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Just a quick question on how to deal with multiple intermediaries in a proposed reaction mechanism.

Here is the overall reaction:

$$\ce{I-(aq) + OCl-(aq) -> IO-(aq) + Cl-(aq)}$$

Here is the proposed mechanism:

$$ \begin{align} \ce{OCl- + H2O &<=>[$k_1$][$k_{-1}$] HOCl + OH-} & &\text{(Fast equilibrium)} \\ \ce{I- + HOCl &->[$k_2$] HOI + Cl-} & &\text{(Slow/RDS, non-reversible)} \\ \ce{HOI + OH- &->[$k_1$] H2O + IO-} & &\text{(Fast)} \end{align} $$

A few things to note, the question does indeed label both the first reaction going forwards and the third reaction going forwards as $k_1$. It also mentions that we should treat the concentration of water as being constant.

Here is my question: I know how to write the rate law for the Rate Determining Step as $k_2[\ce{I-}][\ce{HOCl}]$ and then I know I need to use the equilibrium expression in the first reaction to isolate concentration of $[\ce{HOCl}]$ as

$$[\ce{HOCl}] = \frac{k_1[\ce{OCl-}]}{k_{-1}[\ce{OH-}]}$$

but then I am confused about how to isolate the second intermediary $(\ce{OH-})$ so I can get the proper rate law. Should I use the third reaction?

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    $\begingroup$ You should use steady state on intermediate species HOI, i.e. $\ce{d[HOI]/dt=0=\cdots }$ and the rate of product formation is $\ce{d[IO^-]/dt=k_3[HOI][OH^-]}$. Use these two equations then the equilibrium to get $\ce{OCL^-}$ in the final answer. $\endgroup$ – porphyrin May 13 at 14:47

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