Elementary Kinetics

THE ORIGIN OF K_eq

Look at the general chemical reaction:

A

+

B

C

+

D

When a reaction has reached equilibrium the reaction has gone as far as it’s going to go in terms of making product. For some reactions, that means most or all the reactant has become product, but for others, the reaction stops after only some of the reactants have converted to product. K_eq, the ratio of product concentrations to reactant concentrations when the reaction is at equilibrium, is a measure of how far a reaction will proceed. For the reaction above:

Notice that K_eq (the equilibrium constant) is a totally different number than the k (rate constant) that you learned about in the last section! Although they are related, as you’ll find out a bit later in this tutorial, they are definitely NOT the same thing. To help you understand the meaning of K_eq better, let’s look at a hypothetical example.

The reaction "fruits to nuts" has a K_eq of 2. That means that at equilibrium:

fruits nuts Keq = [nuts] [fruits] = 2 or [nuts] = 2 [fruits]

The reaction "nuts to bolts" has a K_eq of 0.25. That means that at equilibrium:

nuts bolts Keq = [bolts] [nuts] = 0.25 or [bolts] = 0.25 [fruits]

At first, it doesn't seem to make a lot of sense for the "fruit to nut" reaction to just stop when there are twice as many nuts as fruits, or for the "nuts to bolts" reaction to stop when there are still 4 times as many nuts as bolts. To understand K_eq, it is important to remember that the reaction does not really "stop." Rather, once a reaction has reached equilibrium the overall ratio of products to reactants does not change because the rate of the forward reaction

fruits nuts

is exactly balanced by the reverse reaction

muts fruits

If the forward reaction proceeds quickly, but the reverse reaction is slower you end up with more product. This is what happens in the "fruits to nuts" reaction, where there are lots more nuts than fruits at equilibrium. However, if the reverse reaction is faster than the forward reaction, there will always be more reactants. That’s what happens in the "nuts to bolts" reaction, where there are more nuts than bolts at equilibrium.

Another way to understand this explanation of K_eq is to think about a day at the beach. You have a large shovel and are trying to dig a deep hole in the sand. Your friend, always the comedian, has a small shovel and is throwing the sand back into the hole. With your large shovel, you are able to get the sand out faster than your friend can scoop it back in, and the hole gets deeper. However, as the hole gets very deep, your scooping rate slows down as it gets more difficult to get the sand out, but your friend can continue to dump sand back in the hole at the same rate as before. At a certain point, your rate of scooping sand out will exactly balance your friend’s rate of tossing sand back in, and the hole will stay the same size. Thus, your sand pit digging reaction has reached equilibrium.

Relationship between equilibrium and rate constants

The K_eq of a reaction has to do with how fast a reaction goes forward, and how fast it goes backward. In other words, there must be a relationship between K_eq and those rate laws we discussed earlier. We have all the tools to discover this relationship if we proceed one step at a time.

First look at the rate equation for the elementary process:

Then what about the reverse of the reaction?

We know that at equilibrium the rate of the forward reaction exactly balances the rate of the reverse reaction:

Since we know that

then it follows that

Thus, the K_eq is really the ratio of the forward reaction rate constant to the reverse reaction rate constant.

Equilibrium constants and catalysts

Remember that although reaction rates are affected by catalysts, a catalyst does not change the K_eq of a chemical reaction. Although a catalyst speeds up a reaction proceeding toward equilibrium, it does so by speeding up the both the forward and reverse reactions.