Steady-State Approximation and Rate Laws
Consider the following two-step reaction mechanism where X is an intermediate:
Step I: (fast)
Step II: (slow)
When the steady-state approximation is applied to derive the overall rate law for this mechanism, the intermediate X does not appear in the final expression. Which of the following statements provides the best explanation for this?
The overall rate law is determined by the slow step after expressing the intermediate in terms of reactants, so intermediates do not appear in the final rate law.
Intermediates are always included in the overall rate law because they are formed during the reaction.
The steady state approximation requires the concentration of the intermediate to fluctuate rapidly, thereby affecting the rate law directly.
Canceling the intermediate implies that it has no role in the reaction mechanism, making the rate law independent of the initial reactants.
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