# Chemistry - Why are equilibrium constants unitless?

## Solution 1:

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

## Solution 2:

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

## Solution 3:

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.