The rate at which a chemical reaction occurs depends on several factors. Some of these include the concentration of the reactants, the temperature at which the reaction takes place, and the presence of a possible catalyst. For a given chemical reaction the rate typically increases with the concentration of a reactant.
For the reaction xA + yB ® products
the rate law is expressed by the following relationship: Rate = k [A]m [B]n.
In the rate law, [A] and [B] are the molar concentrations of the reactants and k is the rate constant for the reaction, which depends only on the temperature, but not on the concentration of the reactants. The variables m and n are the orders of the reaction with respect to the reactants A and B and are usually whole small numbers. If m is 1, the reaction is said to be first order with respect to A. If n is 2, the reaction is said to be second order with respect to B. The order can also be zero with respect to a reactant. The overall order of the reaction is the sum of n and m.
In today’s experiment we will study the kinetics of the reaction between iodine, I2, and acetone, C3H6O according to:
(1) C3H6O (aq) + I2 (aq) ® C3H5IO (aq) + I- (aq) + H+ (aq)
The rate of this reaction will depend on the concentration of the two reactants, acetone and iodine, but also on the concentration of the hydrogen ion in solution. The rate law for this particular reaction will thus be as follows:
(2) rate = k [acetone]m [iodine]n [H+]p
The rate law in equation (2) is fairly complex, having seven variables. Knowing that the reaction does not depend upon [I2] simplifies matters. That means that the reaction is zero order with respect to iodine, or [I2]o = 1. In this experiment, we will use iodine as the limiting reagent, which is the one that will be used up completely in the reaction with acetone. Since iodine has a yellow color, it will be easy to observe the change in the concentration of iodine by the disappearance of the yellow color. The reaction is complete when the yellow color has completely disappeared. We will measure the time necessary for the solution to become colorless once the iodine has been added to the acetone. If a small concentration of iodine and a large excess of both acetone and hydrogen ions are present, the change of concentrations of iodine and hydrogen ions will be negligible.
Under these conditions, the rate of this reaction can be expressed as the change in the concentration of iodine, D(I2) divided by the time Dt required for the change to take place.
(3) rate = - D(I2) / Dt = initial [I2]/t
Since the change in concentration is negative, we introduce a minus sign to make the rate positive.
In this experiment, you will be running several trials of the reaction, at constant temperature, but changing concentration of one of the reactants, to determine the rate constant.
As an example, suppose we run two trials. In the first run, we have a set of concentrations for all three reactants, while in the second trial, we double the concentration of acetone while keeping all other concentrations constant. We will now look at each rate separately, according to equation (2).
Trial 1: rate1 = k [acetone]m [iodine]n [H+]p
Trial 2 rate2 = k [acetone]m [iodine]n [H+]p
Taking a rate ratio of the two trials, that is dividing the second trial by the first, the k’s cancel and so do the terms in hydrogen ion and iodine concentration. We therefore obtain
rate2/rate1 = (2)m
ln(rate2/rate1) = m x ln2
Having measured both rate 1 and rate 2 by equation (3), we find their ratio, which must be equal to (2)m. We can then solve for m as shown above and so find the order of the reaction with respect to acetone.
By a similar procedure we can measure the order of the reaction with respect to the hydrogen ion concentration, comparing the rates of two runs where only [H+] changes and the other concentration is kept constant. Once m and p are known and recalling that n = 0, equation (2) can be solved to determine the rate constant k for the reaction.
You can find a similar experiment at
Materials and Equipment
4 M acetone
0.005 M iodine solution
1 M HCl
100 mL beakers
Prelaboratory Questions: Rates of Reactions
1. A solution containing 5 mL of each 4 M acetone, 1 M HCl, and 0.005 M iodine is mixed with 10 mL of water.
(a) What is the molarity of acetone in the reaction mixture?
(b) How could you double the molarity of acetone in the reaction mixture, using the solutions above, and keeping the total volume the same (25 mL)?
2. Why is it necessary to keep the total volume of the reagents constant after mixing in all kinetic runs?
3. List four factors that affect the rate of chemical reactions.
4. Nitrogen monoxide reacts with chlorine according to the equation
2NO(g) + Cl2(g) ® 2 NOCl(g)
The following initial rates have been observed for certain reactant concentrations.
Rate, mol/(L hr)
a. What is the rate law? b. Determine the order with respect to each reactant and calculate the rate constant for this reaction.
1. Use two large test tubes. These test tubes will be used as reference. Fill them with deionized water and make sure they look clear when viewed against a white sheet of paper.
2. Use three 100-mL beakers and label them as follows:
I: 4 M acetone
II: 1 M HCl
III: 0.005 M iodine
Using a graduated cylinder pour 50.0 mL of each of the solutions into their appropriate container. Cover the beakers with a watch glass to avoid any evaporation. These are your stock solutions.
3. Run #1:
Obtain a large test tube. With a pipette or a graduated cylinder, measure exactly 5.00 mL of acetone, 5.00 ml of HCl, and 10.0 mL of water into the test tube. Shake the test tube gently to mix. Then measure 5.00 mL of iodine into the test tube and note the time (in seconds) it takes for the solution to become clear. Use one of the test tubes from Step 1 as a color reference. Look from the top, not from the side of the test tube to observe the color change.
Repeat the procedure. The amount of time required for the reaction to take place should agree within about 10 seconds. Record the time on your data sheet.
4. Run #2
To calculate the order of the reaction with respect to acetone we will do a second run in which we will double the concentration of acetone, but keep the total volume the same. Repeat the procedure. Again, the amount of time required for the reaction to take place should agree within about 10 seconds. Record the time on your data sheet.
5. Run #3
In run #3, we will calculate the order with respect to H+. The concentration of H+ should be doubled from run #1, but the other concentrations are held constant. Repeat the procedure. The amount of time required for the reaction to take place should agree within about 10 seconds. Record the time on your data sheet.
6. Run #4
In run #4, we will calculate the order with respect to iodine. Choose an appropriate concentration for iodine, but keep the total volume the same. Repeat the procedure. The amount of time required for the reaction to take place should agree within about 10 seconds. Record the time on your data sheet.
Acetone should be disposed of in the organic waste container.
Data and Analysis: Rates of Reaction
Reaction Rate Data
Record the reaction rate and the temperature for each of the four runs (time for reaction in seconds).
Calculate the molarity (initial concentration) of each I2, acetone, and H+ in your 4 runs.
The total volume of each run is 25 mL.
Calculate the average reaction rate for each run according to equation (3).
Substitute the initial concentration from Table II into equation (2) to obtain the rate for all four runs. Then find the orders m, n, and p with respect to each reactant.
Given the values for m, n, and p, calculate the rate constant k for each run by substituting the orders you obtained, the initial concentrations, and the observed rate into equation (2). Then calculate the average value for k.