Reaction Mechanisms and Rate Laws – Comprehensive Notes

  • Key ideas in reaction mechanisms and kinetics

    • Reaction mechanism = the sequence of elementary steps that describes what individual molecules do to convert reactants to products. It goes beyond the overall observed stoichiometry and rate law.
    • Elementary step = a single molecular event (e.g., a collision between two specific molecules) with its own rate law. For a bimolecular elementary step, the rate depends on the concentrations of the two reactants involved.
    • Rate law vs. mechanism
    • The observed (macroscopic) rate law comes from the combination of all steps and, in particular, the rate-determining step.
    • You cannot judge the order in the overall reaction simply from the overall stoichiometry; the rate law reflects the elementary steps and the slowest step.
    • Rate-determining step (RDS) = the slowest elementary step that controls the overall rate of the reaction. The overall rate cannot exceed the pace set by this step.
    • Intermediates = species formed in one step and consumed in a later step; they do not appear in the overall balanced equation.
    • Law of mass action = for an elementary step aA + bB → products, the rate law is proportional to the concentrations raised to the stoichiometric powers: rate ∝ [A]^a[B]^b.
    • Equilibrium concept in kinetics = forward and reverse rates can be equal at equilibrium; this leads to definitions of the equilibrium constant K_eq and relationships between rate constants.
    • Pre-equilibrium approximation = when the first steps of a mechanism rapidly reach equilibrium (fast) relative to a slower subsequent step, the intermediate can be eliminated from the rate law to yield a rate law in terms of reactants only.
    • Practical approach to mechanisms = propose a mechanism, test two things: (i) do the steps sum to the observed overall reaction, and (ii) does the predicted rate law match experimental data? Experimental data is essential to validate any proposed mechanism.

  • Example 1: Two-step mechanism for 2A + C → 2D

    • Proposed mechanism
    • Step 1 (elementary, forms intermediate): 2A → I (slower, i.e., the rate-determining step in this example)
    • Step 2 (elementary): C + I → 2D
    • Overall reaction implied by combining steps
    • Sum: 2A + C → 2D
    • The intermediate I cancels between the steps when adding them.
    • Rate laws for the steps (for reference)
    • Step 1 rate: r1 = k1 [A]^2
    • Step 2 rate: r2 = k2 [I][C]
    • From the mechanism, the rate of the overall reaction is governed by the slow (rate-determining) step, which is Step 1 here, giving the observed rate law:
    • r = k_1 [A]^2 = k [A]^2
    • Test criteria for an acceptable mechanism (as applied here)
    • The sum of elementary steps must equal the observed overall reaction: achieved, giving 2A + C
      ightarrow 2D.
    • The predicted rate law from the mechanism must match the experimental rate law: here, predicted r = k_1 [A]^2 matches the observed second-order dependence on A and no dependence on C.
    • Notes on experimentation and interpretation
    • The rate law is derived from the slowest elementary step, not from the overall stoichiometry.
    • It is possible to conceive other mechanisms that fit the same rate law, but this example shows a mechanism that is consistent with both stoichiometry and the rate law.
    • Practical takeaway
    • When proposing a mechanism, you should ensure both the stoichiometric sum and the predicted rate law align with experimental data; you may need to consider alternative steps, especially if the slow step could be different.

  • Example 2: NO₂ + CO mechanism and rate law validation

    • Proposed mechanism (two steps)
    • Step 1 (slow): 2 NO₂ → NO₃ + NO
    • Step 2 (fast): NO₃ + CO → NO₂ + CO₂
    • Net reaction from steps
    • Sum: 2 NO₂ + CO → NO + CO₂
    • Predicted rate law from mechanism
    • Since Step 1 is the slow (rate-determining) step, the rate is:
    • r = k1 [ ext{NO}2]^2
    • Experimental check
    • The observed rate law is second-order in NO₂ and independent of CO, which is consistent with the proposed mechanism where the RDS involves only NO₂ collisions.
    • Additional discussion
    • If the second step were the RDS, the rate law would involve the intermediate NO₃ and CO and would require a different derivation (e.g., via steady-state or pre-equilibrium approximations).
    • Takeaway
    • The RDS dominates the form of the rate law; intermediates can appear in intermediate steps, but the final rate law may avoid explicit intermediates if justified by the mechanism and data.

  • Important concepts about equilibrium in kinetics

    • Simple reversible elementary step example
    • A + B ⇌ AB with forward rate rf = k1[A][B] and reverse rate rr = k−1[AB]
    • At equilibrium: rf = rr, so k1[A][B] = k{-1}[AB]
    • Equilibrium constant (law of mass action):
      • K{eq} = rac{[AB]}{[A][B]} = rac{k1}{k_{-1}}
    • Observations
      • The ratio of product to reactants at equilibrium is determined by the rate constants (forward and reverse).
      • The equilibrium constant is related to the rate constants but is a separate quantity (uppercase K_eq vs lowercase k).
    • Practical role of equilibrium in mechanism analysis
    • When a step in a mechanism reaches equilibrium rapidly relative to slower steps, you can use the equilibrium relationship to eliminate intermediates from rate laws (pre-equilibrium approximation).

  • Pre-equilibrium approximation with a three-step example (NO-atom-like mechanism)

    • General setup
    • Mechanism with an intermediate I formed in a fast pre-equilibrium step and consumed in a slower second step:
      • Step 1: A + B ⇌ I (fast forward and fast reverse)
      • Step 2: I → products (slow, rate-determining)
    • Rate law implications
    • The overall rate is determined by the slow Step 2:
      • r = k_2 [I]
    • Since I is an intermediate, you aim to express [I] in terms of reactants using the pre-equilibrium condition for Step 1:
      • Forward rate for Step 1: r{f1} = k1 [A][B]
      • Reverse rate for Step 1: r{-1} = k{-1} [I] (assuming the reverse involves I and the appropriate reactant concentrations; the exact expression depends on the step)
      • At pre-equilibrium: r{f1} \,=\ r{-1} \Rightarrow [I] = \frac{k1}{k{-1}} \,[A][B] / [ ext{(something involving a reactant/condition depending on the mechanism)}]
    • Substituting back yields a rate law in terms of reactants only.
    • Ozone decomposition as a concrete pre-equilibrium example
    • Mechanism (illustrative):
      • Step 1: ext{O}3 ightleftharpoons ext{O} + ext{O}2 (fast, equilibrium)
      • Step 2: ext{O} + ext{O}3 ightarrow 2 ext{O}2 (slow, rate-determining)
    • Predicted rate law (without pre-equilibrium):
      • r = k2 [ ext{O}][ ext{O}3]
      • This contains the intermediate [O], which is not desirable in a rate law.
    • Using the pre-equilibrium relation for Step 1:
      • At equilibrium: k1 [ ext{O}3] = k{-1} [ ext{O}][ ext{O}2]
      • Solve for [O]: [ ext{O}] = rac{k1}{k{-1}} rac{[ ext{O}3]}{[ ext{O}2]}
    • Substitute into the rate law for Step 2:
      • r = k2 [ ext{O}3] [ ext{O}] = k2 igg( rac{k1}{k{-1}} rac{[ ext{O}3]}{[ ext{O}2]} igg) [ ext{O}3] = rac{k2 k1}{k{-1}} rac{[ ext{O}3]^2}{[ ext{O}_2]}
    • Define an effective rate constant for the overall process:
      • k \,\equiv\ rac{k2 k1}{k_{-1}}
      • Therefore, the observed rate law is:
      • r = k \frac{[ ext{O}3]^2}{[ ext{O}2]}
    • Key takeaway: Pre-equilibrium allows elimination of the intermediate to yield a rate law in terms of reactants only, when the first steps rapidly equilibrate and the second step is slow.
    • Important caveats about pre-equilibrium
    • The pre-equilibrium approximation requires justification from rate data: the first steps must be fast relative to the slow step.
    • If Step 2 were not slow, or if the equilibrium assumption did not hold, the derived rate law would be unreliable.

  • Equilibrium concepts revisited for clarity

    • Forward and reverse rates and equilibrium
    • For a generic A + B ⇌ C reaction with two rate constants k1 and k−1, the forward and reverse rates are:
      • Forward: rf = k1 [A][B]
      • Reverse: rr = k{-1} [C]
    • At equilibrium: rf = rr
    • Equilibrium constant from rate constants:
      • K{eq} = rac{[C]}{[A][B]} = rac{k1}{k_{-1}}
    • Relationship between rate constants and equilibrium
    • The equilibrium position is determined by the relative sizes of the forward and reverse rate constants. A faster forward reaction (larger k1) generally pushes the equilibrium toward products (larger [C]), reflected in a larger K_eq.
    • Practical note on notation
    • Lowercase k usually denotes a rate constant for a particular elementary step.
    • Uppercase K usually denotes an equilibrium constant (dimensionless in ideal conditions, based on activities or concentrations).

  • Ongoing questions and discussion points (conceptual clarifications)

    • If there are more than two steps, there can be a step before the rate-determining step that contributes to the observed rate law.
    • If the second step were the slow one in the NO₂/CO example, the rate law would involve the intermediate NO₃ and CO, and further analysis (steady-state or pre-equilibrium) would be required to express the rate in terms of reactants.
    • In many cases, reactions go to equilibrium rather than to completion; forward and reverse rates become equal, and concentrations stop changing although the individual rates remain nonzero.

  • Quick recap: how to evaluate a proposed mechanism

    • Check 1: Do the elementary steps sum to the experimentally observed overall reaction?
    • Check 2: Does the predicted rate law from the mechanism (considering the RDS and any applicable approximations like pre-equilibrium) match the experimentally observed rate law?
    • If both checks pass, the mechanism is acceptable as a plausible description of the process; others may exist, but this one is consistent with data.

  • Practical exam-style tips

    • Always identify the rate-determining step to infer the form of the rate law from an elementary-step mechanism.
    • Use the sum of steps to verify the overall reaction; intermediates should cancel in the net equation.
    • When intermediates appear in a rate law, consider whether a pre-equilibrium or steady-state approximation can eliminate them to give a rate law in terms of reactants.
    • Distinguish between rate constants for elementary steps (lowercase k) and equilibrium constants (uppercase K) and know how they relate via the law of mass action.

  • Summary of key equations to memorize (LaTeX)

    • Overall reaction from a two-step mechanism: 2A + C
      ightarrow 2D
    • Rate law for the NO₂ + CO mechanism (RDS on Step 1): r = k1 [ ext{NO}2]^2
    • O3 decomposition via pre-equilibrium (illustrative):
    • Step 1: ext{O}3 ightleftharpoons ext{O} + ext{O}2 \text{forward: } r1 = k1 [ ext{O}3], ext{ reverse: } r{-1} = k{-1} [ ext{O}][ ext{O}2]
    • Step 2: ext{O} + ext{O}3 ightarrow 2 ext{O}2 \ r2 = k2 [ ext{O}][ ext{O}_3]
    • Pre-equilibrium result: [ ext{O}] \,=\ rac{k1}{k{-1}} \frac{[ ext{O}3]}{[ ext{O}2]}
    • Resulting rate law (eliminating the intermediate): r = k \, rac{[ ext{O}3]^2}{[ ext{O}2]}, \text{where } k = rac{k2 k1}{k_{-1}}
    • Equilibrium constant for a simple reversible step: A + B
      ightleftharpoons AB, \ rf = k1 [A][B], \ rr = k{-1} [AB], \ k1 [A][B] = k{-1} [AB] \ K{eq} = rac{[AB]}{[A][B]} = rac{k1}{k_{-1}}
    • Net rate constant in a pre-equilibrium scenario (example): k{obs} = rac{k2 k1}{k{-1}}

  • End of notes: these concepts connect the microscopic (molecular) steps to the macroscopic (measured) rate laws, showing how experimental data informs which mechanism is plausible and how to manipulate rate laws to remove intermediates for practical analysis.