Introduction to Reaction Mechanisms - AP Chemistry

1. For the net reaction $\mathrm{2\,H_2O_2(aq) \rightarrow 2\,H_2O(l) + O_2(g)}$, a mechanism is proposed: Step 1: $\mathrm{H_2O_2(aq) + I^-(aq) \rightarrow H_2O(l) + IO^-(aq)}$ Step 2: $\mathrm{H_2O_2(aq) + IO^-(aq) \rightarrow H_2O(l) + O_2(g) + I^-(aq)}$ Which statement correctly identifies a catalyst in the mechanism?
   Answer: $\mathrm{I^-(aq)}$ is a catalyst because it is consumed in Step 1 and regenerated in Step 2. This question assesses the introduction to reaction mechanisms. Mechanisms involve steps where catalysts are consumed and then regenerated, appearing in the steps but not in the overall reaction. Adding the steps here yields 2 H₂O₂(aq) + I⁻(aq) + IO⁻(aq) → 2 H₂O(l) + IO⁻(aq) + O₂(g) + I⁻(aq), with I⁻(aq) and IO⁻(aq) canceling to give 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g). Thus, I⁻(aq) is the catalyst as it is used in Step 1 and reformed in Step 2. Choice A fails because IO⁻(aq) is an intermediate, not a catalyst, as it is produced and then consumed without regeneration at the start. When analyzing mechanisms, distinguish catalysts by checking if they are regenerated after being consumed, unlike intermediates which are not.
2. A proposed mechanism for the overall reaction $2\text{H}_2\text{O}_2(aq)\rightarrow 2\text{H}_2\text{O}(l)+\text{O}_2(g)$ is: Step 1: $\text{H}_2\text{O}_2(aq)+\text{I}^-(aq)\rightarrow \text{H}_2\text{O}(l)+\text{IO}^-(aq)$ Step 2: $\text{H}_2\text{O}_2(aq)+\text{IO}^-(aq)\rightarrow \text{H}_2\text{O}(l)+\text{O}_2(g)+\text{I}^-(aq)$ Which species is a catalyst in the mechanism?
   Answer: $\text{I}^-(aq)$ because it is consumed in Step 1 and regenerated in Step 2. This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, steps combine to give the overall reaction, where catalysts are consumed in one step and regenerated in another, not appearing in the net. Here, I- is consumed in Step 1 to form IO- and regenerated in Step 2, acting as a catalyst. The steps sum to 2H2O2 → 2H2O + O2 after canceling IO-, confirming the role. Choice A fails because IO- is produced in Step 1 and consumed in Step 2, making it an intermediate, not a catalyst. A transferable strategy is that catalysts appear in steps but not in the overall reaction, as they are regenerated.
3. A proposed mechanism for the reaction $\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}$ is shown below: Step 1: $\mathrm{NO(g) + NO(g) \rightarrow N_2O_2(g)}$ Step 2: $\mathrm{N_2O_2(g) + O_2(g) \rightarrow 2NO_2(g)}$ Which species is an intermediate in this mechanism?​
   Answer: $\mathrm{N_2O_2(g)}$ This question tests understanding of introduction to reaction mechanisms. In a reaction mechanism, intermediates are species that are produced in one step and consumed in another step, never appearing in the overall reaction. Looking at the mechanism, N₂O₂(g) is produced in Step 1 when two NO molecules combine, and then consumed in Step 2 when it reacts with O₂ to form the final product. When we add the two steps together and cancel species that appear on both sides, N₂O₂ cancels out, confirming it's an intermediate. Choice E is incorrect because it lists reactants from the overall equation, not an intermediate species. Remember: intermediates appear in the mechanism steps but not in the overall reaction equation.
4. A proposed mechanism for the overall reaction $\text{O}_3(g)+\text{O}(g)\rightarrow 2\text{O}_2(g)$ is: Step 1: $\text{O}_3(g)+\text{Cl}(g)\rightarrow \text{ClO}(g)+\text{O}_2(g)$ Step 2: $\text{ClO}(g)+\text{O}(g)\rightarrow \text{Cl}(g)+\text{O}_2(g)$ Which species acts as a catalyst in the mechanism?
   Answer: $\text{Cl}(g)$ because it is consumed in Step 1 and regenerated in Step 2. This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, steps combine to yield the overall, with catalysts consumed and regenerated, not in the net. Here, Cl is consumed in Step 1 with O3 and regenerated in Step 2 from ClO. The steps add to O3 + O → 2O2 after canceling ClO and Cl. Choice A fails because ClO is produced in Step 1 and consumed in Step 2, making it an intermediate, not a catalyst. A transferable strategy is that catalysts appear in steps but not in the overall reaction, as they are regenerated.
5. A mechanism is proposed for the reaction $\mathrm{2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)}$ in the presence of iodide ion: Step 1: $\mathrm{H_2O_2(aq) + I^-(aq) \rightarrow H_2O(l) + IO^-(aq)}$ Step 2: $\mathrm{H_2O_2(aq) + IO^-(aq) \rightarrow H_2O(l) + O_2(g) + I^-(aq)}$ Which species acts as a catalyst in this mechanism?​
   Answer: $\mathrm{I^-(aq)}$ This question tests understanding of introduction to reaction mechanisms, specifically identifying catalysts. A catalyst is a species that is consumed in one step and regenerated in a later step, appearing unchanged in the overall reaction. In this mechanism, I⁻(aq) is consumed in Step 1 when it reacts with H₂O₂ to form IO⁻, and then I⁻ is regenerated in Step 2 when IO⁻ reacts with another H₂O₂ molecule. Since I⁻ is present at both the beginning and end of the reaction without being permanently consumed, it acts as a catalyst. Choice A (IO⁻) is incorrect because it's an intermediate that's produced then consumed, not a catalyst. Remember: catalysts are consumed then regenerated, appearing unchanged overall.
6. A proposed mechanism for the net reaction $\mathrm{2\,NO(g) + O_2(g) \rightarrow 2\,NO_2(g)}$ is: Step 1: $\mathrm{NO(g) + NO(g) \rightleftharpoons N_2O_2(g)}$ Step 2: $\mathrm{N_2O_2(g) + O_2(g) \rightarrow 2\,NO_2(g)}$ Which statement correctly identifies the catalyst or intermediate role of $\mathrm{N_2O_2(g)}$ in the mechanism?
   Answer: $\mathrm{N_2O_2(g)}$ is an intermediate because it is formed in Step 1 and consumed in Step 2. This question assesses the introduction to reaction mechanisms. Mechanisms include steps where species like intermediates are formed and consumed, differing from catalysts which are regenerated. The steps sum to NO(g) + NO(g) + N₂O₂(g) + O₂(g) → N₂O₂(g) + 2 NO₂(g), canceling N₂O₂(g) to give 2 NO(g) + O₂(g) → 2 NO₂(g). Thus, N₂O₂(g) is an intermediate as it is produced in Step 1 and consumed in Step 2. Choice A is wrong because N₂O₂(g) is not regenerated, so it is not a catalyst. To differentiate roles, note that intermediates appear in steps but not in the overall reaction, while catalysts are restored at the end.
7. A student proposes the following mechanism for the net reaction $\mathrm{2NO(g) + Cl_2(g) \rightarrow 2NOCl(g)}$: Step 1: $\mathrm{NO(g) + Cl_2(g) \rightarrow NOCl_2(g)}$ Step 2: $\mathrm{NOCl_2(g) \rightarrow NOCl(g) + Cl(g)}$ Step 3: $\mathrm{Cl(g) + NO(g) \rightarrow NOCl(g)}$ Which statement best describes the overall consistency of the mechanism with the net reaction?​
   Answer: The mechanism is consistent because adding the steps cancels $\mathrm{NOCl_2(g)}$ and $\mathrm{Cl(g)}$ to give the net reaction. This question tests understanding of introduction to reaction mechanisms and verifying mechanism consistency. To check consistency, we must add all steps and verify they sum to the overall reaction. Step 1: NO + Cl₂ → NOCl₂; Step 2: NOCl₂ → NOCl + Cl; Step 3: Cl + NO → NOCl. Adding these steps: NO + Cl₂ + NOCl₂ + Cl + NO → NOCl₂ + NOCl + Cl + NOCl. Canceling species that appear on both sides (NOCl₂ and Cl), we get: 2NO + Cl₂ → 2NOCl, which matches the given net reaction. Choice A is incorrect because Cl(g) is an intermediate that properly cancels out, not an inconsistency. The key strategy is to add all steps algebraically and verify that intermediates cancel to yield the net reaction.
8. A proposed mechanism for the overall reaction $\text{2NO}(g)+\text{O}_2(g)\rightarrow 2\text{NO}_2(g)$ is: Step 1: $\text{NO}(g)+\text{O}_2(g)\rightarrow \text{NO}_3(g)$ Step 2: $\text{NO}_3(g)+\text{NO}(g)\rightarrow 2\text{NO}_2(g)$ Which statement is correct about the mechanism?
   Answer: The mechanism is consistent because $\text{NO}_3(g)$ cancels when the steps are added. This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, the steps must sum to the overall reaction after canceling intermediates. Here, NO3 is produced in Step 1 and consumed in Step 2, serving as an intermediate. Adding the steps yields 2NO + O2 → 2NO2 after NO3 cancels, matching the overall. Choice A fails because NO3 does not appear in the net reaction; it is an intermediate that cancels. A transferable strategy is that intermediates appear in steps but not in the overall reaction.
9. A student proposes the following mechanism for the overall reaction $\text{Cl}_2(g)+\text{CH}_4(g)\rightarrow \text{CH}_3\text{Cl}(g)+\text{HCl}(g)$: Step 1: $\text{Cl}_2(g)\rightarrow 2\text{Cl}(g)$ Step 2: $\text{Cl}(g)+\text{CH}_4(g)\rightarrow \text{HCl}(g)+\text{CH}_3(g)$ Step 3: $\text{CH}_3(g)+\text{Cl}_2(g)\rightarrow \text{CH}_3\text{Cl}(g)+\text{Cl}(g)$ Which species is an intermediate in the mechanism?
   Answer: $\text{Cl}(g)$ because it is produced in Step 1 and consumed in Step 2 while also being regenerated in Step 3. This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, steps add to the overall after canceling intermediates, which are produced and consumed within the mechanism. Here, Cl is produced in Step 1, consumed in Step 2, and regenerated in Step 3, functioning as an intermediate in this chain. The propagation steps (2 and 3) sum to CH4 + Cl2 → CH3Cl + HCl, consistent with overall. Choice E fails because CH4 is a reactant in the net reaction, not an intermediate. A transferable strategy is that intermediates appear in steps but not in the overall reaction.
10. A mechanism is proposed for the net reaction $\mathrm{CH_3Br(aq) + OH^-(aq) \rightarrow CH_3OH(aq) + Br^-(aq)}$: Step 1: $\mathrm{CH_3Br(aq) \rightarrow CH_3^+(aq) + Br^-(aq)}$ Step 2: $\mathrm{CH_3^+(aq) + OH^-(aq) \rightarrow CH_3OH(aq)}$ Which species is an intermediate in the mechanism?
    Answer: $\mathrm{CH_3^+(aq)}$ This question assesses the introduction to reaction mechanisms. Reaction steps in a mechanism add up to the overall process, featuring intermediates that are temporary and cancel out. Adding the steps results in CH₃Br(aq) + CH₃⁺(aq) + OH⁻(aq) → CH₃⁺(aq) + Br⁻(aq) + CH₃OH(aq), canceling CH₃⁺(aq) to give CH₃Br(aq) + OH⁻(aq) → CH₃OH(aq) + Br⁻(aq). Hence, CH₃⁺(aq) is the intermediate produced in Step 1 and consumed in Step 2. Choice A fails because Br⁻(aq) is a product in the overall reaction, not an intermediate. Intermediates can be spotted by their appearance in mechanism steps but absence from the overall balanced reaction.
11. A student proposes the following multistep mechanism for the net reaction $\mathrm{2\,O_3(g) \rightarrow 3\,O_2(g)}$: Step 1: $\mathrm{O_3(g) \rightarrow O_2(g) + O(g)}$ Step 2: $\mathrm{O(g) + O_3(g) \rightarrow 2\,O_2(g)}$ Which species is an intermediate in this mechanism?
    Answer: $\mathrm{O(g)}$ This question assesses the introduction to reaction mechanisms. Reaction mechanisms consist of elementary steps that add up to the overall reaction, with intermediates being transient species not present in the net equation. In this case, summing Step 1 and Step 2 results in O₃(g) + O(g) + O₃(g) → O₂(g) + O(g) + 2 O₂(g), and canceling O(g) gives 2 O₃(g) → 3 O₂(g). Therefore, O(g) is the intermediate as it is generated in the first step and consumed in the second. Choice A is incorrect because O₂(g) is a product in the overall reaction, not an intermediate. A useful strategy is to verify mechanisms by ensuring intermediates cancel out when steps are added to match the overall reaction.
12. A proposed mechanism for the formation of $\mathrm{NO_2}$ is shown below: Step 1: $\mathrm{NO + O_2 \rightarrow NO_3}$ Step 2: $\mathrm{NO_3 + NO \rightarrow 2NO_2}$ Which species is an intermediate in the mechanism?​​
    Answer: $\mathrm{NO_3}$ is an intermediate because it is produced in Step 1 and consumed in Step 2. This question tests understanding of introduction to reaction mechanisms, specifically identifying intermediates. An intermediate is produced in one step and consumed in another, never appearing in the overall reaction. In this mechanism, NO₃ is produced in Step 1 (NO + O₂ → NO₃) and consumed in Step 2 (NO₃ + NO → 2NO₂), making it an intermediate. Choice E is incorrect because intermediates are not catalysts—catalysts are consumed then regenerated to their original form, while intermediates are produced then consumed. Adding the steps gives 2NO + O₂ → 2NO₂ as the net reaction, with NO₃ canceling out completely. Remember: intermediates appear in the mechanism steps but cancel out in the overall reaction.
13. A student suggests the following mechanism for forming hydrogen iodide: Step 1: $\mathrm{I_2(g) \rightarrow 2\,I(g)}$ Step 2: $\mathrm{I(g) + H_2(g) \rightarrow HI(g) + H(g)}$ Step 3: $\mathrm{H(g) + I(g) \rightarrow HI(g)}$ The student claims the net reaction is $\mathrm{H_2(g) + I_2(g) \rightarrow 2\,HI(g)}$. Which statement best assesses the consistency of the mechanism with the claimed net reaction?
    Answer: The mechanism is consistent because intermediates cancel to give $\mathrm{H_2(g) + I_2(g) \rightarrow 2\,HI(g)}$. This question assesses the introduction to reaction mechanisms. A consistent mechanism has steps that sum to the overall reaction, with intermediates canceling out appropriately. Adding the steps results in I₂(g) + I(g) + H₂(g) + H(g) + I(g) → 2 I(g) + HI(g) + H(g) + HI(g), and canceling 2 I(g) and H(g) gives H₂(g) + I₂(g) → 2 HI(g). This shows the mechanism is consistent as I(g) and H(g) are intermediates that balance out. Choice C fails because H(g) is produced in Step 2 and consumed in Step 3, so it does cancel properly. To evaluate mechanism consistency, add the steps and ensure all intermediates cancel to match the claimed net reaction.
14. A proposed mechanism for the decomposition of ozone is shown below: Step 1: $\mathrm{O_3 + Cl \rightarrow ClO + O_2}$ Step 2: $\mathrm{ClO + O \rightarrow Cl + O_2}$ Based on the mechanism, which species acts as a catalyst?​​
    Answer: $\mathrm{Cl}$ acts as a catalyst because it is consumed in Step 1 and regenerated in Step 2. This question tests understanding of introduction to reaction mechanisms, specifically identifying catalysts. A catalyst is a species that is consumed in an early step and regenerated in a later step, appearing unchanged after the reaction. In this mechanism, Cl is consumed in Step 1 (O₃ + Cl → ClO + O₂) and regenerated in Step 2 (ClO + O → Cl + O₂), making it a catalyst. Choice B is incorrect because ClO is an intermediate (produced then consumed), not a catalyst—catalysts must be regenerated to their original form. When adding the steps together, Cl cancels out, and the net reaction is O₃ + O → 2O₂, confirming Cl acts catalytically. Remember: catalysts are consumed then regenerated, while intermediates are produced then consumed.
15. A proposed mechanism for the overall reaction $\text{2SO}_2(g)+\text{O}_2(g)\rightarrow 2\text{SO}_3(g)$ is: Step 1: $\text{SO}_2(g)+\text{O}_2(g)\rightarrow \text{SO}_4(g)$ Step 2: $\text{SO}_4(g)+\text{SO}_2(g)\rightarrow 2\text{SO}_3(g)$ Which species is an intermediate in the mechanism?
    Answer: $\text{SO}_4(g)$ because it is produced in Step 1 and consumed in Step 2. This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, the elementary steps add to the overall, with intermediates formed and consumed internally. Here, SO4 is produced in Step 1 from SO2 and O2 and consumed in Step 2 with another SO2. The steps sum to 2SO2 + O2 → 2SO3 after canceling SO4. Choice A fails because SO2 is a reactant in the net reaction, not an intermediate. A transferable strategy is that intermediates appear in steps but not in the overall reaction.
16. A proposed mechanism for the reaction $2\text{NO}(g)+\text{Br}_2(g)\rightarrow 2\text{NOBr}(g)$ is: Step 1: $\text{NO}(g)+\text{Br}_2(g)\rightarrow \text{NOBr}_2(g)$ Step 2: $\text{NOBr}_2(g)+\text{NO}(g)\rightarrow 2\text{NOBr}(g)$ Which species is an intermediate in this mechanism?
    Answer: $\text{NOBr}_2(g)$ because it is produced in Step 1 and consumed in Step 2. This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, the steps must add up to the overall reaction, with intermediates being species that are generated and then depleted, not appearing in the net. Here, NOBr2 is produced in Step 1 from NO and Br2 and consumed in Step 2 with another NO. The steps sum to 2NO + Br2 → 2NOBr after canceling NOBr2. Choice C fails because NO is a reactant in the net reaction, not an intermediate. A transferable strategy is that intermediates appear in steps but not in the overall reaction.
17. A reaction is proposed to occur by the following mechanism: Step 1: $\mathrm{ClO(g) + O_3(g) \rightarrow ClO_2(g) + O_2(g)}$ Step 2: $\mathrm{ClO_2(g) + O(g) \rightarrow ClO(g) + O_2(g)}$ Which statement correctly identifies the catalyst for the overall reaction?
    Answer: $\mathrm{ClO(g)}$ is the catalyst because it is consumed in Step 1 and regenerated in Step 2. This question tests understanding of introduction to reaction mechanisms. A catalyst is a species that is consumed in an early step of a mechanism and regenerated in a later step, appearing unchanged after the reaction is complete. In this mechanism, ClO(g) is consumed as a reactant in Step 1 and regenerated as a product in Step 2, making it the catalyst. When we add the steps together, ClO(g) cancels out and doesn't appear in the overall reaction, which is characteristic of a catalyst. Choice C incorrectly identifies ClO₂(g) as a catalyst, but ClO₂(g) is actually an intermediate because it's produced in Step 1 and consumed in Step 2. To identify catalysts in mechanisms, look for species that are consumed early and regenerated later—they speed up the reaction without being consumed overall.
18. A proposed mechanism for the overall reaction $\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}$ is: Step 1: $\mathrm{NO(g) + O_2(g) \rightarrow NO_3(g)}$ Step 2: $\mathrm{NO_3(g) + NO(g) \rightarrow 2NO_2(g)}$ Which statement best describes the consistency of the mechanism with the net reaction?​​​
    Answer: The mechanism is consistent because adding the steps cancels $\mathrm{NO_3(g)}$ and gives the net reaction. This question tests understanding of introduction to reaction mechanisms, specifically verifying mechanism consistency. To check consistency, we add all steps and cancel species appearing on both sides. Step 1: NO(g) + O₂(g) → NO₃(g); Step 2: NO₃(g) + NO(g) → 2NO₂(g). Adding these gives: 2NO(g) + O₂(g) + NO₃(g) → NO₃(g) + 2NO₂(g). NO₃(g) appears on both sides and cancels, leaving 2NO(g) + O₂(g) → 2NO₂(g), which matches the given overall reaction. Choice B is incorrect because NO₃(g) is an intermediate that cancels out, not a reactant in the net reaction. The strategy for verifying mechanisms is to add all steps algebraically and confirm that intermediates cancel to give the stated overall reaction.
19. A proposed mechanism for the reaction $\mathrm{2NO(g) + Br_2(g) \rightarrow 2NOBr(g)}$ is shown. Step 1: $\mathrm{NO(g) + Br_2(g) \rightarrow NOBr_2(g)}$ Step 2: $\mathrm{NOBr_2(g) + NO(g) \rightarrow 2NOBr(g)}$ Which species is an intermediate in the mechanism?
    Answer: $\mathrm{NOBr_2(g)}$ This question tests understanding of introduction to reaction mechanisms. Intermediates are species that are produced in one step and consumed in another step, not appearing in the overall reaction. In this mechanism, NOBr₂(g) is produced in Step 1 and consumed in Step 2, making it an intermediate. When we add the two steps together and cancel species that appear on both sides, NOBr₂(g) cancels out, confirming it doesn't appear in the overall reaction 2NO(g) + Br₂(g) → 2NOBr(g). Choice A (NOBr) is incorrect because NOBr is the final product that appears in the overall reaction, not an intermediate. To identify intermediates in any mechanism, look for species that are formed in one step and used up in another—they serve as temporary species that facilitate the reaction.
20. A mechanism is proposed for the decomposition of ozone: Step 1: $\mathrm{O_3(g) \rightarrow O_2(g) + O(g)}$ Step 2: $\mathrm{O(g) + O_3(g) \rightarrow 2O_2(g)}$ Which statement correctly identifies a catalyst or intermediate for this mechanism?​
    Answer: $\mathrm{O(g)}$ is an intermediate because it is produced in Step 1 and consumed in Step 2. This question tests understanding of introduction to reaction mechanisms, specifically distinguishing between intermediates and catalysts. An intermediate is produced in one step and consumed in another, while a catalyst is consumed in one step and regenerated in another. In this mechanism, O(g) is produced in Step 1 when O₃ decomposes, and then O(g) is consumed in Step 2 when it reacts with another O₃ molecule. Since O(g) doesn't appear in the overall reaction (2O₃ → 3O₂), it's an intermediate. Choice D incorrectly identifies O(g) as a catalyst, but catalysts must be present at both the beginning and end of the reaction. Remember: intermediates are produced then consumed; catalysts are consumed then regenerated.

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