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A reaction mechanism is a step-by-step sequence of elementary reactions that shows how a chemical reaction actually happens at the molecular level. The rate-determining step (RDS) is the slowest step in this sequence—it acts like a bottleneck, controlling the overall reaction rate. This topic is critical on the AP exam because it connects kinetics (how fast reactions occur) to molecular behavior. For example, imagine a factory assembly line: if one station (like painting) is slower than the others, it limits how fast the entire product can be made—this is the RDS in action. A real-world example is the ozone depletion reaction (O₃ + O → 2O₂), where the slow step determines how quickly ozone breaks down in the atmosphere.
Problem: Given the mechanism below, derive the rate law for the overall reaction: Overall: 2NO + O₂ → 2NO₂ Mechanism:1. NO + NO ⇌ N₂O₂ (fast equilibrium) 2. N₂O₂ + O₂ → 2NO₂ (slow, RDS)
Steps:1. Identify the RDS: The slow step (Step 2) is the RDS. Write its rate law directly from its stoichiometry: - Rate = k₂[N₂O₂][O₂] 2. Express intermediates in terms of reactants: N₂O₂ is an intermediate (produced in Step 1, consumed in Step 2). Since Step 1 is a fast equilibrium, use its equilibrium constant (K): - K = [N₂O₂] / [NO]² → [N₂O₂] = K[NO]² 3. Substitute into the RDS rate law: - Rate = k₂(K[NO]²)[O₂] = k[NO]²[O₂] (where k = k₂K) 4. Check the overall order: The rate law is second-order in NO and first-order in O₂, matching experimental data for this reaction.
Key Takeaway: If the RDS involves an intermediate, always replace it using equilibrium expressions from earlier steps.
Mistake: Assuming the rate law comes from the overall reaction stoichiometry. Correction: Rate laws are determined experimentally or from the RDS. The overall reaction is just a summary—it doesn’t show the mechanism!
Mistake: Forgetting that intermediates cannot appear in the final rate law. Correction: Intermediates must be replaced using equilibrium expressions (e.g., [N₂O₂] = K[NO]²) or the steady-state approximation.
Mistake: Confusing molecularity (for elementary steps) with order (for the overall rate law). Correction: Molecularity applies to individual steps (e.g., a bimolecular step has 2 molecules colliding). Order is the exponent in the final rate law (e.g., rate = k[A]² is second-order in A).
Mistake: Ignoring the fast equilibrium assumption when deriving rate laws. Correction: If the first step is fast and reversible, use its equilibrium constant (K) to substitute for intermediates.
Mistake: Drawing energy profiles with intermediates as peaks (instead of valleys). Correction: Intermediates are stable (valleys), while transition states are unstable (peaks). The RDS has the highest peak (largest Eₐ).
Justify why a step is the RDS (e.g., "Step 2 is slow because it has a high activation energy").
Multiple-Choice Traps:
Trick: A catalyst is included in the mechanism. Remember: catalysts appear in the rate law (e.g., rate = k[A][catalyst]) but aren’t consumed.
Energy Diagrams: You might be asked to:
Explain why a catalyst lowers the Eₐ of the RDS.
Terminology Pitfalls:
Answer: (A) rate = k[A][B] Explanation: The slow step (Step 1) is the RDS, and its rate law is written directly from its stoichiometry.
Answer: (a) Intermediate: NOCl₂ (produced in Step 1, consumed in Step 2). (b) Rate Law: - RDS is Step 2: rate = k₂[NOCl₂][NO] - From Step 1 (fast equilibrium): K = [NOCl₂] / [NO][Cl₂] → [NOCl₂] = K[NO][Cl₂] - Substitute: rate = k₂(K[NO][Cl₂])[NO] = k[NO]²[Cl₂] (where k = k₂K)
Answer: (C) It is always the first step in the mechanism. Explanation: The RDS can be any step in the mechanism—it’s just the slowest one!
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