This page supplements the first handout on the "clock reaction." You should have your copy of the handout available as you look over this page. If you need a spare, or just want an electronic (pdf) version, click the link below.

• Get a pdf version of the first handout: rate law and rate constant.

A "Clock" Reaction: Initial rates

MAIN IDEAS: Stoichiometry and reaction rate measurement

The net reaction has two sulfur oxyanion reactants combining in a 1:2 stoichiometry to produce two different sulfur oxyanions, also in a 1:2 stoichiometry:

None of these four species is particularly easy to detect quantitatively; they are all colorless, they all have the same charge, and they are chemically quite similar. If one had an easy to observe signature (i.e., suppose thiosulfate was green and the other ions were colorless), then we could measure the rate of this reaction through a quantitative measurement of that signature over time. (This is the technique you will use to study a different reaction during the first two weeks of lab.) Here, we will use a clever chemical trick. We will arrange for another species (iodide ion, I^–, and its solution partner, triiodide, I₃^–) to monitor the reaction for us. Even though these species are also colorless by themselves, triiodide forms a deep brown-purple complex with starch molecules. We will arrange the reactions' initial conditions so that the sudden appearance of this color signals a quantitative change in reactant concentration.

Here's how it works. Consider the first solution we will study (called Mixture A in the handout). You should verify that at the start of the reaction (the instant at which all the reacting species are mixed in one solution), the concentrations are as shown to scale in the picture below.

Note that thiosulfate, S₂O₃^2–, has a concentration far less than its reaction partner peroxydisulfate, S₂O₈^2–. This makes thiosulfate the limiting reagent. We will run out of it first. When we do, the solution composition will be as shown below:

Note that the peroxydisulfate concentration is only slightly lower (it was in great excess), and note that the product ions are present in the correct stoichiometric ratio.

Now we turn to the role of the iodide ion, I^–. As the figures indicate, iodide does not directly enter in the net reaction we are studying. It plays the role of a catalyst through the two reactions discussed in the handout:

The first step slowly generates triiodide ion, but the second step quickly turns triiodide back to iodide. This second step operates as long as we have some thiosulfate, S₂O₃^2–, in the solution, but as soon as we run out, I₃^– builds up almost instantly to a significant concentration, the triiodide finds the starch added to the solution at the start, and binds to it, producing a deep brown-purple color throughout the solution.

This is our signal that all the thiosulfate is gone and that we have reached the concentrations shown above for the reaction's end. We measure the time it takes to go from initial mixing (which defines time t = 0) to the colored solution, and we will call this time t. We calculate the rate of the net reaction from the expression

For Mixture A, and using the concentrations in the figures above, you can see that this equation tells us that the rate equals –(79.2 mM – 80.0 mM)/t = 0.8 mM/t. Once we measure the reaction times, we will have a quick and easy route to the reaction rates.

Note that the SLOW reaction above will truck along all by itself whether or not there is thiosulfate in the reaction mixture, and thus it must keep going once all the thiosulfate is gone. This reaction has three iodide ions consumed per S₂O₈^2–. In Mixture A, we start with [S₂O₈^2–] = [I^–] = 80 mM, and the slow reaction's stoichiometry tells us that we will run out of iodide first. (This is a slight simplification of a more complicated situation, because starch eats up some triiodide, and even without starch, iodide and triiodide establish an equilibrium between themselves.) When all the iodide is gone, [S₂O₈^2–] will have fallen to 80 mM – (80 mM)/3 = 53.3 mM, and the solution will be at equilibrium. This is much less than the 79.2 mM value [S₂O₈^2–] falls to when the thiosulfate is consumed and the color changes. This slow approach to equilibrium is shown in the graph below. (Remember: Equilibrium means concentrations no longer change with time.)

In contrast, we will measure the relatively short time, t, needed for [S₂O₈^2–] to fall to only 79.2 mM, the concentration corresponding to loss of all thiosulfate. This time, about 20 seconds, is shown in the graph below, which is just the initial part of the previous graph (and the iodide concentration isn't shown).

Thus, our experiment is a nice example of the method of initial rates; we measure a finite, but small, concentration change over a finite, but short time (short compared to the time required to reach equilibrium). The ratio of these two quantities is the initial rate.

The rates, calculated for various initial concentrations, allow us to deduce the reaction rate law and to calculate the reaction rate constant. Once we know the rate law, we can repeat the experiment at different temperatures to study the temperature dependence of the rate constant. This is covered in the second part of the discussion of this demonstration.

• Go to part 2, temperature dependence.