Questions: Unimolecular Reactions: Lindemann and RRKM Theory

Question 1

A chemist measures the rate of a gas-phase isomerization at various pressures and finds: rate = k[A][M] at very low pressure, but rate = k[A] at high pressure. What causes the rate law to change?

• A. The reaction mechanism changes from unimolecular to bimolecular at low pressure
• B. At low pressure, the activation step (collision with M) is rate-limiting; at high pressure, the unimolecular reaction of A* is rate-limiting
• C. The transition state structure changes at different pressures, altering the stoichiometry
• D. Low-pressure experiments measure a different reaction because impurities become significant

Answer: (B)

The Lindemann mechanism explains this pressure dependence. At high pressure, collisions are frequent enough to maintain a steady concentration of activated molecules A* — the unimolecular step (A* → products, rate constant k₂) is rate-limiting, giving first-order kinetics in [A]. At low pressure, A* reacts before it can be deactivated, so every activation event directly leads to product. The activation step (A + M → A*, rate constant k₁) is now rate-limiting, and it depends on both [A] and [M], giving apparent second-order kinetics.

Question 2

Why does RRKM theory predict the pressure-dependent falloff of unimolecular rate constants more accurately than the simple Lindemann mechanism?

• A. RRKM includes relativistic corrections that Lindemann neglects at high temperatures
• B. RRKM uses quantum statistical mechanics to account for energy redistribution into the specific reaction coordinate, not just whether total energy exceeds the barrier
• C. RRKM assumes that all activated molecules react instantaneously, simplifying the rate expression
• D. RRKM replaces the binary collision model with a field-theoretic treatment of intermolecular forces

Answer: (B)

The Lindemann mechanism treats any molecule with total energy above the barrier as equally likely to react, giving an oversimplified falloff prediction. RRKM theory recognizes that energy must flow into the specific vibrational mode (the reaction coordinate) that leads over the barrier. The rate constant k(E) = σ·W‡(E−E₀)/(h·ρ(E)) expresses this: W‡ counts the quantum states accessible at the transition state (energy channeled into the reaction coordinate), while ρ(E) counts all the ways the reactant molecule can distribute that energy. Molecules with energy spread across many non-reactive modes react slowly; those with energy concentrated in the reaction coordinate react faster. This microcanonical treatment produces quantitatively accurate falloff curves.

Question 3

At sufficiently high pressure, the rate constant for a unimolecular reaction becomes independent of pressure and the reaction exhibits clean first-order kinetics.

• T. True
• F. False

Answer: True

At high pressure (the high-pressure limit), collisions are so frequent that activation and deactivation maintain A* in a pseudo-equilibrium with A. Any A* that reacts is quickly replenished. The concentration of A* is proportional to [A], and since the rate-limiting step is the unimolecular decomposition of A*, the overall rate equals k_uni[A] — purely first-order in A and independent of the concentration of the bath gas M.

Question 4

According to RRKM theory, once a molecule has accumulated enough total energy to exceed the reaction barrier, it reacts immediately, because intramolecular vibrational energy redistribution (IVR) is essentially instantaneous.

• T. True
• F. False

Answer: False

This is the key assumption RRKM refines relative to the older RRK theory. RRKM explicitly accounts for the rate at which energy flows from non-reactive vibrational modes into the reaction coordinate. While RRKM assumes IVR is fast enough to justify a statistical (microcanonical) treatment, the actual rate k(E) depends on how many transition-state states are accessible — which depends on energy being in the right modes. Moreover, the Explainer notes that in some molecules, energy can remain localized long enough to violate RRKM assumptions entirely — this is an active area of chemical physics research.

Question 5

Explain why a unimolecular gas-phase reaction that is first-order at atmospheric pressure becomes second-order at very low pressure.

Model answer: At atmospheric pressure, collisions with bath gas molecules (M) are frequent and maintain a steady-state population of activated reactant molecules (A*). The rate-limiting step is the unimolecular conversion A* → products, giving rate = k₂[A*] ∝ [A], i.e., first-order. At very low pressure, collisions are rare: a molecule that gains enough energy from a collision (A + M → A*) reacts before it can collide again and lose that energy. Now the rate-limiting step is the activation collision itself — rate = k₁[A][M] — which depends on both the reactant concentration and the bath gas concentration, giving second-order kinetics. The crossover between the two regimes is the falloff region, the experimental signature of a Lindemann mechanism.

The key is identifying which step is rate-limiting at each pressure limit. At high pressure, deactivation is fast relative to reaction (k₋₁[M] ≫ k₂), so A* equilibrates with A. At low pressure, reaction is fast relative to deactivation (k₂ ≫ k₋₁[M]), so every A* formed immediately proceeds to products.