Reaction Mechanism Practice Questions with Answers

A reaction mechanism is a step-by-step description of the path by which reactants are converted into products, detailing the individual elementary steps and reactive intermediates involved. Understanding these processes is critical for chemists to predict the rate of a reaction and the influence of different variables, such as concentration and temperature. By breaking down a complex overall reaction into its constituent parts, we can identify the rate-determining step and apply principles from chemical kinetics to optimize industrial and biological processes.

Concept Explanation

A reaction mechanism provides a theoretical framework that explains how chemical bonds break and form at the molecular level. Unlike an overall balanced chemical equation, which only shows the starting materials and final products, a mechanism reveals the "elementary steps"—individual molecular events like collisions or bond dissociations. Each elementary step has its own molecularity, typically classified as unimolecular (one reactant molecule) or bimolecular (two reactant molecules). Termolecular steps are extremely rare because the probability of three molecules colliding simultaneously with the correct orientation is statistically low.

Key components of a reaction mechanism include:

• Intermediates: Species that are produced in one step and consumed in a subsequent step. They do not appear in the overall balanced equation.

• Catalysts: Species that are consumed in an early step and regenerated in a later step. They speed up the reaction without being permanently altered.

• Rate-Determining Step (RDS): The slowest elementary step in a mechanism. The rate law of the overall reaction is governed by the stoichiometry of the reactants in this specific step.

• Transition States: High-energy, unstable arrangements of atoms that exist briefly at the peak of the activation energy barrier for each step.

To validate a proposed mechanism, it must satisfy two conditions: the sum of the elementary steps must equal the overall balanced equation, and the mechanism must predict a rate law that matches experimental data. This often involves techniques like calculating activation energy using the Arrhenius equation or analyzing the geometry of molecules as seen in VSEPR geometry practice questions.

Solved Examples

These examples demonstrate how to derive rate laws and identify intermediates within a reaction mechanism.

Example 1: Identifying Intermediates and the Overall Equation
Given the following mechanism:
Step 1: NO₂ + NO₂ → NO₃ + NO (Slow)
Step 2: NO₃ + CO → NO₂ + CO₂ (Fast)

1. Identify the intermediate: NO₃ is produced in Step 1 and consumed in Step 2.

2. Determine the overall reaction: Add the steps and cancel common species. (NO₂ + NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂). This simplifies to: NO₂ + CO → NO + CO₂.

3. Determine the rate law: Since Step 1 is the rate-determining step, the rate law is Rate = k[NO₂]².

Example 2: Mechanism with a Fast Initial Step
Consider the reaction: 2NO + Br₂ → 2NOBr. Proposed mechanism:
Step 1: NO + Br₂ ⇌ NOBr₂ (Fast, equilibrium)
Step 2: NOBr₂ + NO → 2NOBr (Slow)

1. The RDS is Step 2, so Rate = k₂[NOBr₂][NO].

2. Because NOBr₂ is an intermediate, we use the equilibrium in Step 1 to express it in terms of reactants: K_eq = [NOBr₂] / ([NO][Br₂]), so [NOBr₂] = K_eq[NO][Br₂].

3. Substitute into the rate law: Rate = k₂(K_eq[NO][Br₂])[NO] = k[NO]²[Br₂].

Example 3: Decomposition of Ozone
Reaction: 2O₃ → 3O₂. Mechanism:
Step 1: O₃ ⇌ O₂ + O (Fast equilibrium)
Step 2: O + O₃ → 2O₂ (Slow)

1. Identify the RDS: Step 2. Rate = k₂[O][O₃].

2. Substitute the intermediate [O]: From Step 1, [O] = K_eq[O₃] / [O₂].

3. Final rate law: Rate = k[O₃]²[O₂]⁻¹. This shows that O₂ actually inhibits the reaction.

Practice Questions

Test your understanding of reaction mechanisms with the following problems.

1. The reaction 2A + B → C has a proposed mechanism of A + B → D (slow) and D + A → C (fast). What is the predicted rate law?

2. In the mechanism Step 1: H₂O₂ + I⁻ → H₂O + IO⁻ (slow) and Step 2: H₂O₂ + IO⁻ → H₂O + O₂ + I⁻ (fast), identify the catalyst and the intermediate.

3. A reaction has the rate law Rate = k[X][Y]. If the overall reaction is X + Y + Z → P, is a single-step mechanism possible? Why or why not?

4. Consider the reaction 2NO + 2H₂ → N₂ + 2H₂O. The experimental rate law is Rate = k[NO]²[H₂]. Is the mechanism Step 1: NO + H₂ → N + H₂O (slow) consistent with this?

5. Define molecularity and explain why termolecular reactions are rare in nature.

6. Given Step 1: Cl₂ → 2Cl (fast equilibrium) and Step 2: Cl + CHCl₃ → HCl + CCl₃ (slow), derive the rate law for the formation of HCl.

7. Explain the difference between a transition state and a reaction intermediate using a potential energy diagram.

8. If a reaction rate doubles when the concentration of a reactant is doubled, and the reaction occurs in a single elementary step, what is the molecularity of that step?

9. How does the presence of a catalyst change the reaction mechanism and the activation energy?

10. A reaction mechanism consists of three steps. The first step is the slowest. Does the concentration of reactants in the third step affect the overall rate law?

Answers & Explanations

1. Rate = k[A][B]. The rate law is determined by the stoichiometry of the slow step (Step 1). Since Step 1 involves one molecule of A and one of B, the rate law is first order in both.

2. Catalyst: I⁻; Intermediate: IO⁻. I⁻ is consumed in Step 1 and regenerated in Step 2. IO⁻ is produced in Step 1 and consumed in Step 2.

3. No. A single-step mechanism would require the rate law to be Rate = k[X][Y][Z]. Since Z is missing from the experimental rate law, the reaction must occur in multiple steps where Z is involved after the rate-determining step.

4. No. The proposed slow step would result in Rate = k[NO][H₂]. This does not match the experimental rate law of Rate = k[NO]²[H₂].

5. Molecularity is the number of molecules colliding in an elementary step. Termolecular reactions are rare because the probability of three particles colliding at the exact same time and with the correct orientation is extremely low.

6. Rate = k[Cl₂]¹/²[CHCl₃]. From Step 1, [Cl]² = K[Cl₂], so [Cl] = K¹/²[Cl₂]¹/². Substituting into the slow step (Rate = k₂[Cl][CHCl₃]) gives the final rate law.

7. A transition state is a local maximum on the energy profile (unstable, cannot be isolated), while an intermediate is a local minimum (relatively stable, exists for a finite time).

8. Unimolecular. If doubling the concentration doubles the rate, the reaction is first order. In a single elementary step, the order equals the molecularity, hence it is unimolecular.

9. A catalyst provides an alternative mechanism with a lower activation energy. It does not lower the energy of the existing path but creates a new, faster path. This relates to how electronic structures change, which you can explore in Lewis structure practice questions.

10. No. The rate-determining step is the first step. Any steps occurring after the RDS do not influence the overall rate of the reaction.

Quick Quiz

Frequently Asked Questions

What is the rate-determining step?

The rate-determining step is the slowest elementary step in a reaction mechanism that limits the overall speed of the chemical process. It functions like a bottleneck in a funnel, where the overall flow is restricted by the narrowest point.

Can a reaction mechanism be proven?

A reaction mechanism cannot be proven absolutely; it can only be supported by experimental evidence. If a proposed mechanism predicts a rate law that matches experimental data, it is considered a plausible explanation for how the reaction occurs.

How do intermediates differ from catalysts?

Intermediates are produced in one step and consumed in a later step, while catalysts are present at the start, consumed, and then regenerated by the end. Consequently, neither appears in the final balanced chemical equation, but for different reasons.

Why is the molecularity of an elementary step usually one or two?

Molecularity refers to the number of molecules colliding; unimolecular and bimolecular collisions are statistically common. Termolecular collisions (three molecules) are exceptionally rare because they require simultaneous timing and precise orientation of three independent particles.

How does temperature affect the reaction mechanism?

Temperature increases the kinetic energy of molecules, leading to more frequent and energetic collisions. While it doesn't usually change the mechanism itself, it increases the rate constants of the elementary steps, particularly the rate-determining step, as described by the Arrhenius Law.

What is a reaction profile?

A reaction profile is a graph showing the energy changes during a chemical reaction, plotting potential energy against the reaction coordinate. It visually depicts the activation energy, transition states, intermediates, and the overall change in enthalpy (ΔH) for the process.

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