Reactions that happen in one step
U5.2 said that the rate law of an overall reaction can only be found by experiment. The reason is that a balanced equation often shows only the start and the finish; the actual reaction may take several steps. A reaction that happens in a single step, exactly as written, is an elementary reaction基元反应, or elementary step. It describes one real molecular event: one particle falling apart or rearranging, or several particles colliding at once.
Molecularity
The number of particles that take part in an elementary reaction is its molecularity分子数:
| Molecularity | General form | Example (OpenStax 12.6) | Rate law |
|---|---|---|---|
| Unimolecular | A → products | rate = k[A] | |
| Bimolecular | A + B → products | rate = k[A][B] | |
| Bimolecular | 2A → products | rate = k[A]² | |
| Termolecular | 2A + B → products | (thought to involve termolecular steps) | rate = k[A]²[B] |
Three particles colliding at the same instant is very unlikely, so termolecular steps are rare; elementary steps with four or more particles colliding at once essentially don’t happen.
Why elementary rate laws come from coefficients
Take the bimolecular step A + B → products. For it to happen, an A must collide with a B. If the number of A particles per unit volume doubles, each B is twice as likely to meet an A, so A–B collisions per unit time double and the rate doubles; the same goes for B. So the rate is proportional to [A][B]. For 2A → products, two A particles must collide, so the rate is proportional to [A]².
In a unimolecular step, each molecule has a certain probability of reacting per unit time; doubling the number of molecules doubles the number that react per unit time, so the step is first order.
The conclusion: for an elementary reaction, the orders in the rate law equal the coefficients, and the overall order equals the molecularity.
Overall reactions are different
An overall reaction may be the sum of several elementary steps, so its rate law can only be found by experiment. OpenStax gives a good example, NO₂ + CO → NO + CO₂:
- Above 225 °C: rate = k[NO₂][CO]. This matches a single bimolecular collision, so at high temperature it may be an elementary reaction.
- Below 225 °C: rate = k[NO₂]², independent of CO. This doesn’t match a single collision, which shows the reaction takes two steps at low temperature (U5.7 discusses this mechanism).
Even when an experimental rate law matches the coefficients, it only shows that a single step is possible; it doesn’t prove the reaction is elementary.
Each example below asks: if the first reactant’s concentration doubles, by what factor does the rate change? For an elementary step the coefficients answer it; for an overall reaction only experiment can:
O₃ → O₂ + OElementary step
If [O₃] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k[O₃]
Unimolecular: one particle falls apart or rearranges on its own; first order.
- Doubling [O₃] → rate multiplied by 2
C₄H₈ → 2C₂H₄Elementary step
If [C₄H₈] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k[C₄H₈]
Unimolecular: one particle falls apart or rearranges on its own; first order.
- Doubling [C₄H₈] → rate multiplied by 2
O + O₃ → 2O₂Elementary step
If [O] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k[O][O₃]
Bimolecular: two particles collide; second order.
- Doubling [O] → rate multiplied by 2
- Doubling [O₃] → rate multiplied by 2
NO₂ + NO₂ → NO₃ + NOElementary step
If [NO₂] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k[NO₂]²
Bimolecular: two particles collide; second order.
- Doubling [NO₂] → rate multiplied by 4
NO₃ + CO → NO₂ + CO₂Elementary step
If [NO₃] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k[NO₃][CO]
Bimolecular: two particles collide; second order.
- Doubling [NO₃] → rate multiplied by 2
- Doubling [CO] → rate multiplied by 2
2NO + Cl₂ → 2NOClElementary step
If [NO] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k[NO]²[Cl₂]
Termolecular: three particles collide at once, which is very rare; third order.
- Doubling [NO] → rate multiplied by 4
- Doubling [Cl₂] → rate multiplied by 2
NO₂ + CO → NO + CO₂Overall reaction
If [NO₂] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k[NO₂]²
This is the experimental result (below 225 °C): the rate doesn’t depend on [CO] and is second order in NO₂, unlike the coefficients 1 and 1.
- Doubling [NO₂] → rate multiplied by 4
2H₂O₂ → 2H₂O + O₂Overall reaction
If [H₂O₂] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k[H₂O₂]
This is the experimental result: first order in H₂O₂, not the second order the coefficient 2 would suggest.
- Doubling [H₂O₂] → rate multiplied by 2
2NH₃ → N₂ + 3H₂Overall reaction
If [NH₃] doubles (with everything else the same), the rate is multiplied by what factor?
rate = k
This is the experimental result (on a hot tungsten surface): zero order, so the rate doesn’t depend on [NH₃].
Zero order: changing [NH₃] leaves the rate unchanged.
Example 1: Rate laws of elementary steps
Give the molecularity and rate law of each elementary step: (1) ; (2) ; (3) .
-
Identify: the rule for elementary steps. An elementary step is a single molecular event, so its coefficients are the numbers of each particle in that one collision (or breakup) and can be used directly as orders; the total number of particles is the molecularity.
-
(1) . One O₃ molecule falls apart by itself, so 1 particle is involved: unimolecular.
-
(2) . One O atom collides with one O₃ molecule, 1 + 1 = 2 particles: bimolecular.
-
(3) . Two NO molecules and one Cl₂ molecule collide at once, 2 + 1 = 3 particles: termolecular.
-
Check. The overall order of each rate law equals its molecularity (1, 2, 3). Three-particle collisions are rare, so termolecular steps are rare too.
Example 2: Evaluating a claim
A student says, “The equation for H₂O₂ decomposition is 2H₂O₂ → 2H₂O + O₂, so rate = k[H₂O₂]².” Experiments show the reaction is first order in H₂O₂. Evaluate the claim.
-
Identify what the student did. The student took the coefficient 2 of H₂O₂ in the overall equation 2H₂O₂ → 2H₂O + O₂ and used it as the order.
-
Find the problem. Only the coefficients of an elementary step can be used as orders. This equation is an overall reaction that need not happen in one step, so its orders must be found by experiment.
-
Compare with experiment. Experiments show first order in H₂O₂:
The student’s rate law is second order, so they disagree.
-
Connect to the mechanism. If the reaction really were two H₂O₂ molecules colliding in one step, it would be second order in H₂O₂; first order means the reaction does not happen in one step but goes through several (for example, with a catalyst involved).
-
Conclusion. The claim is incorrect: an overall equation’s coefficients can’t be used to write the rate law, which has to come from experiment.
- Cl₂ → 2Cl
One Cl₂ molecule falls apart. The number of products doesn't affect molecularity.
- H + Cl₂ → HCl + Cl
Two particles collide.
- 2NO₂ → NO₃ + NO
Two NO₂ molecules collide: rate = k[NO₂]².
- C₄H₈ → 2C₂H₄
One cyclobutane molecule falls apart.
- 2NO + O₂ → 2NO₂
Three particles collide at once.
- NO₃ + CO → NO₂ + CO₂
Two particles collide.
- CH₃NC → CH₃CN
One molecule rearranges (isomerizes).
- 2I + H₂ → 2HI
Two I atoms and one H₂ molecule collide at once.
- O₃ → O₂ + O
One O₃ molecule falls apart.
- A + B + C → products
Three particles collide at once, which is very rare.
Common mistakes
Mistake 1: “You can write a rate law from the coefficients of any equation.” Only for elementary steps. An ordinary chemical equation is an overall reaction, and its rate law must be found by experiment; for example, 2H₂O₂ → 2H₂O + O₂ is first order in H₂O₂.
Mistake 2: “If the experimental rate law matches the coefficients, the reaction is elementary.” A match only shows that a single step is possible. The rate law of H₂ + 2NO → N₂O + H₂O, k[NO]²[H₂], matches the coefficients, but that alone doesn’t show the reaction happens in one step.
Mistake 3: “Molecularity counts the products.” Molecularity counts only the reactant particles that take part in the event. Cl₂ → 2Cl forms two atoms but is unimolecular.
Quick check
NO + O₃ → NO₂ + O₂ is an elementary step. What is its rate law?
For an elementary step, the orders equal the coefficients: one NO collides with one O₃, a bimolecular step. C is true only for overall reactions.
If 2NO₂ → 2NO + O₂ were an elementary step, what would its molecularity and rate law be?
Two NO₂ molecules collide, so it's bimolecular and the rate is proportional to [NO₂]².
Which is the best explanation for why termolecular elementary steps are rare?
A collision requires the particles to arrive at the same place at the same time, which is far less likely for three particles than for two.
The experimental rate law for 2A + B → C is rate = k[A][B]. Which statement is correct?
As a single elementary step, the rate law would be k[A]²[B]. The different experimental result shows the reaction takes more than one step.
For the elementary step A + B → products, if [A] doubles and [B] triples, the rate is multiplied by what factor?
rate = k[A][B], so 2 × 3 = 6.
Above 225 °C, NO₂ + CO → NO + CO₂ has the rate law k[NO₂][CO]. Which statement is best?
Below 225 °C the rate law is k[NO₂]², so the same overall reaction can follow different pathways under different conditions. A match only shows a one-step mechanism is possible.
Suppose 2HI → H₂ + I₂ happens as a single elementary step. What is its rate law?
An elementary step’s rate law comes from its coefficients: two HI molecules collide, so it’s bimolecular and second order in HI. B uses the products; C is true only for overall reactions.
A reaction’s experimental rate law is rate = k[X]²[Y]. If the reaction happens in a single step, which elementary step is consistent with it?
For a single step, the exponents in the rate law are that step’s coefficients: two X and one Y collide at once.
AP-style free response
The overall decomposition of ozone is . A two-step process has been proposed:
(a) Give the molecularity of each step.
(b) Write the rate law for each step.
(c) Show that the two steps add up to the overall reaction. What role does the O atom play?
(d) A student says, “The overall reaction is 2O₃ → 3O₂, so rate = k[O₃]².” Evaluate the claim.
Model answer and scoring
(a) Step 1: unimolecular; step 2: bimolecular.
(b) Step 1: rate = k₁[O₃]; step 2: rate = k₂[O][O₃]. Each step is elementary, so its rate law comes from its coefficients.
(c) Adding: O₃ + O + O₃ → O₂ + O + 2O₂; canceling O from both sides gives 2O₃ → 3O₂. O forms in step 1 and is used up in step 2, so it doesn’t appear in the overall reaction: it’s a reaction intermediate (U5.7).
(d) The claim is incorrect: 2O₃ → 3O₂ is an overall reaction, so its rate law can’t be written from its coefficients. The rate law must come from experiment (or from which step of the mechanism is slowest; see U5.8).
Scoring: (a) both molecularities; (b) both rate laws; (c) correct sum, identifying O as an intermediate; (d) states that an overall reaction’s rate law can’t be written from its coefficients.
Summary
- An elementary reaction happens in one step, exactly as written; its molecularity is the number of particles in that one event: unimolecular, bimolecular, or (rarely) termolecular.
- Rate laws of elementary steps come from their coefficients: A → products, rate = k[A]; A + B → products, rate = k[A][B]; 2A → products, rate = k[A]².
- The reason: the collision frequency is proportional to the product of the reactant concentrations.
- An overall reaction’s rate law can only be found by experiment; a match with the coefficients only shows that a single step is possible.
AP tip: Write a rate law directly from coefficients only when the question says “elementary reaction” or “elementary step.” For an ordinary equation, use experimental data (U5.2) or a mechanism (U5.8) to find the rate law.