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I haven’t quite reached the point where I can read a full-fledged text on chemical kinetics and thermodynamics yet, so bear with me, please.

I’m wondering why a value like $K_\text{eq} = \frac{[\ce{NO}]^2[\ce{O2}]}{[\ce{NO2}]^2}$ wouldn't have units of M?

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This source explains it well. It looks like part of class material, but it clearly explains the dimensionlessness of $K_{eq}$.

The resolution of this apparent paradox is that the above equation, while perfectly satisfactory for everyday use, is not technically correct. A more correct version is:

$$K_{eq} = \frac{\frac{\lvert B \rvert_{eq}}{\lvert B \rvert_{ss}} \frac{\lvert C \rvert_{eq}}{\lvert C \rvert_{ss}}}{\frac{\lvert A\rvert_{eq}}{\lvert A\rvert_{ss}}}$$

where the "ss" subscripts refer to the concentration of that species in the standard state. (By this definition, Keq is always unitless.)

It then goes on to state:

Strictly speaking, division by the standard state concentrations is also necessary in every thermodynamics equation in which you take the log of a concentration product, otherwise the units don’t come out right.) We will NEVER use this "correct" version of the equation in this class (well, never except in one problem on this week’s problem set...), and $K_{eq}$ for a reaction with unequal numbers of reactants and products is ALWAYS given with units, even in published papers.

[1] http://www.bio.brandeis.edu/classes/archives/biochem102/102-02stand.pdf

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I goofed up the first time I tried to answer this question, erroneously applying dimensional analysis to your equilibrium expression.

It turns out that Silberberg[1] gives a good explanation of why $K_\text{eq}$ is dimensionless, which is often glossed over as the terms of the equilibrium expression are generally taught as concentrations. In actual fact, the terms are ratios of the concentration or activity of each species with a reference concentration (1 $\mathrm{mol\cdot{L^{-1}}}$ for solutions.) For example, a concentration of 2 $\mathrm{mol\cdot{L^{-1}}}$ divided by a reference of 1 $\mathrm{mol\cdot{L^{-1}}}$ yields a ratio of 2, with no units. As each term has no units, so too does $K_\text{eq}$.

[1] Silberberg, M.E.; Chemistry – The Molecular Nature of Matter and Change 3e; 2003, p. 719

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  • $\begingroup$ I don't quite follow but that's a step in the right direction at least. $\endgroup$
    – readyready15728
    Commented Sep 11, 2012 at 8:08
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    $\begingroup$ This only works if the number of products is equal to the number of reactants. In the example Sadiq gives it is not, and so the equilibrium constant as written has units of $(\text{mol}\cdot\text{L}^{-1})^3/(\text{mol}\cdot\text{L}^{-1})^2 = \text{mol}\cdot\text{L}^{-1}$. $\endgroup$
    – N. Virgo
    Commented Sep 12, 2012 at 13:13
  • $\begingroup$ I understand this much, but I what I don't understand is how it can have significance in equilibrium reactions $\endgroup$
    – Kian
    Commented Feb 11, 2013 at 17:32
  • $\begingroup$ The stated facts is true for the standard equilibrium constant, not for an equilibrium constant. $\endgroup$
    – Martin - マーチン ♦
    Commented Jun 1, 2014 at 14:46
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Dimensional analysis is useless. The correct answer is the one already given involving activities, which are dimensionless.

Activities are defined as ratios. For example a pressure activity (there are many kinds) is defined in terms of the ratio of the actual pressure of a gas divided by the reference pressure, often 1 atm or 1 bar.

In the present example, the activity is the ratio of the molality divided by the reference molality of 1 molal. This assumes ideal solutions, which is good enough if the solutions are dilute. If the solution is not ideal one has to correct the molality for nonideality.

Because of these complications, detailed discussions of activities are usually left for a course in physical chemistry.

So while we use square brackets and molarities, we have to understand that we are really dealing with activities.

By the way, the activity of a pure liquid or solid is 1, which is why $[\ce{H2O}]$, for example, is dropped from equilibrium calculations.

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  • $\begingroup$ Wait a sec. Don't you mean molarity (as opposed to molality)? $\endgroup$
    – readyready15728
    Commented Sep 13, 2012 at 5:03
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    $\begingroup$ The issue of activity vs. concentration (or activity vs. pressure) is subtly different from the issue of dimensional vs. dimensionless. "Activity" can be related to the chemical potential, which does have dimension of energy/particle or energy/mol. It is the inherent definition of energy relative to a standard state that leads to $K_{eq}$ being dimensionless. $\endgroup$
    – Curt F.
    Commented Apr 20, 2015 at 14:12
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Note

This answer is incorrect, but because it is a common pitfall I'll leave it to show. It is wrong 1) because the OP meant $\lvert \ce{NO} \rvert ^2$ in the numerator not $\lvert \ce{NO2} \rvert$ as I took it (edited) and 2) because the $K_{eq}$ formula is incorrect. I will post a better answer.

The concise answer: Because the units cancel. Here I'll use $\lvert \rvert$ = concentration = moles/liter = m/L

$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert^2} (?) units$

$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} (?) units$

$K_{eq} = \frac{\lvert NO_2 \rvert m/L \lvert O_2 \rvert m/L}{\lvert {NO}_2 \rvert m/L \lvert {NO}_2 \rvert m/L}$

$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} \frac{m^2/L^2}{m^2/L^2}$

$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} \frac{m^2}{m^2}$

$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} (unitless)$

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  • $\begingroup$ Unfortunately not so - check the exponent on [NO] in the numerator. Straight dimensional analysis on Sadiq's example results in residual units. $\endgroup$
    – Richard Terrett
    Commented Sep 11, 2012 at 9:38
  • $\begingroup$ So he meant $\lvert NO \rvert ^2$ and not $\lvert NO^2 \rvert$ (which I took to mean $\lvert NO_2 \rvert$)? $\endgroup$
    – Ehryk
    Commented Sep 11, 2012 at 9:44
  • $\begingroup$ Yeah, I read it like that as well. But the stoichiometries don't work out otherwise. $\endgroup$
    – Richard Terrett
    Commented Sep 11, 2012 at 9:57
  • $\begingroup$ I've edited the question. Sorry. Yes I meant $[\ce{NO}]^2$. (A reminder: you can use \ce{ } to format molecular formulae easily in LaTeX.) $\endgroup$
    – readyready15728
    Commented Sep 11, 2012 at 10:00
  • $\begingroup$ Thanks for the tip. What's with the left [ being darker than ] in formulas? I switched to \lvert and \rvert just because it was ugly. $\endgroup$
    – Ehryk
    Commented Sep 11, 2012 at 10:56

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