Reaction Mechanisms and Rate Law - AP Chemistry

1. A proposed mechanism for the decomposition of hydrogen peroxide, $2\text{H}_2\text{O}_2(aq)\rightarrow 2\text{H}_2\text{O}(l)+\text{O}_2(g)$, is: Step 1 (slow): $\text{H}_2\text{O}_2(aq)\rightarrow 2\text{OH}(aq)$ Step 2 (fast): $\text{OH}(aq)+\text{H}_2\text{O}_2(aq)\rightarrow \text{HO}_2(aq)+\text{H}_2\text{O}(l)$ Step 3 (fast): $\text{HO}_2(aq)+\text{OH}(aq)\rightarrow \text{O}_2(g)+\text{H}_2\text{O}(l)$ Based only on the rate-determining step, which rate law is implied?​​
   Answer: $\text{rate}=k[\text{H}_2\text{O}_2]$ This question tests understanding of reaction mechanisms and rate law. The rate-determining step controls the overall reaction rate, meaning the rate law depends only on the reactants in the slowest step. Step 1 is marked as slow: H₂O₂(aq) → 2OH(aq), which has only H₂O₂ as a reactant. Therefore, the rate law must be rate = k[H₂O₂], showing first-order dependence on hydrogen peroxide. Choice A (rate = k[H₂O₂]²) incorrectly assumes the rate law should match the stoichiometric coefficient in the overall reaction (2H₂O₂ → products), which is a common misconception. When determining rate laws from mechanisms, use only the reactants from the slow step with their stoichiometric coefficients in that step, not from the overall equation.
2. A student suggests the following mechanism for the overall reaction $2\text{NO}_2(g)\rightarrow 2\text{NO}(g)+\text{O}_2(g)$. Step 1 (slow): $\text{NO}_2(g)+\text{NO}_2(g)\rightarrow \text{NO}_3(g)+\text{NO}(g)$ Step 2 (fast): $\text{NO}_3(g)\rightarrow \text{NO}(g)+\text{O}_2(g)$ Based only on the rate-determining step, which rate law is implied by this mechanism?
   Answer: $\text{rate}=k[\text{NO}_2]^2$ 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.
3. A proposed mechanism for the reaction $2\text{NO}(g)+\text{O}_2(g)\rightarrow 2\text{NO}_2(g)$ is shown below. Step 1 (slow): $\text{NO}+\text{O}_2 \rightarrow \text{NO}_3$ Step 2 (fast): $\text{NO}_3+\text{NO} \rightarrow 2\text{NO}_2$ Based only on the rate-determining step, which rate law is implied by this mechanism?
   Answer: $\text{rate}=k[\text{NO}][\text{O}_2]$ 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.
4. A proposed mechanism for the reaction $2\text{H}_2\text{O}_2(aq)\rightarrow 2\text{H}_2\text{O}(l)+\text{O}_2(g)$ is shown below. Step 1 (slow): $\text{H}_2\text{O}_2 \rightarrow 2\text{OH}$ Step 2 (fast): $\text{OH}+\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O}+\text{HO}_2$ Step 3 (fast): $\text{HO}_2+\text{OH} \rightarrow \text{H}_2\text{O}+\text{O}_2$ Based only on the rate-determining step, which rate law is implied by this mechanism?
   Answer: $\text{rate}=k[\text{H}_2\text{O}_2]$ 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 $\text{H}_2\text{O}_2 \rightarrow 2\text{OH}$, a unimolecular decomposition, so the rate law is $\text{rate} = k[\text{H}_2\text{O}_2]$. This reflects only the concentration of H2O2 in that step. A tempting distractor is choice A, $\text{rate} = k[\text{H}_2\text{O}_2]^2$, 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.
5. A student proposes the following mechanism for the reaction $\mathrm{2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)}$ in the presence of iodide ion as a catalyst: Step 1 (slow): $\mathrm{H_2O_2 + I^- \rightarrow H_2O + IO^-}$ Step 2 (fast): $\mathrm{H_2O_2 + IO^- \rightarrow H_2O + O_2 + I^-}$ Which rate law is implied by the rate-determining step?
   Answer: $\text{rate}=k[\mathrm{H_2O_2}][\mathrm{I^-}]$ 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: H₂O₂ + I⁻ → H₂O + IO⁻. This step directly shows that the rate depends on [H₂O₂] and [I⁻], each appearing once, giving rate = k[H₂O₂][I⁻]. Note that I⁻ acts as a catalyst (it's consumed in Step 1 and regenerated in Step 2), but it still appears in the rate law because it participates in the rate-determining step. A common error is excluding the catalyst from the rate law (choice B: rate = k[H₂O₂]), incorrectly assuming catalysts never appear in rate expressions. Remember: if a catalyst participates in the slow step, it will appear in the rate law even though it's regenerated later.
6. In aqueous solution, a student proposes the following mechanism for the reaction $\mathrm{S_2O_8^{2-}(aq)+2I^-(aq)\rightarrow I_2(aq)+2SO_4^{2-}(aq)}$: Step 1 (slow): $\mathrm{S_2O_8^{2-}+I^-\rightarrow SO_4^{2-}+SO_4^-+I}$ Step 2 (fast): $\mathrm{SO_4^-+I^-\rightarrow SO_4^{2-}+I}$ Step 3 (fast): $\mathrm{I+I\rightarrow I_2}$ Which rate law is implied by the mechanism?
   Answer: $\text{rate}=k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}]$ 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.
7. A proposed mechanism for the reaction $\text{Br}_2(g)+\text{H}_2(g)\rightarrow 2\text{HBr}(g)$ is shown below. Step 1 (slow): $\text{Br}_2 \rightarrow 2\text{Br}$ Step 2 (fast): $\text{Br}+\text{H}_2 \rightarrow \text{HBr}+\text{H}$ Step 3 (fast): $\text{H}+\text{Br}_2 \rightarrow \text{HBr}+\text{Br}$ Based only on the rate-determining step, which rate law is implied by this mechanism?
   Answer: $\text{rate}=k[\text{Br}_2]$ 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.
8. A proposed mechanism for the reaction $\text{CO}(g)+\text{NO}_2(g)\rightarrow \text{CO}_2(g)+\text{NO}(g)$ is: Step 1 (slow, rate-determining): $\text{NO}_2(g)\rightarrow \text{NO}(g)+\text{O}(g)$ Step 2 (fast): $\text{CO}(g)+\text{O}(g)\rightarrow \text{CO}_2(g)$ Based only on the rate-determining step, which rate law is implied?
   Answer: $\text{rate}=k[\text{NO}_2]$ This question tests understanding of reaction mechanisms and rate law. The rate-determining (slowest) step controls the overall reaction rate, so the rate law includes only the reactants from that step. Step 1 is the slow step: NO₂(g) → NO(g) + O(g), which involves only NO₂ decomposing, giving rate = k[NO₂]. Students often mistakenly use the overall reaction CO + NO₂ → CO₂ + NO to write rate = k[CO][NO₂] (choice B), but CO doesn't participate until the fast step 2. For mechanisms, the rate law always reflects only the reactants in the slow step, regardless of what appears in the overall equation.
9. A proposed mechanism for the reaction $2\text{NO}_2(g)+\text{F}_2(g)\rightarrow 2\text{NO}_2\text{F}(g)$ is shown below. Step 1 (fast): $\text{NO}_2 + \text{F}_2 \rightleftharpoons \text{NO}_2\text{F} + \text{F}$ Step 2 (slow): $\text{F}+\text{NO}_2 \rightarrow \text{NO}_2\text{F}$ Based only on the rate-determining step, which rate law is implied by this mechanism?
   Answer: $\text{rate}=k[\text{F}][\text{NO}_2]$ 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.
10. A mechanism is proposed for the overall reaction $\text{SO}_2(g)+\text{NO}_2(g)\rightarrow \text{SO}_3(g)+\text{NO}(g)$. Step 1 (slow): $\text{NO}_2(g)\rightarrow \text{NO}(g)+\text{O}(g)$ Step 2 (fast): $\text{SO}_2(g)+\text{O}(g)\rightarrow \text{SO}_3(g)$ Based only on the rate-determining step, which rate law is implied by this mechanism?
    Answer: $\text{rate}=k[\text{NO}_2]$ 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 determines how quickly products can form. In this mechanism, the slow step is the unimolecular decomposition of NO2. Therefore, the rate law is solely based on this slow step, resulting in rate = k[NO2]. A tempting distractor is choice A, rate = k[SO2][NO2], 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.
11. A proposed mechanism for the reaction $\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}$ is: Step 1 (slow): $\mathrm{NO + O_2 \rightarrow NO_3}$ Step 2 (fast): $\mathrm{NO_3 + NO \rightarrow 2NO_2}$ Based only on the rate-determining step, which rate law is implied by the mechanism?
    Answer: $\text{Rate}=k[\mathrm{NO}][\mathrm{O_2}]$ This problem focuses on reaction mechanisms and rate law determination. The rate-determining step is Step 1: NO + O₂ → NO₃, which contains one molecule each of NO and O₂ as reactants. Therefore, the rate law is Rate = k[NO][O₂]. A common error would be choosing option A (Rate = k[NO]²[O₂]) by using the coefficient 2 from the overall equation, but coefficients in the overall equation don't determine the rate law—only the molecularity of the slow step matters. The reliable approach is to identify the slow step and write the rate law using only its reactants with exponents equal to their stoichiometric coefficients in that step.
12. A proposed mechanism for the reaction $\text{CH}_3\text{Br}(aq)+\text{OH}^-(aq)\rightarrow \text{CH}_3\text{OH}(aq)+\text{Br}^-(aq)$ is shown below. Step 1 (slow): $\text{CH}_3\text{Br} \rightarrow \text{CH}_3^+ + \text{Br}^-$ Step 2 (fast): $\text{CH}_3^+ + \text{OH}^- \rightarrow \text{CH}_3\text{OH}$ Based only on the rate-determining step, which rate law is implied by this mechanism?
    Answer: $\text{rate}=k[\text{CH}_3\text{Br}]$ 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.
13. A proposed mechanism for the overall reaction $2\text{NO}(g)+\text{O}_2(g)\rightarrow 2\text{NO}_2(g)$ is given. Step 1 (fast): $\text{NO}(g)+\text{O}_2(g)\rightleftharpoons \text{NO}_3(g)$ Step 2 (slow): $\text{NO}_3(g)+\text{NO}(g)\rightarrow 2\text{NO}_2(g)$ Which rate law, written only in terms of the concentrations of reactants in the overall reaction (no intermediates), is consistent with this mechanism?
    Answer: $\text{rate}=k[\text{NO}]^2[\text{O}_2]$ Step 2 is slow, so rate = k₂[NO₃][NO]. To express the rate using reactant concentrations only, eliminate the intermediate NO₃ with the fast pre-equilibrium in Step 1: [NO₃] = K₁[NO][O₂]. Substitution gives rate = k₂K₁[NO]²[O₂] = k[NO]²[O₂], choice C. Choice B describes the slow step but does not meet the reactant-only requirement. Choice A describes Step 1 alone.
14. A proposed mechanism for the reaction $\mathrm{NO_2(g) + CO(g) \rightarrow NO(g) + CO_2(g)}$ is shown below: Step 1 (fast, reversible): $\mathrm{NO_2 + NO_2 \rightleftharpoons N_2O_4}$ Step 2 (slow): $\mathrm{N_2O_4 + CO \rightarrow NO + NO_3 + CO_2}$ Step 3 (fast): $\mathrm{NO_3 + NO_2 \rightarrow NO + O_2 + NO_2}$ Based only on the rate-determining step, which rate law is implied by the mechanism?
    Answer: $\text{Rate}=k[\mathrm{N_2O_4}][\mathrm{CO}]$ This problem tests understanding of reaction mechanisms and rate law. The rate law is determined by the slowest (rate-determining) step, which is Step 2: N₂O₄ + CO → NO + NO₃ + CO₂. The rate law must include all reactants in this slow step, giving Rate = k[N₂O₄][CO]. Students might incorrectly choose option D (Rate = k[NO₂]²) by looking at the overall reaction or Step 1, but this ignores that CO is also involved in the rate-determining step. The key strategy is to write the rate law using only the reactants that appear in the slow step.
15. A proposed mechanism for the reaction $2\text{NO}(g)+2\text{H}_2(g)\rightarrow \text{N}_2(g)+2\text{H}_2\text{O}(g)$ is shown below. Step 1 (fast): $2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$ Step 2 (slow): $\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}$ Step 3 (fast): $\text{N}_2\text{O}+\text{H}_2 \rightarrow \text{N}_2+\text{H}_2\text{O}$ Based only on the rate-determining step, which rate law is implied by this mechanism?
    Answer: $\text{rate}=k[\text{N}_2\text{O}_2][\text{H}_2]$ 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.
16. A proposed mechanism for the reaction $\text{CH}_3\text{Br}(aq)+\text{OH}^-(aq)\rightarrow \text{CH}_3\text{OH}(aq)+\text{Br}^-(aq)$ is: Step 1 (slow, rate-determining): $\text{CH}_3\text{Br}(aq)\rightarrow \text{CH}_3^+(aq)+\text{Br}^-(aq)$ Step 2 (fast): $\text{CH}_3^+(aq)+\text{OH}^-(aq)\rightarrow \text{CH}_3\text{OH}(aq)$ Based only on the rate-determining step, which rate law is implied?
    Answer: $\text{rate}=k[\text{CH}_3\text{Br}]$ This question tests understanding of reaction mechanisms and rate law. The slowest (rate-determining) step controls the overall reaction rate, meaning the rate law reflects only the reactants in that step. Step 1 is the slow step: CH₃Br(aq) → CH₃⁺(aq) + Br⁻(aq), which shows unimolecular decomposition of CH₃Br, giving rate = k[CH₃Br]. A common error is to include OH⁻ from the overall reaction, writing rate = k[CH₃Br][OH⁻] (choice A), but OH⁻ only participates in the fast step 2, not the rate-determining step. For mechanisms, always write the rate law based exclusively on the reactants in the slow step.
17. A mechanism is proposed for the decomposition of ozone in the presence of chlorine atoms: Overall reaction: $\mathrm{O_3(g)+O(g)\rightarrow 2O_2(g)}$ Step 1 (fast): $\mathrm{Cl + O_3 \rightarrow ClO + O_2}$ Step 2 (slow): $\mathrm{ClO + O \rightarrow Cl + O_2}$ Based only on the slow step, what rate law is implied?
    Answer: $\text{rate}=k[\mathrm{ClO}][\mathrm{O}]$ 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.
18. In acidic solution, the following mechanism is proposed for the reaction $\mathrm{BrO_3^-(aq)+5Br^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)}$: Step 1 (slow): $\mathrm{BrO_3^- + Br^- + 2H^+ \rightarrow HBrO_2 + HOBr}$ Step 2 (fast): $\mathrm{HBrO_2 + Br^- + H^+ \rightarrow 2HOBr}$ Step 3 (fast): $\mathrm{HOBr + Br^- + H^+ \rightarrow Br_2 + H_2O}$ Based only on the slow step, which rate law is implied?
    Answer: $\text{rate}=k[\mathrm{BrO_3^-}][\mathrm{Br^-}][\mathrm{H^+}]^2$ 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.
19. A proposed mechanism for the overall reaction $\mathrm{2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)}$ is: Step 1 (slow): $\mathrm{H_2O_2 \rightarrow H_2O + O}$ Step 2 (fast): $\mathrm{O + H_2O_2 \rightarrow H_2O + O_2}$ Based only on the rate-determining step, which rate law is implied by the mechanism?
    Answer: $\text{Rate}=k[\mathrm{H_2O_2}]$ 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.
20. A proposed mechanism for the reaction $\mathrm{NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g)}$ is shown below: Step 1 (fast): $\mathrm{NO_2 + NO_2 \rightleftharpoons N_2O_4}$ Step 2 (slow): $\mathrm{N_2O_4 + CO \rightarrow NO + NO_3 + CO_2}$ Step 3 (fast): $\mathrm{NO_3 + NO_2 \rightarrow NO + O_2 + NO_2}$ 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?
    Answer: $\text{rate}=k[\mathrm{NO_2}]^2[\mathrm{CO}]$ 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.

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