Questions: Elementary Reaction Mechanisms and Catalysis

Question 1

A proposed mechanism has two steps: Step 1 (fast, reversible): A + B ⇌ X, and Step 2 (slow): X → C. What is the rate law for the overall reaction?

• A. rate = k[X]
• B. rate = k[A][B]
• C. rate = k[A]
• D. rate = k[A][B][C]

Answer: (B)

Step 2 is the rate-determining step, so rate = k₂[X]. But X is an intermediate — it cannot appear in the observable rate law. Using the pre-equilibrium approximation (Step 1 is fast and reversible), K_eq = [X]/([A][B]), so [X] = K_eq[A][B]. Substituting gives rate = k₂·K_eq·[A][B] = k_obs[A][B]. Option A is wrong because rate laws must be expressed in terms of reactants, not intermediates.

Question 2

A catalyst doubles the rate of the forward reaction. What happens to the rate of the reverse reaction?

• A. It stays the same — the catalyst only speeds up the forward direction
• B. It doubles — the catalyst lowers the activation energy for both directions equally
• C. It decreases — the catalyst shifts equilibrium toward products
• D. It increases by more than double — the reverse reaction benefits more from the lowered barrier

Answer: (B)

A catalyst provides an alternative mechanism with a lower activation energy. Because the energy difference between reactants and products (ΔG°) is unchanged, lowering the forward barrier by a given amount lowers the reverse barrier by the same amount. Both rate constants increase by the same factor, leaving the equilibrium constant K = k_f/k_r unchanged. Catalysts change kinetics, never thermodynamics.

Question 3

A catalyst can make a thermodynamically unfavorable reaction proceed by lowering the activation energy sufficiently.

• T. True
• F. False

Answer: False

Activation energy determines the rate of a reaction, not whether it is thermodynamically favorable. A catalyst cannot change the Gibbs energy difference between reactants and products (ΔG°), and therefore cannot change the equilibrium constant. If a reaction is thermodynamically unfavorable (ΔG° > 0), it will still favor reactants at equilibrium regardless of the catalyst — it just reaches that equilibrium faster.

Question 4

For an elementary reaction step, you can always write the rate law directly from the balanced stoichiometric equation for that step.

• T. True
• F. False

Answer: True

This is the defining property of an elementary step: it occurs in a single molecular event, so the rate law must reflect the actual collision or decomposition. A bimolecular elementary step A + B → products has rate = k[A][B] by definition. This is NOT true for overall (non-elementary) reactions, whose rate laws must be determined experimentally and can have fractional or unexpected orders.

Question 5

Why must the rate law for an overall reaction exclude intermediates, and how do chemists eliminate them?

Model answer: Intermediates are transient species that cannot be measured or controlled by the experimenter. To eliminate them, chemists use either the steady-state approximation (set d[intermediate]/dt = 0, since it is consumed as fast as it is formed) or the pre-equilibrium approximation (a fast reversible step establishes an equilibrium before the slow step, allowing expression of [intermediate] in terms of reactant concentrations via the equilibrium constant). Both techniques yield an observable rate law in terms of measurable quantities.

The practical need is measurement: you can only measure and control concentrations of stable reactants, not fleeting intermediates. The steady-state applies when the intermediate's rate of formation is slow relative to its consumption; the pre-equilibrium applies when the fast step truly equilibrates before the slow step proceeds. In either case, the goal is an observable rate law expressed in measurable quantities.