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Write the rate laws for the formation and dissociation of [es].

User Teka
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Final answer:

Chemical reaction rates are tied to reactant concentrations via rate laws for both formation and dissociation. Rate laws for elementary reactions can be derived from stoichiometry, and in multi-step reactions, the rate-determining step governs the overall rate law.

Step-by-step explanation:

To describe the rate of a chemical reaction, we use rate laws which relate the rate of reaction to the concentration of the reactants. For a generic reaction where reactant A forms product B, the rate law for the formation of B might be written as rate = k[A], where 'k' is the rate constant and [A] is the molar concentration of reactant A. For the dissociation of B back into A, the rate law would be rate = k[B], assuming a first-order reaction in both directions.

In the case of an elementary reaction, such as A → B, the rate law can be determined directly from the stoichiometry of the balanced chemical equation. For the reaction 2NO₂ → 2NO + O₂, the rate law according to the stoichiometry would be rate = k[NO₂]².

If we consider a more complex reaction that follows an elementary mechanism, we might have a two-step reaction with elementary steps A → I and I → B. The rate law for each step would be based on its stoichiometry; for example: rate1 = k1[A] and rate2 = k2[I]. If the first step is the rate-determining step, then the overall rate law for the reaction would align with rate1, assuming that the concentration of the intermediate I is steady and does not accumulate significantly.

User Shamseer PC
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