U5.4 · Elementary ReactionsPractice · 基元反应

For an elementary step, the rate law comes straight from the reactant coefficients: one $\ce{NO2}$ collides with one $\ce{F2}$, a bimolecular step, so rate = k[$\ce{NO2}$][$\ce{F2}$]. Products don’t appear in the rate law.

Molecularity counts the particles on the reactant side: $\ce{N2O4}$ falls apart by itself, 1 particle, unimolecular. $\ce{NO + O3}$ and $\ce{H + Cl2}$ have 2 (bimolecular) and $\ce{2NO + O2}$ has 3 (termolecular, rare). The number of products doesn’t matter.

If it were a single elementary step, two $\ce{N2O5}$ would collide and the rate law would be k[$\ce{N2O5}$]². Experiment shows first order, so the reaction goes in several steps and the slowest one sets the rate law. Overall coefficients can’t be used as orders; orders come from experiment.

Elementary step: rate = k[A][B]². The rate changes by 2 × 3² = 18. B has a coefficient of 2, so the step is second order in B and tripling [B] multiplies the rate by 9.

An elementary step’s rate law follows the reactant coefficients, and all three particles must collide at once, so M is included: rate = k[O][$\ce{O2}$][M]. M appears on both sides, but it takes part in the collision, so more M makes the step faster.