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5.9Kinetics

Pre-Equilibrium Approximation — AP CHEMISTRY Revision Notes & Topical Questions

5 min read 8 parts Lesson 12 of 14 1 quiz

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Lesson summary

5.9 Pre-Equilibrium Approximation — AP CHEMISTRY notes: Eliminate an intermediate from the rate law when a fast reversible step precedes the rate-limiting step.Situation — a fast reversible first step is followed by a slow rate-limiting step that uses an intermediateThe rate law …

  • Equality of rates
  • Step 1 (fast, reversible)
  • Step 2 (slow)
  • Substitution
  • Step 1 (fast)
  • What the symbols mean

Section 1 of 8When it is needed

  • Situation
    • a fast reversible first step is followed by a slow rate-limiting step that uses an intermediate
    • The rate law of the slow step contains the intermediate, which cannot appear in the experimental rate law.

Section 2 of 8The approximation

1Pre-equilibrium approximation
  • Pre-equilibrium approximation the fast initial step is assumed to reach equilibrium
  • The forward and reverse rates of that step are equal.
2Equality of rates
  • Equality of rates kf[reactants] = kb[intermediate]
  • Rearranging gives [intermediate] = (kf/kb)[reactants].

Section 3 of 8Procedure

1
Write the rate law from the slow step, including the intermediate.
2
Write the forward and reverse rate expressions for the fast step and set them equal.
3
Solve for the intermediate concentration in terms of reactants.
4
Substitute into the slow-step rate law.
5
Combine all constants into one observed rate constant.

Section 4 of 8Pattern: 2NO + Br₂ → 2NOBr

1Step 1 (fast, reversible)
  • NO + Br2 ⇌ NOBr2
2Step 2 (slow)
  • NOBr2 + NO → 2NOBr, rate = k2[NOBr2][NO]
3Substitution
  • [NOBr2] = (kf/kb)[NO][Br2]
4Result
  • Result rate = k′[NO]2[Br2], matching the experimental law
  • A termolecular step is avoided.

Section 5 of 8Pattern: H₂ + I₂ → 2HI

1Step 1 (fast)
  • I2 ⇌ 2I
2Step 2 (slow)
  • H2 + 2I → 2HI, rate = k[H2][I]2
    What the symbols mean: rate (mol dm−3 s−1) · k = rate constant · [ ] = concentration (mol dm−3) · each power is the order with respect to that reactant
    • (units depend on the overall order)
3Using step 1
  • [I]2 = (kf/kb)[I2], so rate = k′[H2][I2]

Section 6 of 8More from the specification

1
Endothermic
  • A reaction that takes in thermal energy from the surroundings.
2
Energy profile
  • Diagram showing energy during a reaction
3
Elementary reaction
  • A single molecular event; one step of a mechanism.
4
Activation energy
  • The minimum energy colliding particles need to react.
5
Elementary step
  • A single stage in a reaction mechanism.
6
Chemical equation
  • A statement using formulas to show reactants and products.
7
Mechanism
  • The sequence of steps by which reactants form products.

Section 7 of 8Key terms

TermMeaning
Pre-equilibrium approximationAssumption that a fast reversible step is at equilibrium.
Fast stepStep whose forward and reverse rates quickly become equal.
IntermediateSpecies eliminated from the rate law using the equilibrium expression.
Observed rate constantCombined constant in the final rate law, k·(kf/kb).

Section 8 of 8Exam technique

Avoid these mistakes
Leaving the intermediate in the final rate law — it must be replaced by reactant concentrations.
Applying the approximation when the first step is the slow step — it is used only when a fast reversible step comes first.
Forgetting that kf/kb is a constant — it is merged into the observed rate constant.

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