A bimolecular elementary step A + B → products has rate law

Answer: rate = k[A][B] Bimolecular collision: first order in each reactant.

For an elementary step, the rate law

Answer: Comes from the balanced step (molecularity) Only for elementary steps do exponents in the rate law match stoichiometric coefficients.

Termolecular elementary steps are rare because:

Answer: Three-body simultaneous collisions are improbable The probability of three molecules colliding simultaneously with proper orientation and energy is very low.

AP Chemistry · Topic 5.4

Elementary Reactions Practice

Part of Kinetics.(TRA-4.A)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$A bimolecular elementary step A + B \to products has rate law$
- A
  $rate = k[A]²[B]²$
- B
  $rate = k[A]$
- C
  $rate = k$
- D
  $rate = k[A][B]$
  check_circle
Why
$Bimolecular collision: first order in each reactant.$
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Sample 2difficulty 2/5
$For an elementary step, the rate law$
- A
  $Always equals overall rate law$
- B
  $Is always first order$
- C
  $Cannot be deduced$
- D
  $Comes from the balanced step (molecularity)$
  check_circle
Why
$Only for elementary steps do exponents in the rate law match stoichiometric coefficients.$
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Sample 3difficulty 3/5
$Termolecular elementary steps are rare because:$
- A
  $They require a catalyst$
- B
  $Three-body simultaneous collisions are improbable$
  check_circle
- C
  $They are impossible by quantum rules$
- D
  $They violate conservation of energy$
Why
$The probability of three molecules colliding simultaneously with proper orientation and energy is very low.$
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