The fundamental driving forces that determine the outcome of every chemical reaction

The mixing of balls of varying densities are used to demonstrate the basic driving force to lower the energy in a reaction.

Ingredients: lead shot, colored gum balls, large test tube, stopper

Procedure: A complete recipe follows.

1. Place white gum balls in bottom of test tube.

2. Add lead shot on top of gum balls.

3. Shake test tube up and down until the system comes to "equilibrium."

4. Observe distribution of lead shot and gum balls.

Understanding: The reaction mixture begins with the more dense lead shot above the less dense white gum balls. This leaves the reactant configuration segregated and top heavy. As the reaction proceeds through the shaking of the test tube, the particles rearrange themselves as they approach a state of equilibrium. The state of equilibrium is a dynamic one. The balls continue to rearrange themselves, but the overall distribution of balls in the test tube does not change with time.

In this reaction, the dominant driving force that will determine the state of equilibrium in the test tube is the drive to lower the energy of the reaction system. The energy of a single particle in the test tube is

E = mgh

where m is the mass of the particle, g is the constant of gravitational acceleration, and h is the height of the particle in the test tube. When the test tube is shaken, the particles move about to explore arrangements of lower energy. Due to the great disparity in the density of the lead shot and the white gum balls, the lead shot all move to the bottom of the container and the white gum balls move to the top of the container. In that way, the potential energy of the system is minimized.

If the moderate shaking of the test tube continues, the particles continue to move about, but the overall distribution in the test tube is not changed. They will remain segregated with the more dense lead shot on the bottom and the less dense gum balls on top. The system has reached a state of equilibrium. The principal driving force that determines the point of equilibrium for this reaction is drive to lower the energy of the system.


The mixing of balls of differing colors but identical densities is used to demonstrate the basic driving force to increase the entropy in a reaction.

Ingredients: colored gum balls, large test tube, stopper

Procedure: A complete recipe follows.

1. Place yellow gum balls in bottom of test tube.

2. Add the purple gum balls on top of the yellow gum balls.

3. Shake test tube up and down until the system comes to "equilibrium."

4. Observe distribution of yellow and purple gum balls.

Understanding: The reaction mixture begins with the gum balls segregated by color. There is no difference in the mass or size of the gum balls, and so no difference in density. That means that every arrangement of the gum balls can be considered to be of equal energy. You might think that the system is already at equilibrium, as the gum balls are not going to rearrange themselves to lower the energy of the system. But the system is not at equilibrium!

As the test tube is shaken, the gum balls do rearrange themselves. After a time, the gum balls are no longer segregated by color, they are mixed. Moreover, as the test tube is shaken further, the gum balls remain mixed. For the modest number of gum balls used, we never see the system revisit the segregated arrangement of gum balls that started the reaction. How can that be?

Every arrangement of gum balls is equal in energy. There is no driving force to lower the energy of the system. Yet there is a clear and powerful driving force that moves the system to equilibrium, an even distribution of yellow and purple gum balls in the test tube.

The driving force in this reaction is to increase the disorder of the system. As the test tube is shaken, we randomly explore different arrangements of gum balls in the test tube. There are many more ways to arrange the gum balls in a mixed state of yellow and purple gum balls than a segregated state of all yellow gum balls on the bottom and all purple gum balls on the top. So when the system is shaken and randomly visits arrangements of gum balls, we usually see the mixed state and rarely would see a segregated state.

Developing a quantitative measure of disorder

How many ways are there to arrange the gum balls in the test tube? First, we will assume that every gum ball is unique - we can tell them apart, even the yellow gum balls from the yellow gum balls. Now let's think of the test tube as being divided into a number of spaces or cells, each of which can contain a gum ball. We have a total of 2N gum balls, where N are purple and N are yellow. So we have a number of cells in the test tube equal to the total number of gum balls, 2N.

In the reactant state, all of the yellow gum balls are on the bottom half of the test tube, occupying N of the spaces in the tube. How many ways can we arrange the N yellow gum balls in the N spaces? Let's see. We have N choices for the first gum ball to place in the first space. In filling the second space, we have (N-1) remaining so we have (N-1) choices for the second gum ball to place in the second space. And so on. The total number of ways to arrange the N yellow gum balls in N spaces is

N x (N-1) x (N-2)... 3 x 2 x 1 = N!

where N! is "N factorial." For example, 4! = 4x3x2x1 = 24, and there are 24 ways of arranging 4 gum balls in 4 spaces. If we have 24 gum balls and 24 spaces, there are 24! = 6.20448402 x 1023 ways of arranging the gum balls. That is approximately Avogadro's number of ways of arranging only 24 gum balls! We are now prepared to measure the disorder in our reactant and product states.

First the reactant state. We have N yellow gum balls on the bottom and N purple gum balls on the top. The total number of ways of arranging the 2N gum balls so that they are segregated in this way will be

Ωreactant = N! x N! = N!2

which is the number of ways of arranging the N yellow gum balls in N spaces times the number of ways of arranging N purple gum balls in N spaces. Why times? Well, for each 1 way of arranging the yellow gum balls, you could have all N! ways of arranging the purple gum balls. If you have two ways of arranging the yellow gum balls and N! ways of arranging the purple gum balls, there are 2 x N! ways of arranging all the gum balls. And so on.

Now the product state. As the reaction proceeds the gum balls can mix together so that we no longer have the yellow gum balls on top and the purple gum balls on the bottom. Every arrangement is equally likely. The total number of ways of arranging the 2N gum balls in 2N spaces is

Ωproduct = 2N!

Now let's compare our results.

The total number of ways of arranging the gum balls in the segregated reactant state is Ωreactant = N!2, and the total number of ways of arranging the gum balls in the product state is Ωproduct = 2N!. Which is bigger? Let's take two examples.

Suppose that N=4, then Ωreactant = (4!)2=242=576, while Ωproduct = (2x4)!=8!=40,320. Even for the case of eight gum balls, we find a much larger number of ways of mixing the gum balls together in a mixed arrangement than having a segregated arrangement with all the yellow gum balls on the bottom and all the purple gum balls on the top.

If we have 24 yellow gum balls and 24 purple gum balls, we find Ωreactant=(24!)2=3.84956219 x 1047 while Ωproduct = (2 x 24)!= 1.24139156 x 1061 so that there are

Ωproduct/Ωreactant=3.22476037 x 1013

times the number of ways of arranging the mixed products as there are of arranging the segregated reactants. As the test tube of 48 gum balls is shaken, the probability of observing the segregated arrangement of all 24 yellow gum balls on the bottom and all 24 purple gum balls on the top is less than 1 in 30 trillion!

Waiting for the improbable

Question: Suppose that you have a test tube with 50 yellow gum balls and 50 purple gum balls. You shake the test tube and observe a new arrangement of gum balls every 2 seconds. How long it will take you to observe, on average, a segregated arrangement of gum balls in the test tube, with all the yellow gum balls on the bottom and all the purple gum balls on top?

You can check your answers here.