# 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. > *"The slowest step in the reaction mechanism, which determines the overall rate of the reaction, is called the **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: $$ \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{CO}_{(\mathrm{g})} \longrightarrow \mathrm{NO}_{(\mathrm{g})}+\mathrm{CO}_{2(\mathrm{~g})} $$ The experimentally determined rate law is: $$ \text{Rate} = k[\mathrm{NO}_{2}]^{2} $$ What information does this provide about the rate-determining step? **Solution:** The rate equation tells us the following: 1. The reaction is **second order** with respect to $NO_2$ and **zero order** with respect to $CO$. This means the reaction rate is independent of the concentration of $CO$. 2. Two molecules of $NO_2$ must be involved in the rate-determining step. 3. Because the stoichiometry of the overall reaction ($1\,NO_2$) does not match the rate law (involving $2\,NO_2$), the reaction must proceed in more than one step. A proposed mechanism consistent with this rate law is: * **Step I (Slow):** $$ \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{NO}_{2(\mathrm{~g})} \xrightarrow{\text{slow}} \mathrm{NO}_{3(\mathrm{~g})}+\mathrm{NO}_{(\mathrm{g})} $$ * **Step II (Fast):** $$ \mathrm{NO}_{3(\mathrm{~g})}+\mathrm{CO}_{(\mathrm{g})} \xrightarrow{\text{fast}} \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})} $$ The first step is slow and involves two $NO_2$ molecules, matching the rate law. Therefore, **Step I is the rate-determining step**. The species $NO_3$ is a **reaction intermediate** as it is produced in Step I and consumed in Step II. ### Example 7.5: Decomposition of Hypochlorite Ion The hypochlorite ion ($ClO^-$) decomposes in aqueous solution: $$ 3 \mathrm{ClO}_{(\text{aq})}^{-} \longrightarrow \mathrm{ClO}_{3(\text{aq})}^{-} + 2 \mathrm{Cl}_{(\text{aq})}^{-} $$ The rate law is: $$ \text{Rate} = k[\mathrm{ClO}^{-}]^{2} $$ The proposed two-step mechanism is: * **Step I:** $\mathrm{ClO}_{(\text{aq})}^{-} + \mathrm{ClO}_{(\text{aq})}^{-} \longrightarrow \mathrm{ClO}_{2(\text{aq})}^{-} + \mathrm{Cl}_{(\text{aq})}^{-}$ * **Step II:** $\mathrm{ClO}_{2(\text{aq})}^{-} + \mathrm{ClO}_{(\text{aq})}^{-} \longrightarrow \mathrm{ClO}_{3(\text{aq})}^{-} + \mathrm{Cl}_{(\text{aq})}^{-}$ Select the rate-determining step. **Solution:** The rate law indicates that two $ClO^-$ ions participate in the rate-determining step. Step I of the mechanism involves the collision of two $ClO^-$ ions. Therefore, **Step I is the rate-determining step**. ### Example 7.6: Reactants in the Rate-Determining Step For the reaction: $$ 2 \mathrm{NO}_{2(\text{g})} + \mathrm{O}_{3(\text{g})} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5(\text{g})} + \mathrm{O}_{2(\text{g})} $$ The rate law is: $$ \text{Rate} = k[\mathrm{NO}_{2}][\mathrm{O}_{3}] $$ This rate law indicates that one molecule of $NO_2$ and one molecule of $O_3$ 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: $$ 2 \mathrm{NO}_{(\text{g})} + 2 \mathrm{H}_{2(\text{g})} \longrightarrow \mathrm{N}_{2(\text{g})} + 2 \mathrm{H}_{2} \mathrm{O}_{(\text{g})} $$ The proposed mechanism is: * **Step I (slow):** $2 \mathrm{NO} + \mathrm{H}_{2} \longrightarrow \mathrm{N}_{2} + \mathrm{H}_{2} \mathrm{O}_{2}$ * **Step II (fast):** $\mathrm{H}_{2} \mathrm{O}_{2} + \mathrm{H}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}$ Write the experimental rate law for this reaction. **Solution:** The rate law is determined by the slow step of the mechanism. In this case, Step I is the slow (rate-determining) step. The reactants in this step are two molecules of $NO$ and one molecule of $H_2$. Therefore, the experimental rate law is: $$ \text{Rate} = k[\mathrm{NO}]^{2}[\mathrm{H}_{2}] $$ ## 7.4.1 Potential Energy Diagrams and Reaction Mechanisms A potential energy diagram visualizes the energy changes during a reaction, providing insights into its mechanism. * Each **peak** on the diagram represents the transition state of an elementary step. * The number of peaks corresponds to the **number of steps** in the mechanism. * The step with the **highest activation energy ($E_a$)** (the highest peak) is the slowest, **rate-determining step**. ### Example 7.8: Proposing a Mechanism from a PE Diagram and Rate Law For the reaction: $$ 2 \mathrm{NO}_{2(\mathrm{~g})} + \mathrm{F}_{2(\text{g})} \longrightarrow 2 \mathrm{NO}_{2} \mathrm{F}_{(\mathrm{g})} $$ The rate law is: $$ \text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}] $$ Using the potential energy diagram (Figure 7.3), propose a reaction mechanism. **Solution:** 1. **Determine the number of steps:** The potential energy diagram shows two peaks, indicating a **two-step mechanism**. 2. **Determine the rate-determining step:** The activation energy for the first step is greater than for the second step. Therefore, **Step 1 is the slow, rate-determining step**. 3. **Use the rate law to identify reactants in the slow step:** The rate law, Rate = $k[NO_2][F_2]$, indicates that one molecule of $NO_2$ and one molecule of $F_2$ are involved in the slow step. 4. **Propose the elementary steps:** * The slow step involves one $NO_2$ and one $F_2$ molecule. They react to form one product molecule ($NO_2F$) and a reaction intermediate (in this case, an F atom). * The fast second step involves the reaction intermediate (F) reacting with the remaining reactant ($NO_2$) to form the other product molecule ($NO_2F$). The proposed mechanism is: * **Step I (slow):** $\mathrm{NO}_{2} + \mathrm{F}_{2} \longrightarrow \mathrm{NO}_{2} \mathrm{F} + \mathrm{F}$ * **Step II (fast):** $\mathrm{NO}_{2} + \mathrm{F} \longrightarrow \mathrm{NO}_{2} \mathrm{F}$ 5. **Verify the overall reaction:** Adding the two steps together and canceling the intermediate (F) gives the overall reaction. $$ \begin{aligned} \mathrm{NO}_{2} + \mathrm{F}_{2} &\longrightarrow \mathrm{NO}_{2} \mathrm{F} + \cancel{\mathrm{F}} \\ \mathrm{NO}_{2} + \cancel{\mathrm{F}} &\longrightarrow \mathrm{NO}_{2} \mathrm{F} \\ \hline 2 \mathrm{NO}_{2} + \mathrm{F}_{2} &\longrightarrow 2 \mathrm{NO}_{2} \mathrm{F} \end{aligned} $$ Since the sum of the elementary steps matches the overall reaction, the proposed mechanism is plausible. ## Possible Questions/Answers **Q:** For the reaction $2 \mathrm{NO}_{2(g)}+\mathrm{F}_{2(g)} \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F}_{(g)}$ with the rate law $\text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}]$, which of the following mechanisms is consistent? (a) $2 \mathrm{NO}_{2}+\mathrm{F}_{2} \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F}$ (single step) (b) $\mathrm{NO}_{2}+\mathrm{F}_{2} \longrightarrow \mathrm{NO}_{2} \mathrm{~F}+\mathrm{F}$ (fast); $\mathrm{NO}_{2}+\mathrm{F} \longrightarrow \mathrm{NO}_{2} \mathrm{~F}$ (slow) (c) $\mathrm{NO}_{2}+\mathrm{F}_{2} \longrightarrow \mathrm{NO}_{2} \mathrm{~F}+\mathrm{F}$ (slow); $\mathrm{NO}_{2}+\mathrm{F} \longrightarrow \mathrm{NO}_{2} \mathrm{~F}$ (fast) (d) $\mathrm{F}_{2} \longrightarrow 2 \mathrm{~F}$ (slow); $2 \mathrm{NO}_{2}+2 \mathrm{~F} \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F}$ (fast) **A:** The correct mechanism is **(c)**. The rate law $\text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}]$ implies that the slow, rate-determining step must involve one molecule of $NO_2$ and one molecule of $F_2$. Mechanism (c) is the only option where the slow step has these reactants.

7.4 The Mechanism Of A Chemical Reaction | Chemical Kinetics | Chemistry 11