How Does Concentration Affect Reaction Time and Rate Law?

[ND3G]

Mixture-----[I-] (mol/L)---[S2O8^2] (mol/L)---Time (s)
1---------------0.10--------------0.050-------------20
2---------------0.075-------------0.050-------------28
3---------------0.050-------------0.050-------------41
4---------------0.025-------------0.050-------------84

5---------------0.10--------------0.038-------------25
6---------------0.10--------------0.025-------------39
7---------------0.10--------------0.013-------------82

what is the effect of doubling the concentration of iodide ions? of doubling the persulphate ions? Write the rate law for the reaction

Answers:

Doubling the concentration of iodide ions cuts the reaction time in half.

Doubling the concentration of persulphate ions also cuts the reaction time in half.

Now the rate law portion is what is giving me problems. I think I understand it (kind of) but I'm not sure...

What I put down was r = k[A]

Does that look right?

If the increase in concentration has no effect then [A] will not be found in the rate-determining step.
If it doubles doubles when [A] doubles then [A]^1 = [A]

If [A] increases by a factor of 2, such as 2^2 or 3^2 then [A]^2 and so on...

 

[gradeaswimr]

the rate is mol/(L*s) and not s right?

 

[Stevedye56]

Did they give you an equation or no?