AP Chemistry Quiz: Introduction To Reaction Mechanisms
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Introduction To Reaction MechanismsQuestion 1 of 20

A student proposes the following mechanism for the overall reaction Cl2(g)+CH4(g)→CH3Cl(g)+HCl(g)\text{Cl}_2(g)+\text{CH}_4(g)\rightarrow \text{CH}_3\text{Cl}(g)+\text{HCl}(g):

Step 1: Cl2(g)→2Cl(g)\text{Cl}_2(g)\rightarrow 2\text{Cl}(g) Step 2: Cl(g)+CH4(g)→HCl(g)+CH3(g)\text{Cl}(g)+\text{CH}_4(g)\rightarrow \text{HCl}(g)+\text{CH}_3(g) Step 3: CH3(g)+Cl2(g)→CH3Cl(g)+Cl(g)\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?

Cl2(g)\text{Cl}_2(g) because it is consumed in Step 1 and Step 3.
CH3Cl(g)\text{CH}_3\text{Cl}(g) because it is produced in Step 3.
Cl(g)\text{Cl}(g) because it is produced in Step 1 and consumed in Step 2 while also being regenerated in Step 3.
HCl(g)\text{HCl}(g) because it is produced in Step 2.
CH4(g)\text{CH}_4(g) because it is consumed in Step 2.
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AP Chemistry Quiz

AP Chemistry Quiz: Introduction To Reaction Mechanisms

Practice Introduction To Reaction Mechanisms in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Introduction To Reaction Mechanisms, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A student proposes the following mechanism for the overall reaction Cl2(g)+CH4(g)→CH3Cl(g)+HCl(g)\text{Cl}_2(g)+\text{CH}_4(g)\rightarrow \text{CH}_3\text{Cl}(g)+\text{HCl}(g):

Step 1: Cl2(g)→2Cl(g)\text{Cl}_2(g)\rightarrow 2\text{Cl}(g) Step 2: Cl(g)+CH4(g)→HCl(g)+CH3(g)\text{Cl}(g)+\text{CH}_4(g)\rightarrow \text{HCl}(g)+\text{CH}_3(g) Step 3: CH3(g)+Cl2(g)→CH3Cl(g)+Cl(g)\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?

  1. Cl2(g)\text{Cl}_2(g) because it is consumed in Step 1 and Step 3.
  2. CH3Cl(g)\text{CH}_3\text{Cl}(g) because it is produced in Step 3.
  3. Cl(g)\text{Cl}(g) because it is produced in Step 1 and consumed in Step 2 while also being regenerated in Step 3. (correct answer)
  4. HCl(g)\text{HCl}(g) because it is produced in Step 2.
  5. CH4(g)\text{CH}_4(g) because it is consumed in Step 2.

Explanation: 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.

Question 2

A proposed mechanism for the reaction H2(g)+Br2(g)→2HBr(g)\text{H}_2(g)+\text{Br}_2(g)\rightarrow 2\text{HBr}(g) is:

Step 1: Br2(g)→2Br(g)\text{Br}_2(g)\rightarrow 2\text{Br}(g) Step 2: Br(g)+H2(g)→HBr(g)+H(g)\text{Br}(g)+\text{H}_2(g)\rightarrow \text{HBr}(g)+\text{H}(g) Step 3: H(g)+Br2(g)→HBr(g)+Br(g)\text{H}(g)+\text{Br}_2(g)\rightarrow \text{HBr}(g)+\text{Br}(g)

Which species is an intermediate in the mechanism?

  1. HBr(g)\text{HBr}(g) because it is formed in Steps 2 and 3.
  2. Br2(g)\text{Br}_2(g) because it is consumed in Steps 1 and 3.
  3. H2(g)\text{H}_2(g) because it is consumed in Step 2.
  4. Br(g)\text{Br}(g) because it is produced in Step 1 and consumed in Step 2 (and regenerated in Step 3). (correct answer)
  5. H(g)\text{H}(g) because it appears in the net reaction.

Explanation: This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, steps add to the overall, with intermediates produced and consumed. Here, Br is produced in Step 1, consumed in Step 2, and regenerated in Step 3, acting as an intermediate. The propagation steps sum to H2 + Br2 → 2HBr after canceling. Choice E fails because H does not appear in the net reaction; it is an intermediate. A transferable strategy is that intermediates appear in steps but not in the overall reaction.

Question 3

A proposed mechanism for the reaction N2O5(g)→2NO2(g)+12O2(g)\text{N}_2\text{O}_5(g)\rightarrow 2\text{NO}_2(g)+\tfrac{1}{2}\text{O}_2(g) is:

Step 1: N2O5(g)→NO2(g)+NO3(g)\text{N}_2\text{O}_5(g)\rightarrow \text{NO}_2(g)+\text{NO}_3(g) Step 2: 2NO3(g)→2NO2(g)+O2(g)2\text{NO}_3(g)\rightarrow 2\text{NO}_2(g)+\text{O}_2(g)

Which statement best describes the overall consistency of the mechanism with the net reaction?

  1. The mechanism is inconsistent because NO2(g)\text{NO}_2(g) is an intermediate and should not appear in the net reaction.
  2. The mechanism is inconsistent because O2(g)\text{O}_2(g) must cancel when steps are added.
  3. The mechanism is consistent because NO3(g)\text{NO}_3(g) cancels and the summed steps yield the net reaction (after scaling). (correct answer)
  4. The mechanism is consistent only if NO3(g)\text{NO}_3(g) is included in the net reaction.
  5. The mechanism is inconsistent because NO3(g)\text{NO}_3(g) appears as a reactant in Step 2.

Explanation: This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, steps may need scaling to match the overall stoichiometry, with intermediates canceling out. Here, NO3 is produced in Step 1 and consumed in Step 2, but scaling Step 1 twice and Step 2 once cancels NO3 properly. The scaled steps sum to 2N2O5 → 4NO2 + O2, or equivalently N2O5 → 2NO2 + 1/2 O2. Choice A fails because NO3 does not appear in the net reaction; it is an intermediate. A transferable strategy is that intermediates appear in steps but not in the overall reaction.

Question 4

A proposed mechanism for the overall reaction 2NO(g)+O2(g)→2NO2(g)\text{2NO}(g)+\text{O}_2(g)\rightarrow 2\text{NO}_2(g) is:

Step 1: NO(g)+O2(g)→NO3(g)\text{NO}(g)+\text{O}_2(g)\rightarrow \text{NO}_3(g) Step 2: NO3(g)+NO(g)→2NO2(g)\text{NO}_3(g)+\text{NO}(g)\rightarrow 2\text{NO}_2(g)

Which statement is correct about the mechanism?

  1. The mechanism is inconsistent because NO3(g)\text{NO}_3(g) is produced and should appear in the net reaction.
  2. The mechanism is consistent because NO3(g)\text{NO}_3(g) cancels when the steps are added. (correct answer)
  3. The mechanism is inconsistent because O2(g)\text{O}_2(g) must be produced in one of the steps.
  4. The mechanism is inconsistent because NO(g)\text{NO}(g) cancels completely and cannot be a reactant overall.
  5. The mechanism is consistent only if NO3(g)\text{NO}_3(g) is treated as a catalyst.

Explanation: 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.

Question 5

A proposed mechanism for the reaction 2NO(g)+Br2(g)→2NOBr(g)\mathrm{2NO(g) + Br_2(g) \rightarrow 2NOBr(g)} is shown.

Step 1: NO(g)+Br2(g)→NOBr2(g)\mathrm{NO(g) + Br_2(g) \rightarrow NOBr_2(g)} Step 2: NOBr2(g)+NO(g)→2NOBr(g)\mathrm{NOBr_2(g) + NO(g) \rightarrow 2NOBr(g)}

Which species is an intermediate in the mechanism?

  1. NOBr(g)\mathrm{NOBr(g)}
  2. NO(g)\mathrm{NO(g)}
  3. Br2(g)\mathrm{Br_2(g)}
  4. NOBr2(g)\mathrm{NOBr_2(g)} (correct answer)
  5. Br(g)\mathrm{Br(g)}

Explanation: 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.

Question 6

A proposed mechanism for the reaction 2NO(g)+O2(g)→2NO2(g)\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)} is shown below:

Step 1: NO(g)+NO(g)→N2O2(g)\mathrm{NO(g) + NO(g) \rightarrow N_2O_2(g)} Step 2: N2O2(g)+O2(g)→2NO2(g)\mathrm{N_2O_2(g) + O_2(g) \rightarrow 2NO_2(g)}

Which species is an intermediate in this mechanism?​

  1. O2(g)\mathrm{O_2(g)}
  2. NO2(g)\mathrm{NO_2(g)}
  3. NO(g)\mathrm{NO(g)}
  4. N2O2(g)\mathrm{N_2O_2(g)} (correct answer)
  5. 2NO(g)+O2(g)\mathrm{2NO(g) + O_2(g)}

Explanation: 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.

Question 7

A student proposes the following mechanism for the net reaction ClO−(aq)+2 I−(aq)+2 H+(aq)→I2(aq)+Cl−(aq)+H2O(l)\mathrm{ClO^-(aq) + 2\,I^-(aq) + 2\,H^+(aq) \rightarrow I_2(aq) + Cl^-(aq) + H_2O(l)}:

Step 1: ClO−(aq)+H+(aq)→HOCl(aq)\mathrm{ClO^-(aq) + H^+(aq) \rightarrow HOCl(aq)} Step 2: HOCl(aq)+I−(aq)→HOI(aq)+Cl−(aq)\mathrm{HOCl(aq) + I^-(aq) \rightarrow HOI(aq) + Cl^-(aq)} Step 3: HOI(aq)+I−(aq)+H+(aq)→I2(aq)+H2O(l)\mathrm{HOI(aq) + I^-(aq) + H^+(aq) \rightarrow I_2(aq) + H_2O(l)}

Which species acts as an intermediate in this mechanism?

  1. HOCl(aq)\mathrm{HOCl(aq)} (correct answer)
  2. I2(aq)\mathrm{I_2(aq)}
  3. ClO−(aq)\mathrm{ClO^-(aq)}
  4. H+(aq)\mathrm{H^+(aq)}
  5. Cl−(aq)\mathrm{Cl^-(aq)}

Explanation: This question assesses the introduction to reaction mechanisms. Elementary steps form the mechanism, with intermediates being produced and consumed across them, not appearing in the net reaction. Summing the steps gives ClO⁻(aq) + H⁺(aq) + HOCl(aq) + I⁻(aq) + HOI(aq) + I⁻(aq) + H⁺(aq) → HOCl(aq) + HOI(aq) + Cl⁻(aq) + I₂(aq) + H₂O(l), canceling HOCl(aq) and HOI(aq) to yield ClO⁻(aq) + 2 I⁻(aq) + 2 H⁺(aq) → I₂(aq) + Cl⁻(aq) + H₂O(l). Therefore, HOCl(aq) is an intermediate formed in Step 1 and used in Step 2. Choice B fails because I₂(aq) is a product in the overall reaction, not an intermediate. A general strategy is to add mechanism steps and identify intermediates as those that cancel out, ensuring the net reaction matches.

Question 8

A student proposes the following mechanism for the overall reaction H2(g)+I2(g)→2HI(g)\text{H}_2(g)+\text{I}_2(g)\rightarrow 2\text{HI}(g):

Step 1: I2(g)→2I(g)\text{I}_2(g)\rightarrow 2\text{I}(g) Step 2: H2(g)+I(g)→HI(g)+H(g)\text{H}_2(g)+\text{I}(g)\rightarrow \text{HI}(g)+\text{H}(g) Step 3: H(g)+I(g)→HI(g)\text{H}(g)+\text{I}(g)\rightarrow \text{HI}(g)

Which species is an intermediate in the mechanism?

  1. HI(g)\text{HI}(g) because it is produced in Step 2 and Step 3.
  2. I2(g)\text{I}_2(g) because it is consumed in Step 1.
  3. I(g)\text{I}(g) because it is produced in Step 1 and consumed in Steps 2 and 3. (correct answer)
  4. H2(g)\text{H}_2(g) because it is consumed in Step 2.
  5. H(g)\text{H}(g) because it appears in the net reaction.

Explanation: This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, the steps combine to yield the overall reaction, with intermediates being species produced in an early step and consumed in later ones, ensuring they cancel out and do not appear in the net equation. Here, I atoms are produced in Step 1 and consumed in Steps 2 and 3, fitting the definition of an intermediate. The steps sum to H2 + I2 → 2HI after canceling I and H, confirming consistency. Choice E fails because H does not appear in the net reaction; it is actually an intermediate produced in Step 2 and consumed in Step 3. A transferable strategy is that intermediates appear in steps but not in the overall reaction.

Question 9

A proposed mechanism for the reaction 2NO(g)+Br2(g)→2NOBr(g)2\text{NO}(g)+\text{Br}_2(g)\rightarrow 2\text{NOBr}(g) is:

Step 1: NO(g)+Br2(g)→NOBr2(g)\text{NO}(g)+\text{Br}_2(g)\rightarrow \text{NOBr}_2(g) Step 2: NOBr2(g)+NO(g)→2NOBr(g)\text{NOBr}_2(g)+\text{NO}(g)\rightarrow 2\text{NOBr}(g)

Which species is an intermediate in this mechanism?

  1. NOBr(g)\text{NOBr}(g) because it is produced in Step 2.
  2. Br2(g)\text{Br}_2(g) because it is consumed in Step 1.
  3. NO(g)\text{NO}(g) because it is consumed in Step 1 and Step 2.
  4. NOBr2(g)\text{NOBr}_2(g) because it is produced in Step 1 and consumed in Step 2. (correct answer)
  5. Br(g)\text{Br}(g) because it is implied by the presence of bromine.

Explanation: 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.

Question 10

A proposed mechanism for the overall reaction O3(g)+O(g)→2O2(g)\text{O}_3(g)+\text{O}(g)\rightarrow 2\text{O}_2(g) is:

Step 1: O3(g)+Cl(g)→ClO(g)+O2(g)\text{O}_3(g)+\text{Cl}(g)\rightarrow \text{ClO}(g)+\text{O}_2(g) Step 2: ClO(g)+O(g)→Cl(g)+O2(g)\text{ClO}(g)+\text{O}(g)\rightarrow \text{Cl}(g)+\text{O}_2(g)

Which species acts as a catalyst in the mechanism?

  1. O3(g)\text{O}_3(g) because it is consumed in Step 1.
  2. O(g)\text{O}(g) because it is consumed in Step 2.
  3. Cl(g)\text{Cl}(g) because it is consumed in Step 1 and regenerated in Step 2. (correct answer)
  4. O2(g)\text{O}_2(g) because it is produced in both steps.
  5. ClO(g)\text{ClO}(g) because it is produced in Step 1 and consumed in Step 2.

Explanation: 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.

Question 11

Consider the proposed mechanism for the reaction CO(g)+NO2(g)→CO2(g)+NO(g)\text{CO}(g)+\text{NO}_2(g)\rightarrow \text{CO}_2(g)+\text{NO}(g):

Step 1: NO2(g)→NO(g)+O(g)\text{NO}_2(g)\rightarrow \text{NO}(g)+\text{O}(g) Step 2: CO(g)+O(g)→CO2(g)\text{CO}(g)+\text{O}(g)\rightarrow \text{CO}_2(g)

Which species is an intermediate in the mechanism?

  1. CO(g)\text{CO}(g) because it is consumed in Step 2.
  2. CO2(g)\text{CO}_2(g) because it is produced in Step 2.
  3. NO(g)\text{NO}(g) because it is produced overall.
  4. NO2(g)\text{NO}_2(g) because it is consumed in Step 1.
  5. O(g)\text{O}(g) because it is produced in Step 1 and consumed in Step 2. (correct answer)

Explanation: This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, steps sum to the overall reaction, with intermediates being transient species formed in one step and used in another, absent from the net equation. Here, O is produced from NO2 in Step 1 and consumed with CO in Step 2 to form CO2. The steps add to CO + NO2 → CO2 + NO after O cancels, matching the overall. Choice E fails because NO is a product in the net reaction, not an intermediate. A transferable strategy is that intermediates appear in steps but not in the overall reaction.

Question 12

A proposed mechanism for the overall reaction 2SO2(g)+O2(g)→2SO3(g)\text{2SO}_2(g)+\text{O}_2(g)\rightarrow 2\text{SO}_3(g) is:

Step 1: SO2(g)+O2(g)→SO4(g)\text{SO}_2(g)+\text{O}_2(g)\rightarrow \text{SO}_4(g) Step 2: SO4(g)+SO2(g)→2SO3(g)\text{SO}_4(g)+\text{SO}_2(g)\rightarrow 2\text{SO}_3(g)

Which species is an intermediate in the mechanism?

  1. SO2(g)\text{SO}_2(g) because it is consumed in Step 1 and Step 2.
  2. SO4(g)\text{SO}_4(g) because it is produced in Step 1 and consumed in Step 2. (correct answer)
  3. SO3(g)\text{SO}_3(g) because it is produced in Step 2.
  4. O2(g)\text{O}_2(g) because it is consumed in Step 1.
  5. No intermediate is present because there are only two steps.

Explanation: 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.

Question 13

A proposed mechanism for the net reaction 2 NO(g)+O2(g)→2 NO2(g)\mathrm{2\,NO(g) + O_2(g) \rightarrow 2\,NO_2(g)} is:

Step 1: NO(g)+NO(g)⇌N2O2(g)\mathrm{NO(g) + NO(g) \rightleftharpoons N_2O_2(g)} Step 2: N2O2(g)+O2(g)→2 NO2(g)\mathrm{N_2O_2(g) + O_2(g) \rightarrow 2\,NO_2(g)}

Which statement correctly identifies the catalyst or intermediate role of N2O2(g)\mathrm{N_2O_2(g)} in the mechanism?

  1. N2O2(g)\mathrm{N_2O_2(g)} is a catalyst because it is present at the start and end of the mechanism.
  2. N2O2(g)\mathrm{N_2O_2(g)} is an intermediate because it is formed in Step 1 and consumed in Step 2. (correct answer)
  3. N2O2(g)\mathrm{N_2O_2(g)} is a reactant because it appears on the reactant side of Step 2.
  4. N2O2(g)\mathrm{N_2O_2(g)} is a product because it appears on the product side of Step 1.
  5. N2O2(g)\mathrm{N_2O_2(g)} is neither because it cancels only if Step 1 is irreversible.

Explanation: 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.

Question 14

A mechanism is proposed for the net reaction CH3Br(aq)+OH−(aq)→CH3OH(aq)+Br−(aq)\mathrm{CH_3Br(aq) + OH^-(aq) \rightarrow CH_3OH(aq) + Br^-(aq)}:

Step 1: CH3Br(aq)→CH3+(aq)+Br−(aq)\mathrm{CH_3Br(aq) \rightarrow CH_3^+(aq) + Br^-(aq)} Step 2: CH3+(aq)+OH−(aq)→CH3OH(aq)\mathrm{CH_3^+(aq) + OH^-(aq) \rightarrow CH_3OH(aq)}

Which species is an intermediate in the mechanism?

  1. CH3Br(aq)\mathrm{CH_3Br(aq)}
  2. CH3OH(aq)\mathrm{CH_3OH(aq)}
  3. OH−(aq)\mathrm{OH^-(aq)}
  4. CH3+(aq)\mathrm{CH_3^+(aq)} (correct answer)
  5. Br−(aq)\mathrm{Br^-(aq)}

Explanation: 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.

Question 15

A mechanism is proposed for the reaction 2H2O2(aq)→2H2O(l)+O2(g)\mathrm{2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)} in the presence of iodide ion:

Step 1: H2O2(aq)+I−(aq)→H2O(l)+IO−(aq)\mathrm{H_2O_2(aq) + I^-(aq) \rightarrow H_2O(l) + IO^-(aq)} Step 2: H2O2(aq)+IO−(aq)→H2O(l)+O2(g)+I−(aq)\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?​

  1. IO−(aq)\mathrm{IO^-(aq)}
  2. H2O2(aq)\mathrm{H_2O_2(aq)}
  3. I−(aq)\mathrm{I^-(aq)} (correct answer)
  4. O2(g)\mathrm{O_2(g)}
  5. H2O(l)\mathrm{H_2O(l)}

Explanation: 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.

Question 16

A proposed mechanism for the overall reaction 2O3(g)→3O2(g)\mathrm{2O_3(g) \rightarrow 3O_2(g)} is:

Step 1: O3(g)→O2(g)+O(g)\mathrm{O_3(g) \rightarrow O_2(g) + O(g)} Step 2: O(g)+O3(g)→2O2(g)\mathrm{O(g) + O_3(g) \rightarrow 2O_2(g)}

Which species is an intermediate in this mechanism?​​​

  1. O3(g)\mathrm{O_3(g)}
  2. O2(g)\mathrm{O_2(g)}
  3. O(g)\mathrm{O(g)} (correct answer)
  4. 2O2(g)\mathrm{2O_2(g)}
  5. 3O2(g)\mathrm{3O_2(g)}

Explanation: This question tests understanding of introduction to reaction mechanisms, specifically identifying intermediates. An intermediate is produced in one step and consumed in another step of the mechanism. In Step 1, O(g) is produced as a product, and in Step 2, O(g) is consumed as a reactant. When we add the steps: O₃(g) → O₂(g) + O(g) and O(g) + O₃(g) → 2O₂(g), we get 2O₃(g) → 3O₂(g) after O(g) cancels out. Choice A (O₃(g)) is incorrect because it's a reactant in the overall reaction, not an intermediate—it appears in the net equation. The key strategy is to identify species that appear in individual steps but cancel out when steps are added together.

Question 17

A mechanism is proposed for the decomposition of ozone:

Step 1: O3(g)→O2(g)+O(g)\mathrm{O_3(g) \rightarrow O_2(g) + O(g)} Step 2: O(g)+O3(g)→2O2(g)\mathrm{O(g) + O_3(g) \rightarrow 2O_2(g)}

Which statement correctly identifies a catalyst or intermediate for this mechanism?​

  1. O2(g)\mathrm{O_2(g)} is a catalyst because it appears in the first step and in the net reaction.
  2. O3(g)\mathrm{O_3(g)} is an intermediate because it is consumed in Step 2.
  3. O(g)\mathrm{O(g)} is an intermediate because it is produced in Step 1 and consumed in Step 2. (correct answer)
  4. O(g)\mathrm{O(g)} is a catalyst because it is produced and consumed and does not appear in any step.
  5. O2(g)\mathrm{O_2(g)} is an intermediate because it cancels when the steps are added.

Explanation: 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.

Question 18

A proposed mechanism for the formation of NO2\mathrm{NO_2} is shown below:

Step 1: NO+O2→NO3\mathrm{NO + O_2 \rightarrow NO_3} Step 2: NO3+NO→2NO2\mathrm{NO_3 + NO \rightarrow 2NO_2}

Which species is an intermediate in the mechanism?​​

  1. NO\mathrm{NO} is an intermediate because it appears in both steps.
  2. O2\mathrm{O_2} is an intermediate because it is consumed in Step 1.
  3. NO2\mathrm{NO_2} is an intermediate because it is produced in Step 2.
  4. NO3\mathrm{NO_3} is an intermediate because it is produced in Step 1 and consumed in Step 2. (correct answer)
  5. NO3\mathrm{NO_3} is a catalyst because it cancels in the net reaction.

Explanation: 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.

Question 19

A proposed mechanism for the overall reaction CH3Br(aq)+OH−(aq)→CH3OH(aq)+Br−(aq)\text{CH}_3\text{Br}(aq)+\text{OH}^-(aq)\rightarrow \text{CH}_3\text{OH}(aq)+\text{Br}^-(aq) is:

Step 1: CH3Br(aq)→CH3+(aq)+Br−(aq)\text{CH}_3\text{Br}(aq)\rightarrow \text{CH}_3^+(aq)+\text{Br}^-(aq) Step 2: CH3+(aq)+OH−(aq)→CH3OH(aq)\text{CH}_3^+(aq)+\text{OH}^-(aq)\rightarrow \text{CH}_3\text{OH}(aq)

Which species is an intermediate in the mechanism?

  1. Br−(aq)\text{Br}^-(aq) because it is produced in Step 1.
  2. CH3OH(aq)\text{CH}_3\text{OH}(aq) because it is produced in Step 2.
  3. OH−(aq)\text{OH}^-(aq) because it is consumed in Step 2.
  4. CH3+(aq)\text{CH}_3^+(aq) because it is produced in Step 1 and consumed in Step 2. (correct answer)
  5. CH3Br(aq)\text{CH}_3\text{Br}(aq) because it is consumed in Step 1.

Explanation: This question tests your understanding of the introduction to reaction mechanisms. In a reaction mechanism, steps combine to give the overall, with intermediates generated and depleted. Here, CH3+ is produced in Step 1 from CH3Br and consumed in Step 2 with OH-. The steps add to CH3Br + OH- → CH3OH + Br- after canceling CH3+. Choice A fails because Br- is a product in the net reaction, not an intermediate. A transferable strategy is that intermediates appear in steps but not in the overall reaction.

Question 20

Consider the following proposed mechanism in acidic solution:

Step 1: HNO2(aq)+H+(aq)⇌H2NO2+(aq)\mathrm{HNO_2(aq) + H^+(aq) \rightleftharpoons H_2NO_2^+(aq)} Step 2: H2NO2+(aq)→NO+(aq)+H2O(l)\mathrm{H_2NO_2^+(aq) \rightarrow NO^+(aq) + H_2O(l)} Step 3: NO+(aq)+I−(aq)→NOI(aq)\mathrm{NO^+(aq) + I^-(aq) \rightarrow NOI(aq)}

A student claims the overall reaction is HNO2(aq)+H+(aq)+I−(aq)→NOI(aq)+H2O(l)\mathrm{HNO_2(aq) + H^+(aq) + I^-(aq) \rightarrow NOI(aq) + H_2O(l)}. Which species is an intermediate in the mechanism?

  1. NOI(aq)\mathrm{NOI(aq)}
  2. I−(aq)\mathrm{I^-(aq)}
  3. H2NO2+(aq)\mathrm{H_2NO_2^+(aq)} (correct answer)
  4. H2O(l)\mathrm{H_2O(l)}
  5. H+(aq)\mathrm{H^+(aq)}

Explanation: This question assesses the introduction to reaction mechanisms. Mechanisms feature steps where intermediates are formed and later consumed, ensuring the overall reaction balances without them. Combining the steps yields HNO₂(aq) + H⁺(aq) + H₂NO₂⁺(aq) + NO⁺(aq) + I⁻(aq) → H₂NO₂⁺(aq) + NO⁺(aq) + H₂O(l) + NOI(aq), canceling H₂NO₂⁺(aq) and NO⁺(aq) to give HNO₂(aq) + H⁺(aq) + I⁻(aq) → NOI(aq) + H₂O(l). Thus, H₂NO₂⁺(aq) is an intermediate produced in Step 1 and consumed in Step 2. Choice C fails because NOI(aq) is a product in the overall reaction, not an intermediate. Remember, intermediates appear in steps but not in the overall reaction, providing a key way to identify them across mechanisms.