Lesson 4 · 30 min

Reaction mechanisms

An overall equation tells you only the start and the finish; the mechanism is the step-by-step story in between. Integrate every elementary step in the simulator and see how the rate-determining step, the steady state and pre-equilibrium turn a mechanism into a testable rate law.

In this lesson

Distinguish elementary steps from the overall reaction and write an elementary step’s rate law from its molecularity
Identify intermediates and catalysts in a mechanism and pick out the rate-determining step
Derive rate laws from mechanisms with the steady-state and pre-equilibrium approximations
Judge whether a mechanism is consistent with the observed rate law

Nitric oxide turns red-brown in air as it becomes nitrogen dioxide: $\ce{2NO} + \ce{O2} \mathrel{\ce{->}} \ce{2NO2}$ . The measured rate law is $r = k[\ce{NO}]^2[\ce{O2}]$ , which looks like three molecules colliding at once — and three-body collisions are extremely rare. Stranger still, this reaction slows down when heated. The answer lies in its mechanism: the overall equation records only the start and the finish, while in between lie a few simple steps.

Predict first

Below 225 °C the rate law of $\ce{NO2} + \ce{CO} \mathrel{\ce{->}} \ce{NO} + \ce{CO2}$ is $r = k[\ce{NO2}]^2$ , independent of CO. This tells you that:

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

CO takes part only in a fast step after the slow one CO is a catalyst The reaction is termolecular The coefficient of NO₂ in the equation should be 2

Elementary steps and molecularity

An elementary step is a single molecular event. The number of particles that collide in it is its molecularity: unimolecular (one molecule falls apart or rearranges), bimolecular (two particles collide — by far the most common) and termolecular (rare). Only for an elementary step can the rate law be read off the equation — its order equals its molecularity: $r = k[\mathrm{A}]$ for $\ce{A} \mathrel{\ce{->}} \ce{P}$ , $r = k[\mathrm{A}][\mathrm{B}]$ for $\ce{A} + \ce{B} \mathrel{\ce{->}} \ce{P}$ .

A mechanism is a set of elementary steps that must add up to the overall reaction. A species formed in one step and used up in a later one is an intermediate (such as $\ce{NO3}$ ); one used up at the start and regenerated at the end is a catalyst. The slowest step acts as a bottleneck on the overall rate: the rate-determining step.

\frac{d[\mathrm{I}]}{dt} \approx 0\ \ \text{(SS)}\qquad\qquad \ce{2NO <=>[k_1][k_{-1}] N2O2}:\ [\ce{N2O2}] = K_1[\ce{NO}]^2 \ \Rightarrow\ r = k_2K_1[\ce{NO}]^2[\ce{O2}]

Left: the steady-state approximation (SS) — a reactive intermediate is consumed as fast as it forms, so its concentration stays low and nearly constant. Right: the pre-equilibrium approximation — a fast, reversible first step stays at equilibrium while a slow second step drains it.

$[\mathrm{I}]$: concentration of the intermediate
$K_1 = k_1/k_{-1}$: equilibrium constant of the first step
$k_2$: rate constant of the slow step $\ce{N2O2} + \ce{O2} \mathrel{\ce{->}} \ce{2NO2}$

Worked example

2NO + O₂: the rate law from the mechanism

Mechanism: ① $\ce{2NO} \mathrel{\ce{<=>}} \ce{N2O2}$ (fast and reversible, $k_1$ , $k_{-1}$ ); ② $\ce{N2O2} + \ce{O2} \mathrel{\ce{->}} \ce{2NO2}$ (slow, $k_2$ ). Derive the rate law and explain why heating slows the reaction.

Integrating every elementary step · testing the approximations

Mechanism simulator

Overall reaction: $\ce{2NO} + \ce{O2} \mathrel{\ce{->}} \ce{2NO2}$

$\ce{2NO} \mathrel{\ce{<=>[k_1][k_{-1}]}} \ce{N2O2}$ bimolecular fast
$\ce{N2O2} + \ce{O2} \mathrel{\ce{->[k_2]}} \ce{2NO2}$ bimolecular slow

k₁ / (M⁻¹ s⁻¹) · 1 k₋₁ / s⁻¹ · 200 k₂ / (M⁻¹ s⁻¹) · 10

All species (solid = full mechanism; dashed = approximate rate law)

NO
O₂
N₂O₂
NO₂
Product from the approximation

The intermediate, magnified (dashed = approximation) — N₂O₂

Pre-equilibrium approximation

r = k_2K_1[\ce{NO}]^2[\ce{O2}],\qquad K_1 = \dfrac{k_1}{k_{-1}}

Condition for the approximation: ratio ≫ 1; now 40

Largest gap between the full mechanism and the approximation: 0.9 %

Step rates at t = 10.0 s

2NO → N₂O₂ 0.502 M s⁻¹
N₂O₂ → 2NO 0.494 M s⁻¹
N₂O₂ + O₂ → 2NO₂ 0.0088 M s⁻¹

The first step is fast and reversible, the second slow: N₂O₂ stays in equilibrium with NO, [N₂O₂] = K₁[NO]². The overall rate is second order in NO and first order in O₂ — it looks like a three-body collision but is two bimolecular steps. Once k₋₁ is no longer much larger than k₂[O₂], the approximation fails.

Start with the defaults: the dashed curve (pre-equilibrium) sits almost on the solid one, and “Measure the order” gives about 2 for NO and 1 for O₂. Now lower k₋₁ and raise k₂ until the ratio is no longer much greater than 1 — the approximation drifts away. Then switch to “Consecutive steps” and watch the intermediate I rise and fall.

Concept check

Iodide-catalysed decomposition of hydrogen peroxide: ① $\ce{H2O2} + \ce{I-} \mathrel{\ce{->}} \ce{H2O} + \ce{IO-}$ (slow); ② $\ce{H2O2} + \ce{IO-} \mathrel{\ce{->}} \ce{H2O} + \ce{O2} + \ce{I-}$ (fast). Here:

\ce{IO-}

is the catalyst and

\ce{I-}

the intermediate

\ce{I-}

is the catalyst and

\ce{IO-}

the intermediate Both are intermediates Both are catalysts

Try the calculation

For consecutive steps $\ce{A} \mathrel{\ce{->}} \ce{I} \mathrel{\ce{->}} \ce{P}$ with $k_1 = 0.50\ \mathrm{s^{-1}}$ and $k_2 = 5.0\ \mathrm{s^{-1}}$ , starting from A only, when does the intermediate I peak? (You can check in the simulator.)

Key takeaways

For an elementary step, order equals molecularity; an overall rate law must come from experiment or a mechanism.
The steps of a mechanism add up to the overall reaction; intermediates are made then used, catalysts used then regenerated.
The slowest step is rate-determining; a fast pre-equilibrium brings the species before it into the rate law.
Steady-state and pre-equilibrium approximations turn mechanisms into testable rate laws, each within its own limits.

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Data and references

R. Atkinson et al., “Evaluated kinetic and photochemical data for atmospheric chemistry: Volume I”, Atmos. Chem. Phys. 4, 1461–1738 (2004) (NO + NO + O₂).
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 17E–17F (reaction mechanisms, steady state, pre-equilibria).
Chemistry LibreTexts, “Ea and catalysts” (H₂O₂ + I⁻ mechanism and barriers).