Consider the following reaction where $\ce{A}$ & $\ce{B}$ reacts to produce $\ce{C}$ & $\ce{D}$.
$$\ce{ A(g) + B(g) <=> C(g) + D(g)}$$
Is it possible for $\ce{A}$ or/and $\ce{B}$ to run out before equilibrium at some given starting moles of $\ce{A}$,$\ce{B}$,$\ce{C}$ & $\ce{D}$.
Answer
No, that is not possible.
Consider the definition of equilibrium from the IUPAC gold book:
chemical equilibrium Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy, $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that: \begin{align} \Delta G_\mathrm{r} &= \Delta G_\mathrm{r}^\circ + R T \ln K &&= 0\\ \Delta G_\mathrm{r}^\circ &= − R T \ln K\\ \end{align} The equilibrium constant, $K$, is given by the mass-law effect.
The important thing is that the reaction proceeds in both directions. It means, that as soon as some of the product is formed, it will react back towards the reactants. The equilibrium is reached when the rates of both of the directions are identical.
From the formula in the definition, you can also see, that the location of the equilibrium, i.e. the concentration of all species or $K$, is only determined by the difference in the Gibbs energy between products and reactants.
In other words, a system that forms an equilibrium will always form an equilibrium, independent of the concentration of the starting materials.
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