it says "$K_{eq}=1$" is that it is so close to $1$ that it's within the size of the fluctuations about equilibrium.

**I say "for some chemical reactions" because in some cases the reactants or products might break down before the required temperature is reached. E.g., suppose you were trying to find a reaction involving glucose at which $K_{eq} =1$, and did a theoretical extrapolation in which you found that this occurred at, say, 10,000 K. Well, glucose doesn't exist at that temperature.

Note #1: $\Delta G = -R T \ln K_{eq}$ applies to processes carred out at constant T and p. If the process were instead at, say, constant T and V, then the equilibrium condition would be $\Delta A = -R T \ln K_{eq}$. [And don't be confused: Constant T and p doesn't mean a specific T and p. T and p can be whatever you please, so long as the reactants and products can exist at those conditions. Rather, constant T and p just means that, whatever T and p are, they stay fixed during the reaction.]

Note #2 (on time to attain equilibrium): The above has nothing to do with whether or not you can get to equilbrium, which is addressed in the question linked in Ivan's comment on your post. That question pertains to any equilibrium, regardless of the value of the equilibrium constant (whether it's 1 or 10.303 or 4.8 x 10^-22, or any other value). Instead, the above is about whether or not the value of the equilibrium constant can be 1.

Note #3 (on enantiomers in an achiral environment): Andrew raises an interesting point that the reaction:

$$\ce{A_L <=> A_D},$$

where $\ce{A_L}$ and $\ce{A_D}$ are two different enantiomers of compound A, will have $\Delta G = 0$, and thus $K_{eq}=1$, in an achiral environment because, in an achiral environment, the two compounds are chemically identical. And as they are chemically identical, this will obtain at all temperatures and pressures at which the compound exists. I.e., in an achiral environment, chemically,

$$\ce{A_L <=> A_D}$$ reduces to: $$\ce{A <=> A}$$.

But: While enantiomers are chemically identical, the weak nuclear force that exists within their nuclei can recognize handedness, and thus the restriction in the first paragraph still applies: $K_{eq}$ won't exactly equal 1, but can be immeasurably (at least with current technology) close to 1. [I haven't done the calculation to determine if the difference in energy caused by the weak nuclear force puts the deviation from 1 within the, say, the 90% fluctation range about equilibrium for a representative pair of enantiomers at room temperature.]

  • 2
    $\begingroup$ Another example is enantiomers in an achiral environment $\endgroup$
    – Andrew
    May 31, 2021 at 23:27
  • 1
    $\begingroup$ @Andrew: Thanks for your comment. See my addition at the end. $\endgroup$
    – theorist
    Jun 1, 2021 at 2:33
  • 4
    $\begingroup$ Couldn't the same be said about any value of K or $\Delta G^\circ$?? I don't see the point in discussing K=1 as a special case. $\endgroup$
    – Buck Thorn
    Jun 1, 2021 at 7:19
  • 1
    $\begingroup$ You mean $\Delta G ^\circ = -RT\ln K_{eq}$? (note the superscript indicating std conditions) $\endgroup$
    – Buck Thorn
    Jun 1, 2021 at 8:46
  • 4
    $\begingroup$ Would you consider an identity reaction, let's say intramolecular proton transfer $$\ce{\color{red}{H}O\color{red}{C}H2\color{blue}{C}H2O\color{blue}{H} <=> \color{blue}{H}O\color{red}{C}H2\color{blue}{C}H2O\color{red}{H}},$$ as still being a valid "reaction"? Apologies for the grotesque colours. This one is a bit contrived, but I think you could still define a rate constant for it, and I'd suspect that there are valid reactions like this which are of genuine interest to someone. $\endgroup$
    – orthocresol
    Jun 1, 2021 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.