A two-step mechanism has a fast equilibrium step followed by a slow step. The rate-determining step is:

Answer: The slow second step The slowest step (highest Ea) is rate-determining.

What is the predicted rate law?

Answer: rate = k[A][B][C] Slow step rate = k2[I][C]. Substituting [I] = K[A][B] gives rate = k2K[A][B][C] = k[A][B][C].

A reactive intermediate I has [I] that quickly rises and then stays roughly constant. This justifies which approximation?

Answer: Steady-state approximation When d[I]/dt ≈ 0 because formation and consumption rates balance, the steady-state approximation applies.

AP Chemistry · Topic 5.9

Steady-State Approximation Practice

Part of Kinetics.(TRA-5.C)

Practice questions

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Sample questions

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Sample 1difficulty 3/5
$A two-step mechanism has a fast equilibrium step followed by a slow step. The rate-determining step is:$
- A
  $The first equilibrium step$
- B
  $The slow second step$
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- C
  $Whichever has more reactants$
- D
  $Both steps equally$
Why
$The slowest step (highest Ea) is rate-determining.$
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Sample 2difficulty 4/5
$Two-step mechanism with a fast pre-equilibrium and a slow second step.$
$What is the predicted rate law?$
- A
  $rate = k[I][C]$
- B
  $rate = k[C]$
- C
  $rate = k[A][B]$
- D
  $rate = k[A][B][C]$
  check_circle
Why
$Slow step rate = k2[I][C]. Substituting [I] = K[A][B] gives rate = k2K[A][B][C] = k[A][B][C].$
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Sample 3difficulty 4/5
$A reactive intermediate I has [I] that quickly rises and then stays roughly constant. This justifies which approximation?$
- A
  $Equilibrium approximation only$
- B
  $Zero-order assumption$
- C
  $Catalytic cycle assumption$
- D
  $Steady-state approximation$
  check_circle
Why
$When d[I]/dt \approx 0 because formation and consumption rates balance, the steady-state approximation applies.$
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Sample 4difficulty 4/5
$In a fast pre-equilibrium followed by a slow step, intermediate [I] is expressed by$
- A
  $Always negligibly small$
- B
  $An equilibrium expression of step 1$
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- C
  $Cannot be expressed$
- D
  $Set equal to [reactants]$
Why
$Pre-equilibrium: K = [I]/[reactants in step 1], allowing [I] to be replaced in the rate-determining step.$
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Sample 5difficulty 4/5
$The steady-state approximation assumes$
- A
  $Eₐ is constant$
- B
  $[Reactants] is constant$
- C
  $[Intermediate] is constant$
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- D
  $T is constant$
Why
$d[I]/dt \approx 0 — formation rate \approx consumption rate of the intermediate.$
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