AP Chemistry 5.7-5.8: Reaction Mechanisms and Rate Law

Reaction mechanism

A proposed sequence of elementary reaction steps describing how an overall chemical reaction occurs at the molecular level. It identifies the individual molecular events and any intermediates formed.

How does an overall chemical equation differ from a reaction mechanism?

The overall equation shows only the net reactants and products, whereas a mechanism gives the individual elementary steps. The overall equation does not indicate the actual molecular pathway or the order in which bonds change.

Elementary reaction (elementary step)

A single molecular event in a reaction mechanism that occurs exactly as written. Its rate law can be inferred directly from the reactant coefficients in its balanced equation.

How is the net reaction obtained from a multistep mechanism?

Add the elementary-step equations and cancel species that are produced in one step and consumed in another. The remaining species and coefficients form the balanced net reaction.

Intermediate

A species formed in one elementary step and consumed in a later step. It appears in the mechanism but cancels out of the net reaction and therefore does not appear in the overall equation.

In the ozone decomposition mechanism $\mathrm{O_3 \rightarrow O_2 + O}$ followed by $\mathrm{O + O_3 \rightarrow 2O_2}$, what is the net reaction and which species is the intermediate?

Adding the steps gives $\mathrm{2O_3 \rightarrow 3O_2}$. Atomic oxygen, $\mathrm{O}$, is the intermediate because it is produced in the first step and consumed in the second.

Molecularity

The number of reactant particles that participate in a single elementary step. Molecularity is defined only for elementary reactions, not for an overall reaction assembled from multiple steps.

Unimolecular elementary reaction

An elementary step involving one reactant entity, such as $\mathrm{A \rightarrow products}$. Its rate law is $\mathrm{rate = k[A]}$, so it is first order in $\mathrm{A}$.

Bimolecular elementary reaction

An elementary step involving two reactant entities, such as $\mathrm{A+B \rightarrow products}$ or $\mathrm{2A \rightarrow products}$. The corresponding rate laws are $\mathrm{rate=k[A][B]}$ and $\mathrm{rate=k[A]^2}$, respectively.

Termolecular elementary reaction

An elementary step involving the simultaneous collision of three particles, such as $\mathrm{2A+B \rightarrow products}$. Its rate law is $\mathrm{rate=k[A]^2[B]}$, but such steps are uncommon because simultaneous three-particle collisions are unlikely.

Why are elementary steps involving four or more simultaneous reactant particles generally not proposed?

The probability of four or more particles colliding with the required orientations and energies at the same instant is extremely small. Mechanisms therefore generally proceed through successive unimolecular or bimolecular steps.

How can the rate law of an elementary step be determined from its equation?

For an elementary step, the reactant coefficients become the exponents in the rate law. For example, $\mathrm{2A+B\rightarrow products}$ gives $\mathrm{rate=k[A]^2[B]}$.

Why can the rate law of an overall reaction usually not be obtained directly from its balanced equation?

An overall equation usually represents several elementary steps, not one molecular event. The experimentally measured rate law depends on the mechanism and must generally be determined from kinetic data.

Rate-determining step (rate-limiting step)

The slowest elementary step in a proposed mechanism. Because the overall process cannot proceed faster than this bottleneck, the slow step often determines the observed rate law.

When does the overall rate law usually match the rate law of the rate-determining step?

When the rate-determining step is the first step and contains only reactants from the overall reaction, its rate law usually gives the overall rate law directly. If it contains an intermediate, additional algebra is needed to eliminate that intermediate.

For the mechanism $\mathrm{2NO_2\rightarrow NO_3+NO}$ (slow) and $\mathrm{NO_3+CO\rightarrow NO_2+CO_2}$ (fast), what rate law is predicted?

The slow elementary step predicts $\mathrm{rate=k[NO_2]^2}$. The intermediate $\mathrm{NO_3}$ does not appear in the rate law, and adding the steps gives $\mathrm{NO_2+CO\rightarrow NO+CO_2}$.

A reaction has the rate law $\mathrm{rate=k[NO_2]^2}$ for the net reaction $\mathrm{NO_2+CO\rightarrow NO+CO_2}$. What does this imply about a possible mechanism?

The rate-determining step likely involves two $\mathrm{NO_2}$ particles and does not involve $\mathrm{CO}$. A consistent possible mechanism is $\mathrm{2NO_2\rightarrow NO_3+NO}$ followed by a fast reaction of $\mathrm{NO_3}$ with $\mathrm{CO}$.

How can a proposed mechanism be tested against an experimentally determined rate law?

Derive the rate law predicted by the mechanism and compare its concentration dependence with the experimental rate law. A mechanism is supported only if it also produces the correct net equation.

Rapid pre-equilibrium approximation

When a reversible step is much faster than a later slow step, the reversible step may be treated as being at equilibrium. Equating its forward and reverse rates allows the concentration of an intermediate to be expressed in terms of measurable reactant concentrations.

For the rapid equilibrium $\mathrm{A+B\rightleftharpoons I}$, how is the intermediate concentration expressed?

At equilibrium, $\mathrm{k_1[A][B]=k_{-1}[I]}$. Therefore, $\mathrm{[I]=(k_1/k_{-1})[A][B]}$.

For the mechanism $\mathrm{NO+Cl_2\rightleftharpoons NOCl_2}$ (fast) followed by $\mathrm{NOCl_2+NO\rightarrow 2NOCl}$ (slow), derive the predicted rate law.

The fast equilibrium gives $\mathrm{[NOCl_2]=(k_1/k_{-1})[NO][Cl_2]}$. Substitution into the slow-step rate law gives $\mathrm{rate=(k_2k_1/k_{-1})[NO]^2[Cl_2]}$, or equivalently $\mathrm{rate=k[NO]^2[Cl_2]}$.

Why must intermediates usually be eliminated from an overall rate law?

Intermediate concentrations are not normally controlled or measured as reactant concentrations in the overall reaction. A useful rate law is written in terms of observable reactants, products, or other experimentally accessible species.

For the rapid equilibrium $\mathrm{F_2\rightleftharpoons 2F}$, what relationship connects $[F]$ and $[F_2]$?

Equating the forward and reverse rates gives $\mathrm{k_1[F_2]=k_{-1}[F]^2}$. Thus, $\mathrm{[F]=(k_1[F_2]/k_{-1})^{1/2}}$.