homework - Equilibrium constant. Can it be reached?

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.