📚 A-Level Chemistry Equilibrium: Core Concepts and Exam Skills | A-Level化学平衡核心考点突破
Chemical equilibrium is one of the highest-yielding topics in any A-Level Chemistry specification, and it is also one of the most reliably examined. Whether you sit CAIE, Edexcel, AQA or OCR, questions on the equilibrium constant, Le Chatelier's principle, acid-base equilibria and solubility products appear year after year, often carrying a large share of the marks in Paper 2 and Paper 4. The difficulty is rarely the mathematics; it is the precision of the language and the accuracy of the expressions you write. This guide walks through every core idea you need, from the meaning of a dynamic equilibrium to the calculation habits that separate an A grade from an A*.
化学平衡是所有 A-Level 化学考纲中分值最高、出现频率最稳定的主题之一，也是命题最规律的主题之一。无论你考的是 CAIE、Edexcel、AQA 还是 OCR，平衡常数、勒夏特列原理、酸碱平衡与溶度积的题目几乎年年出现，并且在 Paper 2 与 Paper 4 中往往占据相当大的分值。真正的难点通常不在数学运算，而在于表述的精确性和表达式书写的准确性。本文将系统梳理全部核心考点，从动态平衡的含义，一直讲到能把 A 与 A* 区分开的计算习惯与应试细节。
1. What a Dynamic Equilibrium Really Means | 动态平衡的真正含义
A reversible reaction in a closed system eventually reaches a state in which the rate of the forward reaction equals the rate of the backward reaction. At that point the concentrations of all species remain constant, even though both reactions continue to occur. This is why we call it dynamic: nothing has stopped, the two opposing processes simply cancel each other out in terms of observable change.
在封闭体系中，可逆反应最终会达到这样一个状态：正反应速率等于逆反应速率。此时所有物质的浓度保持不变，但两个方向的反应其实都在持续进行。这正是"动态"的含义：反应并没有停止，只是两个相反方向的过程在宏观可观察的变化上互相抵消。
The essential features of a dynamic equilibrium are frequently tested as a "state which" or "explain why" question, so learn them as a checklist rather than as a vague idea.
动态平衡的基本特征经常以"说明原因"或"判断正误"的形式考查，因此应当把它当作一份清单来记忆，而不是模糊的印象。
- It can only be established in a closed system, so that no substance escapes. / 只有在封闭体系中才能建立，确保没有任何物质逸出。
- The forward and reverse rates are equal and non-zero. / 正、逆反应速率相等且均不为零。
- The concentrations of reactants and products stay constant, but they are not necessarily equal. / 反应物与生成物的浓度保持不变，但两者浓度不一定相等。
- It is a dynamic state: both reactions continue at the molecular level. / 它是动态的：在分子层面上两个反应都在继续。
- The same equilibrium position is reached whether you start from reactants or from products, provided the temperature is the same. / 只要温度相同，无论从反应物还是从生成物出发，最终都到达同一个平衡位置。
- A catalyst changes the rate at which equilibrium is reached, not the position of equilibrium. / 催化剂只改变达到平衡的快慢，不改变平衡位置。
When you write an equation for a reversible reaction, always use the reversible arrow ⇌ rs deduct marks for this surprisingly often, and it is one of the easiest marks to secure.
书写可逆反应方程式时，务必使用可逆箭头 ⇌，而不是单向箭头 →。阅卷中因这一细节失分的情况出奇地多，而它恰恰是最容易拿到的分数之一。
2. The Equilibrium Constant: Kc and Kp | 平衡常数：Kc 与 Kp
For a general homogeneous equilibrium in solution or in the gas phase, the equilibrium constant expresses the ratio of product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient. For the reaction aA + bB ⇌ cC + dD, the expression is written as follows.
对于溶液或气相中的均相平衡，平衡常数表示生成物浓度与反应物浓度之比，各浓度项的指数等于其化学计量数。对于反应 aA + bB ⇌ cC + dD，表达式如下。
Kc = [C]ᶜ [D]ᵈ ÷ ( [A]ᵃ [B]ᵇ )
If the system is gaseous, you may be asked for Kp instead, which uses partial pressures in place of concentrations. The partial pressure of a gas is its mole fraction multiplied by the total pressure.
如果体系为气态，题目可能要求使用 Kp，即以分压代替浓度。某气体的分压等于其摩尔分数乘以总压。
p(A) = mole fraction of A × total pressure
Kp = p(C)ᶜ p(D)ᵈ ÷ ( p(A)ᵃ p(B)ᵇ )
Three properties of K are examined repeatedly and should be memorised precisely.
K 有三个性质被反复考查，必须准确记忆。
- K depends only on temperature. Changing concentration, pressure or adding a catalyst does not change K. / K 只随温度变化；改变浓度、压强或加入催化剂都不会改变 K。
- A large K (much greater than 1) means the equilibrium lies to the right, favouring products. / K 很大（远大于 1）表示平衡右移，有利于生成物。
- A small K (much less than 1) means the equilibrium lies to the left, favouring reactants. / K 很小（远小于 1）表示平衡左移，有利于反应物。
Units deserve care. Work them out by substituting the powers of concentration into the expression and cancelling, rather than trying to remember a list.
单位问题必须格外小心。正确做法是把各浓度项的幂次代入表达式后逐步约分，而不是死记一堆结果。
| Equilibrium | 平衡 | Units of Kc | Kc 的单位 |
| H₂(g) + I₂(g) ⇌ 2HI(g) | no units | 无单位 |
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | mol⁻² dm⁶ |
| PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) | mol dm⁻³ |
| CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) | mol dm⁻³ |
3. Writing Equilibrium Expressions Correctly | 正确书写平衡常数表达式
Most marks lost on equilibrium expressions come from including the wrong species rather than from getting the arithmetic wrong. Two rules cover almost every case you will meet.
平衡常数表达式的失分，绝大多数来自写入了不该出现的物质，而不是算错数。掌握下面两条规则，几乎可以覆盖你遇到的所有情形。
First, solids and pure liquids are never included, because their concentration is constant by definition. In a heterogeneous equilibrium such as the decomposition of calcium carbonate, only the gaseous species appear.
第一，固体与纯液体永远不写入表达式，因为它们的浓度按定义为常数。对于碳酸钙分解这类多相平衡，只有气态物质出现在表达式中。
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Kc = [CO₂]
Second, when water is the solvent rather than a reactant or product present in significant amount, its concentration is also treated as constant and omitted. This is why the acid dissociation constant of ethanoic acid is written without a term for water.
第二，当水是溶剂而非大量参与反应的物质时，其浓度同样视为常数并省略。这正是乙酸的电离常数表达式中不出现水的原因。
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)
Ka = [CH₃COO⁻][H⁺] ÷ [CH₃COOH]
A practical checklist when writing any expression: check state symbols, cross out every solid and pure liquid, make sure the products are on top, and confirm that each power matches the coefficient in the balanced equation.
书写表达式时的实用检查清单：核对状态符号，划掉所有固体与纯液体，确认生成物写在分子上，并逐一核对每个指数是否与配平方程式中的计量数一致。
4. Le Chatelier's Principle: Concentration and Pressure | 勒夏特列原理：浓度与压强
Le Chatelier's principle states that when a system at equilibrium is subjected to a change, the position of equilibrium shifts in the direction that tends to oppose the change. This principle predicts the direction of shift, but it does not tell you how large the shift is, and it does not change the value of K unless the temperature changes.
勒夏特列原理指出：当处于平衡的体系受到外界改变时，平衡位置会向能够削弱这一改变的方向移动。该原理只能预测移动方向，不能说明移动幅度；除温度外，任何改变都不会改变 K 的数值。
For concentration changes, the reasoning is direct. Adding more of a reactant pushes the equilibrium to the right; removing a product pulls it to the right; adding a product pushes it to the left.
对于浓度变化，推理非常直接。增加反应物使平衡右移；移走生成物使平衡右移；增加生成物使平衡左移。
For pressure changes, only gaseous equilibria are affected, and only when the number of moles of gas differs between the two sides.
对于压强变化，只有气态平衡会受影响，而且仅在两侧气体总物质的量不同时才有影响。
- Increase in pressure shifts the equilibrium towards the side with fewer moles of gas. / 增大压强使平衡向气体物质的量较少的一侧移动。
- Decrease in pressure shifts the equilibrium towards the side with more moles of gas. / 减小压强使平衡向气体物质的量较多的一侧移动。
- If the number of moles of gas is the same on both sides, pressure has no effect on the position of equilibrium. / 若两侧气体物质的量相同，压强对平衡位置没有影响。
- Adding an inert gas at constant volume has no effect, because partial pressures of the reacting gases are unchanged. / 在恒容条件下加入惰性气体没有影响，因为反应气体的分压未变。
- Adding an inert gas at constant pressure increases the volume and lowers partial pressures, so the equilibrium shifts towards the side with more moles of gas. / 在恒压条件下加入惰性气体使体积增大、分压降低，平衡向气体物质的量较多的一侧移动。
When explaining a shift, always link the direction to the change itself. For example: "The equilibrium shifts to the right because the forward reaction reduces the number of gas moles, partially opposing the increase in pressure."
解释平衡移动时，务必把方向与所施加的改变联系起来。例如："平衡右移，因为正反应使气体物质的量减少，从而部分抵消压强的增大。"
5. Temperature, Catalysts and the Value of K | 温度、催化剂与 K 值
Temperature is the only factor that changes the value of the equilibrium constant, and it is also the factor students most often confuse with pressure. The rule is straightforward once you anchor it to enthalpy.
温度是唯一能改变平衡常数数值的因素，也是学生最容易与压强混淆的因素。只要抓住与焓变的关系，规则就非常简单。
- If the forward reaction is exothermic (ΔH negative), raising the temperature decreases K and shifts the equilibrium to the left. / 若正反应放热（ΔH 为负），升温使 K 减小，平衡左移。
- If the forward reaction is endothermic (ΔH positive), raising the temperature increases K and shifts the equilibrium to the right. / 若正反应吸热（ΔH 为正），升温使 K 增大，平衡右移。
- Lowering the temperature has exactly the opposite effect in both cases. / 降温在两种情形下的效果正好相反。
A catalyst provides an alternative route with a lower activation energy, so both the forward and the reverse reactions are accelerated by the same factor. The equilibrium position is unchanged, K is unchanged, and only the time taken to reach equilibrium is shortened.
催化剂提供活化能更低的反应路径，使正、逆反应速率以相同倍数加快。因此平衡位置不变，K 不变，改变的只是达到平衡所需的时间。
The link between thermodynamics and equilibrium is sometimes examined at the top end of the specification. The standard Gibbs free energy change and K are related by the following equation, where R is the gas constant in J mol⁻¹ K⁻¹ and T is the thermodynamic temperature in K.
热力学与平衡之间的联系有时会在考纲的高阶部分考查。标准吉布斯自由能变与 K 的关系如下式所示，其中 R 为气体常数，单位为 J mol⁻¹ K⁻¹，T 为热力学温度，单位为 K。
ΔG° = −RT ln K
ΔG° is positive, K is less than 1 and the reactants are favoured. At ΔG° = 0, K equals 1.
当 ΔG° 为负时，ln K 为正，K 大于 1，有利于生成物；当 ΔG° 为正时，K 小于 1，有利于反应物；当 ΔG° = 0 时，K 等于 1。
6. ICE Tables and Structured Calculation | ICE 表格与结构化计算
Almost every equilibrium calculation can be organised with an ICE table: Initial amounts, Change amounts, Equilibrium amounts. Writing the table first prevents the two most common errors, which are forgetting to convert moles to concentration and mixing up initial and equilibrium values.
几乎所有平衡计算都可以用 ICE 表格整理：初始量（Initial）、变化量（Change）、平衡量（Equilibrium）。先列表格可以避免两个最常见的错误：忘记把物质的量换算成浓度，以及把初始值与平衡值混用。
Consider a worked example. In a 1.00 dm³ closed vessel, 1.00 mol of H₂ and 1.00 mol of I₂ react at a fixed temperature. At equilibrium the concentration of HI is 1.56 mol dm⁻³. Find Kc.
举一个实例。在 1.00 dm³ 的密闭容器中，1.00 mol H₂ 与 1.00 mol I₂ 在恒温下反应，平衡时 HI 的浓度为 1.56 mol dm⁻³，求 Kc。
H₂(g) + I₂(g) ⇌ 2HI(g)
| Species | 物质 | H₂ | I₂ | HI |
| Initial / mol dm⁻³ | 1.00 | 1.00 | 0.00 |
| Change / mol dm⁻³ | −x | −x | +2x |
| Equilibrium / mol dm⁻³ | 1.00 − x | 1.00 − x | 2x |
m⁻³ and [HI] = 1.56 mol dm⁻³. Substituting into the expression gives the following result.
由 2x = 1.56 得 x = 0.78，因此平衡浓度为 [H₂] = [I₂] = 0.22 mol dm⁻³，[HI] = 1.56 mol dm⁻³。代入表达式得如下结果。
Kc = (1.56)² ÷ (0.22 × 0.22) = 2.43 ÷ 0.0484 = 50.3
Notice that Kc has no units here, because the powers of concentration on the top and bottom cancel. If the volume had been different, the same numerical answer would still appear, because the volume term cancels for reactions with equal numbers of gas moles on both sides.
注意此处 Kc 没有单位，因为分子与分母中浓度的幂次相互抵消。若容器体积不同，结果依然相同，因为对于两侧气体物质的量相等的反应，体积项会整体约去。
7. Weak Acids, Ka and pH Calculations | 弱酸、Ka 与 pH 计算
A weak acid only partially dissociates in water, so an equilibrium is established between the undissociated acid and its ions. The acid dissociation constant Ka quantifies the strength of that acid, and pKa is simply the negative logarithm of Ka.
弱酸在水中只部分电离，因此未电离的酸分子与其离子之间会建立平衡。酸电离常数 Ka 定量描述酸的强度，而 pKa 就是 Ka 的负对数。
Ka = [H⁺][A⁻] ÷ [HA] pKa = −log₁₀ Ka
s only slightly dissociated, the approximation [H⁺] ≈ [A⁻] gives a very convenient result.
对于浓度为 c 且电离程度很小的单元弱酸，利用近似 [H⁺] ≈ [A⁻] 可得到非常便捷的结论。
[H⁺] = √Ka × c
pH = −log₁₀[H⁺]
× 10⁻³ mol dm⁻³ and pH = 2.88. The degree of dissociation is [H⁺] ÷ c, which here is 1.32 %.
以 0.100 mol dm⁻³ 的乙酸为例，Ka = 1.75 × 10⁻⁵ mol dm⁻³。则 Ka × c = 1.75 × 10⁻⁶，故 [H⁺] = 1.32 × 10⁻³ mol dm⁻³，pH = 2.88。电离度为 [H⁺] ÷ c，此处为 1.32 %。
Two conditions must be satisfied before you use the square-root approximation: the acid must be weak, and its concentration must not be extremely low. If the calculated degree of dissociation exceeds about 5 %, you should solve the full quadratic expression instead.
使用平方根近似前必须满足两个条件：酸必须是弱酸，且浓度不能极低。若算出的电离度超过约 5 %，则应改用完整的一元二次方程求解。
Ka = x² ÷ (c − x)
o a tenfold change in [H⁺] changes the pH by exactly one unit, and pH + pOH = 14 at 298 K.
同时要记住 pH 是对数标度，[H⁺] 变化十倍，pH 恰好变化一个单位；在 298 K 下 pH + pOH = 14。
8. Buffer Solutions and Salt Hydrolysis | 缓冲溶液与盐类水解
A buffer solution resists changes in pH when small amounts of acid or base are added, or when it is diluted. An acidic buffer is usually made from a weak acid and its conjugate base, such as ethanoic acid and sodium ethanoate, while a basic buffer uses a weak base and its conjugate acid, such as ammonia and ammonium chloride.
缓冲溶液在加入少量酸或碱、或被稀释时能够抵抗 pH 的变化。酸性缓冲液通常由弱酸与其共轭碱组成，例如乙酸与乙酸钠；碱性缓冲液则由弱碱与其共轭酸组成，例如氨与氯化铵。
Because the acid is only slightly dissociated and the salt is fully dissociated, the equilibrium concentrations of HA and A⁻ are approximately the same as the amounts you mixed. This leads to the buffer equation.
由于弱酸电离程度很小而盐完全电离，HA 与 A⁻ 的平衡浓度近似等于配制时的量。由此得到缓冲方程。
pH = pKa + log₁₀( [A⁻] ÷ [HA] )
The buffering action can be explained in two sentences that examiners look for. Added H⁺ ions are removed by the conjugate base A⁻, and added OH⁻ ions are removed by the undissociated acid HA, so the ratio [A⁻] ÷ [HA] changes very little.
缓冲作用可以用阅卷人期待的两句话解释：加入的 H⁺ 被共轭碱 A⁻ 消耗，加入的 OH⁻ 被未电离的酸 HA 消耗，因此 [A⁻] ÷ [HA] 的比值变化极小。
Salts of weak acids and strong bases, such as sodium ethanoate, give alkaline solutions because the ethanoate ion hydrolyses water and releases OH⁻. Salts of weak bases and strong acids, such as ammonium chloride, give acidic solutions for the mirror-image reason.
弱酸强碱盐（如乙酸钠）的水溶液呈碱性，因为乙酸根离子水解产生 OH⁻；弱碱强酸盐（如氯化铵）的水溶液呈酸性，原因正好相反。
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
9. Solubility Product Ksp and Precipitation | 溶度积 Ksp 与沉淀
For a sparingly soluble salt in contact with its saturated solution, an equilibrium exists between the solid and its ions. The solubility product is the equilibrium constant for this process, and like all equilibrium constants it depends only on temperature.
对于难溶盐与其饱和溶液接触的体系，固体与离子之间存在平衡。溶度积就是该过程的平衡常数，与所有平衡常数一样，它只随温度变化。
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰ mol² dm⁻⁶ at 298 K
The solid silver chloride does not appear in the expression, for exactly the same reason that solids never appear in any equilibrium constant. The units here are mol² dm⁻⁶ because two concentration terms are multiplied together.
固体氯化银不出现在表达式中，理由与前述"固体不写入平衡常数"完全相同。此处单位为 mol² dm⁻⁶，因为两个浓度项相乘。
be s mol dm⁻³. Then both ion concentrations equal s, so Ksp = s². For silver chloride, s = √1.8 × 10⁻¹⁰ = 1.34 × 10⁻⁵ mol dm⁻³.
求 1:1 型难溶盐的溶解度时，设溶解度为 s mol dm⁻³，则两种离子浓度均为 s，故 Ksp = s²。对氯化银而言，s = √(1.8 × 10⁻¹
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