Mechanism &
the Rate-Determining Step

Two-step mechanism with the first step slow

Rate law

The rate law is written from the slow step, treated as elementary.

Rate law reflects the rate-determining step

1. The first step is slow, so it is rate-determining.
2. It involves one A and one B colliding.
3. Treating it as elementary, the exponents equal the coefficients: Rate = k[A][B].
4. D appears only in the fast second step and therefore does not enter the rate law.

The two-step mechanism

Overall balanced equation

Adding the steps and cancelling species appearing on both sides gives the overall equation.

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1. Add the two steps: 2H₂O₂ + I⁻ + IO⁻ → 2H₂O + IO⁻ + I⁻ + O₂.
2. Cancel I⁻, which appears on both sides — it is the catalyst.
3. Cancel IO⁻, which also appears on both sides — it is the intermediate.
4. What remains is 2H₂O₂ → 2H₂O + O₂.

Overall: 2H₂O₂ → 2H₂O + O₂ ; Rate = k[H₂O₂][I⁻]

Why prediction was impossible

A catalyst absent from the equation can still control the rate.

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1. The overall equation involves only H₂O₂, H₂O and O₂.
2. Iodide does not appear anywhere in it, being consumed and regenerated.
3. Yet the measured rate depends directly on [I⁻].
4. So no inspection of the balanced equation could have produced this rate law — which is exactly NCERT's point.

Definition question

Name of the species

The order of events defines the role.

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1. It does not exist at the start, being produced during the reaction.
2. It does not exist at the end, being consumed before completion.
3. NCERT calls such a species an intermediate, since it is formed during the course of the reaction but not in the overall balanced equation.
4. A catalyst is the reverse — consumed first and regenerated later.

A curve that dips and returns to its starting level

Identity of the species

Returning to the starting level means the species was regenerated.

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1. Starting above zero means the species was present before the reaction began.
2. Dipping means it is being consumed at some stage.
3. Returning to its original level means it is fully regenerated.
4. That pattern defines a catalyst. An intermediate would instead start at zero and return to zero.

A multi-step mechanism

Which step is rate-determining

The largest barrier gives the smallest rate constant and therefore the slowest step.

k = A e^(−Ea/RT) — larger Ea means smaller k

1. A larger activation energy corresponds to a smaller rate constant.
2. A smaller rate constant means a slower step.
3. The slowest step limits the whole sequence.
4. Position in the sequence is irrelevant — a slow last step is just as limiting as a slow first one.

A complex reaction with a rate-determining step

What determines the overall order

The rate law comes from the slow step, so its molecularity fixes the order.

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1. The rate law is written from the rate-determining step.
2. For that elementary step, order equals molecularity.
3. So the overall observed order equals the molecularity of the slowest step.
4. This is NCERT's conclusion (iii), stated verbatim.

Comparison of activated complex and intermediate

The distinguishing feature

Position on the energy profile determines stability.

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1. An activated complex sits at the top of a barrier — an energy maximum.
2. Anything at a maximum is unstable, since any displacement sends it downhill.
3. So it exists only for an instant and cannot be isolated.
4. An intermediate sits in a dip, a local minimum, which gives it some stability and a measurable lifetime.

Step 1 slow, producing IO⁻; step 2 fast, consuming it

Reason the intermediate's formation rate governs the reaction

A fast step can only process material as quickly as the slow step supplies it.

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1. The slow first step produces IO⁻.
2. The fast second step consumes IO⁻ as quickly as it appears.
3. So the second step is starved — it can never go faster than the supply.
4. NCERT states that the rate of formation of intermediate will determine the rate of this reaction.

Fast first step producing B; slow second step consuming B

Rate law and its problem

Gap content — a rate law containing an intermediate must be rewritten.

Rate law from the slow step, then eliminate the intermediate

1. The slow step is B + C → D, so directly Rate = k[B][C].
2. But B is an intermediate — it does not exist at the start.
3. A valid rate law must contain only species present initially.
4. So [B] must be expressed in terms of [A] using the fast first step, giving a rate law in [A] and [C].

Requirements for a valid rate law

What must be excluded

A rate law must be expressible in terms of species present at the start.

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1. Reactants and catalysts are both present at the start and may legitimately appear.
2. A rate constant must appear in every rate law.
3. An intermediate does not exist at t = 0, so its concentration cannot be measured or controlled.
4. If it appears, it must be eliminated using an earlier fast equilibrium.

Criteria for an acceptable mechanism

A necessary requirement

Mass balance must be preserved across the whole sequence.

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1. The mechanism describes how the overall reaction actually occurs.
2. Adding the steps must therefore reproduce the overall balanced equation once intermediates and catalysts cancel.
3. Steps may be fast or slow in any combination, and may be uni- or bimolecular.
4. Catalysts are permitted and often essential, so option (d) is wrong.

A mechanism with a slow step and fast steps

Effect of accelerating a fast step

Only the bottleneck limits the sequence.

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1. The overall rate is limited entirely by the slowest step.
2. Fast steps already process material faster than it arrives.
3. Making them faster still cannot increase throughput — the material simply is not there.
4. So the overall rate is essentially unchanged, exactly as with the production-line analogy.

H₂O₂ + I⁻ → H₂O + IO⁻ ; H₂O₂ + IO⁻ → H₂O + I⁻ + O₂

Nature of the steps

Count the colliding species in each step.

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1. Step 1 involves two species colliding: H₂O₂ and I⁻.
2. Step 2 also involves two: H₂O₂ and IO⁻.
3. Each is a single elementary event with molecularity 2.
4. NCERT states directly that both the steps are bimolecular elementary reactions.

A conflict between predicted and measured rate law

The correct response

Experiment is the arbiter; a mechanism is a hypothesis tested against it.

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1. A mechanism is a proposal about how the reaction occurs.
2. Its test is whether the rate law it predicts matches the measured one.
3. If they disagree, the proposal has failed and must be rejected or revised.
4. This asymmetry is why mechanisms can be disproved but never proved.

A species present at both start and end but consumed in between

Its identity

Regeneration after consumption is the catalyst signature.

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1. A reactant is present at the start but gone at the end.
2. A product is absent at the start and present at the end.
3. An intermediate is absent at both.
4. Present at both ends while being consumed in between describes a catalyst — I⁻ in the worked mechanism.

A fractional experimental order

Its usual mechanistic origin

Gap content — eliminating an intermediate via an equilibrium can introduce a square root.

e.g. Cl₂ ⇌ 2Cl gives [Cl] ∝ [Cl₂]^½

1. A single collision can never produce a fractional exponent, since molecules come in whole numbers.
2. But if a species dissociates in a fast pre-equilibrium, its concentration goes as the square root of the parent's.
3. Eliminating that intermediate from the rate law then leaves a half-power.
4. This is where the ½ in Rate = k[CHCl₃][Cl₂]^½ comes from.

A three-step mechanism

Number of intermediates

An intermediate sits between each consecutive pair of steps.

intermediates = steps − 1

1. Each elementary step ends by producing something.
2. The products of the first two steps are consumed by later steps, making them intermediates.
3. The product of the final step is the actual product.
4. So three steps normally give two intermediates — equivalently, three peaks with two dips between them.

A multi-peak energy profile

How to measure each barrier

Each step's barrier is measured from where that step begins.

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1. An activation energy is the extra energy needed to get from a starting state to a transition state.
2. For step 2, the starting state is the intermediate, not the original reactants.
3. So its barrier is measured from the dip up to the second peak.
4. Measuring both barriers from the baseline is the standard error and can identify the wrong rate-determining step.