7b: Determining the order of reaction using an initial rates method

Question 1

What is the reaction called

Answer 1

Iodine clock

Question 2

What are the two equations for the iodine clock reaction

Answer 2

H2O2(aq) + 2H+(aq) + 2I-(aq) = I2(aq) + 2H2O(l)

2S2O3(2-)(aq) + I2(aq) = 2I-(aq) + S4O6(2-)(aq)

Question 3

Describe in words what happens during the iodine clock reaction

Answer 3

Hydrogen peroxide and iodide ions react to form iodine. Iodine reacts with thiosulfate producing iodide ions. Thiosulfate is a reactant not a product

Question 4

What is the procedure of this experiment

Answer 4

Prepare the experiment by adding 10 cm³ of a 0.1M of hydrogen peroxide. Start the timer. Record the time it takes to get from a clear solution to a blue/black solution.

Question 5

what is the purpose of starch in this experiment

Answer 5

thiosulfate immediately reacts with iodine. However when there is no more thiosulfate left no iodine can react. Therefore no more iodide ions are produced. The iodine will change the solution from colorless to a blue black solution. This indicates when the thiosulfate has run out.

Question 6

what is the purpose of water

Answer 6

to keep the total volume of the solution the same and this ensures the concentration of the other reactants don’t change apart from potassium iodide.

[Increasing the concentration of potassium iodide means decreasing the concentration of water/volume of water]

Question 7

How do we find the order of the rate of reaction with respect to potassium iodide [KI]

Answer 7

X axis: log [the volume of potassium iodide]
Y axis: log [1/time taken]

We plot these points and if a straight line forms the order with respect to potassium iodide is 1