Reaction mechanisms — elementary steps, rate-determining step: Question bank

This page is a workout, not a lecture. Each line below is a claim or question with the answer hidden after the :::. Cover the right side, commit to an answer out loud with a reason, then reveal. Bare "yes/no" answers score zero — the goal is to catch the exact misconceptions this topic is famous for. If a term feels shaky, jump back to the parent note the topic note or a prerequisite like Rate Laws and Reaction Order before continuing.

True or false — justify

For each: decide true or false, then give the reason — the reason is the whole point.

True/False: The rate law of an overall reaction can always be read off from its balanced coefficients.

False — only elementary steps let you read exponents off stoichiometry; an overall equation is a summary of many steps, so its order must come from the mechanism (via the RDS) or from experiment.

True/False: The rate-determining step is always the first step written in a mechanism.

False — the RDS is the slowest step (highest activation-energy barrier), and it can sit anywhere in the sequence; "first" and "slowest" are unrelated.

True/False: Molecularity and reaction order are the same thing.

False — molecularity is the count of molecules colliding in one elementary step (a theoretical integer 1, 2, or 3); order is an experimentally measured exponent for the overall reaction. They coincide only for a single elementary step.

True/False: An intermediate is allowed to appear in the final rate law.

False — an intermediate is produced and consumed inside the mechanism, so you can't measure its concentration in the lab; it must be eliminated using a fast-equilibrium or steady-state substitution.

True/False: A catalyst appears in the overall balanced equation.

False — a catalyst is consumed in an early step and regenerated in a later step, so it cancels out of the overall equation (see Activation Energy and Catalysts); it can legitimately appear in the rate law.

True/False: For an elementary step

2 A \to products

, the rate law is

k [A]^{2}

True — because it is elementary, the exponents equal the coefficients; two

A

molecules must collide, so the collision probability scales as

[A]^{2}

True/False: The sum of all elementary steps in a valid mechanism must equal the overall balanced equation.

True — this is the primary consistency check; if the steps don't add up to the observed overall reaction, the mechanism is wrong regardless of how nicely it predicts the rate.

True/False: A reaction that is third-order overall must proceed by a single termolecular step.

False — third-order kinetics usually comes from a fast pre-equilibrium feeding a bimolecular slow step; genuine three-body simultaneous collisions are extremely rare (see Collision Theory).

True/False: If Step 2 is the slow step, reactants pile up before it and products form only as fast as Step 2 runs.

True — the slow step is the bottleneck; everything upstream equilibrates or accumulates, and the overall throughput is capped by the slowest gate.

Spot the error

Each line states a piece of (flawed) student reasoning. Name the mistake.

2NO_{2} + F_{2} \to 2NO_{2} F

, so Rate

= k [NO_{2}]^{2} [F_{2}]

by inspection."

Error — treating the overall equation as elementary. The real slow step is

NO_{2} + F_{2} \to NO_{2} F + F

, giving Rate

= k [NO_{2}] [F_{2}]

(first order in each).

"The RDS is

N_{2} O_{2} + O_{2} \to 2 NO_{2}

, so the answer is Rate

= k_{2} [N_{2} O_{2}] [O_{2}]

Error — leaving the intermediate

[N_{2} O_{2}]

in the answer. Use the fast equilibrium

[N_{2} O_{2}] = K [NO]^{2}

to get Rate

= k_{2} K [NO]^{2} [O_{2}]

"Adding two elementary rate laws gives the overall rate law."

Error — you don't add rate laws; you take the rate of the RDS and substitute out intermediates. Rate laws of separate steps are never summed.

"Molecularity of 4 is fine because four molecules appear on the reactant side."

Error — molecularity above 3 is essentially never observed for a single step; a "4-molecule reaction" must break into elementary steps of molecularity 1, 2, or (rarely) 3.

"The equilibrium constant

K = k_{1} / k_{- 1}

has units of a rate constant."

Error —

K

is a ratio of forward and reverse rate constants, so their units partly cancel;

K

generally does not carry a rate constant's units, and

k_{obs} = k_{2} K

absorbs the difference.

"Since Step 1 is fast, it can't affect the rate law at all."

Error — a fast pre-equilibrium step absolutely shapes the rate law, because it fixes the intermediate concentration that feeds the slow step (that's how

[NO]^{2}

enters).

"An intermediate and a catalyst are the same kind of species."

Error — an intermediate is made then consumed (appears first as a product, later as a reactant); a catalyst is consumed then regenerated (appears first as a reactant, later as a product). Neither is in the overall equation, but their roles are opposite.

Why questions

"Why?" is where understanding lives. Answer before revealing.

Why can we write the rate law of an elementary step directly from its coefficients, but not for an overall reaction?

Because an elementary step is the physical collision event, so its rate is literally proportional to how often those exact molecules meet; the overall equation is a bookkeeping sum with no single collision behind it.

Why does a fast reversible first step reach equilibrium before the slow step matters?

Its forward and reverse rates are both large, so it rapidly settles to a balance; the slow step drains the intermediate so gently that the equilibrium is barely disturbed and holds throughout.

Why does the overall rate never exceed the rate of the slowest step?

Every product molecule must pass through that bottleneck; upstream steps can only supply material to the gate, never bypass it, so the gate caps total throughput.

Why must we eliminate intermediates from the final rate law?

Intermediates are short-lived and present in tiny, hard-to-measure amounts; a rate law is only useful if it's written in terms of species (reactants, sometimes products) whose concentrations we can actually control and measure.

Why is molecularity restricted to 1, 2, or rarely 3?

Because it counts particles that must collide simultaneously in one event; the chance of three particles meeting at once with the right geometry is already small, and four-at-once is effectively negligible (this is Collision Theory logic).

Why does the mechanism for

2 NO + O_{2}

give second order in NO even though the slow step contains no NO?

Two NO molecules combine in the fast equilibrium to form the intermediate

N_{2} O_{2}

; substituting

[N_{2} O_{2}] = K [NO]^{2}

carries that squared dependence into the final rate law.

Why do chemists bother deriving mechanisms if the experimental rate law already tells them the order?

The rate law constrains but doesn't prove a mechanism; the mechanism explains the molecular pathway, predicts intermediates and catalysts, and connects to activation energy and Reaction Coordinate Diagrams.

Edge cases

The scenarios the tidy examples skip.

Edge case: What if two steps have almost identical (comparable) rates — is there still a single RDS?

No clean single RDS; when barriers are similar you cannot approximate with one slow step and must use the full Steady-State Approximation treatment instead.

Edge case: What if the slow step is the very last step and every earlier step is a fast equilibrium?

The rate law is still the slow step's rate, but you may need to chain several equilibrium substitutions to express each intermediate in terms of reactants before you arrive at measurable species.

Edge case: A mechanism whose steps sum correctly but predicts the wrong experimental order — is it valid?

No — passing the summation check is necessary but not sufficient; a valid mechanism must reproduce both the overall equation and the observed rate law. Failing the rate law rules it out.

Edge case: What happens to the rate law if the "slow" step actually becomes fast (e.g., a catalyst lowers its barrier)?

The bottleneck shifts to whatever step is now slowest, so the RDS — and therefore the entire form of the rate law — can change; the RDS is not a fixed label but a comparison of current barriers.

Edge case: A unimolecular elementary step

A \to products

— what is its order and molecularity?

Both are 1: molecularity is 1 (one molecule decomposing), and since it's elementary the order equals the molecularity, so Rate

= k [A]

Edge case: If an intermediate is also fed by a second pathway, can you still use a simple fast-equilibrium substitution?

Often not — multiple production/consumption routes for one intermediate usually demand the Steady-State Approximation, which balances all formation and removal terms rather than assuming a single equilibrium.

Edge case: Zero-order behaviour in a reactant that clearly appears in the overall equation — contradiction?

No contradiction — if that reactant only enters after the RDS (or saturates a surface/enzyme site as in Enzyme Kinetics), changing its concentration doesn't change the bottleneck's rate, giving zero order.

Recall One-line self-test before you leave

Cover everything and answer: "What two questions defuse almost every mechanism trap?" ::: (1) Is this line an elementary step or a summary? (2) Does an intermediate survive into the final rate law when it shouldn't?