[ILoveISO]

x y and k, do they always have to be a WHOLE number like 1, 2? For example I'm trying to find x so I'm using this equation

rate1 = k[s2o8]^x[I-]^y
______________________
rate2 = k[s2o8]^x[I-]^y

Cancelled the K and I's to get

rate1 = [S2O8]^x
_______________
rate2 = [S2O8]^x

x = 0.72 is what I got is this possible or am I supposed to get near 1 or 2 etc?

[Teeny]

I would just round up to 1.

[cliverlong]

x y and k, do they always have to be a WHOLE number like 1, 2?

It appears orders do not have to be integers - but generally are:

http://en.wikipedia.org/wiki/Order_of_reaction (if you trust it)

For example I'm trying to find x so I'm using this equation

rate1 = k[s2o8]^x[I-]^y
______________________
rate2 = k[s2o8]^x[I-]^y

For which reaction? Where is your chemical equation?

Cancelled the K and I's to get

rate1 = [S2O8]^x
_______________
rate2 = [S2O8]^x

If (and I only write if) what you have done until now is valid, then why not cancel the terms above? From what you have written they are identical and should cancel, leaving the number 1.

x = 0.72 is what I got is this possible or am I supposed to get near 1 or 2 etc?

And how did you arrive at that number?
You have this habit of only giving fragments of information in your posts (I have read them before) and even less information on the methods you use - so it becomes almost impossible to work out what you are doing or where you are going wrong.

Also, please use the provided super-script and sub-script options provided on the forum editor. They take barely a second to use. Element symbols are written starting with upper-case.

Which of the following is easier to read?

rate1 = k[s2o8]^x[I-]^y

Rate₁ = k[S₂O₈]^x[I^-]^y

Clive

[ILoveISO]

Well I have two equations and I combined them to one S2O8 + 2I --> I2 + 2SO4 and  2S2O8 + I2 --> 2I + S4O6 so I got the ratio 1:2

here is what I plugged into the equation

rate1 = k[S2O8]^x[I-]^y
______________________________________
rate2 = k[S2O8]₂^x[I-]₂^y

5.75x10^-6 = [0.383]^x
____________________________
1.50x10^-5 = [0.500]₂^x

x = 1.38

I used Logs a = b^x

[ILoveISO]

Is what I did correct? Also how would I solve for K, not sure how to isolate it since it's on top and bottom...

[cliverlong]

Well I have two equations and I combined them to one S2O8 + 2I --> I2 + 2SO4 and  2S2O8 + I2 --> 2I + S4O6 so I got the ratio 1:2

I find it extremely difficult to read what you have typed because of your refusal to use sub-scripts and superscripts  provided by the editor when you type in polyatomic ions. Also you casually ignore charges on ions

Do you mean by your equations the following

S₂O₈^2- + 2I^- -> I₂ + 2SO₄ ^2-

2S₂O₈^2- + I₂ --> 2I^- + S₄O₆ ^2-

If that is what you meant, then what doesn't make sense to me is in the first equation

S₂O₈^2- is oxidizing I^- to I₂

Whereas in your second equation

S₂O₈^2- is reducing  I₂ to I^-

and I don't think both can happen without something else going on.

so I got the ratio 1:2

I don't understand what you mean by that at all.
Maybe if you clarify the earlier questions the meaning of the ratio statement will become clear.
Remember, rate equation indices have very little to do with the stoichiometry of the chemical equation that relates to them

here is what I plugged into the equation

rate1 = k[S2O8]^x[I-]^y
______________________________________
rate2 = k[S2O8]₂^x[I-]₂^y

5.75x10^-6 = [0.383]^x
____________________________
1.50x10^-5 = [0.500]₂^x

x = 1.38

I used Logs a = b^x

I don't understand what you mean by the subscripts in

[S2O8]₂^x
and
[I-]₂^y

I don't understand where you get your data from.

Clive

[cliverlong]

Have a look at the following thread

http://www.chemicalforums.com/index.php?topic=40038.0

are those the reactions your wish to discuss?

Clive