Consider the following three-step representation of a reaction mechanism

Explanation

The reason why the rate law for the overall reaction is not the same as the rate equation for the rate-determining step is rooted in the nature of the reaction mechanism. In the given scenario, the first step is a fast equilibrium reaction between A and B to form AB. Because this step is fast and reversible, it can quickly establish an equilibrium concentration of AB, which is used in the rate-determining step. The rate law for the overall reaction is not simply the rate law of the rate-determining step because it must take into account the concentration of AB generated in the fast equilibrium step.

When the rate-determining step is not the first step, deriving the rate law for the overall reaction becomes more complex because the concentration of intermediates, such as AB in this case, is not necessarily directly proportional to the concentration of the reactants as they appear in the balanced equation of the overall reaction. Instead, the rate law for the overall reaction depends on the mechanism as a whole and how the concentration of intermediates is related to the concentration of reactants.

In summary, the rate-determining step dictates the rate at which the overall reaction can proceed, but the presence of a fast initiatory step can alter the relationship between reactant concentrations and the overall rate law.

Final answer

The rate law for the overall reaction differs from that of the rate-determining step because the initial fast equilibrium affects the concentration of intermediates which, in turn, affects the overall rate law.

Which explains why the rate law for the overall equation is not the same as the rate equation for the rate-determining step?

When the rate-determining step is preceded by a step involving a rapidly reversible reaction, the rate law for the overall reaction may be more difficult to derive.

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