Question #22d95

Well, no. If you read your own question again, look for the key phrases that allow you to answer your own question:

• "a disturbance is introduced"
• "to a system at equilibrium"

A system at dynamic equilibrium is in a state where it's doing its own thing, and nothing is disrupting its current state. It will continue to be like that if nothing disturbs it.

ANALOGY:
Imagine a person walking backwards on one of those airport conveyor belts while the conveyor belt is moving forward. If they move backwards at the same speed as the conveyor belt is moving forwards, they are in dynamic equilibrium.

By analogy, in this dynamic equilibrium state, the rate of the forward reaction is equal to the rate of the reverse reaction.

A disturbance introduced is pretty much the disruption of that equilibrium. Hence, the rate of the forward reaction is momentarily no longer equal to the rate of the reverse reaction.

(Or you could think of it has pushing the person so they fall onto the conveyor belt and then they actually do move forward!)

According to Le Chatelier's principle, the equilibrium will try to re-establish itself; i.e. it will try to re-achieve the state where the rate of the forward reaction is equal to the rate of the reverse reaction.

So, as an example:

#"HA" + "B" rightleftharpoons "A"^(-) + "BH"^(+)#

A simple equilibrium between a weak acid #"HA"# and a weak base #"B"# for instance, can be disturbed by adding more reactive acid #"HA"#.

That acid would cause the equilibrium to shift to the right and make more product (#"A"^(-)# and #"BH"^(+)#) because that extra acid is reactive. It's not going to just sit there doing nothing. It's not inert. It's going to react to make more product.