Reaction Mechanisms and Rate-Determining Step Revision Notes | A Level AQA Chemistry

1. Why the Slowest Step Rules Everything

Imagine a factory assembly line where one worker takes ten times longer than everyone else. No matter how fast the others work, the finished product can only come out as fast as that one slow worker allows. Reaction mechanisms work exactly the same way.

Most reactions do not happen in a single collision. They proceed through a sequence of elementary steps, and one of those steps is always slower than the rest. That step — the rate-determining step — sets the ceiling on how fast the overall reaction can go.

Here is the crucial insight: the rate equation you determine experimentally only includes the species that are present at or before the rate-determining step. Species that only appear in steps after the slow step have no influence on the rate, because by the time they react, the bottleneck has already passed.

This connection between the rate equation and the mechanism is what makes kinetics so powerful at A-Level. You can use rate data collected in a laboratory to make genuine deductions about molecular-level events that you cannot observe directly.

2. Reading the Rate Equation for Mechanistic Clues

The rate equation is your window into the mechanism. Each order tells you something specific: if a reactant appears in the rate equation, it must be involved in the rate-determining step or in a step that occurs before it.

Consider a general reaction between A, B, and C. Suppose experiments give the rate equation: $\text{rate} = k[A]^2[B]$. For the purposes of AQA A-level, the orders are interpreted as the number of molecules of each species consumed up to and including the rate-determining step: two molecules of A and one molecule of B are involved up to and including the slow step. (Strictly, this interpretation applies to elementary steps; for non-elementary mechanisms, orders are empirical exponents determined experimentally.) C does not appear — so C either reacts after the RDS, or it is not involved in the mechanism at all up to that point.

3. Linking Mechanism Steps to the Rate Equation

Now for the practical skill: given a proposed mechanism, you need to decide whether it is consistent with the experimentally determined rate equation. The rule is straightforward — count the species that appear in the RDS and all steps before it, and check whether those species match the rate equation.

Work through this systematically. Identify which step is labelled as the slow step, then list every species that has been consumed (used up) in that step and in any steps that precede it. Those species — and only those species — should appear in the rate equation.

Worked example: Checking a Proposed Mechanism Against a Rate Equation

A reaction between NO(g) and Cl2(g) has the experimentally determined rate equation: rate = k[NO]^2[Cl2]. A student proposes the following two-step mechanism: Step 1 (slow): NO + Cl2 → NOCl2 Step 2 (fast): NOCl2 + NO → 2NOCl Determine whether this mechanism is consistent with the rate equation.

1. Step 1 is labelled slow, so Step 1 is the RDS. — The slow step controls the rate. Only species involved in Step 1 and any steps before it will appear in the rate equation.
2. Step 1 consumes: 1 molecule of NO and 1 molecule of Cl2. No steps occur before Step 1. — These are the only species that have been used up by the time the slow step occurs.
3. Predicted rate equation: rate = k[NO][Cl2] — One NO and one Cl2 appear in the RDS, giving first order in each.
4. Experimental: rate = k[NO]^2[Cl2] Predicted: rate = k[NO][Cl2] — The predicted equation is first order in NO, but the experimental equation is second order in NO. These do not match.
5. The proposed mechanism is NOT consistent with the experimental rate equation. — The mechanism predicts first order in NO, but the data shows second order in NO. The mechanism must be revised.

Answer: The proposed mechanism is inconsistent with the rate equation because it predicts rate = k[NO][Cl2], whereas the experimental rate equation is rate = k[NO]^2[Cl2].

Notice that Step 2 involves NO, but because Step 2 is fast (after the RDS), the NO in Step 2 does not appear in the rate equation. This is a key point that examiners test regularly.

4. When a Pre-Equilibrium Step Precedes the RDS

Some mechanisms include a fast reversible step before the slow step. This is called a pre-equilibrium, and it adds a layer of complexity — but the logic remains the same. Any species consumed in the fast step before the RDS still counts towards the rate equation, because those species must be present for the slow step to proceed.

Here is a classic example. Suppose the mechanism is:

• Step 1 (fast, reversible): A + B ⇌ AB (pre-equilibrium, where AB represents the intermediate)
• Step 2 (slow): AB + A → products (rate-determining step)

To reach the RDS, one A and one B have already been consumed in Step 1 to form intermediate AB. Then in the RDS itself, another A is consumed. In total, two A molecules and one B molecule have been used up by the time the slow step completes. The predicted rate equation is therefore: $\text{rate} = k[A]^2[B]$.

The intermediate AB does not appear in the rate equation because it is formed from A and B — you substitute back in terms of the original reactants. At A-Level, AQA expects you to track which original reactants are consumed, not to write rate equations in terms of intermediates.

5. Deducing the Mechanism from the Rate Equation

The reverse skill is equally important: given a rate equation, deduce what the rate-determining step must look like. This is where you work backwards from the kinetic data to propose a plausible mechanism.

The approach is to read the orders directly as the number of molecules of each species involved up to the RDS. If the rate equation is $\text{rate} = k[A][B]^2$, then the RDS (or the combination of steps leading to it) must involve one molecule of A and two molecules of B.

Worked example: Deducing the Rate-Determining Step from a Rate Equation

The reaction between hydrogen and nitrogen monoxide has the overall equation: 2H2(g) + 2NO(g) → N2(g) + 2H2O(g) The experimentally determined rate equation is: rate = k[H2][NO]^2 Propose a two-step mechanism consistent with this rate equation and identify the rate-determining step.

1. Order in H2 = 1 (one H2 molecule involved up to RDS) Order in NO = 2 (two NO molecules involved up to RDS) — The orders tell us the stoichiometry of species consumed up to and including the slow step.
2. RDS (slow): H2 + 2NO → N2O + H2O — This step consumes one H2 and two NO, matching the rate equation exactly. It is a plausible elementary step.
3. Step 2 (fast): N2O + H2 → N2 + H2O — This step uses the intermediate N2O formed in the RDS, plus the remaining H2 from the overall equation. It is fast, so it does not affect the rate equation.
4. Step 1: H2 + 2NO → N2O + H2O Step 2: N2O + H2 → N2 + H2O Overall: 2H2 + 2NO → N2 + 2H2O — Add the two steps together. N2O is an intermediate — it appears as a product in Step 1 and a reactant in Step 2, so it cancels. The overall equation matches the given equation.
5. RDS involves 1 H2 and 2 NO → predicted rate = k[H2][NO]^2 This matches the experimental rate equation. — The mechanism is consistent. Step 2 involves H2, but because it is fast (after the RDS), H2 in Step 2 does not appear in the rate equation.

Answer: A consistent two-step mechanism is: Step 1 (slow, RDS): H2 + 2NO → N2O + H2O Step 2 (fast): N2O + H2 → N2 + H2O The rate-determining step is Step 1, which involves one H2 and two NO molecules, consistent with rate = k[H2][NO]^2.

One important check: always verify that the elementary steps add up to give the correct overall equation. Intermediates must cancel when you sum the steps. If they do not cancel, the mechanism is internally inconsistent.

6. A Worked Mechanism: SN1 and SN2 as a Case Study

Nucleophilic substitution reactions of halogenoalkanes provide a textbook illustration of how the rate equation reveals the mechanism. Two distinct pathways exist — $S_N1$ and $S_N2$ — and their rate equations are fundamentally different.

In an $S_N2$ reaction, the nucleophile attacks the carbon at the same time as the leaving group departs. This is a single concerted step — both the nucleophile and the halogenoalkane are involved simultaneously. The rate equation is therefore: $\text{rate} = k[\text{halogenoalkane}][\text{nucleophile}]$. Both species appear because both are present in the one and only step, which is also the RDS.

In an $S_N1$ reaction, the mechanism has two steps. The slow step is the ionisation of the halogenoalkane to form a carbocation intermediate — the nucleophile plays no part in this step. The fast step is the nucleophile attacking the carbocation. Because the nucleophile only appears after the RDS, it does not feature in the rate equation: $\text{rate} = k[\text{halogenoalkane}]$. The reaction is first order overall, and the nucleophile concentration has no effect on the rate.

The rate equation $\text{rate} = k[(CH_3)_3CBr]$ is entirely consistent with this mechanism. The hydroxide ion only reacts in Step 2, which is fast. Doubling the concentration of hydroxide has no effect on the rate — a result that would be impossible to explain if the reaction occurred in a single step.

7. Exam Technique and Common Pitfalls

AQA questions on this topic follow predictable patterns. You will typically be given either a rate equation and asked to comment on the mechanism, or a proposed mechanism and asked to check its consistency. Knowing the exact language the mark scheme expects will make a significant difference to your marks.

When asked to explain what the rate equation tells you about the mechanism, structure your answer in two parts: first state which species are involved (and how many molecules of each), then explicitly link this to the rate-determining step. A response such as "the reaction is second order in NO, so two molecules of NO are involved in the rate-determining step or in a step before it" will score the mark; writing "NO is involved" will not.

When asked to suggest a mechanism, always verify two things: the proposed RDS must involve the correct species in the correct numbers to match the rate equation, and the steps must sum to give the correct overall equation with all intermediates cancelling.

The key principle running through this entire topic is that the rate equation is a fingerprint of the rate-determining step: the species and their stoichiometry in the slow step (and any steps before it) are directly encoded in the orders of reaction, giving you a powerful tool for deducing and evaluating reaction mechanisms from experimental data.