Consider the reaction below:

$$\ce{A + B <=> C}$$

Suppose that the equilibrium constant for this reaction is $K = 10$.

I then prepare a reaction vessel with volume of $1~\mathrm{dm^{-3}}$ which contains $1$ molecule of $\ce{A}$, $500$ molecules of $\ce{B}$ and $1$ molecule of $\ce{C}$. Thus the initial concentration of each species (in $\mathrm{mol~dm^{-3}}$) will be $1/N_\mathrm{A}$, $500/N_\mathrm{A}$, and $1/N_\mathrm{A}$ respectively, where $N_\mathrm{A}$ is the Avogadro constant.

As a result the initial reaction quotient is $N_\mathrm{A}/500$.

Now, am I right in saying that this system cannot reach equilibrium, since the only possible values of the reaction quotient are $+\infty$, $0$, or $N_\mathrm{A}/500$?

Does this arise because there simply is not enough of each reactant/product in order to reach equilibrium? I feel like this system must reach equilibrium but I'm struggling to see how the value of $K$ can be reached in this closed system.

In a bimolecular reaction step (such as the one shown above), the rate constant for the forward reaction has units $\mathrm{M^{-1}~s^{-1}}$. I have learnt that if the concentration of $\ce{B}$ is increased significantly, then the Gibbs free energy of the $\ce{A + B}$ mixture will increase because the activation energy has a concentration dependence. Is this true, and is equilibrium only reached once this Gibbs free energy decreases again?

This is confusing as I am used to the Gibbs free energies of the reactants and products remaining constant while the relative amounts of each change throughout the course of the reaction.


Dave's mathematical answer is perfectly valid, but I feel like another, perhaps more chemical point of view is based on the the law of large numbers, i.e., the "thermodynamic" limit.

"Equilibrium" is a statistical concept that only applies to large numbers of molecules. Another way to say it is that equilibrium is a thermodynamic concept and thus only applies in the thermodynamic limit. What is "large" depends on the particular system of interest. In particular it depends on the equilibrium constant. For your reaction of interest, $K_{eq} = 10$. Thus the amount of $A$ and/or $B$ at equilibrium will be smaller than the amount of $C$. For the equilibrium concept to apply to a system of $A$, $B$, and $C$ molecules, the calculated equilibrium amount of $A$ or $B$ molecules must be very large, say tens of thousands or so for precise work, or perhaps as low as hundreds for the equilibrium concept to be "approximately" correct.

If the equilibrium constant were enormously large, say $K_{eq} = 10^{100}$, then by the same logic, the system would need to contain on the order of $10^{102}$ total molecules for the equilibrium to be meaningful. (And by the way that is larger than the number of particles in the universe!)

Speaking loosely, if $K_{eq}$ is $10^{100}$ and there aren't close to that many molecules, the reaction is irreversible.

A probabilistic view of equilibrium for small numbers of particles, such as that mentioned in DavePhD's answer, is occasionally useful but not always valid. For example, consider if a second reaction consumed $A$ irreversibly, converting it to $D$ at some rate. If you take DavePhD's probabilistic view of the $\ce{A + B <=> C}$ equilibrium, you would assume that the probability distribution of persistence times for $A$ is exactly the same as in the thermodynamic limit. However, that is not necessarily true with small systems. Statistical fluctuations in the $\ce{A + B <=> C}$ system will mean that some of the time, $A$ persists far longer than in the thermodynamic case because due to a fluctuation, more than the "expected" amount of the molecules were in the $C$ form. Some other times, $A$ will resist degradation to $D$ for far shorter than the thermodynamic equilibrium calculation would suggest. To treat these cases, more complex equations such as the chemical master equation are required.


If you pick an arbitrary real number, there is zero probability that it is a rational number. Therefore, there is zero probability that a finite sample of molecules A,B,C can exactly represent a arbitrary real number equilibrium constant.

However, the more molecules you have in the system, the closer a given real number equilibrium constant could be represented by the finite number of molecules.

In your example, the system would spend some of the time in the initial state, and some of the time in A=0, B=499, C=2, and some of the time in A=2, B=501, C = 0.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.