Elementary Reactions - AP Chemistry

Formed in one step and consumed in a later step. Intermediates don't appear in reactants or products of overall reaction.

Unimolecular. One molecule participates in the elementary step.

A single-step molecular event with no intermediates. Elementary steps occur in one collision without breaking into smaller steps.

$\text{rate}=k[A]$. For elementary steps, rate law exponents equal stoichiometric coefficients.

$\text{rate}=k[A][B]$. Each reactant appears to the first power in bimolecular steps.

$\text{rate}=k[A]^2$. Coefficient 2 becomes exponent 2 in elementary rate laws.

$\text{rate}=k[A]^2[B]$. Exponents match stoichiometric coefficients for elementary steps.

$3$. Sum all exponents: $2 + 1 = 3$.

$\text{s}^{-1}$. First-order reactions have units of inverse time.

$\text{M}^{-1}\text{s}^{-1}$. Second-order overall requires one inverse concentration unit.

$\text{M}^{-2}\text{s}^{-1}$. Third-order overall requires two inverse concentration units.

Consumed early and regenerated later; not in net equation. Catalysts appear in mechanism but not in overall equation.

The slowest step that limits the overall rate. All other steps must wait for the slowest step to complete.

Observed rate law often matches the RDS rate law. The RDS controls the overall reaction rate.

Intermediates (and catalysts) cancel from the net equation. Species produced and consumed in mechanism don't appear in net equation.

$I$. Produced in step 1, consumed in step 2.

$\text{rate}=k[A][B]$. Elementary step rate laws directly reflect stoichiometry.

Termolecular (rare). Three-body collisions are statistically unlikely.

Bimolecular. Two molecules collide in the elementary step.

$A+B+D \rightarrow E$. Add steps and cancel intermediate C on both sides.

The number of reactant species that collide in that step. Counts molecules/atoms participating in the elementary step.

Unimolecular, bimolecular, or termolecular. 1, 2, or 3 molecules colliding; higher is too improbable.

$\text{rate}=k[A]$. One molecule decomposes, so rate depends on [A] only.

$\text{rate}=k[A][B]$. Two different molecules collide, each contributes to rate.

$\text{rate}=k[A]^2$. Two A molecules collide, so [A] appears squared.

$\text{rate}=k[A][B][C]$. Three molecules must collide simultaneously.

$\text{rate}=k[A]^2[B]$. Two A's and one B collide; A contributes squared term.

For an elementary step, exponents equal reactant coefficients. Unlike overall reactions, elementary steps follow this rule.

No; only an elementary step directly determines its rate law. Overall rate laws must be determined experimentally.