基元反应

AP 化学· AP Chemistry CED — Kinetics· 14 分钟阅读

1. 基元反应的定义与核心性质★★☆☆☆⏱ 2 min

基元反应（当它是更大反应机理的一部分时也称为基元步骤）是一步完成的单步反应，完全按书写形式发生，反应物和产物之间没有中间子步骤。与仅描述多个反应事件净结果的总配平反应不同，基元反应代表反应物粒子之间的单次碰撞事件。

基元反应的核心特点是：每个反应物的化学计量系数直接等于其在速率定律中的反应级数。该规则不适用于总反应，因此基元步骤是AP考试中所有多步反应机理问题的基础。

📘 定义

基元反应

仅通过反应物粒子之间一次碰撞事件一步完成的化学反应，不存在中间子步骤。每个反应物的反应级数等于其化学计量系数。

例:

单分子分解反应，书写形式和实际发生过程均为 $A \to products$ ，属于基元反应。

2. 基元反应的反应分子数★★☆☆☆⏱ 3 min

反应分子数定义为单个基元步骤中发生碰撞并反应的反应物粒子数目。由于它计数的是离散粒子，因此反应分子数只能是较小的正整数：1（单分子反应）、2（双分子反应）或 3（三分子反应）。三分子基元步骤非常罕见，因为三个粒子同时以正确取向、足够能量发生碰撞的概率极低。

常见误区是混淆反应分子数和反应级数。反应级数描述速率对浓度的依赖关系，对于总反应可以是零、分数或负数。反应分子数仅对基元步骤有定义，且只能是 1、2 或 3。要确定反应分子数，只需数配平后的基元步骤左侧的反应物粒子总数即可。

📐 例题

确定下列每个基元反应的反应分子数：1. $C_{4} H_{8} \to 2 C_{2} H_{4}$ 2. $C l + C H_{4} \to H C l + C H_{3}$ 3. $B r + B r + A r \to B r_{2} + A r$

1
数每个基元方程式左侧的反应物粒子总数。
2
反应 1 只有 1 个反应物粒子（ $C_{4} H_{8}$ ），所以分子数为 1（单分子反应）。
3
反应 2 有两个不同的反应物粒子（ $C l$ 和 $C H_{4}$ ），所以分子数为 2（双分子反应）。
4
反应 3 共有三个反应物粒子（两个 $B r$ 和一个 $A r$ ），所以分子数为 3（三分子反应）。

Exam tip:

AP选择题经常会设置非整数反应分子数的干扰选项。如果某个选项给出的反应分子数是 0、1.5 或任何非整数，直接排除即可。

3. 基元反应的速率定律★★☆☆☆⏱ 4 min

对于总配平反应，无法从配平方程式直接得到反应级数——必须通过实验测量。但由于基元反应是单次碰撞事件，反应速率与每个反应粒子浓度的化学计量系数次幂成正比。这是因为所有所需反应物粒子同时碰撞的概率与它们各自浓度的乘积成正比。

\text{rate} = k [A]^a [B]^b \quad \text{for a general elementary reaction } aA + bB \rightarrow \text{products}

where $k$ is the rate constant for the elementary step, $a$ is the reaction order with respect to $A$ , and $b$ is the reaction order with respect to $B$ . The overall order of the elementary step is simply the sum of $a$ and $b$ .

📐 例题

Write the rate law for the elementary reaction $2 H I \to H_{2} + I_{2}$ , and state the overall order of the reaction.

1
Confirm the reaction is explicitly stated to be elementary, so we can use stoichiometric coefficients as reaction orders.
2
The only reactant is HI, with a stoichiometric coefficient of 2.
3
Write the rate law directly from the coefficients:
4
$latex": "\text{rate} = k [HI]^2$
5
Sum the reaction orders to get the overall order of 2.

Exam tip:

In AP FRQ questions asking for a rate law for a forward elementary step, never include product concentrations. Only reactants appear in the rate law for forward elementary steps, which is what you will be asked for 99% of the time on the exam.

4. Elementary Steps in Multi-Step Reaction Mechanisms★★★☆☆⏱ 5 min

Nearly all overall reactions are not single elementary steps — they proceed via a sequence of multiple elementary steps called a reaction mechanism. When you add all elementary steps in a mechanism together, you get the balanced overall reaction.

📘 定义

Key Mechanism Species

Two types of species do not appear in the overall balanced reaction: Reaction intermediates are produced in an early elementary step and consumed in a later step. Catalysts are consumed in an early step and regenerated in a later step.

例:

In the ozone decomposition mechanism, $C l O$ is an intermediate and $C l$ is a catalyst.

For a mechanism to be valid, two conditions must hold: 1) the sum of elementary steps matches the experimental overall reaction, and 2) the rate law derived from the mechanism matches the experimentally determined rate law for the overall reaction.

📐 例题

A reaction mechanism for the conversion of ozone to oxygen is given below: Step 1 (fast): $C l + O_{3} \to C l O + O_{2}$ Step 2 (slow): $C l O + O \to C l + O_{2}$ Identify the reaction intermediate and the catalyst, and confirm the sum of elementary steps gives the overall reaction $O_{3} + O \to 2 O_{2}$ .

1
Track where each non-overall species is produced and consumed: Cl is consumed in step 1 and produced in step 2; ClO is produced in step 1 and consumed in step 2.
2
By definition: a catalyst is consumed first then produced, so Cl is the catalyst. An intermediate is produced first then consumed, so ClO is the reaction intermediate.
3
Add the two steps to get total reactants and products: Left side: $C l + O_{3} + C l O + O$ , Right side: $C l O + O_{2} + C l + O_{2}$ .
4
Cancel species that appear on both sides: Cl and ClO cancel, leaving $O_{3} + O \to 2 O_{2}$ , which matches the given overall reaction.

Exam tip:

Always double-check the order of production/consumption to avoid confusing intermediates and catalysts on FRQ questions — this is one of the most commonly missed points on mechanism problems.

5. AP-Style Practice Problems★★★★☆⏱ 5 min

📐 例题

Which of the following correctly gives the rate law and molecularity for the elementary reaction $2 N O + O_{2} \to 2 N O_{2}$ ?
A) Rate = $k [N O] [O_{2}]$ , unimolecular
B) Rate = $k [N O]^{2} [O_{2}]$ , bimolecular
C) Rate = $k [N O]^{2} [O_{2}]$ , termolecular
D) Rate = $k [N O_{2}]^{2}$ , termolecular

1
For elementary reactions, reaction orders equal the stoichiometric coefficients of reactants, so the rate law here must be $k [N O]^{2} [O_{2}]$ . This eliminates options A and D.
2
Molecularity counts the total number of reactant particles: 2 $N O$ + 1 $O_{2}$ = 3 total particles, so molecularity is termolecular. This eliminates option B.
3
The correct answer is C.

📐 例题

The overall reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ follows the three-step mechanism below:
Step 1 (fast): $N_{2} O_{5} ⇌ N O_{2} + N O_{3}$
Step 2 (slow): $N O_{2} + N O_{3} \to N O + N O_{2} + O_{2}$
Step 3 (fast): $N O + N O_{3} \to 2 N O_{2}$

(a) Identify all reaction intermediates in this mechanism.
(b) Write the rate law for the overall reaction consistent with this mechanism.
(c) What is the overall order of the reaction based on this mechanism?

1
(a) Track production and consumption: $N O_{3}$ is produced in step 1 and consumed in steps 2 and 3; $N O$ is produced in step 2 and consumed in step 3. Both are produced early and consumed late, so $N O_{3}$ and $N O$ are the reaction intermediates.
2
(b) The slow (rate-determining) step is elementary, so its rate law is:
3
$latex": "\text{rate} = k_2 [NO_2][NO_3]$
4
From the equilibrium in step 1, forward rate equals reverse rate: $k_{1} [N_{2} O_{5}] = k_{- 1} [N O_{2}] [N O_{3}]$ , so rearranged, $[N O_{2}] [N O_{3}] = \frac{k _{1}}{k _{- 1}} [N_{2} O_{5}]$ .
5
Substitute into the rate law for the slow step:
6
$latex": "\text{rate} = \frac{k_1 k_2}{k_{-1}} [N_2O_5] = k [N_2O_5]$
7
(c) The only reaction order is 1 for $N_{2} O_{5}$ , so the overall order of the reaction is 1.

📐 例题

The radioactive decay of carbon-14, used for radiocarbon dating, is a unimolecular elementary first-order process with a rate constant $k = 1.21 \times 1 0^{- 4} year^{- 1}$ . A 10 g sample of ancient wood has an initial carbon-14 concentration of $1.65 \times 1 0^{- 10} mol/g$ . What is the initial rate of decay of carbon-14 in this sample, in moles per year?

1
Since decay is an elementary unimolecular reaction, the rate law is directly derived from stoichiometry: $rate = k [^{14} C]$ .
2
First find the total initial moles of C-14 in the 10 g sample: $10 g \times 1.65 \times 1 0^{- 10} mol/g = 1.65 \times 1 0^{- 9} mol$ .
3
Substitute into the rate law to get the initial rate:
4
$latex": "\text{rate} = (1.21 \times 10^{-4}\ \text{year}^{-1})(1.65 \times 10^{-9}\ \text{mol}) = 2.00 \times 10^{-13}\ \text{mol/year}$
5
This result matches the expected slow decay of ancient carbon-14 samples.

6. 常见陷阱

错误做法:

Using stoichiometric coefficients from an overall reaction to write the rate law, the same way you do for an elementary reaction.

原因:

Students generalize the rule for elementary reactions to all reactions, forgetting that only elementary steps have orders matching coefficients.

正确做法:

Always confirm the reaction is explicitly labeled as elementary before using coefficients to get reaction orders; for overall reactions, only use experimentally derived orders.

错误做法:

Assigning a non-integer or zero molecularity to an elementary reaction.

原因:

Students mix up the definitions of molecularity (count of particles) and reaction order (can be any value).

正确做法:

Remember molecularity is only 1, 2, or 3 for elementary steps; eliminate any MCQ option with non-integer molecularity immediately.

错误做法:

Leaving reaction intermediates in the final overall rate law derived from a mechanism.

原因:

Students forget intermediates are not stable species and must be substituted out using equilibrium expressions for fast pre-steps.

正确做法:

Always substitute out intermediate concentrations using expressions from fast equilibrium steps before writing the final rate law.

错误做法:

Counting product particles to determine the molecularity of an elementary reaction.

原因:

Students count all particles in the equation instead of only reactants.

正确做法:

Only count the number of reactant particles on the left-hand side of the elementary step to find molecularity.

错误做法:

Labeling a catalyst as a reaction intermediate in a multi-step mechanism.

原因:

Students mix up the order of production and consumption for the two species types.

正确做法:

Follow the rule: catalysts = consumed first, produced later; intermediates = produced first, consumed later.

7. 速查表

Category	Formula / Rule	Notes
Elementary Step Rate Law	$rate = k \prod [X_{i}]^{n_{i}}, n_{i} = stoichiometric coefficient$	Only valid for elementary reactions; does not apply to overall reactions
Unimolecular Elementary Step	$A \to products, rate = k [A]$	Molecularity = 1, overall order = 1
Bimolecular (A + B)	$A + B \to products, rate = k [A] [B]$	Molecularity = 2, overall order = 2
Bimolecular (2A)	$2 A \to products, rate = k [A]^{2}$	Molecularity = 2, overall order = 2
Termolecular Elementary Step	$a A + b B + c C \to products, rate = k [A]^{a} [B]^{b} [C]^{c}$	Molecularity = 3, rare in mechanisms, overall order = 3
Molecularity	Count of reactant particles in an elementary step	Always 1, 2, or 3; never zero, negative, or fractional
Reaction Intermediate	Produced in early step, consumed in later step	Never included in final overall rate law
Catalyst	Consumed in early step, produced in later step	Not consumed overall; can appear in the rate law

真题中的出现

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2023 · MCQ
识别基元步骤的反应分子数
2022 · FRQ
从反应机理推导总速率定律
2021 · MCQ
区分中间体和催化剂

下一步

Mastering elementary reactions is the foundational prerequisite for working with full reaction mechanisms, the next core topic in AP Chemistry Unit 5 Kinetics. Without understanding how to write rate laws for elementary steps and identify intermediates, you cannot derive the overall rate law for a multi-step mechanism, which is a common high-weight FRQ question on the AP exam. Beyond kinetics, understanding elementary steps helps you interpret collision theory and activation energy, because each elementary step has its own activation energy and Arrhenius behavior. This topic also builds the foundation for understanding biological and industrial catalysis, where catalysts work by providing a new sequence of elementary steps with lower activation energy.