ESAT Chemistry: Reaction Rates and Chemical Equilibrium | 反应速率与化学平衡

📚 ESAT Chemistry: Reaction Rates and Chemical Equilibrium | 反应速率与化学平衡

Reaction kinetics and chemical equilibrium are two cornerstones of physical chemistry, frequently tested in the ESAT chemistry section. This article offers a structured review of the key concepts, equations, and problem-solving strategies you need to master.

反应动力学与化学平衡是物理化学的两大基石，也是 ESAT 化学部分的高频考点。本文系统梳理核心概念、必备方程与解题策略，助你高效备考。

1. What Is Reaction Rate? | 什么是反应速率？

The rate of a chemical reaction measures how quickly the concentration of a reactant or product changes per unit time. It is typically expressed in mol dm⁻³ s⁻¹.

化学反应速率衡量的是反应物或产物浓度随时间变化的快慢，常用单位是 mol dm⁻³ s⁻¹。

For a reaction A → B, the average rate over a time interval Δt is calculated as:

Rate = –Δ[A]/Δt = +Δ[B]/Δt

For reactants, the concentration decreases, so a negative sign is introduced to keep the rate positive. On a concentration–time graph, the instantaneous rate at any point is given by the gradient of the tangent at that point.

对于反应 A → B，某段时间 Δt 内的平均速率可表示为：

速率 = –Δ[A]/Δt = +Δ[B]/Δt

由于反应物浓度随时间减少，因此加负号使速率保持正值。在浓度–时间图上，任意时刻的瞬时速率等于该点切线的斜率。

2. Collision Theory and Activation Energy | 碰撞理论与活化能

For a reaction to occur, reactant particles must collide with sufficient energy and with the correct orientation. This is the essence of collision theory.

反应发生的前提是反应物粒子发生碰撞，并且碰撞具有足够的能量和正确的取向，这就是碰撞理论的核心内容。

Successful collision: a collision that leads to product formation, requiring energy ≥ activation energy (Eₐ) and proper geometry.
有效碰撞：能够生成产物的碰撞，要求能量不低于活化能 Eₐ，且空间取向合适。
Activation energy: the minimum energy that colliding particles must possess to break existing bonds and initiate reaction.
活化能：碰撞粒子必须拥有的最低能量，用以断裂原有化学键并启动反应。

The Maxwell–Boltzmann distribution curve shows the spread of molecular energies at a given temperature. The area under the curve represents the total number of particles; the shaded region beyond Eₐ indicates the fraction of particles with sufficient energy to react.

麦克斯韦–玻尔兹曼分布曲线展示了某一温度下分子能量的分布情况。曲线下面积代表粒子总数，Eₐ 右侧的阴影区域则表示具有足够能量发生反应的粒子比例。

3. Factors Affecting Reaction Rates | 影响反应速率的因素

Several factors can change the rate of a reaction by affecting the frequency of successful collisions. These include concentration, pressure, temperature, surface area, and catalysts.

众多因素通过改变有效碰撞频率来影响反应速率，包括浓度、压强、温度、表面积和催化剂。

Factor 因素	Effect on Rate 对速率的影响
Concentration 浓度	Higher concentration increases collision frequency, hence rate increases. 浓度升高，碰撞频率增大，速率加快。
Pressure (gases) 压强（气体）	Higher pressure compresses gas, increasing concentration and collision frequency. 压强增大使气体体积压缩，浓度升高，碰撞更频繁。
Temperature 温度	Raises average kinetic energy and the fraction of particles exceeding Eₐ; rate increases sharply. 平均动能增大，超过 Eₐ 的粒子比例升高，速率显著加快。
Surface area 表面积	Larger surface area exposes more particles, increasing collision frequency. 表面积增大使更多粒子暴露，碰撞频率增加。
Catalyst 催化剂	Provides an alternative pathway with lower activation energy; catalyst is not consumed. 提供活化能更低的替代路径；催化剂本身不被消耗。

4. Rate Equations and Orders of Reaction | 速率方程与反应级数

A rate equation expresses the reaction rate as a function of reactant concentrations. For a general reaction aA + bB → products, the rate equation often takes the form:

速率方程将反应速率表示为反应物浓度的函数。对一般反应 aA + bB → 产物，速率方程常写为：

Rate = k[A]ᵐ[B]ⁿ

Here, k is the rate constant, m is the order with respect to A, and n is the order with respect to B. The overall order is m + n. Note that the orders m and n are not necessarily equal to the stoichiometric coefficients a and b — they must be determined experimentally.

其中 k 为速率常数，m 是 A 的反应级数，n 是 B 的反应级数，总反应级数为 m + n。注意 m 和 n 不一定等于化学计量数 a 和 b，必须通过实验测定。

Zero order: rate is independent of concentration; [A]⁰ = 1.
零级反应：速率与浓度无关；[A]⁰ = 1。
First order: rate ∝ [A]; half-life is constant.
一级反应：速率正比于 [A]；半衰期为常数。
Second order: rate ∝ [A]².
二级反应：速率正比于 [A]²。

For a first-order reaction, the integrated rate law is:

ln[A]ₜ = –kt + ln[A]₀

A plot of ln[A] against time gives a straight line with gradient –k. The half-life t₁/₂ = 0.693/k.

对一级反应，积分速率定律为：

ln[A]ₜ = –kt + ln[A]₀

以 ln[A] 对时间作图可得一条直线，斜率为 –k。半衰期 t₁/₂ = 0.693/k。

5. Determining Order from Experimental Data | 由实验数据确定反应级数

ESAT questions often provide a table of initial rates at different concentrations. To determine the order with respect to a reactant, compare two experiments where only that reactant’s concentration changes.

ESAT 常给出不同浓度下的初始速率数据表。要确定某一反应物的级数，只需比较仅改变该反应物浓度的两组实验。

Example: For the reaction X + Y → products, experiments show:

示例：对于反应 X + Y → 产物，实验数据如下：

Experiment 实验	[X] / mol dm⁻³	[Y] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.1	0.1	2 × 10⁻³
2	0.2	0.1	8 × 10⁻³
3	0.1	0.2	2 × 10⁻³

Doubling [X] (experiments 1→2) quadruples the rate, so the order with respect to X is 2. Doubling [Y] (experiments 1→3) leaves the rate unchanged, so the order with respect to Y is 0. The rate equation is therefore:

Rate = k[X]²

[X] 浓度加倍（实验 1→2），速率变为原来的 4 倍，因此 X 的级数为 2；[Y] 浓度加倍（实验 1→3），速率不变，因此 Y 的级数为 0。所以速率方程为：

速率 = k[X]²

6. The Rate Constant k and the Arrhenius Equation | 速率常数 k 与阿伦尼乌斯方程

The rate constant k is temperature-dependent. It increases with temperature because a larger fraction of molecules possesses energy greater than Eₐ. The Arrhenius equation describes this relationship:

速率常数 k 依赖于温度。温度升高时，超过 Eₐ 的分子比例增大，因此 k 增大。阿伦尼乌斯方程描述了这种关系：

k = A e^(–Eₐ/RT)

In logarithmic form:

ln k = ln A – Eₐ/(RT)

A plot of ln k against 1/T gives a straight line with gradient –Eₐ/R, from which the activation energy can be determined. A catalyst lowers Eₐ, which significantly increases k at a given temperature.

其对数形式为：

ln k = ln A – Eₐ/(RT)

以 ln k 对 1/T 作图，得到斜率为 –Eₐ/R 的直线，由此可求出活化能。催化剂降低了 Eₐ，使给定温度下的 k 显著增大。

7. Dynamic Equilibrium | 动态平衡

A reaction reaches dynamic equilibrium when the forward and reverse rates are equal, and the concentrations of reactants and products remain constant over time. Equilibrium is only possible in a closed system.

当正反应速率与逆反应速率相等时，体系达到动态平衡，各物质浓度不再随时间变化。只有在封闭体系中才可能建立平衡。

Dynamic: both forward and reverse reactions continue to occur at equal rates.
动态：正逆反应仍在进行，但速率相等。
Constant concentrations: although particles keep reacting, macroscopic concentrations do not change.
浓度恒定：尽管微观粒子持续反应，宏观浓度不再改变。
Closed system: no matter can enter or leave the container.
封闭体系：物质不能进出容器。

It is important to remember that equilibrium is not a static state — it is a balance between two opposing processes occurring simultaneously.

需要注意的是，平衡并非静止状态，而是两个相反过程同时进行时的相互平衡。

8. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s principle states that if a dynamic equilibrium is disturbed by changing conditions, the position of equilibrium will shift to counteract the change. This principle allows us to predict how a system responds to changes in concentration, pressure, and temperature.

勒夏特列原理指出：当动态平衡受到条件改变的干扰时，平衡位置将朝着减弱该改变的方向移动。该原理可用于预测体系对浓度、压强和温度变化的响应。

Concentration change: Increasing a reactant’s concentration shifts equilibrium to the right (more products); decreasing a product’s concentration also shifts right.
浓度变化：增加反应物浓度，平衡向右移动（生成更多产物）；降低产物浓度，平衡同样右移。
Pressure change (gases): Increasing pressure shifts equilibrium toward the side with fewer moles of gas; decreasing pressure shifts toward more gas moles.
压强变化（气体）：增大压强，平衡向气体物质的量较少的方向移动；减小压强则向气体物质的量较多的方向移动。
Temperature change: For an exothermic forward reaction, raising temperature shifts equilibrium left (endothermic direction); lowering temperature shifts right.
温度变化：若正向反应放热，升温使平衡向左（吸热方向）移动；降温则向右移动。

A catalyst does not shift the position of equilibrium; it only speeds up the rate at which equilibrium is reached.

催化剂不改变平衡位置，只加快到达平衡的速率。

9. The Equilibrium Constant Kc | 平衡常数 Kc

For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc at a given temperature is defined as:

对于一般反应 aA + bB ⇌ cC + dD，在给定温度下的平衡常数 Kc 定义为：

Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

Note that Kc has units that depend on the stoichiometry. Pure solids and pure liquids do not appear in the expression for Kc. The value of Kc is constant at a fixed temperature and is unaffected by changes in concentration or pressure.

注意 Kc 的单位取决于化学计量数；纯固体和纯液体不写入 Kc 表达式。Kc 只与温度有关，不受浓度或压强变化的影响。

Interpreting Kc:

Kc 的数值含义：

If Kc >> 1, the equilibrium lies far to the right — products are favored.
若 Kc >> 1，平衡强烈偏向右边——产物占优势。
If Kc << 1, the equilibrium lies far to the left — reactants are favored.
若 Kc << 1，平衡强烈偏向左边——反应物占优势。
If Kc ≈ 1, both reactants and products are present in significant amounts.
若 Kc ≈ 1，反应物和产物均有可观含量。

10. Calculating Kc from Equilibrium Concentrations | 由平衡浓度计算 Kc

ESAT questions often require calculating Kc from initial concentrations and a known equilibrium concentration of one species. A stoichiometric table is an effective tool.

ESAT 题目常要求根据初始浓度和某一物质的平衡浓度计算 Kc。使用化学计量数列表是高效的方法。

Example: For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), initially 0.50 mol of H₂ and 0.50 mol of I₂ are placed in a 1.0 dm³ vessel. At equilibrium, 0.80 mol of HI is present.

示例：对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，初始时在 1.0 dm³ 容器中加入 0.50 mol H₂ 和 0.50 mol I₂。平衡时生成 0.80 mol HI。

Species 物质	Initial / mol 初始	Change / mol 变化	Equilibrium / mol dm⁻³ 平衡浓度
H₂	0.50	–0.40	0.10
I₂	0.50	–0.40	0.10
HI	0	+0.80	0.80

Since producing HI requires consuming H₂ and I₂ in a 1:1:2 ratio, the change in H₂ is –0.40 mol. Using a 1 dm³ volume, concentrations equal moles. Thus:

Kc = (0.80)² / (0.10 × 0.10) = 64

由于生成 HI 时 H₂ 与 I₂ 按 1:1:2 的化学计量比消耗，H₂ 变化量为 –0.40 mol。体积为 1 dm³，因此平衡浓度在数值上等于物质的量。于是：

Kc = (0.80)² / (0.10 × 0.10) = 64

11. Kp for Gaseous Equilibria | 气体平衡的 Kp

For reactions involving gases, the equilibrium constant Kp is expressed in terms of partial pressures. Partial pressure is the pressure that the gas would exert if it alone occupied the container.

对于涉及气体的反应，平衡常数 Kp 以分压表示。分压是指该气体单独占据整个容器时所产生的压强。

Kp = (p_C)ᶜ (p_D)ᵈ / ((p_A)ᵃ (p_B)ᵇ)

The partial pressure of each gas is calculated as:

p_A = (mole fraction of A) × total pressure

每种气体的分压按以下公式计算：

p_A =（A 的摩尔分数）× 总压

The mole fraction of A is n_A / n_total. Remember to use only equilibrium amounts when calculating mole fractions.

A 的摩尔分数为 n_A / n_total。计算摩尔分数时必须使用平衡时的物质的量。

12. Temperature and Equilibrium: van ‘t Hoff Insight | 温度与平衡：范特霍夫视角

At constant pressure, the equilibrium constant changes with temperature according to the van ‘t Hoff equation:

ln K₂/K₁ = –ΔH°/R × (1/T₂ – 1/T₁)

在恒压条件下，平衡常数随温度的变化由范特霍夫方程描述：

ln K₂/K₁ = –ΔH°/R × (1/T₂ – 1/T₁)

Here, ΔH° is the standard enthalpy change of the forward reaction. If ΔH° is negative (exothermic), increasing temperature reduces K — consistent with Le Chatelier’s principle. If ΔH° is positive (endothermic), increasing temperature raises K.

其中 ΔH° 为正向反应的标准焓变。若 ΔH° 为负（放热反应），升高温度使 K 减小，与勒夏特列原理一致；若 ΔH° 为正（吸热反应），升高温度使 K 增大。

On a typical exam question, you may be asked to predict whether K increases or decreases with temperature, or to calculate K at a new temperature given ΔH° and two temperatures. Both skills are important for ESAT.

考试中常要求判断 K 随温度升高而增大还是减小，或给出 ΔH° 和两个温度计算新温度下的 K。这两种能力对 ESAT 均很重要。

13. Catalysts and Equilibrium: A Crucial Distinction | 催化剂与平衡：至关重要的区分

A catalyst increases the rate of both the forward and reverse reactions equally by providing an alternative route with lower activation energy. Therefore, it does not change the value of Kc or Kp, nor does it shift the position of equilibrium.

催化剂通过提供活化能更低的替代路径，同等程度地加快正逆反应速率。因此，它不改变 Kc 或 Kp 的数值，也不改变平衡位置。

However, a catalyst does help the system reach equilibrium faster. In industrial processes such as the Haber process and the Contact process, catalysts are essential for making reactions economically viable at moderate temperatures.

然而，催化剂确实能帮助体系更快地到达平衡。在哈伯法、接触法之类的工业流程中，催化剂对于在温和温度下实现经济可行的反应至关重要。

14. Common Pitfalls in ESAT Questions | ESAT 常见失分点

Even strong students lose marks on subtle but predictable mistakes. Here are the most frequent pitfalls:

即使是优秀的考生也可能在隐蔽而可预测的错误上失分。以下是最常见的失分点：

Using initial concentrations instead of equilibrium concentrations in Kc expressions.
在 Kc 表达式中误用初始浓度而非平衡浓度。
Forgetting that pure solids and liquids are omitted from Kc and Kp expressions.
忘记纯固体和纯液体不写入 Kc、Kp 表达式。
Confusing the order of a reaction with stoichiometric coefficients.
将反应级数与化学计量数混淆。
Applying Le Chatelier’s principle incorrectly to temperature: some students think increased temperature always favors the endothermic direction, which is correct, but they also forget that K changes accordingly — both are true.
对温度变化误用勒夏特列原理：升温总是有利于吸热方向，这是正确的，但同时 K 也随之改变——两者密不可分。
Using the total pressure instead of partial pressure in Kp calculations.
在 Kp 计算中误用总压代替分压。
Forgetting that a catalyst does not alter the equilibrium position.
忘记催化剂不改变平衡位置。

Always double-check the units of Kc and Kp — many questions award a mark for correct units.

务必检查 Kc 和 Kp 的单位——很多题目会单独给单位分。

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ESAT Chemistry: Reaction Rates and Chemical Equilibrium | 反应速率与化学平衡