Rate Law problem

  • Thread starter jools111
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  • #1
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I am working on a lab that deals with ionic concentration and rate of reaction. The data is as follows:

Mixture----- 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 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 [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.
 

Answers and Replies

  • #2
TeethWhitener
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Something must have messed up when you typed this in, because your reasoning is spot on, but the answers as given are wrong. Whatever the first column of concentration data was supposed to represent didn't get displayed properly.

The answer is that the rate is first order in both reagents, which is what I think you tried to type in the first place. As for the something that doesn't sit well, I'm afraid I don't quite understand what the concern is.
 
  • #3
mjc123
Science Advisor
Homework Helper
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A few years late, perhaps, but I think the OP should have typed [I-], but typed [I ] (only without the space), which is wrong chemically, and also is the code for italics.
 

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