When the slow step isn’t first
In U5.8 the slow step came first, so its rate law contained only reactants. When the first step is fast and the second is slow, there’s a problem: the second step’s reactants usually include an intermediate formed in the first step, and intermediates can’t appear in a rate law.
The mechanism for 2NO + Cl₂ → 2NOCl in OpenStax Example 12.14 is like this:
Step 2 is rate-determining: rate = k₂[NOCl₂][NO]. NOCl₂ is an intermediate, so it has to be replaced.
Replacing the intermediate with a fast equilibrium
Step 1 is fast and reversible: its forward and reverse reactions are both much faster than step 2, so step 1 stays essentially at equilibrium (2.11.3). This is called a pre-equilibrium预平衡. At equilibrium the forward and reverse rates are equal:
Substituting into the rate law of the rate-determining step:
The steps, in summary:
- Write the rate law of the rate-determining step.
- Identify the intermediate in it.
- For the earlier fast equilibrium, set forward rate = reverse rate (the equilibrium-constant expression) and solve for the intermediate’s concentration.
- Substitute, and combine the constants into k.
AP files this method under the steady-state approximation稳态近似: an intermediate is used up soon after it forms, so its concentration stays small and nearly constant. What the AP exam tests is the “fast equilibrium + slow step” substitution above.
Fractional and negative orders
After substitution, orders aren’t always positive whole numbers:
- Fractional order. If the first step is A₂ ⇌ 2A (fast), then [A]² = K[A₂] and [A] = (K[A₂])^½, so [A₂]^½ appears in the rate law.
- Negative order. The other product of a fast equilibrium ends up in the denominator. With HX ⇌ H⁺ + X⁻ (fast) and X⁻ + Y → XY⁻ (slow), [X⁻] = K[HX]/[H⁺], so rate = k[HX][Y]/[H⁺]. The higher [H⁺], the further the equilibrium lies toward HX, the less X⁻ there is, and the slower the reaction.
Step through the derivations for the mechanisms below:
Overall reaction: 2NO + Cl₂ → 2NOCl
- NO + Cl₂ ⇌ NOCl₂(fast equilibrium)
- NOCl₂ + NO → 2NOCl(slow)
- Step 2 is slowest, so it is the rate-determining step, and the overall rate equals its rate: rate = k₂[NOCl₂][NO]
- But NOCl₂ is an intermediate (formed in step 1 and used up in step 2); its concentration is tiny and hard to measure, so it can’t appear in the rate law.
- Step 1 is a fast equilibrium, so its forward and reverse rates are equal: k₁[NO][Cl₂] = k₋₁[NOCl₂]
- So [NOCl₂] = K₁[NO][Cl₂], where K₁ = k₁/k₋₁
- Substituting into the rate-determining step: rate = k[NO]²[Cl₂], where k = k₂K₁
This is the derivation in OpenStax Example 12.14. The result is second order in NO and first order in Cl₂, matching the experiment.
Overall reaction: 2NO + O₂ → 2NO₂
- NO + NO ⇌ N₂O₂(fast equilibrium)
- N₂O₂ + O₂ → 2NO₂(slow)
- Step 2 is slowest, so it is the rate-determining step, and the overall rate equals its rate: rate = k₂[N₂O₂][O₂]
- But N₂O₂ is an intermediate (formed in step 1 and used up in step 2); its concentration is tiny and hard to measure, so it can’t appear in the rate law.
- Step 1 is a fast equilibrium, so its forward and reverse rates are equal: k₁[NO]² = k₋₁[N₂O₂]
- So [N₂O₂] = K₁[NO]², where K₁ = k₁/k₋₁
- Substituting into the rate-determining step: rate = k[NO]²[O₂], where k = k₂K₁
The result is the same as the rate law for a single termolecular step 2NO + O₂ → 2NO₂: the rate law alone can’t tell these two mechanisms apart.
Overall reaction: A₂ + 2B → 2AB
- A₂ ⇌ 2A(fast equilibrium)
- A + B → AB(slow), occurs 2 times
- Step 2 is slowest, so it is the rate-determining step, and the overall rate equals its rate: rate = k₂[A][B]
- But A is an intermediate (formed in step 1 and used up in step 2); its concentration is tiny and hard to measure, so it can’t appear in the rate law.
- Step 1 is a fast equilibrium, so its forward and reverse rates are equal: k₁[A₂] = k₋₁[A]²
- So [A]² = K₁[A₂], where K₁ = k₁/k₋₁, that is, [A] = (K₁[A₂])^½
- Substituting into the rate-determining step: rate = k[A₂]^½[B], where k = k₂K₁^½
Orders can be fractions: A₂ splits into two A atoms, so [A] is proportional to the square root of [A₂], and the reaction is order 1/2 in A₂.
Overall reaction: HX + Y → H⁺ + XY⁻
- HX ⇌ H⁺ + X⁻(fast equilibrium)
- X⁻ + Y → XY⁻(slow)
- Step 2 is slowest, so it is the rate-determining step, and the overall rate equals its rate: rate = k₂[X⁻][Y]
- But X⁻ is an intermediate (formed in step 1 and used up in step 2); its concentration is tiny and hard to measure, so it can’t appear in the rate law.
- Step 1 is a fast equilibrium, so its forward and reverse rates are equal: k₁[HX] = k₋₁[H⁺][X⁻]
- So [X⁻] = K₁[HX] / [H⁺], where K₁ = k₁/k₋₁
- Substituting into the rate-determining step: rate = k[HX][Y] / [H⁺], where k = k₂K₁
H⁺ is the other product of the fast equilibrium, so it appears in the denominator: the higher [H⁺], the further the equilibrium lies toward HX, the less X⁻ there is, and the slower the rate (order −1 in H⁺).
Example 1: Deriving a rate law
This mechanism has been proposed for 2NO + O₂ → 2NO₂: NO + NO ⇌ N₂O₂ (fast); N₂O₂ + O₂ → 2NO₂ (slow). Derive the rate law.
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Rate law of the rate-determining step. Step 2 is slow, and an elementary step’s coefficients are its orders:
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Spot the intermediate. forms in step 1 and is used up in step 2; its concentration can’t be measured directly, so it must be replaced with reactant concentrations.
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Replace it using the fast equilibrium. Step 1 is fast, so its forward and reverse rates are equal:
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Substitute into the slow step.
where k = k₂K₁.
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Check. Only reactants remain in the rate law (no intermediate), third order overall.
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Think about it. This is the same rate law as for the single termolecular step 2NO + O₂ → 2NO₂, so the rate law alone can’t tell these two mechanisms apart. A rate law can rule a mechanism out, but it can’t prove one.
Example 2: What a negative order means
The reaction HX + Y → H⁺ + XY⁻ has this mechanism: HX ⇌ H⁺ + X⁻ (fast); X⁻ + Y → XY⁻ (slow). Derive the rate law, and predict the effect of lowering the pH of the solution on the rate.
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Rate law of the rate-determining step. Step 2 is slow:
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Replace the intermediate X⁻ using the fast equilibrium. The equilibrium constant of step 1:
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Substitute to get the rate law.
[H⁺] is in the denominator, so the reaction is order −1 in H⁺.
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Make sense of the negative order. H⁺ is a product of step 1; the larger [H⁺], the farther step 1’s equilibrium shifts left, the less X⁻ there is, and the slower the slow step.
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Predict the effect of lowering the pH. Lowering the pH raises [H⁺]. For example, a pH drop of 1 multiplies [H⁺] by 10:
The rate decreases to 1/10 of its value.
- A + B ⇌ C (fast); C + B → D (slow). Order in B
rate = k₂[C][B] = k₂K₁[A][B]².
- Same mechanism; order in A
[A] appears once in [C] = K₁[A][B].
- X₂ ⇌ 2X (fast); X + Y → XY (slow). Order in X₂
[X] = (K₁[X₂])^½.
- Same mechanism; order in Y
Y appears once, in the slow step.
- HA ⇌ H⁺ + A⁻ (fast); A⁻ + B → AB⁻ (slow). Order in H⁺
[A⁻] = K₁[HA]/[H⁺], so H⁺ is in the denominator.
- Same mechanism; order in HA
rate = k₂K₁[HA][B]/[H⁺].
- 2A ⇌ A₂ (fast); A₂ + B → P (slow). Order in A
[A₂] = K₁[A]².
- A ⇌ B + C (fast); B + D → P (slow). Order in C
[B] = K₁[A]/[C].
- Same mechanism; order in D
D appears once, in the slow step.
- Br₂ ⇌ 2Br (fast); Br + H₂ → HBr + H (slow). Order in Br₂
[Br] = (K₁[Br₂])^½.
Common mistakes
Mistake 1: “Whatever is in the rate-determining step goes into the rate law.” If the rate-determining step contains an intermediate, it must first be replaced using the fast equilibrium; a rate law can contain only concentrations that can be measured.
Mistake 2: “Fast steps don’t affect the rate.” A fast equilibrium before the rate-determining step controls how much intermediate there is, and its equilibrium constant ends up in the overall k (k = k₂K₁). Only fast steps after the rate-determining step leave the rate unaffected.
Mistake 3: “Orders are always positive whole numbers.” After substitution, an order can be 1/2 (when a diatomic molecule first splits apart) or −1 (when a product of the fast equilibrium ends up in the denominator).
Quick check
Mechanism: NO + Cl₂ ⇌ NOCl₂ (fast); NOCl₂ + NO → 2NOCl (slow). What rate law does it predict?
[NOCl₂] = K₁[NO][Cl₂], substituted into k₂[NOCl₂][NO]. B contains an intermediate; C is the rate law of step 1, which isn’t the rate-determining step.
For the fast equilibrium A + B ⇌ I, how can the intermediate’s concentration [I] be written?
k₁[A][B] = k₋₁[I], so [I] = (k₁/k₋₁)[A][B].
A reaction has the rate law rate = k[A][B]/[C], where C is a product. If extra C is added at the start, how does the rate change?
The order in C is −1, so raising [C] lowers the rate. In terms of the mechanism, C shifts the fast equilibrium to the left and reduces the intermediate.
The first step of a mechanism is X₂ ⇌ 2X (fast), and the second, X + Y → XY, is slow. How is [X] related to [X₂]?
K = [X]²/[X₂], so [X]² = K[X₂]; take the square root.
Two mechanisms both give rate = k[NO]²[O₂]: one goes through the intermediate N₂O₂, the other through NO₃. Which statement is best?
With identical rate laws, kinetics can’t choose between the two.
A fast equilibrium has k₁ = 4.0 × 10³ and k₋₁ = 2.0 × 10⁴, and the rate-determining step has k₂ = 50 (units omitted). What is the overall rate constant k = k₂K₁?
K₁ = k₁/k₋₁ = 4.0 × 10³ ÷ 2.0 × 10⁴ = 0.20, so k = 50 × 0.20 = 10. A is K₁; B uses k₂/K₁.
AP-style free response
This mechanism has been proposed for 2NO(g) + Br₂(g) → 2NOBr(g):
(a) Show that the steps add up to the overall reaction, and identify the intermediate.
(b) Write the rate law of the rate-determining step, and explain why it can’t be used directly as the overall rate law.
(c) Derive the overall rate law, and express k in terms of the constants of the individual steps.
(d) Suppose the experimental rate law is rate = k[NO]²[Br₂]. Is this mechanism consistent with the experiment?
(e) A single termolecular step, 2NO + Br₂ → 2NOBr, is also consistent with this rate law. Propose a piece of evidence that could distinguish the two mechanisms.
Model answer and scoring
(a) NO + Br₂ + NOBr₂ + NO → NOBr₂ + 2NOBr; canceling NOBr₂ gives 2NO + Br₂ → 2NOBr. NOBr₂ forms in step 1 and is used up in step 2: it is the intermediate.
(b) rate = k₂[NOBr₂][NO]. It contains the concentration of the intermediate NOBr₂, which is tiny and can’t be controlled or measured, so it must be replaced first.
(c) Step 1 is a fast equilibrium: k₁[NO][Br₂] = k₋₁[NOBr₂], so [NOBr₂] = K₁[NO][Br₂], with K₁ = k₁/k₋₁. Substituting: rate = k₂K₁[NO]²[Br₂], so k = k₂K₁ = k₂k₁/k₋₁.
(d) Yes: the derived rate law matches the experiment, and the steps add up to the overall reaction.
(e) For example, use spectroscopy to look for the intermediate NOBr₂ in the reaction mixture. Detecting it would support the two-step mechanism; the one-step mechanism has no such intermediate.
Scoring: (a) the sum and the intermediate; (b) notes the intermediate; (c) the fast-equilibrium relation, the substitution, and the expression for k; (d) a comparison and conclusion; (e) sensible evidence, such as detecting the intermediate.
Summary
- When a fast reversible first step is followed by a slow step, the rate law of the rate-determining step usually contains an intermediate.
- Solve for the intermediate from the fast equilibrium (forward rate = reverse rate): [intermediate] = K × [reactants] / [other products], then substitute into the rate-determining step. The overall k = k₂K₁.
- Substitution can give order 1/2 (a diatomic molecule splitting first) or order −1 (a product of the fast equilibrium in the denominator).
- Different mechanisms can lead to the same rate law; a rate law can support a mechanism but can’t prove it.
AP tip: Lay out the four steps of a derivation clearly: the rate law of the rate-determining step → identify the intermediate → solve for it from the fast equilibrium → substitute and combine constants. Finish by checking that the rate law contains only concentrations of reactants (or catalysts, or products of the fast equilibrium).