What is the rate law for elementary step A + B - D?

Answer: rate = k[A][B] For an elementary bimolecular step, rate = k × product of reactant concentrations.

A proposed mechanism is acceptable if

Answer: Steps sum to the overall reaction AND predicted rate law matches experiment Both: stoichiometry sum AND consistent predicted rate law are required.

Which step is rate-determining?

Answer: Step 2 The step with highest activation energy (highest TS relative to its starting valley) determines the rate. Here TS2 is highest.

AP Chemistry · Topic 5.8

Reaction Mechanisms and Rate Law Practice

Part of Kinetics.(TRA-5.B)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$Rate laws for elementary steps come directly from the molecularity (stoichiometric coefficients).$
$What is the rate law for elementary step A + B -> D?$
- A
  $rate = k[D]$
- B
  $rate = k[A]+[B]$
- C
  $rate = k[A]^2[B]$
- D
  $rate = k[A][B]$
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Why
$For an elementary bimolecular step, rate = k \times product of reactant concentrations.$
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Sample 2difficulty 3/5
$A proposed mechanism is acceptable if$
- A
  $Only matches the rate law$
- B
  $Only contains elementary steps$
- C
  $Just sums to overall reaction$
- D
  $Steps sum to the overall reaction AND predicted rate law matches experiment$
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Why
$Both: stoichiometry sum AND consistent predicted rate law are required.$
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Sample 3difficulty 3/5
$A three-step mechanism's energy profile shows TS2 highest, then TS3, then TS1.$
$Which step is rate-determining?$
- A
  $Step 1$
- B
  $All equal$
- C
  $Step 3$
- D
  $Step 2$
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Why
$The step with highest activation energy (highest TS relative to its starting valley) determines the rate. Here TS2 is highest.$
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Sample 4difficulty 3/5
$The overall rate of a multi-step mechanism is determined by$
- A
  $Sum of all steps$
- B
  $The fastest step$
- C
  $The slowest (rate-determining) step$
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- D
  $The exothermic step$
Why
$The slowest step bottlenecks the entire mechanism.$
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Sample 5difficulty 3/5
$A reaction A + B \to C is proposed to proceed by mechanism: (i) A + A ⇌ A₂ fast; (ii) A₂ + B \to 2C slow.$
$What is the predicted rate law?$
- A
  $rate = k[A₂][B]$
- B
  $rate = k[A]²[B]$
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- C
  $rate = k[A][B]$
- D
  $rate = k[A][B]²$
Why
$The slow step gives rate = k₂[A₂][B]. Substituting [A₂] = K[A]² gives rate = k₂K[A]²[B] = k[A]²[B].$
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