Question

Consider the reaction of peroxydisulfate ion (S2O8^2-)(Aq) with iodide ion (I-) in aqueous solution: (S2O8^2-)(Aq)+(3I-)(Aq)---->2SO4^2-+(I3-)(Aq)
.
At a particular temperature the rate of disappearance of S2O8^2 varies with reactant concentrations in the following manner:

Experiment S2O8^2- I- Initial Rate
1 0.018 0.036 2.6x10^-6
2 0.027 0.036 3.9x10^-6
3 0.036 0.054 7.8x10^-6
4 0.050 0.072 1.4x10^-5


1) Determine the rate law for the reaction.

Apparently the rate law is Rate=[S2O8^2-][I-]
How did they come up with this?

2)How is the rate of disappearance ofS2O8^2- related to the rate of disappearance of [I-] ?
The Answer is the rate of disappearance of of S2O8^2- is one-third the rate of disappearance of I-
How did they come up with this?

3)What is the average value of the rate constant for the disappearance of S2O8^2- based on the four sets of data? (answer is in form Rate= M^-1 S^-1
Please show me how to come up with this answer.

4)What is the rate of disappearance of S2O8^2- when S2O8^2- = 2.3×10-2 M and I-4.0×10-2 M? (answer is in form Rate= M/s)

Answer 1

Answer 2

Hope this helps.