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
S
2O
82- + 2I
- -> I
2 + 2SO
4 2-2S
2O
82- + I
2 --> 2I
- + S
4O
6 2-If that is what you meant, then what doesn't make sense to me is in the first equation
S
2O
82- is oxidizing I
- to I
2Whereas in your second equation
S
2O
82- is reducing I
2 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]2x[I-]2y
5.75x10^-6 = [0.383]x
____________________________
1.50x10^-5 = [0.500]2x
x = 1.38
I used Logs a = bx
I don't understand what you mean by the subscripts in
[S2O8]
2xand
[I-]
2yI don't understand where you get your data from.
Clive