Introduction to Reaction Mechanisms
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AP Chemistry › Introduction to Reaction Mechanisms
A student proposes the following mechanism for the overall reaction $\text{Cl}_2(g)+\text{CH}_4(g)\rightarrow \text{CH}_3\text{Cl}(g)+\text{HCl}(g)$:
Step 1: $\text{Cl}_2(g)\rightarrow 2\text{Cl}(g)$
Step 2: $\text{Cl}(g)+\text{CH}_4(g)\rightarrow \text{HCl}(g)+\text{CH}_3(g)$
Step 3: $\text{CH}_3(g)+\text{Cl}_2(g)\rightarrow \text{CH}_3\text{Cl}(g)+\text{Cl}(g)$
Which species is an intermediate in the mechanism?
$\text{HCl}(g)$ because it is produced in Step 2.
$\text{Cl}_2(g)$ because it is consumed in Step 1 and Step 3.
$\text{CH}_4(g)$ because it is consumed in Step 2.
$\text{CH}_3\text{Cl}(g)$ because it is produced in Step 3.
$\text{Cl}(g)$ because it is produced in Step 1 and consumed in Step 2 while also being regenerated in Step 3.
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.
A proposed mechanism for the reaction $\text{H}_2(g)+\text{Br}_2(g)\rightarrow 2\text{HBr}(g)$ is:
Step 1: $\text{Br}_2(g)\rightarrow 2\text{Br}(g)$
Step 2: $\text{Br}(g)+\text{H}_2(g)\rightarrow \text{HBr}(g)+\text{H}(g)$
Step 3: $\text{H}(g)+\text{Br}_2(g)\rightarrow \text{HBr}(g)+\text{Br}(g)$
Which species is an intermediate in the mechanism?
$\text{HBr}(g)$ because it is formed in Steps 2 and 3.
$\text{Br}(g)$ because it is produced in Step 1 and consumed in Step 2 (and regenerated in Step 3).
$\text{H}_2(g)$ because it is consumed in Step 2.
$\text{Br}_2(g)$ because it is consumed in Steps 1 and 3.
$\text{H}(g)$ 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.
A proposed mechanism for the reaction $\text{N}_2\text{O}_5(g)\rightarrow 2\text{NO}_2(g)+\tfrac{1}{2}\text{O}_2(g)$ is:
Step 1: $\text{N}_2\text{O}_5(g)\rightarrow \text{NO}_2(g)+\text{NO}_3(g)$
Step 2: $2\text{NO}_3(g)\rightarrow 2\text{NO}_2(g)+\text{O}_2(g)$
Which statement best describes the overall consistency of the mechanism with the net reaction?
The mechanism is consistent only if $\text{NO}_3(g)$ is included in the net reaction.
The mechanism is inconsistent because $\text{O}_2(g)$ must cancel when steps are added.
The mechanism is consistent because $\text{NO}_3(g)$ cancels and the summed steps yield the net reaction (after scaling).
The mechanism is inconsistent because $\text{NO}_3(g)$ appears as a reactant in Step 2.
The mechanism is inconsistent because $\text{NO}_2(g)$ is an intermediate and should not appear in the net reaction.
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.
A proposed mechanism for the overall reaction $\text{2NO}(g)+\text{O}_2(g)\rightarrow 2\text{NO}_2(g)$ is:
Step 1: $\text{NO}(g)+\text{O}_2(g)\rightarrow \text{NO}_3(g)$
Step 2: $\text{NO}_3(g)+\text{NO}(g)\rightarrow 2\text{NO}_2(g)$
Which statement is correct about the mechanism?
The mechanism is consistent because $\text{NO}_3(g)$ cancels when the steps are added.
The mechanism is inconsistent because $\text{NO}(g)$ cancels completely and cannot be a reactant overall.
The mechanism is inconsistent because $\text{O}_2(g)$ must be produced in one of the steps.
The mechanism is consistent only if $\text{NO}_3(g)$ is treated as a catalyst.
The mechanism is inconsistent because $\text{NO}_3(g)$ is produced and should appear in the net reaction.
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.
A proposed mechanism for the reaction $\mathrm{2NO(g) + Br_2(g) \rightarrow 2NOBr(g)}$ is shown.
Step 1: $\mathrm{NO(g) + Br_2(g) \rightarrow NOBr_2(g)}$
Step 2: $\mathrm{NOBr_2(g) + NO(g) \rightarrow 2NOBr(g)}$
Which species is an intermediate in the mechanism?
$\mathrm{Br(g)}$
$\mathrm{NO(g)}$
$\mathrm{NOBr_2(g)}$
$\mathrm{NOBr(g)}$
$\mathrm{Br_2(g)}$
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.
A proposed mechanism for the reaction $\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}$ is shown below:
Step 1: $\mathrm{NO(g) + NO(g) \rightarrow N_2O_2(g)}$
Step 2: $\mathrm{N_2O_2(g) + O_2(g) \rightarrow 2NO_2(g)}$
Which species is an intermediate in this mechanism?
$\mathrm{NO(g)}$
$\mathrm{NO_2(g)}$
$\mathrm{O_2(g)}$
$\mathrm{2NO(g) + O_2(g)}$
$\mathrm{N_2O_2(g)}$
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.
A student proposes the following mechanism for the net reaction $\mathrm{ClO^-(aq) + 2,I^-(aq) + 2,H^+(aq) \rightarrow I_2(aq) + Cl^-(aq) + H_2O(l)}$:
Step 1: $\mathrm{ClO^-(aq) + H^+(aq) \rightarrow HOCl(aq)}$
Step 2: $\mathrm{HOCl(aq) + I^-(aq) \rightarrow HOI(aq) + Cl^-(aq)}$
Step 3: $\mathrm{HOI(aq) + I^-(aq) + H^+(aq) \rightarrow I_2(aq) + H_2O(l)}$
Which species acts as an intermediate in this mechanism?
$\mathrm{ClO^-(aq)}$
$\mathrm{H^+(aq)}$
$\mathrm{Cl^-(aq)}$
$\mathrm{HOCl(aq)}$
$\mathrm{I_2(aq)}$
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.
A student proposes the following mechanism for the overall reaction $\text{H}_2(g)+\text{I}_2(g)\rightarrow 2\text{HI}(g)$:
Step 1: $\text{I}_2(g)\rightarrow 2\text{I}(g)$
Step 2: $\text{H}_2(g)+\text{I}(g)\rightarrow \text{HI}(g)+\text{H}(g)$
Step 3: $\text{H}(g)+\text{I}(g)\rightarrow \text{HI}(g)$
Which species is an intermediate in the mechanism?
$\text{HI}(g)$ because it is produced in Step 2 and Step 3.
$\text{H}(g)$ because it appears in the net reaction.
$\text{I}(g)$ because it is produced in Step 1 and consumed in Steps 2 and 3.
$\text{I}_2(g)$ because it is consumed in Step 1.
$\text{H}_2(g)$ because it is consumed in Step 2.
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.
A proposed mechanism for the reaction $2\text{NO}(g)+\text{Br}_2(g)\rightarrow 2\text{NOBr}(g)$ is:
Step 1: $\text{NO}(g)+\text{Br}_2(g)\rightarrow \text{NOBr}_2(g)$
Step 2: $\text{NOBr}_2(g)+\text{NO}(g)\rightarrow 2\text{NOBr}(g)$
Which species is an intermediate in this mechanism?
$\text{Br}(g)$ because it is implied by the presence of bromine.
$\text{NOBr}(g)$ because it is produced in Step 2.
$\text{NO}(g)$ because it is consumed in Step 1 and Step 2.
$\text{NOBr}_2(g)$ because it is produced in Step 1 and consumed in Step 2.
$\text{Br}_2(g)$ because it is consumed in Step 1.
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.
A proposed mechanism for the overall reaction $\text{O}_3(g)+\text{O}(g)\rightarrow 2\text{O}_2(g)$ is:
Step 1: $\text{O}_3(g)+\text{Cl}(g)\rightarrow \text{ClO}(g)+\text{O}_2(g)$
Step 2: $\text{ClO}(g)+\text{O}(g)\rightarrow \text{Cl}(g)+\text{O}_2(g)$
Which species acts as a catalyst in the mechanism?
$\text{O}(g)$ because it is consumed in Step 2.
$\text{O}_3(g)$ because it is consumed in Step 1.
$\text{ClO}(g)$ because it is produced in Step 1 and consumed in Step 2.
$\text{Cl}(g)$ because it is consumed in Step 1 and regenerated in Step 2.
$\text{O}_2(g)$ because it is produced in both steps.
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.