AP Chem 5.7: Introduction to Reaction Mechanisms

How a reaction proceeds, step by step

U5.4 noted that a balanced equation often shows only the start and the finish. The set of elementary steps a reaction actually goes through is its reaction mechanism反应机理. Ozone decomposition, for example, appears to take two steps (OpenStax 12.6):

\begin{aligned} &\text{Step 1: }\ce{O3(g) -> O2(g) + O(g)} \\ &\text{Step 2: }\ce{O(g) + O3(g) -> 2O2(g)} \end{aligned}

Each step is an elementary reaction that happens exactly as written, and the steps must add up to the overall reaction 2O₃ → 3O₂. Two O₃ molecules never collide directly: one O₃ falls apart, and the O atom it forms then reacts with another O₃.

Adding the steps

Add up the reactants and the products of all the steps, and cancel species that appear on both sides:

If a step must happen more than once for things to balance, multiply that step first.
Cancel identical species on the two sides, number for number.
What remains is the overall reaction, which must match the given equation.

Intermediates and catalysts

Two kinds of species cancel when the steps are added:

An intermediate中间体 is formed in an earlier step and used up in a later one. The O atom above is an intermediate. Intermediates are used up soon after they form, so their concentrations are usually very low and they don’t last long; they don’t appear in the overall reaction.
A catalyst催化剂 is used up in an earlier step and re-formed in a later one. It is present from the start, its amount is the same at the end, and it doesn’t appear in the overall reaction either.

Take the cycle by which Cl atoms destroy ozone, from OpenStax:

\begin{aligned} &\ce{Cl + O3 -> ClO + O2} \\ &\ce{ClO + O -> Cl + O2} \end{aligned}

Cl is used up and then re-formed, so it is a catalyst; ClO is formed and then used up, so it is an intermediate. The overall reaction is O₃ + O → 2O₂. Because Cl can go around the cycle again and again, one Cl atom can destroy thousands of O₃ molecules.

Classify the species in each mechanism below, then see the result of adding the steps:

Try it: add up the steps and find the intermediates and catalysts

O₃ → O₂ + OO₃ → O₂ + O
O + O₃ → 2O₂O + O₃ → 2O₂

What role does each species play in this mechanism?

O₃

O₂

Is a mechanism plausible?

A plausible mechanism must meet these conditions:

each step is an elementary reaction (usually unimolecular or bimolecular; termolecular steps are rare);
the steps add up to the overall reaction;
the rate law it predicts agrees with the experimental rate law (U5.8, U5.9).

Even when all of these hold, a mechanism is only “consistent with the data”; that doesn’t prove it is the real mechanism. Other evidence can support a mechanism further, such as detecting a proposed intermediate by spectroscopy, or finding that adding a proposed catalyst really does speed the reaction up while its own amount stays the same.

Example 1: Adding steps and finding intermediates

The mechanism below has been proposed for $\ce{2NO(g) + 2H2(g) -> N2(g) + 2H2O(g)}$ (illustrative). Show that the steps add up to the overall reaction, and identify the intermediates.

\begin{aligned} &\text{Step 1: }\ce{2NO <=> N2O2} \\ &\text{Step 2: }\ce{N2O2 + H2 -> N2O + H2O} \\ &\text{Step 3: }\ce{N2O + H2 -> N2 + H2O} \end{aligned}

Step-by-step solution

Add the left sides of the three steps.
$\ce{2NO + N2O2 + H2 + N2O + H2}$
Add the right sides.
$\ce{N2O2 + N2O + H2O + N2 + H2O}$
Cancel what appears on both sides and combine. One $\ce{N2O2}$ and one $\ce{N2O}$ appear on each side and cancel; the two $\ce{H2}$ and two $\ce{H2O}$ combine:
$\boxed{\ce{2NO + 2H2 -> N2 + 2H2O}}$
This matches the overall reaction, so the mechanism passes the “steps must add up” test.
Find the intermediates. An intermediate is formed first and used up later, and doesn’t appear in the overall reaction:
- $\ce{N2O2}$ : formed in step 1, used in step 2
- $\ce{N2O}$ : formed in step 2, used in step 3
Both are intermediates.
Is there a catalyst? A catalyst is used up first and re-formed later, and must be present before the reaction starts. Nothing in this mechanism behaves that way, so there is no catalyst.

Example 2: Which is the catalyst?

This mechanism has been proposed for the decomposition of H₂O₂ in the presence of I⁻:

\begin{aligned} &\text{Step 1: }\ce{H2O2 + I- -> H2O + IO-} \\ &\text{Step 2: }\ce{H2O2 + IO- -> H2O + O2 + I-} \end{aligned}

Write the overall reaction, identify the catalyst and the intermediate, and justify your answers.

Step-by-step solution

Add the two steps.
$\begin{aligned} &\ce{2H2O2 + I- + IO-} \\ &\ce{-> 2H2O + IO- + O2 + I-} \end{aligned}$
Cancel what appears on both sides. $\ce{I-}$ and $\ce{IO-}$ are on both sides and cancel, leaving the overall reaction:
$\boxed{\ce{2H2O2 -> 2H2O + O2}}$
Tell the two kinds of species apart by order of appearance.
- Catalyst: used up first, re-formed later (added before the reaction)
- Intermediate: formed first, used up later (absent at the start)
$\ce{I-}$ . Used up in step 1 and re-formed in step 2, and the reaction is run in the presence of $\ce{I-}$ (it is added beforehand): $\ce{I-}$ is the catalyst.
$\ce{IO-}$ . Formed in step 1 and used up in step 2, absent at the start: $\ce{IO-}$ is the intermediate.

Practice: in the mechanism given, what is this species?

O₃ → O₂ + O; O + O₃ → 2O₂. The O
NO + O₃ → NO₂ + O₂; O₃ → O₂ + O; NO₂ + O → NO + O₂. The NO
H₂O₂ + I⁻ → H₂O + IO⁻; H₂O₂ + IO⁻ → H₂O + O₂ + I⁻. The IO⁻
The I⁻ in the same mechanism
The O₂ in the same mechanism
2NO ⇌ N₂O₂; N₂O₂ + O₂ → 2NO₂. The O₂
The N₂O₂ in the same mechanism
A + M → AM; AM + B → AB + M. The M
The AM in the same mechanism
Cl + O₃ → ClO + O₂; ClO + O → Cl + O₂. The O₃

Common mistakes

Mistake 1: “Intermediates and catalysts both cancel, so they’re the same thing.” Look at the order: an intermediate is formed first and used up later, so it isn’t there at the start; a catalyst is used up first and re-formed later, so it must be added before the reaction starts.

Mistake 2: “A catalyst doesn’t take part in the reaction.” A catalyst does react (it is used up in some step); it is simply re-formed in a later step, so its total amount doesn’t change.

Mistake 3: “If the steps add up to the overall reaction, the mechanism is correct.” That’s only a necessary condition. The mechanism must also agree with the experimental rate law, and even then it is only a possible mechanism.

Quick check

What overall reaction do these three steps add up to? 2NO ⇌ N₂O₂; N₂O₂ + H₂ → N₂O + H₂O; N₂O + H₂ → N₂ + H₂O

Below 225 °C, the mechanism for NO₂ + CO → NO + CO₂ is NO₂ + NO₂ → NO₃ + NO (slow); NO₃ + CO → NO₂ + CO₂ (fast). What is the intermediate?

Cl atoms destroy ozone: Cl + O₃ → ClO + O₂; ClO + O → Cl + O₂. Which statement is correct?

During a reaction, the concentration of an intermediate is usually very low. Which explanation is best?

Which observation best supports the claim that Cl atoms catalyze O₃ + O → 2O₂?

How do the products of a reaction compare with and without a catalyst?

Mechanism: X₂ ⇌ 2X (fast); X + Y → XY (slow). For the steps to add up to X₂ + 2Y → 2XY, how many times must step 2 occur?

AP-style free response

This mechanism has been proposed for the decomposition of H₂O₂ in the presence of I⁻:

\begin{aligned} &\text{Step 1: }\ce{H2O2(aq) + I-(aq) -> H2O(l) + IO-(aq)}\quad\text{(slow)} \\ &\text{Step 2: }\ce{H2O2(aq) + IO-(aq) -> H2O(l) + O2(g) + I-(aq)}\quad\text{(fast)} \end{aligned}

(a) Show that the steps add up to the overall reaction 2H₂O₂ → 2H₂O + O₂.

(b) Identify the catalyst and the intermediate, and justify your answers.

(c) If this mechanism is correct, write the rate law. (Hint: the slow step sets the overall rate; see U5.8.)

(d) [I⁻] can appear in the rate law, but [IO⁻] can’t. Explain why.

(e) If more KI is added (with the same amount of H₂O₂), how do the reaction rate and the final amount of O₂ produced change?

Model answer and scoring

(a) Left: 2H₂O₂ + I⁻ + IO⁻; right: 2H₂O + IO⁻ + O₂ + I⁻. Canceling I⁻ and IO⁻ on both sides gives 2H₂O₂ → 2H₂O + O₂.

(b) I⁻ is the catalyst: it is used up in step 1, re-formed in step 2, and present before the reaction starts. IO⁻ is the intermediate: it forms in step 1 and is used up in step 2.

(c) rate = k[H₂O₂][I⁻].

(d) I⁻ is an added catalyst whose concentration can be controlled and measured, and it stays constant during the reaction. IO⁻ is an intermediate, at a very low concentration that is hard to measure; a rate law can only use concentrations of species that can be measured.

(e) The rate increases, because [I⁻] increases and the rate is proportional to [I⁻]. The amount of O₂ is unchanged, because a catalyst doesn’t change the overall reaction, and the amount of O₂ depends on the amount of H₂O₂.

Scoring: (a) correct addition and cancellation; (b) both species, justified by the order of formation and use; (c) the rate law; (d) why an intermediate can’t appear; (e) both conclusions with reasons.

Summary

A reaction mechanism is a series of elementary steps; the steps (multiplied where needed) must add up to the overall reaction.
An intermediate is formed and then used up, at low concentration; a catalyst is used up and then re-formed, and is present from the start. Neither appears in the overall reaction.
A plausible mechanism: every step is elementary, the steps add up to the overall reaction, and it agrees with the experimental rate law. Agreement shows the mechanism is possible, not that it is proven.
A catalyst doesn’t change the overall reaction or its products, only the pathway.

AP tip: To classify a species as an intermediate or a catalyst, state the step where it forms and the step where it is used up; that order is your justification. When writing the overall reaction, check whether any step needs to be multiplied.

Introduction to Reaction Mechanisms反应机理入门

Learning goals

Before you start

How a reaction proceeds, step by step

Adding the steps

Intermediates and catalysts

Is a mechanism plausible?

Example 1: Adding steps and finding intermediates

Example 2: Which is the catalyst?

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem