Chemical Equilibrium Revision for AQA A-Level Chemistry | AQA A-Level 化学：化学平衡考点精讲

📚 Chemical Equilibrium Revision for AQA A-Level Chemistry | AQA A-Level 化学：化学平衡考点精讲

In AQA A-Level Chemistry, the topic of chemical equilibrium bridges the gap between reaction kinetics and thermodynamics. It explains how reversible reactions reach a state of balance, how we quantify the position of that balance using the equilibrium constant Kc, and how external changes shift the equilibrium position according to Le Chatelier’s principle. Mastering this topic is essential for tackling both structured questions and data-analysis problems that frequently appear in Paper 1 and Paper 2.

在 AQA A-Level 化学中，化学平衡主题将反应动力学与热力学联系起来。它解释了可逆反应如何达到平衡状态，我们如何用平衡常数 Kc 定量描述平衡位置，以及外界变化如何根据勒夏特列原理使平衡发生移动。掌握这一主题对攻克试卷一和试卷二中常见的结构化问题与数据分析题至关重要。

1. Dynamic Equilibrium | 动态平衡

A reversible reaction reaches dynamic equilibrium in a closed system when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant, but both reactions continue to occur at the molecular level.

在封闭体系中，当正反应速率与逆反应速率相等时，可逆反应达到动态平衡。此时，反应物和产物的浓度保持不变，但两种反应在分子水平上仍在继续进行。

Dynamic equilibrium can only be established in a closed system where no matter enters or leaves. If the system is open, products or reactants may escape, preventing the rates from ever becoming equal.

动态平衡只能在封闭体系中建立，即没有物质进出。若体系开放，产物或反应物可能逃逸，使得正逆反应速率永远无法相等。

For the general reversible reaction aA + bB ⇌ cC + dD, the double arrow indicates that the reaction proceeds in both directions. The equilibrium position describes the relative amounts of reactants and products at equilibrium; it can lie to the left (more reactants) or to the right (more products).

对于一般可逆反应 aA + bB ⇌ cC + dD，双箭头表示反应可双向进行。平衡位置描述了平衡时反应物与产物的相对含量；平衡位置可以偏左（反应物较多）或偏右（产物较多）。

2. The Equilibrium Constant Kc | 平衡常数 Kc

The equilibrium constant Kc is a mathematical expression that relates the concentrations of products and reactants at equilibrium. For the reaction aA + bB ⇌ cC + dD, the expression is:

平衡常数 Kc 是一个将平衡时产物和反应物浓度联系起来的数学表达式。对于反应 aA + bB ⇌ cC + dD，表达式为：

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

The square brackets denote concentrations in mol dm⁻³. Only substances that have a concentration that can vary are included: gases and aqueous solutions. Pure solids and pure liquids are omitted because their concentrations are essentially constant and are incorporated into the Kc value.

方括号表示浓度，单位为 mol dm⁻³。只有浓度可变的物质才被纳入表达式：气体和水溶液。纯固体和纯液体被省略，因为它们浓度基本恒定，已并入 Kc 值中。

Kc is temperature dependent. For a given reaction, changing the temperature alters the value of Kc. If a change in condition other than temperature causes the equilibrium position to shift, the value of Kc remains unchanged — the reaction simply adjusts concentrations to restore the same Kc.

Kc 与温度有关。对于给定的反应，改变温度会改变 Kc 值。若温度以外的条件变化引起平衡移动，Kc 值保持不变——反应只是通过调整浓度来重新满足同一个 Kc。

3. Homogeneous vs Heterogeneous Equilibria | 均相与非均相平衡

A homogeneous equilibrium is one in which all reactants and products are in the same physical state. Common examples are all gases (e.g., 2SO₂ + O₂ ⇌ 2SO₃) or all in aqueous solution (e.g., esterification of ethanoic acid and ethanol). In such systems, all species appear in the Kc expression.

均相平衡是指所有反应物和产物处于同一相态。常见的例子有全部是气体（如 2SO₂ + O₂ ⇌ 2SO₃）或全部在溶液中（如乙酸与乙醇的酯化反应）。在这类体系中，所有物种都出现在 Kc 表达式中。

A heterogeneous equilibrium involves substances in different physical states. For example, in the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g), the two solids are omitted, so Kc = [CO₂]. This dramatically simplifies the expression and means that the equilibrium position depends only on the concentration (or partial pressure) of the gaseous component.

非均相平衡涉及不同相态的物质。例如反应 CaCO₃(s) ⇌ CaO(s) + CO₂(g)，两种固体被省略，故 Kc = [CO₂]。这大大简化了表达式，也意味着平衡位置仅取决于气体组分的浓度（或分压）。

When calculating Kc for heterogeneous systems, students must recognise which species to exclude. Only gases and aqueous species are included; liquids are also omitted unless the reaction takes place in a non-aqueous solvent where the liquid is also a reactant or product in solution.

在计算非均相体系的 Kc 时，学生必须识别哪些物种应当排除。只有气体和溶液中的物种才纳入；液体也被省略，除非反应发生在非水溶剂中且该液体本身是溶液中的反应物或产物。

4. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s principle states that if a system at dynamic equilibrium is subjected to a change in concentration, pressure, or temperature, the position of equilibrium will shift to partially oppose the change. This principle allows us to predict qualitatively how a system responds to external disturbances.

勒夏特列原理指出，如果一个处于动态平衡的体系受到浓度、压力或温度的变化，平衡位置将发生移动，以部分抵消这种变化。该原理使我们能够定性预测体系如何响应外界扰动。

It is crucial to understand that the shift ‘opposes’ the change, but does not completely nullify it. For instance, increasing the concentration of a reactant shifts equilibrium to the right to use up some of the added reactant, but the final concentration of that reactant will still be higher than before the addition.

关键是要理解，移动是“抵消”这种变化，但并不能完全消除它。例如，增加一种反应物的浓度会使平衡右移以消耗部分加入的反应物，但该反应物的最终浓度仍会比加入前高。

Le Chatelier’s principle applies to all reversible reactions at equilibrium. However, it does not predict the rate at which the new equilibrium is established. For AQA exams, you must be able to apply the principle to unfamiliar reactions, including those in industrial contexts.

勒夏特列原理适用于所有处于平衡的可逆反应。但它不能预测建立新平衡的速率。在 AQA 考试中，你必须能够将该原理应用于陌生反应，包括工业背景下的反应。

5. Effect of Concentration Changes | 浓度变化的影响

If the concentration of a reactant is increased, the system shifts to the right (towards products) to reduce the concentration of that reactant. Conversely, increasing the concentration of a product shifts equilibrium to the left (towards reactants).

若增加一种反应物的浓度，体系会向右移动（趋向产物）以降低该反应物的浓度。反之，增加产物的浓度会使平衡向左移动（趋向反应物）。

Decreasing the concentration of a substance causes the equilibrium to shift in the direction that produces more of that substance. For example, removing a product continuously (e.g., by distillation or precipitation) drives the reaction to completion by constantly pulling the equilibrium to the right.

降低某物质的浓度会使平衡向生成更多该物质的方向移动。例如，不断移走产物（如通过蒸馏或沉淀）会通过持续拉动平衡右移，使反应趋于完全。

Adding a solid or pure liquid does not affect the equilibrium position because its concentration does not change. Similarly, adding a catalyst does not shift the equilibrium position, a point often tested in multiple-choice questions.

加入固体或纯液体不会影响平衡位置，因为其浓度不变。同样地，加入催化剂不会移动平衡位置，这是选择题中常考的点。

6. Effect of Pressure Changes | 压力变化的影响

Pressure changes only affect equilibria involving gases, and only when there is a change in the total number of gaseous moles between reactants and products. Increasing pressure shifts the equilibrium towards the side with fewer gas molecules, as this reduces the pressure.

压力变化只影响涉及气体的平衡，且仅当反应物与产物之间的气体总摩尔数发生变化时。增大压力会使平衡向气体分子数较少的一侧移动，因为这样可以降低压力。

For the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), there are 4 moles of gas on the left and 2 moles on the right. High pressure shifts equilibrium to the right, increasing the yield of ammonia. In contrast, for H₂(g) + I₂(g) ⇌ 2HI(g), both sides have 2 moles of gas, so pressure has no effect on the equilibrium position.

对哈伯法合成氨，N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，左边有 4 mol 气体，右边有 2 mol。高压使平衡右移，提高氨的产率。对比之下，H₂(g) + I₂(g) ⇌ 2HI(g) 两边气体摩尔数均为 2，所以压力对平衡位置无影响。

Decreasing pressure shifts equilibrium to the side with more gas moles. This is rarely used in industry because low pressures often lead to slower rates, but it is a valid prediction for analysis questions.

减小压力会使平衡向气体摩尔数较多的一侧移动。这在工业中很少使用，因为低压常导致速率降低，但在分析题中这是一个合理的预测。

7. Effect of Temperature Changes | 温度变化的影响

Temperature is the only external condition that changes the numerical value of Kc. If the forward reaction is exothermic (ΔH negative), increasing the temperature adds heat, so the equilibrium shifts in the endothermic (backward) direction to absorb the extra heat. This decreases Kc.

温度是唯一能改变 Kc 数值的外部条件。若正反应放热（ΔH 为负），升温相当于加入热量，平衡向吸热方向（逆反应）移动以吸收多余热量，导致 Kc 减小。

If the forward reaction is endothermic (ΔH positive), increasing temperature shifts equilibrium to the right, and Kc increases. Always associate Kc changes with temperature: an increase in Kc means the equilibrium position shifts right, and a decrease means it shifts left.

若正反应吸热（ΔH 为正），升温使平衡右移，Kc 增大。要始终将 Kc 的变化与温度联系起来：Kc 增大意味着平衡位置右移，Kc 减小意味着左移。

For the AQA specification, you must be able to deduce whether a reaction is exothermic or endothermic from the effect of temperature on Kc or equilibrium yield. For instance, in the Contact process 2SO₂ + O₂ ⇌ 2SO₃ (ΔH = –197 kJ mol⁻¹), a higher temperature reduces Kc and lowers the equilibrium yield of SO₃, but a compromise temperature is used to achieve a reasonable rate.

根据 AQA 大纲，你必须能从温度对 Kc 或平衡产率的影响判断反应是放热还是吸热。例如，接触法制硫酸 2SO₂ + O₂ ⇌ 2SO₃（ΔH = –197 kJ mol⁻¹），升高温度会减小 Kc，降低 SO₃ 的平衡产率，但工业上采用折中温度以保证适当速率。

8. Catalysts and Equilibrium | 催化剂与平衡

Adding a catalyst has no effect on the position of equilibrium or the value of Kc. A catalyst provides an alternative reaction pathway with a lower activation energy, which increases the rate of both the forward and backward reactions equally. This means equilibrium is reached faster, but the composition of the equilibrium mixture remains unchanged.

加入催化剂对平衡位置和 Kc 值均无影响。催化剂通过提供较低活化能的替代路径，同等程度地加快正、逆反应速率。这意味着平衡能更快达到，但平衡混合物的组成保持不变。

A common exam trap is to claim that a catalyst increases the yield of a product; it does not. While it is true that in time-limited industrial processes a catalyst can increase the amount of product obtained per unit time, it does not shift the equilibrium position.

常见考试陷阱是声称催化剂提高产物产率；它不提高。虽然在有时间限制的工业过程中，催化剂能提高单位时间所得产物量，但它不会移动平衡位置。

Catalysts are essential in processes like the Haber process (iron catalyst) and the Contact process (vanadium(V) oxide) to allow equilibrium to be reached rapidly at moderate temperatures, thus optimising both rate and yield in a compromise.

催化剂在哈伯法（铁催化剂）和接触法（五氧化二钒）等工艺中至关重要，可使体系在中等温度下迅速达到平衡，从而在速率与产率之间实现最优妥协。

9. Calculating Equilibrium Constants | 平衡常数计算

A typical Kc calculation requires you to determine equilibrium concentrations from initial amounts and the amount of one species at equilibrium. Constructing an ICE table (Initial, Change, Equilibrium) is the recommended approach. All concentrations must be in mol dm⁻³, so you must divide moles by the volume of the container.

典型的 Kc 计算要求你根据初始量和某物质在平衡时的量确定平衡浓度。推荐使用 ICE 表（初始—变化—平衡）的方法。所有浓度必须以 mol dm⁻³ 为单位，因此需将摩尔数除以容器体积。

For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), if initially 1.0 mol of H₂ and 1.0 mol of I₂ are placed in a 1 dm³ vessel, and at equilibrium 0.4 mol of H₂ remains, then the change for H₂ is –0.6 mol, so equilibrium amounts are: H₂ = 0.4, I₂ = 0.4, HI = 1.2 mol. Concentrations are the same because V = 1 dm³. Then Kc = (1.2)² / (0.4 × 0.4) = 9.0 (units cancel).

对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，若初始在 1 dm³ 容器中放入 1.0 mol H₂ 和 1.0 mol I₂，平衡时剩余 0.4 mol H₂，则 H₂ 的变化量为 –0.6 mol，平衡时各物质摩尔数：H₂ = 0.4，I₂ = 0.4，HI = 1.2 mol。体积为 1 dm³，浓度相同。则 Kc = (1.2)² / (0.4 × 0.4) = 9.0（单位消去）。

When volumes are not 1 dm³, always convert moles to concentrations. Also, Kc may have units depending on the stoichiometry. Derive units from the expression: e.g., for N₂ + 3H₂ ⇌ 2NH₃, Kc units = (mol dm⁻³)² / (mol dm⁻³)⁴ = mol⁻² dm⁶.

当体积不为 1 dm³ 时，务必把摩尔数换算为浓度。此外，Kc 可能有单位，取决于化学计量数。从表达式推导单位：例如对于 N₂ + 3H₂ ⇌ 2NH₃，Kc 单位 = (mol dm⁻³)² / (mol dm⁻³)⁴ = mol⁻² dm⁶。

Kc calculations can involve quadratic equations if the change is not directly given, but at AQA A-Level, data are usually chosen so that taking square roots or simple algebra suffices. Always check that your final Kc value is reasonable and matches the expected direction of the equilibrium.

如果没有直接给出变化量，Kc 计算可能会涉及二次方程，但在 AQA A-Level 中，数据的选取通常使得开平方或简单代数就能完成。务必检查最终 Kc 值是否合理并与预期的平衡方向一致。

10. Industrial Applications: Haber and Contact Processes | 工业应用：哈伯法与接触法

The Haber process synthesises ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = –92 kJ mol⁻¹. The forward reaction is exothermic and reduces the number of gas molecules. According to Le Chatelier, high pressure and low temperature favour the forward reaction, giving a higher equilibrium yield of NH₃.

哈伯法合成氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = –92 kJ mol⁻¹。正反应放热且气体分子数减少。根据勒夏特列原理，高压和低温有利正反应，获得更高的氨平衡产率。

In practice, a compromise temperature of about 400–450 °C is used: low enough to give a reasonable Kc, but high enough to achieve an acceptable rate. A pressure of around 200 atm is chosen to shift equilibrium right while keeping plant costs manageable. An iron catalyst speeds up the attainment of equilibrium. The ammonia is continuously liquefied and removed, pulling the equilibrium to the right.

实际采用约 400–450 °C 的折中温度：足够低以得到合理的 Kc，又足够高以实现可接受的速率。压力约 200 atm 以推动平衡右移，同时使设备成本可控。铁催化剂加速达到平衡。氨被不断液化移走，拉动平衡右移。

The Contact process produces sulfuric acid via the oxidation of SO₂: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = –197 kJ mol⁻¹. Again, low temperature and high pressure favour SO₃. Industrial conditions are 1–2 atm (only a small gain from higher pressure) and around 450 °C with a V₂O₅ catalyst.

接触法通过氧化 SO₂ 生产硫酸：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，ΔH = –197 kJ mol⁻¹。同样，低温和高压有利 SO₃ 生成。工业条件采用 1–2 atm（更高压力收益不大）和约 450 °C，并用 V₂O₅ 催化剂。

Both processes illustrate the balance between thermodynamic equilibrium and kinetic practicality. Exam questions often ask you to justify the chosen conditions using Le Chatelier’s principle and to explain why even lower temperatures or higher pressures are not used, referencing rate, cost, and safety.

这两种工艺都体现了热力学平衡与动力学可行性之间的权衡。考题常要求你引用勒夏特列原理解释所选条件的合理性，并说明为什么不采用更低温度或更高压力，需涉及速率、成本和安全。

11. Relationship Between Kc and Equilibrium Position | Kc 与平衡位置的关系

A large Kc value (Kc >> 1) indicates that at equilibrium the concentration of products is much greater than that of reactants: the equilibrium lies to the right. A very small Kc (Kc << 1) means the equilibrium lies to the left, with reactants dominating.

Kc 值很大（Kc >> 1）表明平衡时产物的浓度远大于反应物：平衡位置偏右。Kc 值很小（Kc << 1）意味着平衡偏左，反应物为主。

Comparing Kc values at different temperatures allows you to determine the sign of ΔH. If Kc increases with temperature, the forward reaction is endothermic. If Kc decreases with temperature, the forward reaction is exothermic. This is a direct application of Le Chatelier’s principle.

比较不同温度下的 Kc 值可判断 ΔH 的符号。若 Kc 随温度升高而增大，则正反应吸热；若 Kc 随温度升高而减小，则正反应放热。这是勒夏特列原理的直接应用。

Note that Kc only tells you about the position of equilibrium, not the rate at which equilibrium is achieved. A reaction might have a very large Kc but still be impractically slow without a catalyst.

注意 Kc 只能说明平衡位置，不能反映达到平衡的速率。一个反应可能具有很大的 Kc，但若无催化剂，仍可能慢得不切实际。

12. Common Misconceptions and Exam Tips | 常见误解与应试技巧

Many students mistakenly think that a catalyst increases the yield at equilibrium. Remember: a catalyst only shortens the time to reach equilibrium; it does not change Kc or the equilibrium position.

许多学生误以为催化剂提高平衡产率。请记住：催化剂只能缩短达到平衡的时间，不改变 Kc 或平衡位置。

Another common error is including solids or pure liquids in Kc expressions. Unless a solid is in solution or the liquid is a solvent involved in a homogeneous equilibrium, they are omitted. Always check the state symbols carefully.

另一个常见错误是把固体或纯液体纳入 Kc 表达式。除非固体在溶液中或液体是参与均相平衡的溶剂，否则一概省略。务必仔细审查状态符号。

When using Le Chatelier’s principle, avoid saying the system ‘wants’ or ‘tries’ to do something; use precise language: ‘the equilibrium shifts to the right/left to oppose the increase/decrease in…’ This is the level of scientific phrasing expected by AQA examiners.

使用勒夏特列原理时，不要说体系“想”或“试图”做什么；要用精确语言：“为抵消……的增加/减少，平衡向正/逆方向移动”。这是 AQA 考官期望的科学表述水平。

Finally, in Kc calculations, always state the units if they are not dimensionless. Even if they cancel, state ‘no units’ to show you have considered it. This simple habit can secure marks often lost through careless omission.

最后，在 Kc 计算中，若单位并非无量纲，一定要写出单位。即使单位消去，也写上“无单位”以表明你已考虑。这个简单习惯能帮你拿到经常因粗心遗漏而丢失的分数。

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Chemical Equilibrium Revision for AQA A-Level Chemistry | AQA A-Level 化学：化学平衡 考点精讲