General Safety Guidelines
Introduction
Your Challenge
Some Math
Materials and Apparatus
Pre-lab Exercise
Procedure
Data and Calculations
Results and Discussion

Introduction

The ozone hole would not exist if each chlorofluorocarbon molecule only destroyed one or two ozone molecules. The problem exists because each chlorofluorocarbon molecule leads to the destruction of large numbers of ozone molecules. The way the molecules are destroyed is described under the link titled Making and Destroying Ozone. In a critical step of the reaction, a chlorine atom reacts with ozone, as you see below:

Notice that at the end of this cycle, the chlorine atom is regenerated, and it's ready to destroy another ozone molecule. No matter how many times this cycle repeat, the supply of chlorine atoms is never going to be used up. When an atom or molecule participates in a reaction but isn't used up in that reaction, we call it a catalyst. A catalyst will speed up the reaction rate without being consumed by the reaction. Here we will consider this phenomenon, called catalysis, that enables one molecule to have such a large effect.

We've been talking about how catalysts are not consumed in a reaction, but we haven't said much about how they speed up reactions. CFCs speed up the reactions which destroy ozone, so much that natural processes can't replenish it fast enough to replace that which is destroyed. This in the sinister side of catalysis. We should point out that catalysis can be helpful as well as harmful. Catalysts are used industrially to make useful materials faster and more cheaply, while special catalysts called enzymes help your body carry out an untold number of reactions that keep you alive.

Hydrogen peroxide H₂O₂, that you may use to clean your contact lenses or to sterilize cuts, decomposes to water and to molecular oxygen. The bottle of hydrogen peroxide can remain in your medicine cabinet for weeks without losing any of its strength, but addition of small amounts of a catalyst will cause it to decompose quickly.

This is similar to the way chlorine atoms act as catalysts to break down ozone. Since it is hard to study the reaction that break down ozone using the kind of equipment found in most high school laboratories, we're going to learn a little more about catalysis by taking a look at the use of catalysts to break down hydrogen peroxide.

Your Challenge

Determine if manganese dioxide (MnO₂) or potassium iodide (KI) is the better catalyst, for the decomposition of a 3% solution of hydrogen peroxide. To do this you are going to calculate how long it takes for half of the hydrogen peroxide to decompose using each catalyst. We call this time the half-life of the decomposition reaction.

The question is, how do we know that half of the hydrogen peroxide has decomposed? We know that oxygen gas is produced when the hydrogen peroxide decomposes. Using your knowledge of stoichiometry and the gas laws, it is possible to predict how much oxygen gas should be produced when a given amount of hydrogen peroxide decomposes. Then, all we have to do is measure how long it takes for the decomposition of hydrogen peroxide to produce half of the total oxygen gas the reaction is capable of producing.

This brings us to a second problem. How do we measure the oxygen gas that bubbles up in the reaction mixture, then escapes freely into the atmosphere? To measure it, we have to trap it somehow. Our oxygen trap looks something like the drawing you see below:

Schematic diagram of the gas collection appartus.

The decomposition of hydrogen peroxide takes place in the test tube. The oxygen gas forms bubbles out of the reaction mixture. But the tube is sealed, and the oxygen can't escape. The only place it can go is through the rubber tube, into the trough, and into the inverted graduated cylinder. There the oxygen gas floats to the top of the cylinder as bubbles. The oxygen gas collects at the top of the graduated cylinder, and we can easily measure the volume of the oxygen that has been produced at a given moment.

By measuring the volume of oxygen being produced, we can monitor how fast the hydrogen peroxide is decomposing. The question to be settled is this: Which catalyst, KI or MnO₂, will give the shortest half-life for the decomposition of hydrogen peroxide?

Some Math

Measuring reaction rates means doing a little bit of math. Most of the math involved in studying reaction rates involves what we call the rate law of the reaction. A rate law is simply an equation that describes how fast a particular chemical reaction will take place. For example, the rate law for the decomposition of hydrogen peroxide is:

Sometimes the rates law is more complicated. Let's look at the reactions that destroy ozone. In one step of this reaction cycle, a chlorine atom reacts with an ozone molecule:

For this reaction the rate law is:

The important difference between this rate law and the rate law for the decomposition is that here, two reactants are involved in the reaction, the chlorine atom and the ozone molecule. So, the reaction rate is dependent on the concentrations of both. Since two reactants are involved in the reaction, we call it a second order reaction. Likewise, the decomposition of hydrogen peroxide involves only one reactant, so we call it a first order reaction.

Materials and Apparatus

3% hydrogen peroxide (H₂O₂) aqueous solution
manganese dioxide (MnO₂)
potassium iodide (KI)
test tubes, 20 × 150 mm
rubber stopper with a hole
glass tube
plastic tubing
100 ml graduated cylinder
50 ml beakers (2)
clamp
clamp stand
test tube rack

Prelab Exercise

1. Place a small spatula full of each catalyst and place into a 50 ml beaker half-filled with water. How does each catalyst behave? On the basis of your observation, predict which catalyst will be faster when used to facilitate the decomposition of hydrogen peroxide.

2. Predict how changing the amount of catalyst will affect the rate of the reaction.

Procedure

1. Assemble the apparatus as shown above.

2. Measure 5 ml of 3% hydrogen peroxide solution using a graduated cylinder. Remove the rubber stopper from the test tube. Pour the solution into the test tube.

3. Calculate the mass of 0.002 moles of the catalyst you will use, and wiegh out that amount of the material.

4. Place your 0.002 moles of catalyst in the test tube with the hydrogen peroxide solution and quickly close the tube with the rubber stopper.

5. Measure and record the time it takes for 5 ml of gas to collect at the top of the cylinder. Continue to record the time needed for each additional 5 ml of gas to collect.

6. Repeat the procedure using each catalyst.

7. Using the faster catalyst, repeat the procedure using only 0.001 moles of catalyst.

Data and Calculations

1. For each run, make a graph showing volume of gas produced on the y-axis and the time in seconds on the x-axis.

2. How many milliliters of oxygen gas should 5 ml of 3 % hydrogen peroxide be able to produce when it decomposes?

3. Using the graph you made in question #1, how long was the half-life of the reaction when 0.002 moles of each catalyst were used? Which catalyst did you observe to be the faster of the two?

4. When you repeated the procedure using 0.001 moles of the faster catalyst, what was the half-life of the reaction?

5. What was the total amount of gas produced in each run using the faster catalyst at the time you stopped taking measurements?

Results and Discussion

1. When the amount of the faster catalyst used in a reaction is decreased, does the reaction take place faster or more slowly?

2. How did the amount of catalyst affect the total amount of gas finally produced? Explain.

3. In each of your test runs, did the rate of your reaction increase or decrease at the reaction progressed? Explain.

4. When you stopped taking measurements during your test of the slower catalyst, how much gas had been produced? If you had let the reaction continue running all night, how much gas do you think would be produced? Explain.

5. Why do you think you were told to measure the time needed for half of the hydrogen peroxide to decompose, rather than measuring the time it takes for the reaction to reach equilibrium?

6. Look again at the rate law for the decomposition of hydrogen peroxide:
   rate = k[H₂O₂]

7. How does a catalyst affect the rate constant k? Is it higher or lower for one catalyst than the other, if equal molar amounts of both catalysts are used? How does the concentration of the catalyst affect k?

History of Catalysis — an online exhibit from the University of Kentucky.

References

Silberman, R., and Eubanks, L. ACS Small-Scale Laboratory Assessment Activities. Clemson SC: ACS DivCHED Examinations Institute, 1996.