Questions: Reaction Mechanisms and Elementary Steps

Question 1

The overall balanced equation for a reaction is: 2NO(g) + O₂(g) → 2NO₂(g). A student writes the rate law as rate = k[NO]²[O₂]. What is the problem with this reasoning?

• A. The rate law should use molar concentrations, not partial pressures
• B. The rate law for an overall reaction cannot be written from its stoichiometry — it must be determined experimentally or derived from the mechanism's rate-determining step
• C. The exponents should equal 1 for each reactant in any rate law
• D. Only products can appear in a rate law, not reactants

Answer: (B)

Rate laws for overall reactions must be determined experimentally or derived from the proposed mechanism's rate-determining step. Only for elementary steps can you write the rate law directly from stoichiometry. The overall stoichiometry reflects the net result of multiple steps and gives no direct information about the reaction mechanism or rate law.

Question 2

In a proposed two-step mechanism, Step 1 (fast, reversible) produces intermediate X, and Step 2 (slow) converts X into product. The experimentally measured rate law involves only original reactants. How is this consistent?

• A. Intermediates never appear in rate laws because they have zero concentration
• B. The pre-equilibrium approximation expresses [X] in terms of original reactants using the equilibrium constant of Step 1, which is then substituted into the rate law for Step 2
• C. The intermediate X is treated as a reactant because it is consumed in Step 2
• D. Only the fast step determines the rate law, so X's concentration is irrelevant

Answer: (B)

Since Step 1 is fast and reversible, it reaches equilibrium before the slow step proceeds. Setting the forward and reverse rates of Step 1 equal gives an expression for [X] in terms of the original reactants and the equilibrium constant. Substituting this into the rate expression for the slow step eliminates the intermediate, producing a rate law in terms of measurable concentrations.

Question 3

For an elementary bimolecular reaction A + B → C, you can always write the rate law as rate = k[A][B] directly from the stoichiometry.

• T. True
• F. False

Answer: True

Elementary steps are single molecular-level events. For a bimolecular step, two specific molecules must collide, so the rate depends on the probability of both being present — which is proportional to [A][B]. This is the one case where stoichiometry directly gives the rate law. This rule applies only to elementary steps, not to overall reactions.

Question 4

If the overall balanced equation for a reaction is first order in A and first order in B, the reaction is expected to proceed through a single bimolecular elementary step.

• T. True
• F. False

Answer: False

The experimentally observed rate law tells you about the rate-determining step, not the overall mechanism. A second-order rate law could arise from a multi-step mechanism in which the slow step happens to involve one molecule of A and one of B — even if the overall equation has different stoichiometry or if earlier steps involve other species. The mechanism cannot be read off from the overall rate law alone.

Question 5

Why can't the concentration of a reaction intermediate appear in the final rate law, and how does the pre-equilibrium approximation solve this problem?

Model answer: Intermediates are produced and consumed within the mechanism — they are not present at the start of the reaction, so their concentration cannot be directly measured or controlled. The pre-equilibrium approximation solves this by using the equilibrium constant of a fast reversible step that produces the intermediate: setting forward rate = reverse rate gives an algebraic expression for [intermediate] in terms of original reactant concentrations. Substituting this expression into the slow-step rate law replaces the unmeasurable intermediate with measurable quantities.

The pre-equilibrium approximation is valid when the step producing the intermediate is much faster than the step consuming it, so the intermediate concentration stays at its quasi-equilibrium value. The result is a rate law written entirely in terms of species present at the start of the reaction.