Reaction Mechanisms and Rate Law
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AP Chemistry › Reaction Mechanisms and Rate Law
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?
$\text{rate}=k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}]^2$
$\text{rate}=k[\mathrm{SO_4^-}][\mathrm{I^-}]$
$\text{rate}=k[\mathrm{I^-}]$
$\text{rate}=k[\mathrm{S_2O_8^{2-}}]$
$\text{rate}=k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}]$
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.
A proposed mechanism for the reaction $\mathrm{H_2(g)+I_2(g)\rightarrow 2HI(g)}$ is:
Step 1 (slow): $\mathrm{I_2 \rightarrow 2I}$
Step 2 (fast): $\mathrm{I + H_2 \rightarrow HI + H}$
Step 3 (fast): $\mathrm{H + I \rightarrow HI}$
Which rate law is implied by the mechanism?
$\text{rate}=k[\mathrm{I_2}]$
$\text{rate}=k[\mathrm{I}]^2$
$\text{rate}=k[\mathrm{H_2}]$
$\text{rate}=k[\mathrm{H_2}][\mathrm{I_2}]$
$\text{rate}=k[\mathrm{HI}]^2$
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.
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?
$\text{rate}=k[\mathrm{NO_2}]^2$
$\text{rate}=k[\mathrm{NO_2}][\mathrm{CO}]$
$\text{rate}=k[\mathrm{N_2O_4}][\mathrm{CO}]$
$\text{rate}=k[\mathrm{NO_2}]^2[\mathrm{CO}]$
$\text{rate}=k[\mathrm{CO}]$
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.
A proposed mechanism for the gas-phase 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): $\mathrm{NO_3 + NO \rightarrow 2NO_2}$
Which rate law is implied by the mechanism?
$\text{rate}=k[\mathrm{NO}]$
$\text{rate}=k[\mathrm{NO_3}][\mathrm{NO}]$
$\text{rate}=k[\mathrm{O_2}]$
$\text{rate}=k[\mathrm{NO}][\mathrm{O_2}]$
$\text{rate}=k[\mathrm{NO}]^2[\mathrm{O_2}]$
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: 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.
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?
$\text{rate}=k[\mathrm{Br^-}][\mathrm{H^+}]^2$
$\text{rate}=k[\mathrm{BrO_3^-}][\mathrm{Br^-}][\mathrm{H^+}]^2$
$\text{rate}=k[\mathrm{BrO_3^-}][\mathrm{H^+}]$
$\text{rate}=k[\mathrm{BrO_3^-}][\mathrm{Br^-}]^5[\mathrm{H^+}]^6$
$\text{rate}=k[\mathrm{HBrO_2}][\mathrm{Br^-}][\mathrm{H^+}]$
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.
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?
$\text{rate}=k[\mathrm{Cl}][\mathrm{O_3}]$
$\text{rate}=k[\mathrm{ClO}][\mathrm{O}]$
$\text{rate}=k[\mathrm{O}]$
$\text{rate}=k[\mathrm{O_3}]$
$\text{rate}=k[\mathrm{O_3}][\mathrm{O}]$
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.
A mechanism is proposed for the overall reaction $\text{H}_2(g)+\text{I}_2(g)\rightarrow 2\text{HI}(g)$.
Step 1 (slow): $\text{I}_2(g)\rightarrow 2\text{I}(g)$
Step 2 (fast): $\text{I}(g)+\text{H}_2(g)\rightarrow \text{HI}(g)+\text{H}(g)$
Step 3 (fast): $\text{H}(g)+\text{I}(g)\rightarrow \text{HI}(g)$
Based only on the rate-determining step, which rate law is implied by this mechanism?
$\text{rate}=k[\text{I}_2]$
$\text{rate}=k[\text{H}_2]$
$\text{rate}=k[\text{H}_2][\text{I}_2]$
$\text{rate}=k[\text{I}]^2$
$\text{rate}=k[\text{HI}]^2$
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 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.
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?
$\text{rate}=k[\text{NO}]^2[\text{O}_2]$
$\text{rate}=k[\text{NO}]$
$\text{rate}=k[\text{NO}_2]^2$
$\text{rate}=k[\text{NO}_3]$
$\text{rate}=k[\text{NO}_2]$
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.
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?
$\text{rate}=k[\text{O}_2]$
$\text{rate}=k[\text{NO}][\text{O}_2]$
$\text{rate}=k[\text{NO}_3][\text{NO}]$
$\text{rate}=k[\text{NO}]$
$\text{rate}=k[\text{NO}]^2[\text{O}_2]$
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.
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?
$\text{rate}=k[\text{CH}_3^+][\text{OH}^-]$
$\text{rate}=k[\text{CH}_3\text{OH}]$
$\text{rate}=k[\text{Br}^-]$
$\text{rate}=k[\text{CH}_3\text{Br}][\text{OH}^-]$
$\text{rate}=k[\text{CH}_3\text{Br}]$
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.