This experiment was developed by Profs LH Berka and NK Kildahl, WPI, based on a commercially available experiment.
1 lab period; work in pairs. Complete the Preparation before laboratory.
Goals
Background
Kinetics and Thermodynamics. The distinction between kinetics and thermodynamics is an important one. Kinetics is the study of the rate at which (and the path by which) a chemical reaction occurs. Thermodynamics deals with the extent to which the reaction occurs. There is no necessary connection between the rate, measured by a rate constant k (see below), and the extent, measured by an equilibrium constant Keq. Thus, a reaction in which 99.9% of the reactant molecules are eventually consumed may take years to reach that stage. Conversely, it may take just milliseconds for 0.1% of a reactant to be converted to products, but with no further reaction over an extended period of time. The former is an example of a slow reaction (small k) that occurs essentially to completion (large Keq); the latter is a fast reaction of limited extent.
The Differential Rate Law. For the general reaction (1), the rate is expressed as the change of concentration with respect to time, t, of any reactant or product.
Since A is consumed and F is produced by (1), their derivatives have opposite signs. Thus we introduce a minus sign in front of d[A]/dt to ensure a positive rate. Also, unless a = f, -d[A]/dt and d[F]/dt are different. In general, the rate of reaction is a function of the concentrations of reactants and products, as indicated in (3).
k is the rate constant of the reaction, and [A], [B], etc., are the molar concentrations of reactants and products in (1). The exponents m,n... appearing in (3) are usually either positive or negative integers or zero.
Because (3) is a differential equation, it is called the differential rate law for (1). The goal of a kinetics study is to determine the differential rate law for the reaction (i.e., the values of the exponents m, n, s, and v) and the numerical value of the rate constant k at each of several temperatures. Suppose that appropriate kinetics experiments carried out on (1) give m = l, n = 2, s = 0, and v = -1. The differential rate law for the reaction is then
The exponents m, n, etc. are called orders. We say that the reaction is first order in A, second order in B, and inverse first order in F. It is important to realize that there is no necessary connection between the reaction orders m, n, s, and v, in (3) and the stoichiometric coefficients a, b, d, f, in (l).
The Method of Initial Rates. A straightforward way to determine experimentally the specific form of the rate law (3) is to measure the initial rate, -(d[A]/dt)0, as a function of each of the initial concentrations [A]0, [B]0, etc, and from the data deduce the reaction orders. Suppose that we found that -(d[A]/dt)0 quadrupled when we doubled the [A]0 while keeping the other concentrations constant. We would conclude that m = 2. Similarly, if -(d[A]/dt)0 decreased by a factor of 2 when [B]0 was cut in half while keeping other concentrations the same, we would conclude that n = 1.
This Experiment. We will study the initial rate at which hydrogen peroxide, H2O2, reacts with iodide ion, I- , and hydrogen ion, H+, to give triodide ion, I3-, and water in aqueous solution. By varying the initial concentrations of the participants, we will deduce the differential rate law for reaction (8).
(8): H2O2 + 3I- + 2H+ ---> I3- + 2H2O
To obtain the rate, we will measure the time required for a definite, known amount of H2O2 to be consumed. This can be accomplished by adding a known amount of S2O32- to the system. As reaction (8) proceeds, the I3- produced will react with S2O32- according to (9) and will be converted back to I-.
(9): I3- + 2S2O32- ---> 3I- + S4O62-
As long as (9) occurs very rapidly relative to (8), [I-] will not change until the S2O32- is gone. How do we signal that S2O32-is gone? I3- forms a blue complex with starch. So we put a little starch in the solution, too. As soon as S2O32- is used up, I3- will accumulate and immediately produce a blue color with the starch. If we know how much S2O32- we put in, we can calculate the amount of H2O2 used up in the time it took for the blue color to appear, because the combined stoichiometries of (8) and (9) tell us that 1 mole H2O2 is equivalent to 2 moles S2O32-. Once we know how much H2O2 was consumed, we can estimate the initial reaction rate as -D[H2O2]/Dt.
A sketch of our procedure follows.
Calculate the initial rate, -D[H2O2]/Dt, of the reaction.
The above procedure requires calculation of concentrations in the reaction mixtures. Each reaction mixture will be prepared by mixing aliquots of stock solutions of I- and S2O32- with buffer solution and water to give a total volume of at least 17 and at most 19 mL. Reaction is then initiated by adding enough H2O2 stock solution to give a final total volume of 20.0 mL. If [A]stock symbolizes the concentration of the stock solution of reagent A, VA the volume of the aliquot of that stock solution used, and VT the total volume of reaction mixture, the concentrations of reagents in the reaction mixture can be calculated using the following equations.
In these equations, [H2O2]0 and [I-]0 are the initial concentrations of these reactants in the reaction mixture (t = t0 = 0).
Focus Questions
Equipment and Materials
Note to instructor: Click here for recipes for preparation of solutions.
Safety
Hydrogen peroxide is an oxidizing agent; avoid contact with skin and clothing.
Experimental
Record all data in your notebook. Obtain required glassware and a stopwatch. Clean the glassware thoroughly using brushes and Alconox detergent (pipets cannot be cleaned this way, but can be rinsed). Remove all soap with tap water, then rinse with distilled water and dry thoroughly.
Preliminary Activities
| Run | Solution 1: 0.200 M KIa | Solution
2: Bufferb | Solution 3: 0.2% Starchd | Solution 4: 0.02 M Na2S2O3d | Solution
5: Waterc | Solution 6: 0.4 M H2O2c | VT |
|---|---|---|---|---|---|---|---|
| 1 | 2.00 | 11.5 | 0.50 | 1.00 | 4 | 1.00 | 20.00 |
| 2 | 4.00 | 11.5 | 0.50 | 1.00 | 2 | 1.00 | 20.00 |
| 3 | 6.00 | 11.5 | 0.50 | 1.00 | 0 | 1.00 | 20.00 |
| 4 | 2.00 | 11.5 | 0.50 | 1.00 | 3 | 2.00 | 20.00 |
| 5 | 2.00 | 11.5 | 0.50 | 1.00 | 2 | 3.00 | 20.00 |
Kinetics Runs
Five kinetic runs will be carried out using the volumes of stock solutions in Table 1. The procedure for carrying out Run 1 will be outlined in detail. The others are carried out in a similar manner.
Run 1:
Carry out Run 1 in triplicate. Then carry out runs 2-5 in similar fashion, also in triplicate.
If time allows, design a procedure to determine the order in H+.
If time allows, design a procedure to determine the temperature dependence of the rate.
Clean-up. When you have finished all of your work:
Disposal Methods
Discard left-over solutions in the sink; flush down the drain with water.
Preparation
Kinetics of Reaction of I- with H2O2 Using
Initial Rate Methods