Rate-Determining Step

Spec Mapping — OCR H432 Module 5.1.1 — How fast? The rate-determining step (RDS) of a multi-step mechanism, the link between the experimentally determined rate equation and the molecularity of the RDS, the role of intermediates and catalysts in mechanisms, and the use of rate-equation data to support or reject proposed mechanisms (refer to the official OCR H432 specification document for exact wording).

This lesson is where kinetics turns mechanistic. Most reactions on paper appear to be a single step — $\text{A} + \text{B} \rightarrow \text{products}$A+B→products — but in reality almost all proceed by a sequence of elementary collisions, each with its own activation energy. One step is invariably much slower than the rest, and that step controls the rate at which products appear. The slow step is the rate-determining step (RDS), and the experimentally determined rate equation is its fingerprint: only species involved up to and including the RDS can appear in the rate equation, and the orders count how many of each species the RDS consumes. The lesson develops the conceptual link between mechanism and rate equation through three worked examples — $S_N1/S_N2$SN​1/SN​2 hydrolysis of haloalkanes, $\text{NO}_2 + \text{CO}$NO2​+CO via an $\text{NO}_3$NO3​ intermediate, and the third-order $2\text{NO} + \text{O}_2$2NO+O2​ system mediated by a fast pre-equilibrium — and closes by addressing the subtleties of catalysts and intermediates in the rate equation.

Key Equation: for any multi-step mechanism, $\text{overall rate} = \text{rate of the slowest (rate-determining) step}$overall rate=rate of the slowest (rate-determining) step The species and orders in the experimentally observed rate equation correspond to the molecules consumed up to and including the RDS — not to the overall balanced equation.

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The idea of a rate-determining step

Almost all chemical reactions proceed via a series of elementary steps, not in a single concerted act. Each elementary step has its own activation energy and rate constant. The rate of product formation is set by the slowest step in the sequence — the bottleneck. This is the rate-determining step (RDS).

Production-line analogy: imagine a manufacturing line where stage 3 takes ten times longer than all the others. The rate at which finished products emerge is set entirely by stage 3, no matter how fast stages 1, 2, 4, 5 could operate in isolation. Improving any of the fast stages does nothing for throughput; only making the slow stage faster increases the rate at which goods leave the factory.

In chemistry: the slowest elementary step controls the rate at which products appear. Fast steps after the RDS just keep up; fast steps before the RDS supply intermediates as fast as the RDS can consume them.

Linking the rate equation to the mechanism

A crucial insight: only species involved in or before the rate-determining step appear in the experimentally observed rate equation. Species that participate only in fast steps after the RDS are invisible to the rate law — they cannot influence the rate they themselves do not gate. This gives us a powerful tool: read the rate equation to constrain the mechanism.

The reasoning logic:

1. Identify the species and orders in the rate equation. rate = $k[\text{A}]^{2}[\text{B}]$k[A]2[B] implies two molecules of A and one of B are involved up to and including the RDS.
2. Count how many of each. Order with respect to A tells you how many As are consumed by the time the RDS finishes.
3. Check candidate mechanisms against the observed orders.

The rate equation cannot prove a mechanism, but it can eliminate mechanisms that contradict it.

Worked Example 1 — $S_N1$SN​1 versus $S_N2$SN​2 hydrolysis

A classic application is the hydrolysis of haloalkanes by hydroxide ion: $\text{R-X} + \text{OH}^{-} \rightarrow \text{R-OH} + \text{X}^{-}$R-X+OH−→R-OH+X−. Two mechanisms compete.

$S_N1$SN​1 (unimolecular): two-step mechanism with a slow ionisation followed by fast nucleophilic capture.

Slow: R-X -> R+ + X-
Fast: R+ + OH- -> R-OH

The RDS involves only R-X. Predicted: $\text{rate} = k[\text{RX}]$rate=k[RX], first order in R-X, zero order in OH$^{-}$−.

$S_N2$SN​2 (bimolecular): single concerted step.

Slow: R-X + OH- -> R-OH + X-

The RDS (the only step) involves R-X and OH$^{-}$−. Predicted: $\text{rate} = k[\text{RX}][\text{OH}^{-}]$rate=k[RX][OH−], first order in each.

Experimentally:

• $(\text{CH}_3)_3\text{C-Br}$(CH3​)3​C-Br (tertiary) hydrolyses by $S_N1$SN​1: rate $= k[(\text{CH}_3)_3\text{CBr}]$=k[(CH3​)3​CBr]. The tertiary carbocation is stabilised by three alkyl groups (inductive electron donation), making ionisation feasible.
• $\text{CH}_3\text{-Br}$CH3​-Br (primary) hydrolyses by $S_N2$SN​2: rate $= k[\text{CH}_3\text{Br}][\text{OH}^{-}]$=k[CH3​Br][OH−]. A primary cation would be hopelessly unstable, so the concerted bimolecular pathway is the only option.

The rate equation distinguishes between mechanisms — a single rate-law experiment classifies the haloalkane.

Worked Example 2 — $\text{NO}_2 + \text{CO}$NO2​+CO with hidden intermediate

For $\text{NO}_2\text{(g)} + \text{CO(g)} \rightarrow \text{NO(g)} + \text{CO}_2\text{(g)}$NO2​(g)+CO(g)→NO(g)+CO2​(g), the experimentally determined rate equation is $\text{rate} = k[\text{NO}_2]^{2}$rate=k[NO2​]2. Note: CO does not appear.

A proposed two-step mechanism:

Slow:  2 NO2 -> NO + NO3
Fast:  NO3 + CO -> NO2 + CO2

Check:

• Slow step has $2 \times \text{NO}_2$2×NO2​; expected rate $\propto [\text{NO}_2]^{2}$∝[NO2​]2 — matches experiment.
• CO appears only in the fast step; expected to be absent from the rate equation — matches.
• Adding the two steps: $\text{NO}_2 + \text{NO}_2 + \text{NO}_3 + \text{CO} \rightarrow \text{NO} + \text{NO}_3 + \text{NO}_2 + \text{CO}_2$NO2​+NO2​+NO3​+CO→NO+NO3​+NO2​+CO2​. Cancelling $\text{NO}_3$NO3​ and one $\text{NO}_2$NO2​: $\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2$NO2​+CO→NO+CO2​. Overall equation correct.

The mechanism is consistent. $\text{NO}_3$NO3​ is an intermediate — formed and consumed within the mechanism, present neither as a reactant nor a product of the overall equation.

Worked Example 3 — fast pre-equilibrium and apparent third-order kinetics

For $2\text{NO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{NO}_2\text{(g)}$2NO(g)+O2​(g)→2NO2​(g), the experimental rate equation is $\text{rate} = k[\text{NO}]^{2}[\text{O}_2]$rate=k[NO]2[O2​], overall order 3. A single three-body collision (three molecules meeting simultaneously) is so improbable that the true mechanism is almost certainly not a one-step termolecular step. Instead, a fast pre-equilibrium followed by a bimolecular slow step:

Step 1 (fast equilibrium):  2 NO  <->  N2O2
Step 2 (slow, RDS):         N2O2 + O2 -> 2 NO2

The slow step involves $\text{N}_2\text{O}_2$N2​O2​ (an intermediate) and $\text{O}_2$O2​. From step 1's equilibrium constant $K_1 = [\text{N}_2\text{O}_2]/[\text{NO}]^{2}$K1​=[N2​O2​]/[NO]2, so $[\text{N}_2\text{O}_2] = K_1[\text{NO}]^{2}$[N2​O2​]=K1​[NO]2. Substituting into the rate of step 2:

$\text{rate} = k_2 [\text{N}_2\text{O}_2][\text{O}_2] = k_2 K_1 [\text{NO}]^{2}[\text{O}_2]$rate=k2​[N2​O2​][O2​]=k2​K1​[NO]2[O2​]

Identifying the effective rate constant $k = k_2 K_1$k=k2​K1​, this gives the observed $\text{rate} = k[\text{NO}]^{2}[\text{O}_2]$rate=k[NO]2[O2​]. The order-3 rate equation does not require a three-body collision — it arises from a bimolecular RDS preceded by a fast pre-equilibrium.

This is one of the most beautiful applications of mechanism reasoning: the rate equation fingerprints the participation of intermediates that never appear in the balanced equation.

Mermaid — mechanism-to-rate-equation logic

graph LR
  A[Multi-step mechanism] --> B[Identify slowest step RDS]
  B --> C[List species in RDS]
  C --> D[Plus species in any fast pre-equilibrium feeding RDS]
  D --> E[Write expected rate equation]
  E --> F{Match experiment?}
  F -->|Yes| G[Mechanism consistent]
  F -->|No| H[Reject mechanism]

Orders and the molecularity of the RDS

• Order with respect to A = 0 → A is involved only after the RDS (or its concentration does not affect the rate of the RDS — e.g. enzyme-saturated regime, where active sites are full).
• Order with respect to A = 1 → one A molecule is involved up to and including the RDS.
• Order with respect to A = 2 → either two As collide in the RDS, or one A appears in a fast pre-equilibrium that is squared (as in worked example 3).

The overall order = molecularity of the RDS (unimolecular = 1, bimolecular = 2). A "termolecular" overall order = 3 is almost always an effective rate equation arising from a pre-equilibrium plus a bimolecular RDS, not a genuine three-body collision.

Worked Example 4 — matching a mechanism to a rate equation

A reaction $\text{X} + \text{Y} + \text{Z} \rightarrow \text{products}$X+Y+Z→products has experimental rate equation $\text{rate} = k[\text{X}][\text{Y}]$rate=k[X][Y]. Test two candidate mechanisms.

Mechanism A — single step.

Slow: X + Y + Z -> products

Predicted: $\text{rate} = k[\text{X}][\text{Y}][\text{Z}]$rate=k[X][Y][Z]. Does not match (Z should not appear). Rejected.

Mechanism B — two-step.

Slow:  X + Y -> XY
Fast:  XY + Z -> products

Slow step involves X and Y; rate $= k[\text{X}][\text{Y}]$=k[X][Y]. Z appears only in the fast step, consistent with its absence from the rate equation. Consistent with experiment.

Worked Example 5 — three-step mechanism with rate equation derivation

For the gas-phase reaction $2\text{NO(g)} + 2\text{H}_2\text{(g)} \rightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O(g)}$2NO(g)+2H2​(g)→N2​(g)+2H2​O(g) the experimentally determined rate equation is $\text{rate} = k[\text{NO}]^{2}[\text{H}_2]$rate=k[NO]2[H2​]. A three-step mechanism has been proposed:

Step 1 (fast eq):    2 NO  <->  N2O2
Step 2 (slow, RDS):  N2O2 + H2 -> N2O + H2O
Step 3 (fast):       N2O + H2 -> N2 + H2O

Deriving the rate equation from the mechanism. The rate of product formation is governed by the rate of the slowest step:

$\text{rate}_{\text{step 2}} = k_2 [\text{N}_2\text{O}_2][\text{H}_2]$ratestep 2​=k2​[N2​O2​][H2​]

The intermediate $\text{N}_2\text{O}_2$N2​O2​ is generated and consumed entirely within the mechanism — it does not appear in any rate equation we can measure. We eliminate it using step 1's fast equilibrium:

$K_1 = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^{2}} \quad\Rightarrow\quad [\text{N}_2\text{O}_2] = K_1 [\text{NO}]^{2}$K1​=[NO]2[N2​O2​]​⇒[N2​O2​]=K1​[NO]2

Substituting into the step-2 rate:

$\text{rate} = k_2 K_1 [\text{NO}]^{2}[\text{H}_2] = k_{\text{obs}}[\text{NO}]^{2}[\text{H}_2]\quad\text{with}\quad k_{\text{obs}} = k_2 K_1$rate=k2​K1​[NO]2[H2​]=kobs​[NO]2[H2​]withkobs​=k2​K1​

This recovers the experimentally observed form. Step 3 is invisible to the rate equation because it lies entirely after the RDS — the $\text{N}_2\text{O}$N2​O produced in step 2 is consumed faster than it accumulates, so step 3's rate "keeps up" with whatever step 2 delivers, contributing nothing to the overall expression. The chemistry of step 3 is essential to the stoichiometry of the overall reaction (it accounts for the second mole of water and the conversion of $\text{N}_2\text{O}$N2​O to $\text{N}_2$N2​), but it is kinetically silent.

The takeaway: in a mechanism with $N$N elementary steps, the observed rate equation reflects only the first $r$r steps up to and including the RDS at position $r$r, with all fast pre-equilibria substituted in. Everything after step $r$r is kinetically transparent.

Steady-state approximation — foreshadowing undergraduate kinetics

The fast pre-equilibrium assumption used in worked examples 3 and 5 is a special case of a more general undergraduate tool, the steady-state approximation (SSA). Under SSA we assume the concentration of any reactive intermediate (a species generated and consumed within the mechanism, present only at low concentration) stays essentially constant once the reaction is underway:

$\frac{d[\text{intermediate}]}{dt} \approx 0$dtd[intermediate]​≈0

This is not the same as saying the intermediate concentration is zero — only that its rate of formation equals its rate of consumption at every instant during the reaction's main phase. SSA then provides an algebraic equation to solve for the intermediate concentration in terms of the (measurable) reactants and the elementary rate constants. The fast-pre-equilibrium reasoning emerges as the SSA limit in which one of the consumption pathways (the reverse of step 1) is much faster than the other (step 2). When the two are comparable, the SSA gives a more nuanced rate equation than fast-pre-equilibrium reasoning predicts — the textbook example is the Lindemann–Hinshelwood treatment of unimolecular dissociation, where the apparent order with respect to the dissociating species varies from 1 at high pressure to 2 at low pressure. This is undergraduate material; OCR will not test SSA directly. But knowing that fast pre-equilibrium is a limit of a more general technique is exactly the kind of mental scaffolding that distinguishes A* candidates from A candidates when faced with unfamiliar mechanism questions.