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I am dealing with the iodine clock reaction, which has the half-reactions: $$\begin{alignat}{2}\ce{H2O2 (aq) + 2I- (aq) &-> I2 (s) + 2 OH- (aq)} &&\quad (\text{slow}) \\ \ce{I2 (s) + 2S2O3^2- (aq) &-> 2I- (aq) + S4O6^2- (aq)} &&\quad (\text{fast}) \end{alignat}$$

Now, I have to find the rate law. What I did was 4 solutions – one with standard amounts, another with double the $\ce{I-}$, one with double the $\ce{H2O2}$, and one with double the $\ce{S2O3^2-}$. I recorded the amount of time for the ppt ($\ce{I2}$) to appear. Now I have to calculate the rate law.

The thing is, I am not quite sure how. Usually you take the "slow" step and set up an equation based on that, like $\text{rate}=\ce{[H2O2]$^x$ [I- ]$^y$}$. However, the reaction is not just dependent on those chemicals, it's also dependent on the $\ce{S2O3^2-}$. When it was doubled, the reaction was significantly slower (for obvious reasons).

How do I set up the rate law for this reaction? Or, if I can't, what else can I do with my data?

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IMO, the only reaction that you are studying is the first reaction: the conversion of $I^-$ to $I_2$ by hydrogen peroxide. The thiosulphate reaction is simply a dramatic way to indicate when a particular amount of iodine has been produced.

The reaction was not slower when you doubled the amount of thiosulphate. Twice as much iodine needed to be produced (and reconverted to iodide ion) before the thiosulphate was consumed and the first permanent iodine changed the starch indicator blue.

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  • $\begingroup$ Although, doesn't the fact that $\ce{[I^{-}]}$ stays constant as a result of the fast reaction mean that it has an effect (not necessarily slower) on the reaction rate? $\endgroup$ – Shahar Jun 8 '14 at 1:04

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