AP Chemistry Quiz: Reaction Mechanisms And Rate Law
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Reaction Mechanisms And Rate LawQuestion 1 of 20

In aqueous solution, a student proposes the following mechanism for the reaction S2O82−(aq)+2I−(aq)→I2(aq)+2SO42−(aq)\mathrm{S_2O_8^{2-}(aq)+2I^-(aq)\rightarrow I_2(aq)+2SO_4^{2-}(aq)}:

Step 1 (slow): S2O82−+I−→SO42−+SO4−+I\mathrm{S_2O_8^{2-}+I^-\rightarrow SO_4^{2-}+SO_4^-+I}

Step 2 (fast): SO4−+I−→SO42−+I\mathrm{SO_4^-+I^-\rightarrow SO_4^{2-}+I}

Step 3 (fast): I+I→I2\mathrm{I+I\rightarrow I_2}

Which rate law is implied by the mechanism?

rate=k[I−]\text{rate}=k[\mathrm{I^-}]
rate=k[S2O82−][I−]\text{rate}=k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}]
rate=k[S2O82−]\text{rate}=k[\mathrm{S_2O_8^{2-}}]
rate=k[S2O82−][I−]2\text{rate}=k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}]^2
rate=k[SO4−][I−]\text{rate}=k[\mathrm{SO_4^-}][\mathrm{I^-}]
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AP Chemistry Quiz

AP Chemistry Quiz: Reaction Mechanisms And Rate Law

Practice Reaction Mechanisms And Rate Law 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 Reaction Mechanisms And Rate Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

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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

In aqueous solution, a student proposes the following mechanism for the reaction S2O82−(aq)+2I−(aq)→I2(aq)+2SO42−(aq)\mathrm{S_2O_8^{2-}(aq)+2I^-(aq)\rightarrow I_2(aq)+2SO_4^{2-}(aq)}:

Step 1 (slow): S2O82−+I−→SO42−+SO4−+I\mathrm{S_2O_8^{2-}+I^-\rightarrow SO_4^{2-}+SO_4^-+I}

Step 2 (fast): SO4−+I−→SO42−+I\mathrm{SO_4^-+I^-\rightarrow SO_4^{2-}+I}

Step 3 (fast): I+I→I2\mathrm{I+I\rightarrow I_2}

Which rate law is implied by the mechanism?

  1. rate=k[I−]\text{rate}=k[\mathrm{I^-}]
  2. rate=k[S2O82−][I−]\text{rate}=k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}] (correct answer)
  3. rate=k[S2O82−]\text{rate}=k[\mathrm{S_2O_8^{2-}}]
  4. rate=k[S2O82−][I−]2\text{rate}=k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}]^2
  5. rate=k[SO4−][I−]\text{rate}=k[\mathrm{SO_4^-}][\mathrm{I^-}]

Explanation: This question tests your understanding of reaction mechanisms and rate law. The rate law is determined by the slow (rate-determining) step, which is Step 1: S₂O₈²⁻ + I⁻ → SO₄²⁻ + SO₄⁻ + I. This step directly involves the reactants S₂O₈²⁻ and I⁻, each appearing once, so the rate law is rate = k[S₂O₈²⁻][I⁻]. The fast steps that follow do not affect the rate law since the overall reaction rate is limited by the slowest step. A common error is using the stoichiometry of the overall reaction (choice B: rate = k[S₂O₈²⁻][I⁻]²), which would incorrectly suggest the rate depends on [I⁻]² because two I⁻ ions appear in the balanced equation. Remember: only the slow step determines the rate law, regardless of the overall reaction stoichiometry.

Question 2

A mechanism is proposed for the decomposition of ozone in the presence of chlorine atoms:

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

Step 1 (fast): Cl+O3→ClO+O2\mathrm{Cl + O_3 \rightarrow ClO + O_2}

Step 2 (slow): ClO+O→Cl+O2\mathrm{ClO + O \rightarrow Cl + O_2}

Based only on the slow step, what rate law is implied?

  1. rate=k[O3]\text{rate}=k[\mathrm{O_3}]
  2. rate=k[ClO][O]\text{rate}=k[\mathrm{ClO}][\mathrm{O}] (correct answer)
  3. rate=k[O3][O]\text{rate}=k[\mathrm{O_3}][\mathrm{O}]
  4. rate=k[O]\text{rate}=k[\mathrm{O}]
  5. rate=k[Cl][O3]\text{rate}=k[\mathrm{Cl}][\mathrm{O_3}]

Explanation: This question tests your understanding of reaction mechanisms and rate law. The rate law is determined by the slow (rate-determining) step, which is Step 2: ClO + O → Cl + O₂. This step shows the rate depends on [ClO] and [O], giving rate = k[ClO][O]. Even though ClO is an intermediate formed in Step 1, we don't need to substitute it because the question asks for the rate law based only on the slow step. A common error is using the overall reaction (choice A: rate = k[O₃][O]), which incorrectly ignores the mechanism and assumes a single-step process. Remember: when asked for the rate law based only on the slow step, write it directly from that step without substituting intermediates.

Question 3

A mechanism is proposed 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 (slow): I2(g)→2I(g)\text{I}_2(g)\rightarrow 2\text{I}(g)

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

Step 3 (fast): H(g)+I(g)→HI(g)\text{H}(g)+\text{I}(g)\rightarrow \text{HI}(g)

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[HI]2\text{rate}=k[\text{HI}]^2
  2. rate=k[H2]\text{rate}=k[\text{H}_2]
  3. rate=k[I]2\text{rate}=k[\text{I}]^2
  4. rate=k[H2][I2]\text{rate}=k[\text{H}_2][\text{I}_2]
  5. rate=k[I2]\text{rate}=k[\text{I}_2] (correct answer)

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step in a reaction mechanism is the rate-determining step, which controls the overall rate of the reaction because subsequent fast steps occur quickly once it completes. In this mechanism, the slow step is the unimolecular dissociation of I2 into 2I. Therefore, the rate law reflects only this slow step, resulting in rate = k[I2]. A tempting distractor is choice A, rate = k[H2][I2], which incorrectly uses the overall reaction instead of the rate-determining step. Always remember that only the slow step governs the rate law, and its elementary rate expression should be used directly.

Question 4

A proposed mechanism for the reaction 2NO(g)+2H2(g)→N2(g)+2H2O(g)2\text{NO}(g)+2\text{H}_2(g)\rightarrow \text{N}_2(g)+2\text{H}_2\text{O}(g) is shown below.

Step 1 (fast): 2NO⇌N2O22\text{NO} \rightleftharpoons \text{N}_2\text{O}_2

Step 2 (slow): N2O2+H2→N2O+H2O\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}

Step 3 (fast): N2O+H2→N2+H2O\text{N}_2\text{O}+\text{H}_2 \rightarrow \text{N}_2+\text{H}_2\text{O}

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[N2O]\text{rate}=k[\text{N}_2\text{O}]
  2. rate=k[NO][H2]\text{rate}=k[\text{NO}][\text{H}_2]
  3. rate=k[N2O2][H2]\text{rate}=k[\text{N}_2\text{O}_2][\text{H}_2] (correct answer)
  4. rate=k[NO]2[H2]2\text{rate}=k[\text{NO}]^2[\text{H}_2]^2
  5. rate=k[NO]2[H2]\text{rate}=k[\text{NO}]^2[\text{H}_2]

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step controls the overall rate by acting as the rate-limiting factor, with preceding and following steps adjusting accordingly. In this mechanism, the slow step is N2O2 + H2 → N2O + H2O, bimolecular, so rate = k[N2O2][H2]. This rate law is derived solely from these concentrations, with N2O2 as an intermediate. A tempting distractor is choice D, rate = k[NO]^2[H2], which is wrong because it substitutes for the intermediate without considering the slow step directly, often a misconception of approximating equilibria prematurely. A useful strategy is to write the rate law using only the slow step's reactants, regardless of whether they are intermediates.

Question 5

A proposed mechanism for the reaction 2NO2(g)+F2(g)→2NO2F(g)2\text{NO}_2(g)+\text{F}_2(g)\rightarrow 2\text{NO}_2\text{F}(g) is shown below.

Step 1 (fast): NO2+F2⇌NO2F+F\text{NO}_2 + \text{F}_2 \rightleftharpoons \text{NO}_2\text{F} + \text{F}

Step 2 (slow): F+NO2→NO2F\text{F}+\text{NO}_2 \rightarrow \text{NO}_2\text{F}

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[NO2]2[F2]\text{rate}=k[\text{NO}_2]^2[\text{F}_2]
  2. rate=k[NO2][F2]\text{rate}=k[\text{NO}_2][\text{F}_2]
  3. rate=k[F][NO2]\text{rate}=k[\text{F}][\text{NO}_2] (correct answer)
  4. rate=k[F]\text{rate}=k[\text{F}]
  5. rate=k[NO2F]\text{rate}=k[\text{NO}_2\text{F}]

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step controls the overall rate as it is the bottleneck, with fast steps not influencing the rate significantly. Here, the slow step is F + NO2 → NO2F, bimolecular, so the rate law is rate = k[F][NO2]. This is based exclusively on the species in this step, including the intermediate F. A tempting distractor is choice A, rate = k[NO2]^2[F2], which is wrong because it derives from the overall reaction instead of the rate-determining step. A transferable strategy is to always derive the rate law from the reactants in the slowest step of the mechanism.

Question 6

A proposed mechanism for the reaction NO2(g)+CO(g)→NO(g)+CO2(g)\text{NO}_2(g)+\text{CO}(g)\rightarrow \text{NO}(g)+\text{CO}_2(g) is shown below.

Step 1 (fast): NO2+NO2⇌NO3+NO\text{NO}_2 + \text{NO}_2 \rightleftharpoons \text{NO}_3 + \text{NO}

Step 2 (slow): NO3+CO→NO2+CO2\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[NO3][CO]\text{rate}=k[\text{NO}_3][\text{CO}] (correct answer)
  2. rate=k[NO2]2[CO]\text{rate}=k[\text{NO}_2]^2[\text{CO}]
  3. rate=k[NO2][CO]\text{rate}=k[\text{NO}_2][\text{CO}]
  4. rate=k[NO2]2\text{rate}=k[\text{NO}_2]^2
  5. rate=k[CO]\text{rate}=k[\text{CO}]

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step in a reaction mechanism is the rate-determining step, which controls the overall speed of the reaction because it is the bottleneck where the reaction proceeds most slowly. In this mechanism, the slow step is NO3 + CO → NO2 + CO2, which is a bimolecular elementary reaction involving one molecule of NO3 and one of CO. Therefore, the rate law reflects only this step, expressed as rate = k[NO3][CO], even though NO3 is an intermediate. A tempting distractor is choice D, rate = k[NO2]^2, which is incorrect because it mistakenly uses the stoichiometry of the fast equilibrium step instead of the rate-determining step. To determine the rate law from a mechanism, always focus on the concentrations of the species directly involved in the slowest step.

Question 7

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

Step 1 (slow): NO2+NO2→NO3+NO\mathrm{NO_2 + NO_2 \rightarrow NO_3 + NO}

Step 2 (fast): NO3→NO+O2\mathrm{NO_3 \rightarrow NO + O_2}

Based only on the rate-determining step, which rate law is implied by the mechanism?

  1. Rate=k[NO2]\text{Rate}=k[\mathrm{NO_2}]
  2. Rate=k[NO2]2\text{Rate}=k[\mathrm{NO_2}]^2 (correct answer)
  3. Rate=k[NO3]\text{Rate}=k[\mathrm{NO_3}]
  4. Rate=k[NO][O2]\text{Rate}=k[\mathrm{NO}][\mathrm{O_2}]
  5. Rate=k[NO2][NO3]\text{Rate}=k[\mathrm{NO_2}][\mathrm{NO_3}]

Explanation: This problem tests knowledge of reaction mechanisms and rate law relationships. The rate-determining step is Step 1: NO₂ + NO₂ → NO₃ + NO, which involves two NO₂ molecules colliding. Therefore, the rate law is Rate = k[NO₂]². Students might incorrectly choose option A (Rate = k[NO₂]) by counting NO₂ only once, but when two molecules of the same species react in an elementary step, the concentration appears squared in the rate law. The fundamental principle is that the exponent in the rate law equals the number of molecules of each species participating in the rate-determining step.

Question 8

A proposed mechanism for the overall reaction 2NO(g)+2H2(g)→N2(g)+2H2O(g)\mathrm{2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(g)} is:

Step 1 (slow): 2NO→N2O2\mathrm{2NO \rightarrow N_2O_2}

Step 2 (fast): N2O2+H2→N2O+H2O\mathrm{N_2O_2 + H_2 \rightarrow N_2O + H_2O}

Step 3 (fast): N2O+H2→N2+H2O\mathrm{N_2O + H_2 \rightarrow N_2 + H_2O}

Based only on the rate-determining step, which rate law is implied by the mechanism?

  1. Rate=k[NO]\text{Rate}=k[\mathrm{NO}]
  2. Rate=k[NO]2\text{Rate}=k[\mathrm{NO}]^2 (correct answer)
  3. Rate=k[NO]2[H2]2\text{Rate}=k[\mathrm{NO}]^2[\mathrm{H_2}]^2
  4. Rate=k[N2O2][H2]\text{Rate}=k[\mathrm{N_2O_2}][\mathrm{H_2}]
  5. Rate=k[NO][H2]\text{Rate}=k[\mathrm{NO}][\mathrm{H_2}]

Explanation: This question tests understanding of reaction mechanisms and rate law for complex multi-step processes. The rate-determining step is Step 1: 2NO → N₂O₂, which involves two NO molecules combining. Therefore, the rate law is Rate = k[NO]². A common error would be choosing option C (Rate = k[NO]²[H₂]²) by including all reactants from the overall equation, but H₂ doesn't participate in the slow step—it only appears in the subsequent fast steps. The critical insight is that only reactants in the rate-determining step appear in the rate law, regardless of what happens in other steps.

Question 9

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

Step 1 (slow): Br2→2Br\text{Br}_2 \rightarrow 2\text{Br}

Step 2 (fast): Br+H2→HBr+H\text{Br}+\text{H}_2 \rightarrow \text{HBr}+\text{H}

Step 3 (fast): H+Br2→HBr+Br\text{H}+\text{Br}_2 \rightarrow \text{HBr}+\text{Br}

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[Br2][H2]\text{rate}=k[\text{Br}_2][\text{H}_2]
  2. rate=k[Br]2\text{rate}=k[\text{Br}]^2
  3. rate=k[H2]\text{rate}=k[\text{H}_2]
  4. rate=k[Br2]\text{rate}=k[\text{Br}_2] (correct answer)
  5. rate=k[HBr]\text{rate}=k[\text{HBr}]

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step dictates the overall rate as it is the slowest, controlling the pace at which the entire reaction occurs. Here, the slow step is Br2 → 2Br, unimolecular, resulting in rate = k[Br2]. The rate law reflects only this step's concentration. A tempting distractor is choice A, rate = k[Br2][H2], which is incorrect because it bases the rate on the overall reaction instead of the rate-determining step. Always remember that only the slow step governs the rate law in multi-step mechanisms.

Question 10

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

Step 1 (slow): H2O2→H2O+O\mathrm{H_2O_2 \rightarrow H_2O + O}

Step 2 (fast): O+H2O2→H2O+O2\mathrm{O + H_2O_2 \rightarrow H_2O + O_2}

Based only on the rate-determining step, which rate law is implied by the mechanism?

  1. Rate=k[O]\text{Rate}=k[\mathrm{O}]
  2. Rate=k[H2O2]\text{Rate}=k[\mathrm{H_2O_2}] (correct answer)
  3. Rate=k[O][H2O2]\text{Rate}=k[\mathrm{O}][\mathrm{H_2O_2}]
  4. Rate=k[O2]\text{Rate}=k[\mathrm{O_2}]
  5. Rate=k[H2O2]2\text{Rate}=k[\mathrm{H_2O_2}]^2

Explanation: This question examines reaction mechanisms and rate law determination. The slow step (Step 1: H₂O₂ → H₂O + O) contains only H₂O₂ as a reactant, so the rate law is Rate = k[H₂O₂]. A common mistake would be selecting option B (Rate = k[H₂O₂]²) by looking at the coefficient 2 in the overall equation, but the overall stoichiometry doesn't determine the rate law—only the slow step does. Since only one H₂O₂ molecule participates in the rate-determining step, the order with respect to H₂O₂ is 1. Remember: the rate law reflects the molecular composition of the slowest elementary step.

Question 11

A proposed mechanism for the reaction H2(g)+I2(g)→2HI(g)\mathrm{H_2(g)+I_2(g)\rightarrow 2HI(g)} is:

Step 1 (slow): I2→2I\mathrm{I_2 \rightarrow 2I}

Step 2 (fast): I+H2→HI+H\mathrm{I + H_2 \rightarrow HI + H}

Step 3 (fast): H+I→HI\mathrm{H + I \rightarrow HI}

Which rate law is implied by the mechanism?

  1. rate=k[H2][I2]\text{rate}=k[\mathrm{H_2}][\mathrm{I_2}]
  2. rate=k[I2]\text{rate}=k[\mathrm{I_2}] (correct answer)
  3. rate=k[I]2\text{rate}=k[\mathrm{I}]^2
  4. rate=k[H2]\text{rate}=k[\mathrm{H_2}]
  5. rate=k[HI]2\text{rate}=k[\mathrm{HI}]^2

Explanation: This question tests your understanding of reaction mechanisms and rate law. The rate law is determined by the slow (rate-determining) step, which is Step 1: I₂ → 2I. This step shows that the rate depends only on [I₂], giving rate = k[I₂]. The subsequent fast steps involving H₂ do not affect the rate law because they occur after the rate-determining step. A common error is using the overall reaction stoichiometry (choice A: rate = k[H₂][I₂]), which incorrectly assumes both reactants appear in the rate law based on the balanced equation. Remember: only the species involved in the slow step appear in the rate law, regardless of what happens in subsequent fast steps.

Question 12

A proposed mechanism for the reaction NO2(g)+CO(g)→NO(g)+CO2(g)\mathrm{NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g)} is shown below:

Step 1 (fast): NO2+NO2⇌N2O4\mathrm{NO_2 + NO_2 \rightleftharpoons N_2O_4}

Step 2 (slow): N2O4+CO→NO+NO3+CO2\mathrm{N_2O_4 + CO \rightarrow NO + NO_3 + CO_2}

Step 3 (fast): NO3+NO2→NO+O2+NO2\mathrm{NO_3 + NO_2 \rightarrow NO + O_2 + NO_2}

Based only on the rate-determining step and expressing the rate in terms of reactants in the overall reaction, which rate law is implied by this mechanism?

  1. rate=k[NO2][CO]\text{rate}=k[\mathrm{NO_2}][\mathrm{CO}]
  2. rate=k[N2O4][CO]\text{rate}=k[\mathrm{N_2O_4}][\mathrm{CO}]
  3. rate=k[NO2]2[CO]\text{rate}=k[\mathrm{NO_2}]^2[\mathrm{CO}] (correct answer)
  4. rate=k[CO]\text{rate}=k[\mathrm{CO}]
  5. rate=k[NO2]2\text{rate}=k[\mathrm{NO_2}]^2

Explanation: This question tests your understanding of reaction mechanisms and rate law. The rate law for a multi-step mechanism is determined by the slow (rate-determining) step, which is Step 2: N₂O₄ + CO → NO + NO₃ + CO₂. This step shows the rate depends on [N₂O₄] and [CO], but N₂O₄ is an intermediate formed in the fast equilibrium Step 1: NO₂ + NO₂ ⇌ N₂O₄. Since Step 1 is fast and at equilibrium, we can express [N₂O₄] in terms of [NO₂]² using the equilibrium relationship, giving rate = k[NO₂]²[CO]. A common error is using the overall reaction stoichiometry (choice A: rate = k[NO₂][CO]), which incorrectly assumes a single-step mechanism. Remember: for multi-step mechanisms, write the rate law from the slow step, then substitute any intermediates using fast pre-equilibrium steps.

Question 13

In acidic solution, the following mechanism is proposed for the reaction BrO3−(aq)+5Br−(aq)+6H+(aq)→3Br2(aq)+3H2O(l)\mathrm{BrO_3^-(aq)+5Br^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)}:

Step 1 (slow): BrO3−+Br−+2H+→HBrO2+HOBr\mathrm{BrO_3^- + Br^- + 2H^+ \rightarrow HBrO_2 + HOBr}

Step 2 (fast): HBrO2+Br−+H+→2HOBr\mathrm{HBrO_2 + Br^- + H^+ \rightarrow 2HOBr}

Step 3 (fast): HOBr+Br−+H+→Br2+H2O\mathrm{HOBr + Br^- + H^+ \rightarrow Br_2 + H_2O}

Based only on the slow step, which rate law is implied?

  1. rate=k[BrO3−][Br−][H+]2\text{rate}=k[\mathrm{BrO_3^-}][\mathrm{Br^-}][\mathrm{H^+}]^2 (correct answer)
  2. rate=k[BrO3−][Br−]5[H+]6\text{rate}=k[\mathrm{BrO_3^-}][\mathrm{Br^-}]^5[\mathrm{H^+}]^6
  3. rate=k[HBrO2][Br−][H+]\text{rate}=k[\mathrm{HBrO_2}][\mathrm{Br^-}][\mathrm{H^+}]
  4. rate=k[BrO3−][H+]\text{rate}=k[\mathrm{BrO_3^-}][\mathrm{H^+}]
  5. rate=k[Br−][H+]2\text{rate}=k[\mathrm{Br^-}][\mathrm{H^+}]^2

Explanation: This question tests your understanding of reaction mechanisms and rate law. The rate law is determined by the slow (rate-determining) step, which is Step 1: BrO₃⁻ + Br⁻ + 2H⁺ → HBrO₂ + HOBr. This step directly shows that the rate depends on [BrO₃⁻], [Br⁻], and [H⁺]², giving rate = k[BrO₃⁻][Br⁻][H⁺]². The subsequent fast steps do not affect the rate law since the overall reaction rate is limited by this slowest step. A common error is using the stoichiometry of the overall reaction (choice B: rate = k[BrO₃⁻][Br⁻]⁵[H⁺]⁶), which incorrectly assumes the rate law reflects the total number of each reactant consumed. Remember: the rate law comes only from the slow step, not from the overall reaction stoichiometry.

Question 14

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

Step 1 (slow): NO+O2→NO3\mathrm{NO + O_2 \rightarrow NO_3}

Step 2 (fast): NO3+NO→2NO2\mathrm{NO_3 + NO \rightarrow 2NO_2}

Which rate law is implied by the mechanism?

  1. rate=k[NO]2[O2]\text{rate}=k[\mathrm{NO}]^2[\mathrm{O_2}]
  2. rate=k[NO3][NO]\text{rate}=k[\mathrm{NO_3}][\mathrm{NO}]
  3. rate=k[NO][O2]\text{rate}=k[\mathrm{NO}][\mathrm{O_2}] (correct answer)
  4. rate=k[O2]\text{rate}=k[\mathrm{O_2}]
  5. rate=k[NO]\text{rate}=k[\mathrm{NO}]

Explanation: This question tests your understanding of reaction mechanisms and rate law. The rate law is determined by the slow (rate-determining) step, which is Step 1: NO + O₂ → NO₃. This step directly shows that the rate depends on [NO] and [O₂], each appearing once, giving rate = k[NO][O₂]. The fast Step 2 that follows does not affect the rate law since the overall reaction rate is limited by the slowest step. A common error is using the stoichiometry of the overall reaction (choice A: rate = k[NO]²[O₂]), which incorrectly suggests the rate depends on [NO]² because two NO molecules appear in the balanced equation. Remember: the rate law reflects only the molecularity of the slow step, not the overall reaction stoichiometry.

Question 15

A student suggests the following mechanism for the overall reaction 2NO2(g)→2NO(g)+O2(g)2\text{NO}_2(g)\rightarrow 2\text{NO}(g)+\text{O}_2(g).

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

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

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[NO]\text{rate}=k[\text{NO}]
  2. rate=k[NO3]\text{rate}=k[\text{NO}_3]
  3. rate=k[NO2]2\text{rate}=k[\text{NO}_2]^2 (correct answer)
  4. rate=k[NO]2[O2]\text{rate}=k[\text{NO}]^2[\text{O}_2]
  5. rate=k[NO2]\text{rate}=k[\text{NO}_2]

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step in a reaction mechanism is the rate-determining step, which controls the overall rate of the reaction as it limits the speed of the entire process. In this mechanism, the slow step is the bimolecular collision between two NO2 molecules. Therefore, the rate law is determined solely by the reactants in this slow step, yielding rate = k[NO2]^2. A tempting distractor is choice D, rate = k[NO]^2[O2], which incorrectly incorporates the products' stoichiometry instead of focusing on the rate-determining step. Always remember that only the slow step governs the rate law, and its elementary rate expression should be used directly.

Question 16

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

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

Step 2 (fast): NO3+NO→2NO2\text{NO}_3+\text{NO} \rightarrow 2\text{NO}_2

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[NO]2[O2]\text{rate}=k[\text{NO}]^2[\text{O}_2]
  2. rate=k[NO3][NO]\text{rate}=k[\text{NO}_3][\text{NO}]
  3. rate=k[NO][O2]\text{rate}=k[\text{NO}][\text{O}_2] (correct answer)
  4. rate=k[O2]\text{rate}=k[\text{O}_2]
  5. rate=k[NO]\text{rate}=k[\text{NO}]

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step controls the overall rate because it is the rate-limiting process, meaning the subsequent fast steps occur quickly once the slow step is completed. Here, the slow step is NO + O2 → NO3, a bimolecular reaction, so the rate depends on the concentrations of NO and O2. Thus, the rate law is rate = k[NO][O2], based solely on this step's reactants. A tempting distractor is choice A, rate = k[NO]^2[O2], which is wrong because it uses the overall reaction stoichiometry instead of the rate-determining step. Remember, only the slow step governs the rate law, providing a strategy to identify it first in any mechanism.

Question 17

A proposed mechanism for the 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 shown below.

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

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

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[CH3Br]\text{rate}=k[\text{CH}_3\text{Br}] (correct answer)
  2. rate=k[CH3+][OH−]\text{rate}=k[\text{CH}_3^+][\text{OH}^-]
  3. rate=k[CH3Br][OH−]\text{rate}=k[\text{CH}_3\text{Br}][\text{OH}^-]
  4. rate=k[CH3OH]\text{rate}=k[\text{CH}_3\text{OH}]
  5. rate=k[Br−]\text{rate}=k[\text{Br}^-]

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step controls the overall rate because it is the rate-determining process, with fast steps following rapidly. The slow step is CH3Br → CH3+ + Br-, unimolecular, so rate = k[CH3Br]. This rate law is based only on the concentration of CH3Br in that step. A tempting distractor is choice A, rate = k[CH3Br][OH-], which is wrong because it uses the overall reaction stoichiometry instead of the rate-determining step. A transferable strategy is to identify the slow step and use its elementary rate law directly.

Question 18

A proposed mechanism for the reaction 2H2O2(aq)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(aq)\rightarrow 2\text{H}_2\text{O}(l)+\text{O}_2(g) is shown below.

Step 1 (slow): H2O2→2OH\text{H}_2\text{O}_2 \rightarrow 2\text{OH}

Step 2 (fast): OH+H2O2→H2O+HO2\text{OH}+\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O}+\text{HO}_2

Step 3 (fast): HO2+OH→H2O+O2\text{HO}_2+\text{OH} \rightarrow \text{H}_2\text{O}+\text{O}_2

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[H2O2]2\text{rate}=k[\text{H}_2\text{O}_2]^2
  2. rate=k[OH][H2O2]\text{rate}=k[\text{OH}][\text{H}_2\text{O}_2]
  3. rate=k[H2O2]\text{rate}=k[\text{H}_2\text{O}_2] (correct answer)
  4. rate=k[HO2][OH]\text{rate}=k[\text{HO}_2][\text{OH}]
  5. rate=k[O2]\text{rate}=k[\text{O}_2]

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step governs the overall rate because it proceeds at a much slower pace than the fast steps, determining the reaction's speed. In this mechanism, the slow step is H2O2→2OH\text{H}_2\text{O}_2 \rightarrow 2\text{OH}, a unimolecular decomposition, so the rate law is rate=k[H2O2]\text{rate} = k[\text{H}_2\text{O}_2]. This reflects only the concentration of H2O2 in that step. A tempting distractor is choice A, rate=k[H2O2]2\text{rate} = k[\text{H}_2\text{O}_2]^2, which is incorrect because it uses the overall reaction stoichiometry instead of the rate-determining step. For any mechanism, only the slow step governs the rate law, so prioritize identifying it.

Question 19

A proposed mechanism for the reaction H2(g)+I2(g)→2HI(g)\text{H}_2(g)+\text{I}_2(g)\rightarrow 2\text{HI}(g) is shown below.

Step 1 (fast): I2⇌2I\text{I}_2 \rightleftharpoons 2\text{I}

Step 2 (slow): H2+I→HI+H\text{H}_2 + \text{I} \rightarrow \text{HI} + \text{H}

Step 3 (fast): H+I2→HI+I\text{H}+\text{I}_2 \rightarrow \text{HI}+\text{I}

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[H2][I]\text{rate}=k[\text{H}_2][\text{I}] (correct answer)
  2. rate=k[H2][I2]\text{rate}=k[\text{H}_2][\text{I}_2]
  3. rate=k[I2]\text{rate}=k[\text{I}_2]
  4. rate=k[H2][I2]2\text{rate}=k[\text{H}_2][\text{I}_2]^2
  5. rate=k[HI]2\text{rate}=k[\text{HI}]^2

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step dictates the overall reaction rate as it is the slowest part of the mechanism, limiting how fast products can form. In this case, the slow step is H2 + I → HI + H, involving H2 and I, so the rate law is rate = k[H2][I]. This reflects only the reactants in that step, including the intermediate I from the fast equilibrium. A tempting distractor is choice B, rate = k[H2][I2], which is incorrect because it mistakenly bases the rate on the overall reaction instead of the rate-determining step. Always identify the slowest step and use its reactant concentrations to derive the rate law.

Question 20

Consider the following proposed mechanism for the overall reaction 2NO(g)+Cl2(g)→2NOCl(g)2\text{NO}(g)+\text{Cl}_2(g)\rightarrow 2\text{NOCl}(g).

Step 1 (fast): NO(g)+Cl2(g)⇌NOCl2(g)\text{NO}(g)+\text{Cl}_2(g)\rightleftharpoons \text{NOCl}_2(g)

Step 2 (slow): NO(g)+NOCl2(g)→2NOCl(g)\text{NO}(g)+\text{NOCl}_2(g)\rightarrow 2\text{NOCl}(g)

Based only on the rate-determining step, which rate law is implied by this mechanism?

  1. rate=k[NO]2[Cl2]\text{rate}=k[\text{NO}]^2[\text{Cl}_2]
  2. rate=k[NO][NOCl2]\text{rate}=k[\text{NO}][\text{NOCl}_2] (correct answer)
  3. rate=k[Cl2]\text{rate}=k[\text{Cl}_2]
  4. rate=k[NOCl]2\text{rate}=k[\text{NOCl}]^2
  5. rate=k[NO]2\text{rate}=k[\text{NO}]^2

Explanation: This question tests the skill of reaction mechanisms and rate law. The slow step in a reaction mechanism is the rate-determining step, which controls the overall rate of the reaction because it is the bottleneck in the process. In this mechanism, the slow step involves NO and the intermediate NOCl2. Therefore, the rate law is based solely on the concentrations of the species in this slow step, giving rate = k[NO][NOCl2]. A tempting distractor is choice A, rate = k[NO]^2[Cl2], which incorrectly uses the overall reaction stoichiometry instead of the rate-determining step. Always remember that only the slow step governs the rate law, and its elementary rate expression should be used directly.