[jools]

I am working on a lab that deals with ionic concentration and rate of reaction. The data is as follows:

Mixture-----(I) mol/L-----(S2O8) 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

The question asks me to write the rate law for the reaction. I know from the data that when the concentration of [ i] doubles, the reaction time gets cut in half. The same is true for the persulphate. So I am thinking that the rate law would be:

r = k (I) (S2O8)

Something doesn't sit well with this answer though. When the persulphate is being varies and the iodide remains constant, the reactions take longer, even though they are proportionately rougly the same. Am I missing something here? Thanks.

[mwarner89]

Your right that they are both first order to give an overall order of 2. The iodine and persulphate have different rate constants though which you missed out