In an elementary rate law, the reaction orders are identical to the stoichiometric coefficients. This implies what about the rate law?

In an elementary rate law, the reaction orders are identical to the stoichiometric coefficients. This implies what about the rate law?

• A. It is followed when the reaction orders equal stoichiometric coefficients
• B. It requires a multi-step mechanism
• C. It depends on mass transfer limitations
• D. It is independent of temperature

Answer: (B)

In an elementary reaction, the rate law comes directly from the collision event that defines that step. The rate is proportional to how often the reacting molecules meet, so the rate law takes the form rate = k times the concentrations raised to the stoichiometric powers of the reactants in that step. In other words, the exponents on each reactant’s concentration match its coefficient in that single elementary step. If you observe that the reaction orders align with those stoichiometric coefficients, that suggests you’re looking at, or describing, a single elementary step, for which the rate law is simply the straightforward expression based on that step’s stoichiometry. This means there’s no need to invoke a complex, multi-step mechanism with intermediates to explain the rate law.

In an elementary reaction, the rate law comes directly from the collision event that defines that step. The rate is proportional to how often the reacting molecules meet, so the rate law takes the form rate = k times the concentrations raised to the stoichiometric powers of the reactants in that step. In other words, the exponents on each reactant’s concentration match its coefficient in that single elementary step. If you observe that the reaction orders align with those stoichiometric coefficients, that suggests you’re looking at, or describing, a single elementary step, for which the rate law is simply the straightforward expression based on that step’s stoichiometry. This means there’s no need to invoke a complex, multi-step mechanism with intermediates to explain the rate law.