Elementary Steps and Rate Laws
المؤلف:
University of Missouri System
المصدر:
Introductory chemistry
الجزء والصفحة:
.................
19-12-2020
1368
Elementary Steps and Rate Laws
A complete chemical reaction may occur in one or more elementary steps, each having its own rate law. The rate of a single elementary step can be derived directly from its stoichiometric equation, since it is an individual effective collision describing a single bond-breaking or bond-forming event. This explains why the rate law of an overall reaction, potentially involving several steps, does not necessarily correlate to the stoichiometry of its balanced chemical equation. For example:
5 Br–(aq) + BrO3–(aq) + 6 H+(aq) → 3 Br2(ℓ) + 3 H2O(ℓ)
Rate = k[Br–][BrO3–][H+]2
However, the rate law for an elementary step is determined from its molecularity: the number of molecules involved in the single effective collision (Table 17.2 “Elementary Steps and Their Rate Laws”). For the following elementary step:
A → B
Rate = k[A]
This is a unimolecular step, and as the concentration of reactant A molecules increases, the number of effective collisions also increases.
Table 17.2 Elementary Steps and Their Rate Laws
| |
| Elementary Step |
Molecularity |
Rate Law |
| A → B |
Unimolecular |
Rate = k[A] |
| 2 A → C |
Bimolecular |
Rate = k[A]2 |
| 2 A + D → E |
Termolecular |
Rate = k[A]2[D] |
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