Determining the law from a mechanism with an initial fast, equilibrium step: The oxidation of iodide ion by hydrochlorite ion : `CIO^(-)(aq.)+I^(-)(aq.)rarrCI^(-)(aq.)+IO^(-)(aq.)` has been postulated to occur by the two step mechanism : 1. `CIO^(-)(aq.)+H_(2)O(l)hArrHCIO(aq.)+OH^(-)(aq.)` fast, equilibrium 2. `I^(-)(aq.)+HCI(aq.)overset(k_(2))rarr(HIO(aq.)+CI^(-)(aq.)` slow What rate law is predicted by this mechanism ? Strategy : Write the rate equation for the rate determining (slow) step. However, in this case the equation contains a species that does not appear in the overall eqwuation for the reaction. We need to eliminate it from the final from of the rate law. For this purpose we use the fast and equilibrium step.

Determining the law from a mechanism with an initial fast, equilibrium

1.	Determining the law from a mechanism with an initial fast, equilibrium step: The oxidation of iodide ion by hydrochlorite ion : `CIO^(-)(aq.)+I^(-)(aq.)rarrCI^(-)(aq.)+IO^(-)(aq.)` has been postulated to occur by the two step mechanism : 1. `CIO^(-)(aq.)+H_(2)O(l)hArrHCIO(aq.)+OH^(-)(aq.)` fast, equilibrium 2. `I^(-)(aq.)+HCI(aq.)overset(k_(2))rarr(HIO(aq.)+CI^(-)(aq.)` slow What rate law is predicted by this mechanism ? Strategy : Write the rate equation for the rate determining (slow) step. However, in this case the equation contains a species that does not appear in the overall eqwuation for the reaction. We need to eliminate it from the final from of the rate law. For this purpose we use the fast and equilibrium step.
Answer» The rate - determining step is step 2. Its law is `Rate = k_(2)[I^(-)][HCIO]` This rate law includes the concentration of HCIO, which is not part of the overall reaction as it is not one of the orginial reactants. To convert the rate law to a more usful form, we must find the relationship between the concentration of HCIO and the concentration of substances originally present in the solution. Step 1, which is fast and reaches equilibrium, controls the [HCIO]. To eliminate [HCIO] from the rate law, we use the fact than step 1 is described by the equilibium constant `k_(eq.)= ([HCIO][OH^(-)])/([CIO^(-)])` Note that `[H_(2)O]` is included in the value of `k_(eq.)` as it is constant on according of the fact that `H_(2)O` (sovent) is present in excess. Solving for `[HCIO]` gives `[HCIO] = k_(eq.) ([CIO^(-)])/([OH^(-)])` Substituting this expression into the rate law gives `Rate = k_(2)k_(eq.) ([I^(-)][CIO^(-)])/([OH^(-)])` This is the overall rate law for the reaction. The product `k_(2)k_(eq.)` is aconstant. The reaction is first order in iodide ion, i.e., as the concentration of `OH^(-)` increases, the rate of the reaction decrases. This decrease in rate can be attributed to the decrease in the concentration of HCIO as the solution becomes more alkaline.

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