Questions: Molecularity vs Reaction Order: Elementary and Complex Reactions

Question 1

A reaction has the stoichiometry 2NO + O₂ → 2NO₂. A student writes the rate law as rate = k[NO]²[O₂]. When is this guaranteed to be correct?

• A. Always — stoichiometric coefficients always equal the reaction orders
• B. Only if the reaction is elementary (occurs in a single step with no intermediates)
• C. Only if the reaction is carried out at high temperature
• D. Never — rate laws must always be determined by experiment regardless of mechanism

Answer: (B)

Stoichiometric coefficients equal the rate law exponents ONLY for elementary reactions. For an elementary step, molecularity determines the rate law mechanistically. For a multi-step mechanism, the overall rate law depends on which step is rate-limiting and how intermediates relate — the result can be fractional, zero, or even negative order in a species, with no necessary connection to the balanced equation. The student's rate law may happen to be correct if the reaction is elementary, but you cannot assume it is without mechanistic evidence.

Question 2

Ozone decomposes via 2O₃ → 3O₂ with the experimentally measured rate law: rate = k[O₃]²[O₂]⁻¹. What does the negative order in O₂ indicate?

• A. O₂ is a reactant being consumed, which always produces negative order
• B. The reaction proceeds through a multi-step mechanism where a fast equilibrium produces O₂ as a product that inhibits the rate-limiting step
• C. The experimenter made an error — orders cannot be negative
• D. O₂ has a molecularity of −1 in the rate-determining step

Answer: (B)

Negative reaction order is impossible to rationalize from stoichiometry but makes perfect sense mechanistically. In the ozone mechanism, the fast pre-equilibrium O₃ ⇌ O₂ + O produces an oxygen atom intermediate. The slow step is O + O₃ → 2O₂. Applying the pre-equilibrium approximation yields rate = k[O₃]²[O₂]⁻¹ — the O₂ produced by the fast step accumulates and drives it backward, reducing the concentration of the intermediate and slowing the overall reaction. This is only possible in a multi-step mechanism; no elementary step can have negative molecularity.

Question 3

For an elementary reaction, molecularity and reaction order are the same thing.

• T. True
• F. False

Answer: True

True. For an elementary reaction — one that occurs in a single molecular event with no intermediates — the rate law follows directly from molecularity. A unimolecular step A → products has rate = k[A] (first order); a bimolecular step A + B → products has rate = k[A][B] (second order overall). This is not empirical but mechanistically necessary: if two molecules must collide for the reaction to happen, the rate must depend on the concentration of both. This identity between molecularity and order breaks down entirely for multi-step (complex) reactions.

Question 4

Termolecular elementary steps are common in gas-phase reactions because three molecules can easily collide with sufficient combined energy.

• T. True
• F. False

Answer: False

False. Genuine termolecular elementary steps are exceedingly rare because they require three molecules to collide simultaneously — a statistically improbable event. The probability of a two-body collision is already concentration-dependent; requiring a third body to be present at exactly the right moment and orientation makes simultaneous three-body collisions extremely infrequent. Most reactions that appear termolecular from their stoichiometry actually proceed through two sequential bimolecular steps. This is why unimolecular and bimolecular steps account for nearly all elementary steps in known mechanisms.

Question 5

Explain why the overall reaction order for a multi-step reaction cannot be read from the stoichiometric coefficients of the balanced equation, and give the key condition under which stoichiometry DOES determine the rate law.

Model answer: For a multi-step reaction, the overall rate law is determined by the mechanism — specifically by the rate-limiting step and the steady-state or pre-equilibrium relationships between intermediates. The stoichiometric coefficients describe the overall change in matter, not which molecules are colliding in any single step. Intermediates can appear in the rate expression even though they are absent from the balanced equation. Stoichiometry determines the rate law only for elementary reactions, where the reaction occurs in a single step and molecularity (the count of reacting molecules) directly dictates the rate law exponents.

The key insight is that molecularity is a mechanistic concept (how many molecules collide in one step), while reaction order is an empirical measurement of the overall kinetics. They coincide only when the reaction has a single elementary step. Any multi-step mechanism can produce any overall order, including fractional and negative orders, depending on the topology of the mechanism and which step limits the rate.