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rates of chemical reactions

using the equation + 2S2O8SO4^2-
how dp write the rate law for this reactio
also knowing that rate= K(S2O8^2-)^X(I^-)^Y, how do I find K, mwaning what values do I plug in for the 2 compunds.

Quote:

Originally Posted by fpolycarpe

using the equation + 2S2O8SO4^2-
how to write the rate law for this reaction
also knowing that rate= K(S2O8^2-)^X(I^-)^Y, how do I find K, meaning, what values do I plug in for the 2 compounds.

$2S_2O_8SO_4^{2-}$ isn't a compound or a reaction

Are you trying to write

$S_2O_8^{2-} \rightarrow 2SO_4^{2-} \, + \, 2e^-$

Then the rate law is

$rate= K(S_2O_8^{2-})^X(I^-)^Y$

Where did the "I" come from? Is this an iodine oxidation?

$S_2O_8^{2-} \, + I_2 \rightarrow 2SO_4^{2-} \, + \, 2I^-$

If so, why didn't you say so?

The rate law should be

$rate\,(K)\,=\,\Large \frac {[S_2O_8^{2-}][I_2]^2}{[SO_4^{2-}]^2[I^-]^2}$

Sorry Perito, I haven't done rate yet, but I'm wondering why you used iodine raised to the power of 2? I saw that there are two sulfates, and two iodides...