In an equilibrium we have two
processes going on at the same time:
We already know that changing concentration will change an equilibrium. This was illustrated in the simulation of the hydrogen iodide equilibrium.
The rate of any chemical reaction actually depends on a factor called the activity of the reacting chemicals. In reasonably dilute solutions, the activity is equal to the concentration. In beginning chemistry we always assume that the two are equal, and so we will here too. In other words, the rate depends on the concentration of the reacting chemicals.
The reaction rate depends on the concentration of the reacting chemicals because, with a higher concentration, there will be more collisions. It also depends on a reaction rate constant, and so can be represented by the equation:
Rate = k[Substance]
In an equilibrium we have two
processes going on at the same time:
where Ratef is the rate in the forward, and Rater the rate in the reverse directions. At equilibrium we know that Ratef = Rater, since the forward and reverse rates are equal at equilibrium.
What would happen to our reaction at equilibrium if we were to increase the concentration of the reactants, R? We would expect the forward rate to increase. So, for a while at least, the rate would go faster in the forward direction. This is no longer an equilibrium, since the rates aren't equal any longer. Of course, as the reaction goes forward faster it will make more products, P. More P will increase the rate in the reverse direction. Eventually, we'll reach an equilibrium, but it won't be the same one as we started with.
Similarly, if we were to increase the concentration of P, then the reaction would initially go faster in the reverse direction. Again though, it would eventually reach a new equilibrium with equal rates.