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16 Laboratory work: Determination of the rate constant for the oxidation of iodine by hydrogen peroxide. Hydrogen peroxide oxidizes iodine ions in an acid medium according to the reaction: 2I – + H 2 O 2 + 2H + ⇄ I 2 + 2H 2 O (1) This reaction can take place in the presence of sodium thiosulfate also, then the resulting iodine is rapidly reduced by the thiosulfate ion: I 2 + 2S 2 O 3 2– ⇄ 2I – + S 4 O 6 2– (2) As a result, reaction (1) will proceed at a constant ion concentration I – . In the presence of an excess of acid in the solution, the concentration of hydrogen ions will also be practically constant. Consequently, the reaction rate (1) will depend only on the concentration of hydrogen peroxide, and the kinetic equation for it can be written as: kС dt dС v = = , where C is [H 2 O 2 ]. Thus, under these conditions reaction (1) proceeds as a first-order reaction. Work way: pour 100 ml of a 0.4% solution of potassium iodide in a 200 ml conical flask and add 5 ml of H 2 SO 4 with a concentration of 1 mol/l. A burette is placed in a tripod and filled with a 0.05 M solution of sodium thiosulfate. Add 1 ml of Na 2 S 2 O 3  solution from the burette to the flask with potassium iodide. Add 5 drops of the starch solution. 10 ml H 2 O 2 0.025 M is poured into a measuring tube and this solution is quickly poured into a flask. At the moment of mixing the solutions, a stopwatch is activated. Observe the state of the solution in the flask and record the first time of appearance of the blue color. Then 1 ml of sodium thiosulfate solution is quickly added from the burettewithout stopping time, mixed (the color disappears). Waite for the next moment the appearance of the blue color. When it appears, the time is again recorded. This operation is repeated 5 times. Then a few drops of 0.05 M ammonium molybdate (catalyst) are added to the flask, and the iodine released is titrated with a solution of sodium thiosulfate until the blue color disappears. Record the total volume of sodium thiosulfate Na 2 S 2 O 3 ( V ) spent on the experiment. The amount of Na 2 S 2 O 3  contained in this volume of solution is equivalent to the total amount of hydrogen peroxide. In subsequent moments, the amount of hydrogen peroxide will be proportional to ( V‑n ), where n  is the number of ml of Na 2 S 2 O 3  solution added to the corresponding moment.