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.

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.

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.

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.

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.

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.

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.

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.

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.

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.