📚 Further Aspects of Equilibria | 化学平衡的深入探讨
This article revises the deeper parts of chemical equilibrium required for Cambridge A Level Chemistry. It covers Kc and Kp, partial pressures, temperature effects, acid-base equilibria, buffer solutions and solubility product.
本文复习剑桥 A Level 化学中化学平衡的深入内容，涵盖 Kc 和 Kp、分压、温度效应、酸碱平衡、缓冲溶液及溶度积。
1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡
A reversible reaction proceeds in both the forward and backward directions. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction, so macroscopic concentrations of reactants and products remain constant.
可逆反应同时向正逆两个方向进行。达到平衡时，正反应速率等于逆反应速率，因此反应物和产物的宏观浓度保持恒定。
aA + bB ⇌ cC + dD
2. Equilibrium Constants Kc and Kp | 平衡常数 Kc 和 Kp
For a homogeneous reaction aA + bB ⇌ cC + dD, the concentration equilibrium constant Kc is given by the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient. Kp uses partial pressures instead of concentrations. Units depend on the stoichiometry of the reaction.
对于均相反应 aA + bB ⇌ cC + dD，浓度平衡常数 Kc 表示为产物浓度与反应物浓度之比，每种物质浓度以化学计量数为指数。Kp 使用分压代替浓度。单位取决于反应的化学计量数。
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Kp = p(C)ᶜ p(D)ᵈ / p(A)ᵃ p(B)ᵇ
3. Mole Fraction and Partial Pressure | 摩尔分数与分压
In a gas mixture, the mole fraction of gas X is n(X) divided by n(total). The partial pressure p(X) is the mole fraction multiplied by the total pressure. Dalton’s law states that the total pressure is the sum of all partial pressures.
混合气体中，气体 X 的摩尔分数为 n(X)/n(总)。分压 p(X) 等于摩尔分数乘以总压。道尔顿定律指出总压等于各分压之和。
p(X) = (n(X) / ntotal) × Ptotal
4. Homogeneous and Heterogeneous Equilibria | 均相与多相平衡
Homogeneous equilibria have all species in the same phase, usually gas or aqueous solution. In heterogeneous equilibria, pure solids and pure liquids have constant concentration and are omitted from the Kc or Kp expression.
均相平衡中所有物质处于同一相，通常为气相或水溶液。多相平衡中，纯固体和纯液体的浓度视为常数，在 Kc 或 Kp 表达式中省略。
Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = p(CO₂) and Kc = [CO₂].
示例：CaCO₃(s) ⇌ CaO(s) + CO₂(g)；Kp = p(CO₂)，Kc = [CO₂]。
5. Kp Calculations with Worked Example | Kp 计算示例
At equilibrium, 1.0 mol of PCl₅ is 50% dissociated in a vessel at 500 K: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Calculate Kp if the total pressure is 200 kPa.
在平衡时，1.0 mol PCl₅ 于 500 K 下解离 50%：PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)。若总压为 200 kPa，计算 Kp。
Since dissociation is 50%, n(PCl₅) = 0.5 mol, n(PCl₃) = n(Cl₂) = 0.5 mol, so total moles = 1.5 mol. Mole fractions are all 0.5/1.5 = 0.333. Partial pressures are each 0.333 × 200 kPa = 66.7 kPa.
解离 50% 时，n(PCl₅) = 0.5 mol，n(PCl₃) = n(Cl₂) = 0.5 mol，总物质的量为 1.5 mol。摩尔分数均为 0.5/1.5 = 0.333。各分压为 0.333 × 200 kPa = 66.7 kPa。
Kp = p(PCl₃) × p(Cl₂) / p(PCl₅) = (66.7 × 66.7) / 66.7 = 66.7 kPa
6. Le Chatelier’s Principle and Equilibrium Response | 勒夏特列原理与平衡响应
If a system at equilibrium is disturbed, the position of equilibrium shifts to oppose the change. Increasing pressure favours the side with fewer gas moles. Increasing temperature favours the endothermic direction. Adding a catalyst does not shift the position but speeds up both forward and backward rates equally.
若平衡体系受到扰动，平衡位置会朝削弱该改变的方向移动。增大压力有利于气体物质的量较少的一侧。升高温度有利于吸热方向。加入催化剂不改变平衡位置，但同等加快正逆反应速率。
7. Temperature Dependence of Equilibrium Constants | 平衡常数与温度的关系
Kc and Kp are only affected by temperature. For an exothermic forward reaction, increasing temperature decreases K; for an endothermic forward reaction, increasing temperature increases K. Concentration changes, pressure changes and catalysts may change the position or rates but not the value of K.
Kc 和 Kp 只受温度影响。正反应放热时，升高温度使 K 减小；正反应吸热时，升高温度使 K 增大。浓度、压力和催化剂可改变平衡位置或速率，但不改变 K 值。
If ΔH° < 0, T increase lowers K; if ΔH° > 0, T increase raises K.
若 ΔH° < 0，升温使 K 减小；若 ΔH° > 0，升温使 K 增大。
8. Acid-Base Equilibria: Ka, pH and pKa | 酸碱平衡：Ka、pH 与 pKa
For a weak acid HA ⇌ H⁺ + A⁻, the acid dissociation constant Ka = [H⁺][A⁻]/[HA]. pH = -log₁₀[H⁺] and pKa = -log₁₀ Ka. A smaller pKa indicates a stronger acid. For a weak acid of concentration c, [H⁺] ≈ √(Ka × c).
对弱酸 HA ⇌ H⁺ + A⁻，酸解离常数 Ka = [H⁺][A⁻]/[HA]。pH = -log₁₀[H⁺]，pKa = -log₁₀ Ka。pKa 越小，酸性越强。对浓度为 c 的弱酸，[H⁺] ≈ √(Ka × c)。
Ka = [H⁺][A⁻] / [HA]
pH = -log₁₀[H⁺]
9. Buffer Solutions and Their Action | 缓冲溶液及其作用
A buffer resists pH changes on adding small amounts of acid or base. An acidic buffer contains a weak acid and its conjugate base, for example CH₃COOH/CH₃COO⁻. The pH of a buffer is given by pH = pKa + log₁₀([A⁻]/[HA]).
缓冲溶液能抵抗少量酸或碱引起的 pH 变化。酸性缓冲液含有弱酸及其共轭碱，如 CH₃COOH/CH₃COO⁻。缓冲溶液的 pH 可用 pH = pKa + log₁₀([A⁻]/[HA]) 计算。
pH = pKa + log₁₀([A⁻] / [HA])
10. Solubility Product Ksp and Common Ion Effect | 溶度积 Ksp 与同离子效应
For a sparingly soluble salt such as AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), Ksp = [Ag⁺][Cl⁻]. Ksp is the product of ion concentrations in a saturated solution. Adding a common ion lowers the solubility because the equilibrium shifts to the left.
对于难溶盐如 AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)，Ksp = [Ag⁺][Cl⁻]。Ksp 是饱和溶液中离子浓度的乘积。加入同离子会降低溶解度，因为平衡左移。
If Ksp = 2.0 × 10⁻¹⁰ mol² dm⁻⁶ for AgCl, then solubility s = √Ksp = 1.4 × 10⁻⁵ mol dm⁻³.
若 AgCl 的 Ksp = 2.0 × 10⁻¹⁰ mol² dm⁻⁶，则溶解度 s = √Ksp = 1.4 × 10⁻⁵ mol dm⁻³。
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