📚 Chemical Equilibria and Acid-Base Equilibria | 化学平衡与酸碱平衡
In Edexcel A-Level Chemistry, chemical equilibria and acid-base equilibria form the quantitative heart of physical chemistry. They connect reversible reactions, the equilibrium constants Kc and Kp, the behaviour of weak and strong acids in water, buffer action, titration curves and solubility equilibria. Mastering these ideas means you can predict the direction of change, calculate pH, and explain experimental observations with confidence.
在爱德思 A-Level 化学中，化学平衡与酸碱平衡构成了物理化学的定量核心。它们把可逆反应、平衡常数 Kc 与 Kp、水溶液中弱酸和强酸的行为、缓冲作用、滴定曲线以及溶解平衡联系起来。掌握这些内容意味着你能够预测变化方向、计算 pH，并自信地解释实验现象。
1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡
Many reactions are reversible: the forward reaction and the backward reaction occur at the same time. When the rate of the forward reaction equals the rate of the backward reaction, the system reaches dynamic equilibrium. At equilibrium, the concentrations of reactants and products remain constant, but both reactions continue at the molecular level.
许多反应是可逆的：正反应和逆反应同时发生。当正反应速率等于逆反应速率时，体系达到动态平衡。达到平衡时，反应物和生成物的浓度保持不变，但在分子水平上两个反应仍在继续。
For the general equilibrium aA + bB ⇌ cC + dD, the system must be closed and the temperature must be constant. A dynamic equilibrium can only be reached in a closed system so that no matter is lost. The equilibrium position describes the relative amounts of reactants and products, while the equilibrium constant describes the extent of the reaction at a given temperature.
对于一般的平衡 aA + bB ⇌ cC + dD，体系必须是封闭的且温度恒定。动态平衡只能在封闭体系中达到，这样才能保证没有物质流失。平衡位置描述反应物和生成物的相对量，而平衡常数描述在给定温度下反应进行的程度。
- The forward and backward reactions continue at equal rates. | 正反应和逆反应以相等的速率持续进行。
- The concentrations of reactants and products stay constant. | 反应物和生成物的浓度保持不变。
- The equilibrium can be approached from either direction. | 平衡可以从任一方向达到。
- A catalyst speeds up the attainment of equilibrium but does not change the equilibrium position. | 催化剂加快达到平衡，但不改变平衡位置。
2. The Equilibrium Constant Kc | 平衡常数 Kc
The equilibrium constant Kc is expressed in terms of equilibrium concentrations. For the reaction aA + bB ⇌ cC + dD, the expression is shown below. The square brackets mean concentration in mol dm⁻³.
平衡常数 Kc 用平衡浓度表示。对于反应 aA + bB ⇌ cC + dD，表达式如下。方括号表示浓度，单位为 mol dm⁻³。
Kc = [C]ᶜ[D]ᵈ ÷ [A]ᵃ[B]ᵇ
se their concentrations are effectively constant. For example, in the equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g), the Kc expression is simply Kc = [CO₂].
只有浓度会变化的物种才写入表达式。纯固体和纯液体的浓度实际上恒定，因此不写入 Kc 表达式。例如，在平衡 CaCO₃(s) ⇌ CaO(s) + CO₂(g) 中，Kc 表达式就是 Kc = [CO₂]。
The units of Kc depend on the stoichiometry of the reaction. You should be able to deduce them from the powers in the expression. An ICE table, showing initial, change and equilibrium amounts, is a reliable way to organise the calculation and avoid mistakes.
Kc 的单位取决于反应的化学计量关系。你应当能够从表达式中的幂次推导出单位。ICE 表展示初始量、变化量和平衡量，是组织计算并避免错误的一种可靠方法。
3. The Equilibrium Constant Kp and Partial Pressures | 平衡常数 Kp 与分压
For gas-phase equilibria, it is often more convenient to use partial pressures instead of concentrations. The partial pressure of a gas A is the pressure it would exert if it alone occupied the whole volume. It is calculated from the mole fraction and the total pressure.
对于气相平衡，使用分压通常比使用浓度更方便。气体 A 的分压是指它单独占据整个体积时所施加的压力。它由摩尔分数和总压计算得到。
p(A) = x(A) × p(total)
For the general gas equilibrium aA + bB ⇌ cC + dD, the Kp expression uses partial pressures in place of concentrations. The partial pressures must be in the same units, often atm or Pa.
对于一般的气体平衡 aA + bB ⇌ cC + dD，Kp 表达式用分压代替浓度。各分压必须使用相同单位，通常为 atm 或 Pa。
Kp = (pC)ᶜ(pD)ᵈ ÷ (pA)ᵃ(pB)ᵇ
The Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), is a classic example. Increasing total pressure shifts the equilibrium to the right because there are fewer moles of gas on the product side. This increases the yield of ammonia, but it does not change the value of Kp at a fixed temperature.
哈伯法合成氨 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 是经典例子。增大总压使平衡向右移动，因为生成物一侧气体摩尔数更少。这会提高氨的产率，但在固定温度下不会改变 Kp 的值。
4. Le Chatelier's Principle and Shifting Equilibrium | 勒夏特列原理与平衡移动
Le Chatelier's principle states that when a system at equilibrium is subjected to a change, the equilibrium position shifts in the direction that opposes the change. This principle helps predict the effect of changes in concentration, pressure and temperature.
勒夏特列原理指出，当处于平衡的体系受到改变时，平衡位置会向抵消该改变的方向移动。该原理有助于预测浓度、压强和温度变化的影响。
| Change | 改变 | Effect on equilibrium | 对平衡的影响 |
| Increase concentration of a reactant | 增大反应物浓度 | Shifts right to use up the added reactant | 向右移动以消耗加入的反应物 |
| Increase total pressure | 增大总压 | Shifts towards the side with fewer gas moles | 向气体摩尔数较少的一侧移动 |
| Increase temperature | 升高温度 | Shifts in the endothermic direction | 向吸热方向移动 |
| Add a catalyst | 加入催化剂 | No shift; equilibrium reached faster | 不移动；更快达到平衡 |
Changing concentration or pressure alters the position of equilibrium, but it does not change the equilibrium constant K. Only a change in temperature changes the value of K. This distinction is a frequent source of exam errors.
改变浓度或压强会改变平衡位置，但不会改变平衡常数 K。只有温度变化才会改变 K 的值。这一区别是考试中常见的错误来源。
5. Temperature, Catalysts and the Value of K | 温度、催化剂与 K 值
Temperature is the only factor that changes the numerical value of an equilibrium constant. For an exothermic forward reaction, increasing temperature decreases K, so the equilibrium position shifts to the left. For an endothermic forward reaction, increasing temperature increases K, so the equilibrium position shifts to the right.
温度是唯一会改变平衡常数数值的因素。对于放热正反应，升高温度会降低 K，因此平衡位置向左移动。对于吸热正反应，升高温度会提高 K，因此平衡位置向右移动。
A catalyst lowers the activation energy of both the forward and backward reactions by the same amount. It therefore increases the rate at which equilibrium is reached but has no effect on the equilibrium position or on K.
催化剂同等程度地降低正反应和逆反应的活化能。因此它提高达到平衡的速率，但对平衡位置或 K 没有影响。
When the value of K is very large, the equilibrium position lies far to the right and the reaction is described as going almost to completion. When K is very small, the equilibrium position lies far to the left and only a small amount of product forms.
当 K 值很大时，平衡位置远在右侧，反应几乎进行完全。当 K 很小时，平衡位置远在左侧，只生成少量产物。
6. Brønsted-Lowry Acids and Bases, Kw and pH | 布朗斯特-劳里酸碱、Kw 与 pH
A Brønsted-Lowry acid is a proton donor, and a Brønsted-Lowry base is a proton acceptor. When an acid donates a proton, it forms its conjugate base. When a base accepts a proton, it forms its conjugate acid. For example, CH₃COOH ⇌ CH₃COO⁻ + H⁺, where CH₃COOH and CH₃COO⁻ are a conjugate acid-base pair.
布朗斯特-劳里酸是质子给体，布朗斯特-劳里碱是质子受体。酸给出质子后形成其共轭碱。碱接受质子后形成其共轭酸。例如，CH₃COOH ⇌ CH₃COO⁻ + H⁺，其中 CH₃COOH 和 CH₃COO⁻ 是一对共轭酸碱。
self-ionisation of water is shown below. At 298 K, Kw has the value 1.0 × 10⁻¹⁴ mol² dm⁻⁶.
水是两性的：它既可以作酸，也可以作碱。水的自电离如下所示。在 298 K 时，Kw 的值为 1.0 × 10⁻¹⁴ mol² dm⁻⁶。
H₂O ⇌ H⁺ + OH⁻ Kw = [H⁺][OH⁻]
water and in neutral solutions at 298 K, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³, so pH = 7.00.
溶液的 pH 定义为氢离子浓度的以 10 为底的负对数。在 298 K 的水和中性溶液中，[H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³，所以 pH = 7.00。
pH = -log₁₀[H⁺] [H⁺] = 10⁻ᵖᴴ
At higher temperatures, Kw increases because the self-ionisation of water is endothermic. Pure water then has a pH slightly below 7, but it remains neutral because [H⁺] and [OH⁻] are still equal.
温度升高时，Kw 增大，因为水的自电离是吸热的。此时纯水的 pH 略低于 7，但它仍然是中性的，因为 [H⁺] 和 [OH⁻] 仍然相等。
7. Strong and Weak Acids: Ka, pKa and Degree of Dissociation | 强酸弱酸：Ka、pKa 与电离度
00 mol dm⁻³ and pH = 1.00. A weak acid only partially dissociates, so an equilibrium is established between the undissociated acid and its ions.
强酸在水溶液中完全电离。例如，HCl 几乎完全电离，因此 0.100 mol dm⁻³ 的 HCl 溶液中 [H⁺] = 0.100 mol dm⁻³，pH = 1.00。弱酸只部分电离，因此未电离的酸与其离子之间建立平衡。
The acid dissociation constant Ka is used to compare weak acids quantitatively. A smaller Ka means a weaker acid, because the equilibrium lies further to the left.
酸电离常数 Ka 用于定量比较弱酸。Ka 越小，酸越弱，因为平衡更偏向左侧。
Ka = [H⁺][A⁻] ÷ [HA] pKa = -log₁₀ Ka
The degree of dissociation, α, is the fraction of acid molecules that have ionised. Diluting a weak acid increases the degree of dissociation, even though the concentration of H⁺ decreases. This is another example of Le Chatelier's principle in action.
电离度 α 是已电离的酸分子所占的比例。稀释弱酸会增大电离度，尽管 H⁺ 浓度降低。这是勒夏特列原理的另一个应用实例。
For a weak monoprotic acid, a useful approximation is that [H⁺] = √Ka × [HA]₀ the acid is not too concentrated and Ka is small. This leads to pH = ½(pKa − log₁₀[HA]₀).
对于一元弱酸，一个有用的近似是 [H⁺] = √Ka × [HA]₀，前提是酸不太浓且 Ka 较小。由此可得到 pH = ½(pKa − log₁₀[HA]₀)。
8. Bases, Kb and the Relationship between Ka and Kb | 碱、Kb 以及 Ka 与 Kb 的关系
A weak base accepts protons only partially in water. For a base B, the equilibrium is B + H₂O ⇌ BH⁺ + OH⁻, and the base dissociation constant Kb is defined in the same way as Ka. A larger Kb means a stronger base.
弱碱在水中只部分接受质子。对于碱 B，平衡为 B + H₂O ⇌ BH⁺ + OH⁻，碱电离常数 Kb 的定义方式与 Ka 相同。Kb 越大，碱越强。
Kb = [BH⁺][OH⁻] ÷ [B]
e related through the ionic product of water. At 298 K, pKa + pKb = 14.00. This means that
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