# Why is the equilibrium constant unitless?

I read that the equilibrium constant is unitless because the molar activities of each of the species are used in the equilibrium expression, not the actual concentrations themselves. I understand that.

But it is taught that

$$K_c = \frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$$

I’m having some difficulty understanding this equation. Consider the chemical equation

$$\ce{A + B <=>[k_\mathrm{fwd}][k_\mathrm{rev}] C}$$

The forward rate constant is a second order reaction and has units $$\pu{M-1 s-1},$$ and the reverse rate constant is a first order reaction and has units $$\pu{s-1}.$$ Therefore

$$\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$$

has units $$\pu{M-1},$$ but it’s supposed to be unitless. Where am I going wrong?

• What they don't tell you is these are not just molar activities of the species. It is activities divided by the activities at standard conditions. The latter are numerically 1, but they eat up the dimension. – Ivan Neretin Jul 3 '19 at 13:05
• @Ivan Surely you meant something else by "molar activity", as activity as such is already dimensionless (or with dimension 1) from the start. – Linear Christmas Jul 4 '19 at 18:51
• Also strongly related (among many others): chemistry.stackexchange.com/questions/85529/…, and chemistry.stackexchange.com/questions/5811/… – Linear Christmas Jul 4 '19 at 18:59

There are a few separate issues here to keep in mind:

1. $$K_c$$ (the equilibrium constant in terms of concentrations) is defined as $$K_c = \prod {c_i}^{\nu_i} \tag{1}$$
2. Agreement between $$K_c$$ and the product of forward/back rate constants ($$k_{\pu{fwd}}/k_{\pu{rev}}$$ in the OP example) is expected only if the mechanism is correct, assuming some proportional rate law, and at equilibrium.
3. When dealing with gases, $$K_c$$ cannot generally be equated with $$K_p$$ (the equilibrium constant in terms of partial pressures). They may be numerically equal under some circumstances. For instance, they are numerically equal in reactions involving only ideal gases, provided the number of gaseous molecular entities remains constant during the reaction.
4. The (standard) equilibrium constant $$K$$, which simplifies to $$K_a$$ (under standard fugacity and some other conditions, agreements) which is defined as $$K_a = \prod a_i^{\nu_i}, \tag{2a}$$ cannot generally be equated with $$K_c$$. They may be numerically equal under some circumstances. $$K$$ itself is actually defined as follows: $$\Delta_\pu{r} G^\circ=-RT\log(K) \tag{3}$$ whereas, strictly speaking, $$K_c$$ can't be used in this relation (but see below). The equilibrium constant $$K$$ cannot have units or this equation will not make sense. In terms of concentrations,$$K_a = \prod \left(\frac{\gamma_ic_i}{c_i^\circ}\right)^{\nu_i} \tag{2b}$$ (To be nitpicky, one should use molality-based activities instead of amount concentration-based $$c$$ when saying that $$K$$ reduces to $$K_a$$). When activity coefficients are all unity (ideal conditions), it might be that $$K_a \approx \frac{K_c}{K_c^\circ}= \prod \left(\frac{c_i}{c_i^\circ}\right)^{\nu_i} \tag{2c}$$ Strictly speaking, the equilibrium constant $$K$$ should remain unchanged with a change in the units used to describe concentrations. Units have no place in the constant because they cancel when ratios to reference standard states are computed (see Eq. (2)). $$K_c$$ and $$K$$ differ by the product $$K_c^\circ = \prod ({c_i^\circ})^{\nu_i}\tag{2d}$$ (ignoring, once again, activity coefficients; standard fugacity). This difference is the source of the dangling units leading to the question in the OP.

You should therefore be careful equating $$K_c$$ with the ratio $$k_{\pu{fwd}}/k_{\pu{rev}}$$. And, assuming they are equal, the concentration equilibrium constant cannot generally be equated with $$K$$. However you can divide $$K_c$$ by $$K_c^\circ$$, and, assuming activity coefficients are unity, the result might be the thermodynamically correct unitless equilibrium constant (if standard fugacity), equivalent to that which appears in usual thermodynamic derivations.

Sloppy or not, associating units with an equilibrium constant is common practice. Attaching units to the equilibrium constant during reporting is a useful mnemonic when performing some computations, but it should be kept in mind that such a constant may not fulfill the requirements for computation of an associated standard free energy change, unless first modified by a ratio product of the standard concentrations.

Note that some of this is also explained in an answer to this post. IUPAC likes to use the convention standard equilibrium constant when referring to the thermodynamic equilibrium constant.

• Comments are not for extended discussion; this conversation has been moved to chat. – andselisk Jul 5 '19 at 19:06

This is a complicated question among chemical educationists- with tons of arguments over arguments. Just search do equilibrium constant have units? I would say the "units" of K are in a grey area (just like in any real world science.). You should read the nice section on Dimensionality of the Equilibrium Constant, and I will just give a background:

Now, after ages, I feel that there are mathematical benefits of sticking to the formalism of unit-less equilibrium constants. For example, you need to take logarithms of K. Quantities in transcendental functions (log, ln, e, sin, cos) cannot have units, just like pH cannot have units. It is then beneficial to keep these quantities dimensionless. Yes, one simple argument is that we use the concept of activities and that do not have units. However, most of the calculations involving equilibria use concentrations. So the idea is to make the concentration of each species dimensionless by dividing it with unit concentration. Now, whatever the exponents be in the equilibrium constant expression, you don't worry about the units.

Regarding your query regarding rate constants, one may try to use the same argument of using dimensionless concentrations in the rate expressions. There we have logs and exponential functions too. Otherwise your dimensional analysis is correct. In general we can never write the rate expression or determine the order without an experimental support until and unless the reaction is a so-called elementary reaction.

• If I use dimensionless concentrations in the rate expressions, then the units of the rate constant will always be molarity per second, which is not correct. – Cyclopropane Jul 3 '19 at 18:12
• Your dimensional analysis is correct in the original post. If someone insists that we have to make K unitless then you make a formalism. You should make the whole rate equation dimensionless. When we raise e to the power t, where t is time, the time has to be made dimensionless. These type of functions, such as e^(-t^2) appear in making Gaussian peaks. Can you take natural log of concentration, where [C] has units of concentration? Nope. ln [C] also appears in kinetic equations. It cannot have units. – M. Farooq Jul 3 '19 at 18:26