Hi guys, I'm doing the hydrogen peroxide/ potassium iodide “clock” reaction, but I did the lab last week and it was a big failure. The time it took for the blue colour to appear was really fast for my lowest concentration of hydrogen peroxide (which is my independent variable). I can only imagine how it will be for my highest concentration!
These are the concentrations of the solutions that I am using:
Potassium iodide, KI, solution (1.000 moldm*‐3) (always 25 cm3)
Hydrogen Peroxide Solutions - 0.100 M, 0.300 M, 0.500 M, 0.700 M, 0.900 M.
(Always 12.5 cm3)
Sodium Thiosulphate solution, (0.050 moldm*‐3) (always 20 cm3)
Sulphuric acid (1.000 moldm-3) always 10 cm3
Starch Solution (2%) Always 5 ml
Why is it that this reaction occurs so fast, which concentrations should I change.
I suspect it's the sodium thiosulphate as it is the reactant that acts as a delaying agent.
Please help me. It's very important to me!
EXPERIMENT 10 Chemical Kinetics: Iodine Clock Reaction Objective: The complete concentration and temperature dependence of the reaction rate for the reaction between peroxydisulfate ion and iodide ion in aqueous solution will be determined. Introduction: Peroxydisulfate ion reacts with iodide ion in aqueous solution to give iodine and sulfate ion: 2 I-+ S 2 O 8 2-→ I 2 + 2 SO 4 2-(1) In this experiment you will determine the rate at which reaction (1) occurs. This rate can be expressed in the form of a rate law, Rate = k I " [ ] x S 2 O 8 2 " [ ] y where x and y are the orders of reaction in iodide and peroxydisulfate ions, respectively, and k is the specific rate constant; the specific rate constant is a function of temperature. The initial rate of this reaction will be determined by measuring the time required to generate a certain amount of iodine, the same amount in every trial, according to reaction (1). Two other reactions will be used to signal when this constant amount of iodine has been produced. The first of these two reactions involves thiosulfate ion, S 2 O 3 2-, a certain, identical amount of which will be added to the reaction mixture in each trial. Thiosulfate reacts with iodine as fast as iodine is produced, converting it back to iodide ion: 2 S 2 O 3 2-+ I 2 → S 4 O 6 2-+ 2 I-(2) Because reaction (2) is so fast, relative to reaction (1), iodine will not have a chance to build up in solution until all of the thiosulfate has been consumed. Any buildup of iodine in the solution indicates that the thiosulfate has been used up, and that, of course, means that the constant amount of iodine has been produced. To visually detect the presence of excess iodine in the solution you will add some starch to one of the solutions before mixing the iodide and peroxydisulfate. As soon as iodine begins to build up in solution it will react to form a dark-blue complex with starch: I 2 + starch → I 2-starch complex (3) The time period as measured from the time of mixing the peroxydisulfate with the iodide in the presence of starch and thiosulfate to the appearance of the dark-blue color is the time it