# Rate law of the iodine clock reaction

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?

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.
• 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? – Shahar Jun 8 '14 at 1:04