# Unit of the equilibrium constant: contradiction of Bridgman's theorem?

The following equation is standard in thermodynamics:

$$\Delta G^\circ=-RT\log(K)$$

where $K$ is the equilibrium constant. In dimensional analysis, Bridgman's theorem tells us that the argument of a transcendental function (like $\log$) must always be dimensionless. But $K$ may have dimensions (depending on the particular equilibrium). Why is this OK?

Note: working out dimensions explicitly gives:

\begin{align} [R]&=EN^{-1}\Theta^{-1}\\ [T]&=\Theta\\ [\Delta G^\circ]&=EN^{-1} \end{align}

where $\Theta$ is the dimensions of temperature, $N$ is the dimensions of number and $E$ = dimensions of energy = $MLT^{-2}$. From these, we see that the quantity $\log(K)$ ought to be dimensionless, implying that $K$ should be dimensionless as well. But it isn't always.

Furthermore, suppose $K$ has dimensions $NL^{-3}$ (for an equilibrium of the form $A+B\leftrightarrow AB$, say). Suppose now that we scale the units for number by a factor $a$. Then we get new values for $K,R,\Delta G^\circ$, given by:

\begin{align} \hat K&=K/a\\ \hat R&=aR\\ \hat{\Delta G^\circ}&=a\Delta G^\circ \end{align}

From our original equation:

\begin{align} \Delta G^\circ&=-RT\log(K)\\ a\Delta G^\circ&=-aRT\log(K)\\ \hat{\Delta G^\circ}&=-\hat RT\log(a\hat K)\\ \hat{\Delta G^\circ}&=-\hat RT\log(\hat K)-\hat RT\log(a) \end{align}

So the new quantities do not satisfy the old equation. Rather, they satisfy the old equation, but with a constant factor of $-\hat RT\log(a)$ added on. What is going on here?

The problem is that people are often sloppy with the definition of quantities. The equilibrium constant $K$ in your first equation is indeed a dimensionless quantity while the equilibrium constant $K_c$ that is usually used to describe an equilibrium in a solution is not. I will take some detour to show where they come from and how they are connected.

From thermodynamics it is known that the Gibbs free energy of reation is given by $$\Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ ,$$

where $\xi$ is the extent of reaction and $\nu_{i}$ and $\mu_{i}$ are the stochiometric coefficient and the chemical potential of the $i^{\text{th}}$ component in the reaction, respectively. Now, imagine the situation for a ideal system consisting of two phases, one purely consisting of component $i$ and the other being a mixed phase comprised of components $1, 2, \dots, k$, in equilibrium. Since the system is in equilibrium and shows ideal behavior we know that the chemical potential of component $i$ in the mixed phase (having the temperature $T$ and the total pressure $p$), $\mu_{i}(p, T)$, must be equal to the chemical potential, $\mu^{*}_{i}(p_{i}, T)$, of the pure phase having the same temperature but a different pressure $p_{i}$, whereby $p_{i}$ is equal to the partial pressure of component $i$ in the mixed phase, namely $$\mu_{i}(p, T)= \mu^{*}_{i}(p_{i}, T).$$ From Maxwell's relations it is known that $$\left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = \left( \frac{\partial}{\partial p} \biggl(\frac{\partial G^{*}}{\partial n_{i}}\biggr) \right)_{T} = \Biggl( \frac{\partial}{\partial n_{i}} \underbrace{\biggl(\frac{\partial G^{*}_{i}}{\partial p}\biggr)}_{=\, V_{i}} \Biggr)_{T} = \left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$$

but since $\mu^{*}_{i}$ is associated with a pure phase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to $$\left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}$$

and one gets $$\left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = v_{i} \ ,$$

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase. Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get $$\int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ ,$$ so that, introducing the mole fraction $x_{i}$, $$\mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .$$ Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$ $$a_{i} = \frac{\varphi_{i} p_{i}}{p^0}$$ where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises $$\mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .$$ Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets $$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,$$ where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that $$\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .$$ So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant $$K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .$$ which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via $$K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .$$ it is more or less proportional to $K$ and thus gives qualitatively the same information.

Edit: If the solution at hand behaves like an ideal solution then by definition it's activity/fugacity coefficient is equal to one. Furthermore the state of ideality is defined with respect to standard states: for an ideal solution this is $c^{\ominus} = 1 \, \text{mol}/\text{L}$. Using this together with the ideal gas law on the relation between $K$ and $K_{c}$ $$K = \prod_i [\underbrace{\varphi_{i}}_{= \, 1}]^{\nu_{i}} \Bigl(\underbrace{\frac{RT}{p^{0}}}_{\substack{= \, 1/c^{\ominus} \, = \, 1 \, \text{L} / \text{mol} \\ \text{per definition}}}\Bigr)^{\sum_i \nu_{i}} K_{c} \qquad \Rightarrow \qquad K = \left(\frac{L}{\text{mol}} \right)^{\sum_i \nu_{i}} K_{c} \ .$$ one sees that for an ideal solution $K_{c}$ is identical to $K$ scaled by a dimensional prefactor.

Edit: I forgot to mention there are also "versions" of the equilibrium constants that are defined in terms of partial pressures or mole fractions which provide a more suitable description for gas equlibria but all those "versions" can be traced back to the original equilibrium constant.

• OK, thanks very much. But why then do I see the equation $\Delta G^\circ=-RT\log(K_C)$ (as I do at en.wikipedia.org/wiki/…)? – John Gowers Aug 7 '13 at 16:51
• @Donkey_2009 As I said $K_{c}$ is easier to work with and since it is more or less proportional to $K$ it gives you qualitatively the same information if the systems behavior is not too far away from the ideal behavior. But, I'd say strictly speaking the equation you quote should be incorrect but another opinion on that matter might be worthwhile. Maybe under normal conditions this equation is good enough. Unfortunately, I lack the practical expertise to say something about that. – Philipp Aug 7 '13 at 17:00
• @Donkey_2009 A little late but... because most chemist do not care about math and formalism. That equation have not mathematical sense since logarithm of anything with units is simply not defined. But normally people does not care and just neglect the units and takes logarithms. It is simply wrong, but in most practical cases works, so nobody cares. – user1420303 Mar 19 '16 at 14:53

(this is a comment to Phillip answer; upon editing it failed to get saved on due time)

Basically $K_c$ is valid as long as the solutions show colligative properties. These "concentration" constants can (and should) be defined dimensionless, as they properly emerge as a limit for the dimensionless activity in "ideal" systems, and the activity $a_j$ measures how the chemical potential $\mu_j$ of the substance $j$ changes as its concentration (or in general the chemical composition of the system) changes from some reference state: $a_j = [c]_j/c^\ominus$ where $[c]_j$ is the molar concentration and $c^\ominus = 1$ M. Therefore to remember that inside the logarithm everything is dimensionless one should not forget to write the reference magnitude $c^\ominus$. These constants can use different units for the concentration too, for instance $m_j/m^\ominus$ where $m_j$ is the molal concentration and the reference magnitude $m^\ominus$ is $1$ mol/kg.

Regarding the limits of validity, $K_c$ fails miserably for electrolytes, where the true thermodynamic constant depends on all the charges in the system (ionic strength of the solution), that is, the activity of the electrolyte is not only proportional to its number (concentration, etc.).

• Thanks. If you change the reference magnitude, then the numerical value for $K_c$ changes by a constant multiplicative factor, which means that the expression for $\Delta G^\circ$ changes by a constant additive factor. Does this mean that $\Delta G^\circ$ is only defined up to addition of some constant related to the choice of $c^\ominus$? – John Gowers Aug 8 '13 at 8:38
• Following my recipe, $K_c$ is dimensionless. Take the chemical reaction: aA $\rightarrow$ bB. Using the molar scale, $K_c = ([B]/c^\ominus)^b / ([A]/c^\ominus)^a$. There is no free multiplicative factor. You can use $K_p$ (based on partial pressures) or $K_x$ (based on molar fractions), etc. These $K$'s are different. The relation is such that $\Delta G^\ominus$ is the same. Because $\Delta G^\ominus$ is an energy difference you cannot have any additive factor. $\Delta G$ (Phillip's answer) is a derivative with respect to the advance of reaction, $\xi$, and there is no additive factor either. – perplexity Aug 8 '13 at 11:21
• I understand that your formula for $K_c$ is dimensionless in the way the term is commonly understood. However, it relies on the (somewhat arbitrary) choice of $c^{\ominus}$ to make it work - choose a different value for $c^{\ominus}$, and we get a different numerical value. What do we call a dimensional quantity with an arbitrary fixed value that we measure other quantities against? Oh yes - a unit. Calling it $c^{\ominus}$ rather than $M$ doesn't change that. – John Gowers Aug 8 '13 at 11:52
• @Donkey_2009 The choice of $c^{\ominus} = 1 \, \text{mol}/\text{L}$ is not entirely arbitrary. In thermodynamics there is a model system called the ideal solution, which is (in essence) defined as a solution having a standard concentration of $c^{\ominus} = 1 \, \text{mol}/\text{L}$ but behaving like an infinite dilute system, i.e. no interactions between the solute molecules. It's the analogue of the ideal gas for solutions and yields analogues equations making it very nice to work with. Though the definition shows that it's a very crude model (as is the ideal gas law) it nevertheless... – Philipp Aug 8 '13 at 13:57
• ...describes the behavior of very dilute solutions quite good. But as @perplexity said this doesn't hold for electrolytes and rather concentrated solutions because there the interaction between the solute molecules is too strong to consider the system as "infinite dilute". But if the conditions for the ideal solution model are nearly fulfilled $K_{c} \approx K$ (see edit in my answer). – Philipp Aug 8 '13 at 14:01

Yes, K is unitless, always, if defined by your equation:

$$\Delta G^\circ=-RT\log(K)$$

There is no way around this, unless you use some strange units for free energy, R or T.

Beyond the obvious mathematical answer, if you dig deeper you find that equilibrium constants are unitless because the expression relating $$K$$ to $$\Delta G^o$$ requires the definition of reference states. This is because $$\Delta G^o$$ is defined as the free energy change for the transformation of reagents to products in the specified reference states. As a result, equilibrium constants contain ratios of activites (which can be expressed as functions of concentration) of reagents and products in their reference states and at equilibrium.

For instance, for a species i in the gas phase terms in the equilibrium constant involving that species consist of ratios of fugacities $$(f_i/f^o)^\nu$$ where $$\nu$$ is a stoichiometric factor, $$f_i$$ is the fugacity of the species, and $$f^o$$ is the fugacity in the reference state (fugacities become equal to pressures in the limit of zero pressure). The resulting equilibrium constant is a product of such unitless ratios, and therefore also unitless.

That equilibrium constants are sometimes reported with units is contrary to accepted conventions, but is convenient: specifying a set of units together with the equilibrium constant implicitly declares the reference state(s).

People really need to look at the fundamental definition of the natural logarithm! Calculus (and mathematics in general) always works correctly with units.

1. The argument to a Log most certainly can have units. People aren't just ignoring the argument's units, and it isn't magic.
2. The result of taking a Log is always unitless. Even if the argument has units the Log is unitless.

To prove 1 and 2 are correct you need to look at the definition of Log (natural or base 10 or any other base). The Log(a) (natural log, but other bases behave exactly the same way) is the definite integral of 1/x dx over the interval from 1 to a. The units returned when you calculate an integral are always the units of the x-axis times the units of the y-axis. This is not magic and is absolute, positively correct 100% of the time. But, in the case of the Log the units of the y-axis are the reciprocal of the units of the x-axis (that's the basic definition of what a "Log" is) and therefore the product is always unitless.

Similar things happen with other functions. For example, trig functions are typically defined as ratios. Sin(a) is equal to the ratio of the length of the "opposite edge" over the hypotenuse. Both those measurements have the same unit so the ratio is unitless. No magic or ignoring of units or fundamental flaws in math.

The units work, they always work and if they don't it is because you made a mistake somewhere.

• Please try and keep the discussion here friendly and constructive. There is no need to shout (all caps). The tone of your answer is rude and dismissive, and your answer does not provide any useful information such as a proof. Please consider editing your answer to show an actual explanation of your point and to improve the tone. – Michael Lautman Jun 4 '19 at 22:18
• The flaw in your argument about integrals is that you say 'Log(a) is the definite integral of 1/x dx over the interval from 1 to a'. But 1 is an dimensionless constant! – John Gowers Jun 5 '19 at 8:42
• No, the "1" is a value on the x-axis hence it must have the same units as the axis. Again, look at the fundamental definition of natural log. It must work with units, as all of calculus must. – James Sluka Jun 5 '19 at 22:09
• Although it is mathematically correct that "The argument to a Log most certainly can have units", the Bridgman's Theorem referred to in the question specifically forbids arguments of log functions to have units $\textbf{in the specific case of dimensional analysis}$, whose rules generally govern the use of units in scientific calculations. I'd suggest reading up on dimensional analysis and then see if you're still of the same opinion as expressed in this answer. – Andrew Jun 6 '19 at 0:17
• No, you won't find that anywhere in the definition of dimensional analysis. DA requires that the the mathematics work with units. All of math works with units. – James Sluka Jun 24 '19 at 0:35