I mean the gold book defines two terms as the equilibrium constant - the standard/thermodynamic equilibrium constant and generic equilibrium constant.

To me, only the thermodynamic one seems correct cause it is derived from thermodynamic factors or activities and Gibbs free energy. The other one just seems like a senseless ratio/pedagogical tool to teach this concept of "Law of Mass Action" which itself doesn't have any thermodynamic basis but only a kinetic one.

So, is the second one useful into some other parts of chemistry, if not why can't we do away with it.

Or is it because kinetic equilibrium without thermodynamic equilibrium is a serious thing.

  • $\begingroup$ Why do you think the law of mass action cannot be derived from thermodynamics factors ? $\endgroup$
    – Maurice
    Commented Apr 2, 2022 at 8:28
  • $\begingroup$ There are cases where the TD K looks like a senseless pedagogical tool looking good in textbooks, while not always applicable in real life. $\endgroup$
    – Poutnik
    Commented Apr 2, 2022 at 10:14
  • $\begingroup$ It is both important when teaching chemistry and also in "real life" to understand that from kinetics we can get equilibrium constants. I have seen how using kinetics data for gold chloride complexes interconverting that we can calculate the stability constants for complexes such as AuCl4-. $\endgroup$ Commented Apr 2, 2022 at 10:47
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    $\begingroup$ if you ever use $\Delta G=-RT\ln(K_{eq})$ then $K_{eq}$ has to be dimensionless, which means that all equilibrium constants have to be dimensionless. It is a historical/convenient thing that units are still given, and we implicitly assume that the $K$ are divided by 1 unit to make them dimensionless. The Gold book definition gives them as dimensionless. $\endgroup$
    – porphyrin
    Commented Apr 3, 2022 at 7:04
  • $\begingroup$ @porphyrin what you are saying is just the end results where we have already defined relation between chemical potential and activities (which are dimensionless), this just strengthen the point that we should do away generic equilibrium constant. I am looking for some strong counter arguments. One good one is by Karsten Theis below. $\endgroup$ Commented Apr 3, 2022 at 17:13

1 Answer 1


The three $K$'s of biochemistry

It is common to use $K$ with a unit in biochemistry. There are three main uses, and the unit customarily has the dimension of a concentration (e.g. $\pu{mol L-1}$).

$K_\mathrm{d}$: Dissociation constant, typically of a ligand bound to a biomolecule. You would say "the ligand has nanomolar affinity" to express that the dissociation constant is about $\pu{1e-9}$. This means that at nanomolar concentration, there would be a significant fraction of ligand-bound biomolecule (with the assumption that the ligand is always at higher concentration than the biomolecule).

$K_\mathrm{i}$: Inhibition constant, essentially the dissociation constant of the inhibitor binding to an enzyme. When the inhibitor concentration is equal to the inhibition constant, more than 50% is bound.

$K_\mathrm{m}$: Michaelis–Menten constant, a combination of three rate constants. This describes a steady state of an enzymatic reaction. When the substrate concentration is equal to $K_\mathrm{m}$, the enzymatic rate is half-maximal (compared to saturating concentrations of substrate).

In all three cases, you assume a bimolecular association event with no cooperativity.

Here is a snapshot from the Table 1. Recommended Symbols and their Units [1, p. 290]:

\begin{array}{lll} \hline \text{Symbol} & \text{Meaning} & \text{Customary unit} \\ \hline K_\mathrm{i} & \text{inhibition constant (inhibition type unspecified)} & \pu{mol dm^-3} \\ K_\mathrm{iA} & \text{inhibition constant for A} & \pu{mol dm^-3} \\ K_\mathrm{ic} & \text{competitive inhibition constant} & \pu{mol dm^-3} \\ k_{ij} & \text{rate constant for step from}~E_i~\text{to}~E_j & \text{as}~k \\ K_\mathrm{iu} & \text{uncompetitive inhibition constant} & \pu{mol dm^-3} \\ K_\mathrm{m} & \text{Michaelis constant (or Michaelis concentration)} & \pu{mol dm^-3} \\ K_\mathrm{mA} & \text{Michaelis constant for A} & \pu{mol dm^-3} \\ K_\mathrm{s} & \text{substrate dissociation constant} & \pu{mol dm^-3} \\ K_\mathrm{sA} & \text{value of}~K_\mathrm{s}~\text{for substrate A} & \pu{mol dm^-3} \\ \hline \end{array}

Relation to thermodynamic equilibrium constant

The relation is loose. For example, even if these are pH-dependent, you would not include the hydronium concentration in the scheme but instead quote the values at a specific pH. You would also combine distinct protonation states in a single constant (for examples for ATP).

Time to wash your hands

As a physical chemist, you might feel sullied by the muddy definitions above. However, they are part of the daily language in biochemical labs and in the pharmaceutical industry. This would be the "A" of IUPAC.

Logarithms and units

The argument and the result of a logarithm do not have units, but measurements do. Somewhere before taking a logarithm, you have to have a step that gets rid of the units. Here are some examples:

$$ \mathrm{pH} = -\log [\ce{H+}]\tag{1}$$

$$ \Delta G^\circ = − R T \ln(K)\tag{2}$$

$$ \ln \frac{k_2}{k_1} = \frac{E_A}{R} (\frac{1}{T_1} - \frac{1}{T_2})\tag{3}$$

Textbooks often are not explicit about what happens to the units. In (1), you have to divide the concentration by the standard state (unless you use activities instead of concentration). In (2), you "dropped" the units when calculating the equilibrium constant. In (3), you get away with rate constants that have units because they cancel out. It would be a fine convention to have equilibrium constants with units, and then divide them by $Q^\circ$, the reaction quotient for the standard state. So whether $K$ is defined as dimensionless or with units is not directly linked to thermodynamic or kinetic definition, it is a convention.

Kinetics vs thermodynamics

$K_\mathrm{m}$ by its nature is defined through kinetics, as is $K_\mathrm{i}$. For determining $K_\mathrm{d}$, however, you typically would do an equilibrium measurement. Combined with the (first-order) rate constant of dissociation, $k_\mathrm{off}$, you can calculate the (second-order) rate constant of association. For non-covalent complexes, this will be fast, near diffusion controlled, in many instances. So there is a connection between kinetics and thermodynamics, but it is not directly related to the choice whether $K$ has units or is dimensionless.


  1. Nomenclature Committee of the International Union of Biochemistry (NC-IUB). Symbolism and Terminology in Enzyme Kinetics (Recommendations 1981). Eur J Biochem 1982, 128 (2–3), 281–291. DOI: 10.1111/j.1432-1033.1982.tb06963.x.
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    $\begingroup$ Other fields have chemistry have similar things, for example Henry's Law constants ($K_\mathrm{H}$) are generally reported with units because there is no single standard definition. $\endgroup$
    – Andrew
    Commented Apr 2, 2022 at 19:57
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    $\begingroup$ This answer seems to me like just a series of facts without an explantion of the chemistry. One should aim to understand chemistry rather than merely memorise it. I would be more happy if the answer showed the relationship between kinetics and the thermodynamic idea of an equilibrium. $\endgroup$ Commented Apr 3, 2022 at 6:12
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    $\begingroup$ @NuclearChemist I gave it a shot. However, I can not be responsible for your happiness. $\endgroup$
    – Karsten
    Commented Apr 3, 2022 at 13:42
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    $\begingroup$ @KarstenTheis to me the logarithm argument doesn't make sense. In fact it will point to flaws in our thermodynamic theory if some variable came under logarithm and all we have to say is look logarithm are unitless , so let's divide by something to make our variable unitless. That's just a bad theory/argument. It is unitless cause activities are unitless and activities are unitless cause they are defined w.r.t standard states. $\endgroup$ Commented Apr 3, 2022 at 17:32
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    $\begingroup$ It is sloppy use of math. DeltaG = -RTlnKeq + RTlnQ = -RT [lnKeq -lnQ] = -RTln[Keq/Q]. The units of Keq and Q are the same so the expression is unitless. At equilibrium Q=Keq so deltaG = 0. Under standard conditions All activities are defined as 1. so deltaG = deltaG[0] = _RTln[Keq/1] still unitless. $\endgroup$
    – jimchmst
    Commented Apr 5, 2022 at 5:29

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