Rate-determining step

CIE A-Level Chemistry· Unit 20: Further reaction kinetics· 25 min read

1. Definition and Core Properties★★☆☆☆⏱ 10 min

📘 Definition

Rate-determining step

The slowest elementary step in a multi-step reaction mechanism. It limits the maximum rate of the overall reaction, because the reaction cannot proceed faster than its slowest step.

Example:

A factory line can only produce finished goods as fast as the slowest station, just like a reaction can only go as fast as its RDS.

Each step in a multi-step mechanism has its own activation energy and reaction rate. The RDS is the step with the highest activation energy, so it proceeds slowest at any given temperature. Key properties are:

Only reactants involved in the RDS or steps before it affect the overall rate
The RDS directly determines the overall rate constant for the reaction
Species that only react after the RDS do not change the overall rate

📐 Worked Example

A reaction has three steps: Step 1 (fast), Step 2 (very slow), Step 3 (fast). Which step is the rate-determining step?

1
Recall that the RDS is by definition the slowest step in any reaction mechanism.
2
Step 2 is explicitly given as the slowest step, so it is the rate-determining step.

2. Linking RDS to Experimental Rate Equations★★★☆☆⏱ 15 min

The order of reaction with respect to a reactant equals the number of molecules of that reactant that are involved in the RDS and any preceding steps. This lets us test if a proposed mechanism is consistent with experimental data.

📘 Definition

Molecularity

The number of reactant particles involved in an elementary step. For any elementary step, the order of reaction equals the molecularity.

📐 Worked Example

The overall reaction $2 N O + O_{2} \to 2 N O_{2}$ has the experimental rate equation $rate = k [N O]^{2} [O_{2}]$ . The proposed mechanism is: Step 1 (fast): $N O + N O ⇌ N_{2} O_{2}$ , Step 2 (slow): $N_{2} O_{2} + O_{2} \to 2 N O_{2}$ . Confirm this RDS matches the rate equation.

1
Step 1: The slow step is always the RDS, so RDS is Step 2.
2
Step 2: The RDS contains 1 $N_{2} O_{2}$ (an intermediate formed from 2 $N O$ molecules in Step 1) and 1 $O_{2}$ molecule.
3
$\text{Total reactant particles: } 2 NO + 1 O_2 \implies \text{order} = 2 \text{ for } NO, 1 \text{ for } O_2$
4
Step 3: The derived rate equation matches the experimental result, so the mechanism and RDS are plausible.

✓ Quick check

Test your understanding:

For the reaction $A + B \to C$ , the mechanism is Step 1 (slow): $A \to X$ , Step 2 (fast): $X + B \to C$ . What is the correct rate equation?
- A: $rate = k [A]$
- B: $rate = k [A] [B]$
- C: $rate = k [B]$
- D: $rate = k [A]^{2}$
Reveal answer
A: $\text{rate} = k[A]$ —
B only reacts after the slow RDS, so it does not affect the overall rate and is zero order.

3. Proposing Mechanisms from Rate Data★★★★☆⏱ 15 min

Given an overall reaction and experimental rate equation, you can propose a plausible mechanism by first identifying the RDS, then adding fast steps that add up to the overall reaction. Always cancel intermediates to check your steps add to the overall equation.

📐 Worked Example

The hydrolysis of 2-bromo-2-methylpropane is $(C H_{3})_{3} C B r + O H^{-} \to (C H_{3})_{3} C O H + B r^{-}$ . The experimental rate equation is $rate = k [(C H_{3})_{3} C B r]$ . Propose a two-step mechanism and identify the RDS.

1
Step 1: Only $(C H_{3})_{3} C B r$ appears in the rate equation, so it must be the only reactant in the RDS. The RDS is the first (slow) step.
2
$(CH_3)_3CBr \rightarrow (CH_3)_3C^+ + Br^- \quad (\text{slow, RDS})$
3
Step 2: Add a fast second step that reacts the carbocation intermediate with $O H^{-}$ to form the product:
4
$(CH_3)_3C^+ + OH^- \rightarrow (CH_3)_3COH \quad (\text{fast})$
5
Step 3: Add the two steps and cancel the intermediate $(C H_{3})_{3} C^{+}$ from both sides. The result matches the overall reaction equation, so the mechanism is plausible.

4. Common Pitfalls

Wrong move:

Assuming all reactants in the overall reaction appear in the rate equation

Why:

Reactants that only participate in steps after the RDS do not affect the overall rate, so they are zero order

Correct move:

Only count reactants involved in the RDS or any steps before it when writing the rate equation

Wrong move:

Including intermediates directly in the rate equation

Why:

Intermediates are not starting reactants, so they are not included in experimental rate equations

Correct move:

Replace any intermediate in the RDS with the starting reactants that form it in preceding steps

Wrong move:

Claiming the RDS is always the first step in the mechanism

Why:

The RDS can be any step, depending on the activation energy of each step

Correct move:

Only assign RDS based on the given rate equation, not its position in the mechanism

Wrong move:

Only counting reactants in the RDS to get the overall order

Why:

Reactants from steps before the RDS produce intermediates that enter the RDS, so they must also be counted

Correct move:

Count all reactant molecules from all steps up to and including the RDS to get the overall order

5. Quick Reference Cheatsheet

Rule	Explanation
The slowest step is the RDS	Overall rate equals the rate of the slowest step
Species after RDS do not affect rate	These are zero order, do not appear in rate equation
Order = total reactant molecules up to RDS	Count molecules from all steps before + including RDS
All steps must add to the overall equation	Intermediates cancel out to give the net reaction
RDS has the highest activation energy	It is slow because fewer molecules have enough energy to react

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Deduce RDS from rate data
2023 · 4
Propose mechanism from RDS
2021 · 1
Identify intermediates from RDS

Going deeper

syllabusCIE 9701 A-Level Chemistry SyllabusSee section 20: Further reaction kinetics

What's Next

Understanding the rate-determining step is a foundational concept for all further kinetics and mechanistic study in A-Level Chemistry. You will use RDS to explain how catalysts work: catalysts provide an alternative reaction mechanism with a lower activation energy for the rate-determining step, increasing overall reaction rate. This concept is also core to organic chemistry, where it explains the different rate laws and stereochemical outcomes of SN1 and SN2 nucleophilic substitution reactions, a common topic in Paper 2 and Paper 4 exams.