Iodine Clock Reaction

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Title: The effect of concentration on reaction rates
Introduction: - The iodine clock reaction involves mixing two clear solutions and producing a new clear solution. Then, after several seconds, the solution turns dark blue. This can be used to measure how fast a reaction occurs which is known as chemical kinetics. The time taken for the reaction mix to turn blue can be measured with a stopwatch.
The reactions that form the basis for the iodine clock reaction are shown below.
Equation 1:
H2O2 + 3 I- + 2 H+ → I3- + 2 H2O
• H2O2 = Hydrogen peroxide
• I- = Iodide ion (from potassium iodide)
• H+ = A proton, from hydrochloric acid (HCL)
• I3- = Triiodide
• H2O = Water
Equation 1 shows that hydrogen peroxide reacts with iodide ions in acid solution
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Though there is starch in the mix of chemicals, the triiodide doesn 't react with it because that starch is immediately consumed in a reaction with the thiosulfate.
Equation 2:
I3S- + 2 S2O32- → 3 I- + S4O62-
• I3S- = Triiodide
• S2O32- = Thiosulfate ion
• 3 I- = Iodide ion
• S4O62- = Tetrathionate ion
Here, the triiodide reacts with thiosulfate to form iodide ions and tetrathionate.
This reaction is so fast that none of the triiodide has time to form a complex with starch, even though the starch is in the reaction mix. The reactions in Equations 1 and 2 are moving along during the lag time between mixing the chemicals and the dramatic appearance of the blue color. Note that iodide ions are regenerated in Equation 2, so they are available to react with the hydrogen peroxide in Equation 1. The thiosulfate, on the other hand, is consumed as it is turned into tetrathionate. The lag period ends when the thiosulfate is all used up. At this time, the triiodide is able to react with the starch.
Equation 3:
I3- + starch → (I3- starch complex)
• I3- = Triiodide
• I3- starch complex, which is blue
This equation says that starch reacts with triiodide to form a blue
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In Equation 1, for example, increasing the amount of hydrogen peroxide will increase the rate at which it reacts with iodide. The concentrations of iodide and acid remain the same, so the rate will depend only on the changes in hydrogen peroxide concentration. (The iodide is recycled between Equations 1 and 2, and the concentration of acid is high enough that the change in its concentration is small. Note the concentrations of the reactants in the Materials and Equipment section). The rate actually depends on the concentration of hydrogen peroxide raised to a power, called the "reaction order."
Equation 5:
Rate = k(H2O2)x
• k = Rate constant, in 1/seconds (s)
• (H2O2) = Concentration of hydrogen peroxide, in moles/liter
• x = Order of the reaction for hydrogen peroxide, unit less
The good news from Equation 5 is that the rate depends on the concentration of hydrogen peroxide, and you will know what the concentration of hydrogen peroxide is when the reaction starts. You will use the number of hydrogen peroxide drops as a measure of its concentration.
*insert aim*
Hence I’ve arrived at the following question:
How does varying the concentration of hydrogen peroxide affect the rate of reaction?
Research Question: How does varying the concentration of hydrogen peroxide affect the rate of reaction?
Independent Variable: Volume of hydrogen peroxide used.
Dependent Variable: Volume of distilled water used.
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