Kinetics of the Reaction of Hydrogen Peroxide with Iodide Ion Lab
Hydrogen peroxide will react with iodide ion in an acidic solution to produce iodine
... [Show More] (I2) and water. If this
reaction is carried out with different concentrations of reactants the rate law for the reaction can be
determined. If it is also carried out at different temperatures the activation energy of the reaction can
also be determined.
H2O2 + 2 I-
+ 2 H+
2 H2O + I2 (Rxn 1)
In this experiment you will view a video that shows a part of this experiment and collects some of the
data you will need to calculate the rate expression for this reaction and the activation energy.
Here is the link for the video:
https://www.youtube.com/watch?v=AOdWANd6gmQ
In this experiment the average initial rate is determined by measuring the time it takes for a specific
amount of I2 to be produced. This is done by having the I2 react with thiosulfate ion (S2O3
2-) as it forms.
Starch is also present in the solution, which acts as an indicator. As soon as I2 forms it will immediately
react with the thiosulfate until all the thiosulfate is consumed. As soon as the thiosulfate is completely
consumed the next I2 that forms will combine with the starch to produce a dark blue color. The same
amount of thiosulfate is placed in each of the reaction mixtures, so the reaction is timed until a specific
amount of I2 is produced.
Each reaction mixture has 10.0 mL of 0.0500 M Na2S2O3 added to the mixture. Each mole of Na2S2O3
produces a mole of S2O3
2-, so there would be (0.0500 moles/L)(0.0100 L) = 0.000500 moles of S2O3
2- in
each reaction mixture. The reaction of thiosulfate with iodine is:
I2 + 2 S2O3
2-
S4O6
2- 2 I-
(Rxn 2)
So there are two thiosulfates that react for each I2. Thus when all the thiosulfate in each mixture reacts
there would have been (0.000500 moles S2O3
2-)( 1 mole I2 / 2 mole S2O3
2-) = 0.000250 moles of I2
produces from the hydrogen peroxide reaction (Rxn 1). [Show Less]