# Reaction Mechanism with a Rate Determining Step

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?

• 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. – porphyrin May 13 '19 at 14:47