If the reaction 2 NO2(g)+F2(g) ⟶2 NO2 F(g) occurred by a one-step process, what would be the ex

step 1 For this question, we're given the reaction 2N02 plus F2 goes to 2, NO2F. If this reaction occur

If the reaction 2 NO2(g)+F2(g) ⟶2 NO2 F(g) occurred by a...

If the reaction $$2 \mathrm{NO}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F}(g)$$ occurred by a one-step process, what would be the expected rate law for the reaction? The actual rate law is rate $=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{F}_{2}\right]$, why is this a better rate law?

Key Concepts

Rate Law

The rate law expresses how the rate of a chemical reaction depends on the concentration of its reactants. It is experimentally determined and can differ from the stoichiometric coefficients given by the overall balanced equation, especially if the reaction occurs in multiple steps.

Reaction Mechanism

A reaction mechanism is a series of elementary steps that lead to the overall reaction. While a one-step mechanism implies the rate law would be directly derived from the molecularity of that single step, discrepancies between the expected and actual rate laws indicate that the reaction likely proceeds through several steps.

Molecularity

Molecularity refers to the number of molecules involved in an elementary reaction step. For a one-step process, the expected rate law would have reaction orders equal to the number of molecules colliding. In this case, if the reaction were elementary, one would expect a second-order dependence on NO2, yet the observed first-order dependence suggests a different, more complex process.

Rate-Determining Step

The rate-determining step is the slowest step in a multi-step reaction that limits the overall reaction rate. The measured rate law reflects the kinetics of this slow step rather than the stoichiometry of the overall reaction, which explains why the experimental rate law shows different orders than those predicted by a simple one-step mechanism.

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