DETERMINATION OF THE RATE LAW
FOR A REACTION
The rate law for the reactionH2O2 (aq) + 2 I1– (aq) + 2 H1+ (aq) I2 (aq) + 2 H2O (l) is
The purpose of this experiment is to determine the values of k, x, y, and z using the method of initial rates. You will not actually measure initial rates, but will measure the time required to consume a fixed amount of another substance, thiosulfate ion. The thiosulfate ion reacts rapidly with the I2 produced in the reaction and reduces it back to I1– :
2 S2O32– (aq) + I2 (aq) 2 I1– (aq) + S4O62– (aq)
The I2produced in the first reaction is removed as quickly as it forms, until the thiosulfate (S2O32–) is used up. Then the I2 concentration increases rapidly and forms a dark blue-black complex with starch. This is the endpoint for the reaction, the sudden change from a colorless solution to dark blue, signaling the consumption of all of thiosulfate ion. Because less than 3% of the original H2O2 is consumed, the rate you measure will be a good approximation of the initial rate of the reaction (see section 13-2 in text).
Procedure
1. Obtain 6 50 mL flasks and 6 small test tubes. All glassware must be scrupulously clean and rinsed with deionized water. Shake out excess water but do not dry anything with a paper towel (it leaves little bits that react with the peroxide). Label the flasks 1 – 5, with a duplicate #1.
0.050 M Sodium acetate is prepared by adding 6.804 g in 1.00L of deionized water. Acetic acid is prepared by 8.33 mL of 6M acetic acid or 2.87 mL of 17.4 M Glacial acetic acid.
0.050 M KIis prepared by adding 8.301 g in 1.00L of deionized water
Ammonium molybdate (VI) tetrahydrate 0.176g per 100mL of water
Starch is made by dissolving 0.100 g in 100mL
0.050 M Sodium thiosulfate is prepared by adding 2.482 g in 200 mL of deionized water
2. Prepare the reaction mixtures by combining these ingredients, in order. Use a pipet for the water, and use the Repipets or Eppendorf pipets for the other ingredients (read the labels and be very careful not to mix them up). Measure as accurately as you can. All volumes are in milliliters. Mix thoroughly.
mixture / H2O / 0.050 M NaAc + 0.050 M HAc buffer / 0.30 M acetic acid / 0.050 M KI / 0.1% starch / 0.050 M Na2S2O3 / Mo (VI) catalyst1 / 7.5 / 3.0 / 0 / 2.5 / 0.5 / 0.5 / 0
2 / 6.5 / 3.0 / 0 / 2.5 / 0.5 / 0.5 / 0
3 / 5.0 / 3.0 / 0 / 5.0 / 0.5 / 0.5 / 0
4 / 3.0 / 3.0 / 4.5 / 2.5 / 0.5 / 0.5 / 0
5 / 7.0 / 3.0 / 0 / 2.5 / 0.5 / 0.5 / 0.5
1 / 7.5 / 3.0 / 0 / 2.5 / 0.5 / 0.5 / 0
3. Dispense the 3.0% (0.88 M) H2O2 solution into the test tubes:
tube / 1 / 2 / 3 / 4 / 5 / 6mL H2O2 / 1.0 / 2.0 / 1.0 / 1.0 / 1.0 / 1.0
4. Add all of the contents of tube 1 to flask 1 and mix thoroughly. Measure in seconds the time from the moment you add the peroxide to the moment the mixture turns blue (expect 2–4 minutes).
5. Do mixtures 2–5, then repeat mixture #1 two times to check agreement with the first time. Mixture #1 is the control for all calculations, so if two of the values do not agree within a few seconds, prepare more #1 mixtures until you are satisfied that the control time is reliable.
Analysis
Two reactions are occurring in the reaction mixture:
H2O2 (aq) + 2 I1– (aq) + 2 H1+ (aq) I2 (aq) + 2 H2O (l)
2 S2O32– (aq) + I2 (aq) 2 I1– (aq) + S4O62– (aq)
For every 1 mol of H2O2 that is consumed in the first reaction, 2 moles of S2O32– are consumed in the second reaction. From it we can deduce that
that is, the rate of the reaction is proportional to the rate of thiosulfate consumption (which is what you actually measured). Thus,
because ∆[S2O32–]2 = ∆[S2O32–]1 (you always used the same amount of thiosulfate ion). The reaction rate is inversely proportional to the reaction time.
1. In this experiment, mixture 1 is the control mixture. To compare mixture 2 to mixture 1, set up the ratio of rate 2 to rate 1 using the rate law expression:
If you examine the ingredients for the reaction mixtures carefully, you will notice that the only difference between mixture 1 and mixture 2 is the [H2O2]; the [I1–] and [H1+] (from the buffer) are the same. Assuming temperature is the same, the values of k are constant, and the ratio reduces to
because the amount of H2O2in mixture 2 is twice that in mixture 1. You will probably be able to determine the value of x by inspection, but if not you can use logarithms to calculate the value of x exactly. See example 15-3 in your textbook (page 512).
2. Do a similar analysis comparing mixture 3 to mixture 1 and mixture 4 to mixture 1 to determine the values of y and z in the rate law expression. Mixture 4 contains ten times as much H1+ as mixture 1.
3. Compare mixture 5 to mixture 1 to determine the effect of catalyst on the reaction rate; this time the values of k are not identical, because the catalyst affects the mechanism and thus the rate constant. Calculate the ratio of k5 to k1.
4. The concentrations of the stock solutions are given in the procedure instructions. Calculate [S2O32–] in mixture #1, accounting for dilution. For example, the concentration of the S2O32–stock solution is 0.050 M, and in mixture 1 you used 0.50 mL of stock S2O32–. The total volume of the mixture was 15.0 mL, so [S2O32–] in the mixture was
Using this [S2O32–] and your measured reaction time, calculate the reaction rate for mixture #1 (see the introduction to the analysis for the relationship between reaction rate and [S2O32–] ).
Accounting for dilution, calculate [H2O2], [I1–], and [HC2H3O2] in mixture #1, and use the rate law (with the exponents you determined) to calculate the value of k.
Discussion
Write the chemical equation for the reaction, then write rate law for the reaction at room temperature, showing the experimentally determined values for k, x, y, and z. Comment on the relationship (if any) between the rate law and the stoichiometry of the reaction. Discuss sources of error.