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Rate Law From Mechanism

Hardchemistry

For a reaction whose slow step involves only one molecule decomposing after a fast equilibrium, how is the overall rate law most likely determined?

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About This Question

Subject
chemistry
Chapter
chemical kinetics
Topic
rate law from mechanism
Difficulty
Hard
Year
2025
Tags
rate-determining stepreaction mechanismrate law derivationfast equilibriumintermediate substitution

Solution

Correct Answer:

By the slowest, rate-determining step

In a multistep reaction the overall rate is governed by the slowest step, called the rate-determining step, because the reaction cannot proceed faster than its slowest stage just as a production line is limited by its slowest worker. The rate law is therefore written from the molecularity of this slow step, with any intermediates expressed in terms of measurable reactants using the preceding fast equilibrium. The option that the fast step alone controls the rate is wrong, since fast steps reach completion quickly and do not limit throughput. Adding all step rates has no kinetic meaning, because rates of sequential steps are not summed. The most exothermic step relates to thermodynamics, not to which step is slowest, so it does not set the rate. Deriving the rate law from the rate-determining step, while substituting for intermediates, is a central JEE Advanced skill highlighted in NCERT. A common JEE pitfall is to ignore the role of rate-determining step, yet it is exactly this factor that distinguishes the correct answer from the tempting alternatives. Plausibility check: the experimentally observed order usually matches the molecularity of the proposed slow step, confirming that the rate-determining step dictates the rate law.

This hard difficulty chemistry question is from the chapter chemical kinetics, covering the topic of rate law from mechanism. It appeared in the 2025 exam.

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