7.4 The Mechanism of a Chemical Reaction

The path that reactants take to form products in a chemical reaction is called the reaction mechanism. A reaction's rate equation is highly informative as it provides crucial details about its mechanism. A reaction can occur in a single step or, more commonly, in several elementary steps.

When a reaction proceeds in two or more steps, one of them is inevitably the slowest. The rate of this slowest step dictates the overall reaction rate because it establishes a speed limit for the entire process. No reaction can proceed faster than its rate-determining step.

Understanding the Rate-Determining Step

The rate law for an overall reaction is determined by the stoichiometry of its rate-determining step. The concentration terms in the rate equation correspond to the reactants involved in this slow step.

• Multi-step Reactions: Complex reactions often occur through a sequence of simpler, elementary steps.
• Slowest Step Dominates: The overall rate is limited by the slowest step, much like traffic flow is limited by the slowest car in a single lane.
• Fast Steps: All other steps in the mechanism are usually much faster and do not limit the overall rate.
• Reaction Intermediate: A species that is formed in one step of the mechanism and consumed in a subsequent step is called a reaction intermediate. It does not appear in the overall balanced chemical equation.

Important Note: A balanced chemical equation often provides no information about the reaction mechanism. The mechanism must be determined experimentally.

Worked Examples

Example 7.4: Deducing Mechanism from Rate Law

For the reaction: ${\mathrm{NO}}_{2(g)}+{\mathrm{CO}}_{(g)}⟶{\mathrm{NO}}_{(g)}+{\mathrm{CO}}_{2(g)}$ The experimentally determined rate law is: $\text{Rate}=k[{\mathrm{NO}}_2{]}^2$ What information does this provide about the rate-determining step?

Solution:

1. The reaction is second order with respect to $\mathrm{N}{\mathrm{O}}_2$ and zero order with respect to $\mathrm{CO}$. The rate is independent of $[\mathrm{CO}]$.
2. Two molecules of $\mathrm{N}{\mathrm{O}}_2$ must be involved in the rate-determining step.
3. Because the overall stoichiometry (1 $\mathrm{N}{\mathrm{O}}_2$) does not match the rate law (2 $\mathrm{N}{\mathrm{O}}_2$), the reaction must proceed in more than one step.

A proposed mechanism consistent with this rate law:

• Step I (Slow — Rate-Determining): ${\mathrm{NO}}_{2(g)}+{\mathrm{NO}}_{2(g)}\stackrel{\text{slow}}{\to }{\mathrm{NO}}_{3(g)}+{\mathrm{NO}}_{(g)}$
• Step II (Fast): ${\mathrm{NO}}_{3(g)}+{\mathrm{CO}}_{(g)}\stackrel{\text{fast}}{\to }{\mathrm{NO}}_{2(g)}+{\mathrm{CO}}_{2(g)}$

Step I involves two $\mathrm{N}{\mathrm{O}}_2$ molecules, matching the rate law. The species $\mathrm{N}{\mathrm{O}}_3$ is a reaction intermediate — produced in Step I and consumed in Step II.

Example 7.5: Decomposition of Hypochlorite Ion

The hypochlorite ion ($\mathrm{Cl}{\mathrm{O}}^{-}$) decomposes in aqueous solution: $3{\mathrm{ClO}}_{(aq)}^{-}⟶{\mathrm{ClO}}_{3(aq)}^{-}+2{\mathrm{Cl}}_{(aq)}^{-}$ The rate law is: $\text{Rate}=k[{\mathrm{ClO}}^{-}{]}^2$ The proposed two-step mechanism is:

• Step I: ${\mathrm{ClO}}_{(aq)}^{-}+{\mathrm{ClO}}_{(aq)}^{-}⟶{\mathrm{ClO}}_{2(aq)}^{-}+{\mathrm{Cl}}_{(aq)}^{-}$
• Step II: ${\mathrm{ClO}}_{2(aq)}^{-}+{\mathrm{ClO}}_{(aq)}^{-}⟶{\mathrm{ClO}}_{3(aq)}^{-}+{\mathrm{Cl}}_{(aq)}^{-}$

Solution:

The rate law indicates two $\mathrm{Cl}{\mathrm{O}}^{-}$ ions participate in the rate-determining step. Step I involves the collision of two $\mathrm{Cl}{\mathrm{O}}^{-}$ ions. Therefore, Step I is the rate-determining step. The species $\mathrm{Cl}{\mathrm{O}}_2^{-}$ is a reaction intermediate.

Example 7.6: Reactants in the Rate-Determining Step

For the reaction: $2{\mathrm{NO}}_{2(g)}+{\mathrm{O}}_{3(g)}⟶{{\mathrm{N}}_2\mathrm{O}}_{5(g)}+{\mathrm{O}}_{2(g)}$ The rate law is: $\text{Rate}=k[\mathrm{N}{\mathrm{O}}_2][{\mathrm{O}}_3]$

This rate law indicates that one molecule of $\mathrm{N}{\mathrm{O}}_2$ and one molecule of ${\mathrm{O}}_3$ participate in the rate-determining step. The rate law directly reflects the reactants and their coefficients in the slow step.

Example 7.7: Writing the Rate Law from a Mechanism

Nitric oxide reacts with hydrogen according to the equation: $2{\mathrm{NO}}_{(g)}+2{\mathrm{H}}_{2(g)}⟶{\mathrm{N}}_{2(g)}+2{{\mathrm{H}}_2\mathrm{O}}_{(g)}$ The proposed mechanism is:

• Step I (slow): $2\mathrm{NO}+{\mathrm{H}}_2⟶{\mathrm{N}}_2+{\mathrm{H}}_2{\mathrm{O}}_2$
• Step II (fast): ${\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{H}}_2⟶2{\mathrm{H}}_2\mathrm{O}$

Write the experimental rate law for this reaction.

Solution:

The rate law is determined by the slow (rate-determining) step. Step I is the slow step and involves two molecules of NO and one molecule of ${\mathrm{H}}_2$. Therefore, the rate law is:

$\text{Rate}=k[\mathrm{NO}{]}^2[{\mathrm{H}}_2]$

The species ${\mathrm{H}}_2{\mathrm{O}}_2$ is a reaction intermediate — it is produced in Step I and consumed in Step II, and does not appear in the overall balanced equation.