Every system will be in the state of equilibrium, just not neesarilly every equilibrium will be present
Imagine some amount of water in a closed container. If teh container is small enough, there are both water and vapor - so there is gas/liquid equilibrium. Not take larger container (or heat the syste up) - water disappears, there is no liquid/gas equilibrium. But does it means system is not at equilibrium?
Equilibrium occurs when the rate of forward process equals the rate of reverse process, correct? So if only a vapor is present, wouldn't equilibrium not exist because there is no forward or reverse process in the system? A closed system such as the one in this question would be at equilibrium when the container is saturated with water vapor and the rate of condenstation equals the rate of evaporation. Basically, a system will only reach equilibrium if it is saturated, because if not, all of the water will evaporate into water vapor, correct?
Now, looking at the original problem, if I were to add a drop of water to a closed system that is saturated with water vapor and has a certain volume of liquid water present, the drop of water would increase the volume of liquid water present in the closed container and therefore decrease the volume of available for water vapor. Therefore, the system will shift towards condenstation because it is supersaturated and eventually reestablish equilibrium where the rate of condensation equals the rate of evaporation, correct? Now, if I were to add a drop of water to unsaturated volume of water vapor, the liquid water would simply evaporate and no state of dynamic equilibrium would result since there is no forward/reverse process present. However, if I were to add a drop of water to a closed container saturated with water vapor, a state of equilibrium will be established as the volume of the drop of water decreases the volume available for the water vapor, resulting in supersaturation and shift in the direction of condensation until the system is saturated (i.e. evaporation equals condensation).
So, am I on the right track?