📚 Edexcel A-Level Chemistry: Combined Topic 109 – Equilibrium, Energetics and Redox | Edexcel A-Level 化学：综合专题 109——平衡、能量学与氧化还原
This revision guide brings together the equilibrium, energetics and redox ideas that frequently appear in Edexcel A-Level Chemistry examinations. Topic 109-style questions often mix calculation, explanation and practical methods, so a strong grasp of the links between ΔH, ΔS, Kc, Kp, pH and E° is essential. The aim is not simply to memorise separate facts, but to move confidently between a numerical answer, a chemical explanation and a judgement about feasibility.
本复习指南汇总了 Edexcel A-Level 化学考试中经常出现的平衡、能量学和氧化还原知识点。109 类综合题通常融合计算、解释和实验方法，因此牢固掌握 ΔH、ΔS、Kc、Kp、pH 与 E° 之间的联系至关重要。目标不只是记住零散事实，而是能在数值结果、化学解释与可行性判断之间自如切换。
1. What Combined Topic 109 Tests | 综合专题 109 考查什么
Combined Topic 109 questions usually ask you to link three core ideas: enthalpy and entropy changes, equilibrium constants and Le Chatelier's principle, and redox feasibility from standard electrode potentials. The examiner often provides data in tables and expects a calculation followed by a chemical explanation. A typical question may ask you to calculate ΔH from calorimetry, use ΔS to predict whether a reaction becomes feasible at a given temperature, then calculate Kc or Kp for the same system and finally explain why an apparently feasible redox reaction does not occur in practice.
综合专题 109 的题目通常要求你联系三大核心概念：焓变与熵变、平衡常数与勒夏特列原理，以及由标准电极电势得出的氧化还原可行性。考官常以表格形式给出数据，要求先完成计算，再给出化学解释。典型题目可能要求你由量热实验计算 ΔH，用 ΔS 判断某温度下反应是否可行，再计算同一体系的 Kc 或 Kp，最后解释为何一个看似可行的氧化还原反应在实践中并未发生。
- Energetics: Hess's law, Born-Haber cycles, entropy, Gibbs free energy and the temperature at which a reaction becomes feasible.
能量学：盖斯定律、玻恩-哈伯循环、熵、吉布斯自由能以及反应变为可行的温度。 - Equilibrium: dynamic equilibrium, Le Chatelier's principle, Kc, Kp, partial pressures, pH, buffers and titration curves.
平衡：动态平衡、勒夏特列原理、Kc、Kp、分压、pH、缓冲液与滴定曲线。 - Redox: oxidation states, half-equations, disproportionation, standard electrode potentials and electrochemical cells.
氧化还原：氧化态、半反应式、歧化反应、标准电极电势与电化学电池。
The command words matter. "Calculate" requires clear working and correct units. "Explain" requires a chemical reason, not a restatement of the result. "Suggest" usually means applying familiar principles to an unfamiliar context, such as a new catalyst or an industrial process.
题目中的指令词非常重要。"Calculate"要求清晰的解题步骤和正确单位；"Explain"要求给出化学原因，而不是重复结果；"Suggest"通常意味着把熟悉原理应用到陌生情境中，例如一种新催化剂或一个工业流程。
| Topic area | Typical task | 常见任务 |
| Energetics | Construct a Hess cycle, calculate ΔG, predict feasibility | 构建盖斯循环，计算 ΔG，预测可行性 |
| Equilibrium | Calculate Kc or Kp, deduce the effect of changing T or p | 计算 Kc 或 Kp，推断改变温度或压强的影响 |
| Redox | Combine half-equations, calculate E°cell, assess feasibility | 合并半反应式，计算 E°cell，判断可行性 |
2. Enthalpy Changes and Hess's Law | 焓变与盖斯定律
Enthalpy change ΔH is the heat energy transferred at constant pressure. Standard conditions are 100 kPa, a stated temperature usually 298 K, and 1 mol dm⁻³ for solutions. The sign convention is crucial: exothermic reactions have negative ΔH because the system loses energy to the surroundings, while endothermic reactions have positive ΔH.
焓变 ΔH 是在恒压条件下传递的热能。标准条件为 100 kPa、指定温度（通常 298 K）以及溶液浓度 1 mol dm⁻³。符号约定至关重要：放热反应 ΔH 为负，因为体系向环境释放能量；吸热反应 ΔH 为正。
Hess's law states that the total enthalpy change of a reaction depends only on the initial and final states, not on the route taken. This allows you to build cycles in which an unknown enthalpy change is found from known values. When using standard enthalpies of formation, apply the general relationship below.
盖斯定律指出，反应的总焓变只取决于始态和终态，与路径无关。因此可以构建循环，由已知值求出未知焓变。使用标准生成焓时，可应用下面的通用关系式。
ΔH°reaction = Σ ΔH°f(products) − Σ ΔH°f(reactants)
ts) − Σ ΔH°c(products). Remember that the standard enthalpy of formation of an element in its standard state is zero, and that coefficients in the equation multiply the enthalpy values.
使用燃烧焓数据时，循环方向相反：ΔH°reaction = Σ ΔH°c(反应物) − Σ ΔH°c(生成物)。记住，元素在标准状态下的标准生成焓为零，且方程式中的系数要乘以相应的焓值。
sually 4.18 J g⁻¹ K⁻¹ for water, and ΔT is the temperature change. The enthalpy change per mole is then ΔH = −q ÷ n, with the negative sign making an exothermic temperature rise give a negative ΔH.
常见的实验方法是量热法。放出的热量由 q = mcΔT 计算，其中 m 为溶液质量，c 为比热容（水通常取 4.18 J g⁻¹ K⁻¹），ΔT 为温度变化。每摩尔的焓变则为 ΔH = −q ÷ n；负号使放热导致的升温对应负的 ΔH。
Typical sources of error in calorimetry include heat loss to the surroundings, incomplete combustion, evaporation of the fuel, and assuming that the solution has the same specific heat capacity and density as pure water. These errors usually make the experimental ΔH less exothermic than the standard value.
量热法中的典型误差来源包括向环境散热、燃烧不完全、燃料蒸发，以及假设溶液与纯水具有相同的比热容和密度。这些误差通常使实验测得的 ΔH 比标准值放热更少。
3. Born-Haber Cycles and Lattice Enthalpy | 玻恩-哈伯循环与晶格焓
Lattice enthalpy of formation is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions. It is always exothermic for a stable ionic solid. Lattice enthalpy of dissociation is the reverse process and is endothermic. The magnitude of lattice enthalpy depends mainly on the charges of the ions and their radii.
晶格生成焓是在标准条件下由气态离子生成 1 摩尔离子化合物时的焓变。对于稳定离子固体，它总是放热的。晶格解离焓是逆过程，为吸热。晶格焓的大小主要取决于离子电荷和离子半径。
ΔH°latt ≈ constant × (z⁺ × z⁻) ÷ (r⁺ + r⁻)
Higher charges and smaller radii give a more negative lattice enthalpy. This explains why MgO has a much more exothermic lattice enthalpy than NaCl and therefore a much higher melting point. The perfect ionic model assumes ions are point charges with no polarisation; in reality, polarisation and covalent character make the experimental lattice enthalpy more exothermic than the theoretical value for compounds such as AgI.
电荷越高、半径越小，晶格焓越负。这解释了为何 MgO 的晶格焓比 NaCl 放热得多，因而熔点也高得多。完美离子模型假设离子为无极化作用的点电荷；实际上，极化和共价性使 AgI 等化合物的实验晶格焓比理论值更放热。
A Born-Haber cycle for an ionic compound such as NaCl links the standard enthalpy of formation to a sequence of steps. A standard sequence is shown below.
以 NaCl 为例，玻恩-哈伯循环把标准生成焓与一系列步骤联系起来。标准顺序如下所示。
ΔH°f = ΔH°at(Na) + IE(Na) + ½ΔH°diss(Cl₂) + EA(Cl) + ΔH°latt
In this equation, ΔH°at is the enthalpy of atomisation, IE is the first ionisation energy, ΔH°diss is the bond dissociation enthalpy of chlorine, EA is the first electron affinity, and ΔH°latt is the lattice enthalpy of formation. The second ionisation energy is needed for a 2+ ion such as Mg²⁺, and the second electron affinity for an ion such as O²⁻ is endothermic.
式中 ΔH°at 为原子化焓，IE 为第一电离能，ΔH°diss 为氯的键解离焓，EA 为第一电子亲和能，ΔH°latt 为晶格生成焓。对于 Mg²⁺ 等 2+ 离子需要第二电离能；对于 O²⁻ 等离子，第二电子亲和能是吸热的。
Exam questions often provide all values except one and ask you to calculate it. Rearrange the cycle carefully, keep the signs consistent, and check that the final answer has the correct units, usually kJ mol⁻¹. A common mistake is to forget the ½ before the bond dissociation enthalpy of Cl₂ or to use the wrong number of ionisation energies.
考试题常给出除一个值外的所有数据，要求你计算该值。要仔细重排循环，保持符号一致，并检查最终答案的单位，通常为 kJ mol⁻¹。常见错误是忘记 Cl₂ 键解离焓前的 ½，或使用了错误数目的电离能。
4. Entropy, Gibbs Free Energy and Feasibility | 熵、吉布斯自由能与可行性
Entropy S is a measure of the number of ways particles and energy can be arranged, often described as disorder or dispersal of energy. Solids have low entropy, liquids and gases have higher entropy, and gases have much higher entropy than liquids. Entropy increases when a solid dissolves, when a gas is produced, or when temperature rises.
熵 S 是粒子与能量排布方式数目的量度，常被描述为无序度或能量分散程度。固体熵低，液体和气体熵较高，而气体的熵远高于液体。固体溶解、生成气体或温度升高时，熵都会增加。
ΔS°total = ΔS°system + ΔS°surroundings
The entropy change of the surroundings is related to the enthalpy change and temperature.
环境的熵变与焓变和温度有关。
ΔS°surroundings = −ΔH° ÷ T
A reaction is feasible when ΔS°total is positive. Multiplying by −T gives the Gibbs free energy equation, where a negative ΔG° corresponds to a feasible reaction under standard conditions.
当 ΔS°total 为正时，反应可行。乘以 −T 后得到吉布斯自由能方程；在标准条件下，ΔG° 为负对应反应可行。
ΔG° = ΔH° − TΔS°
r standard conditions. If both have the same sign, the reaction becomes feasible above or below a certain temperature. The boundary temperature is found by setting ΔG° = 0.
若 ΔH° 为负且 ΔS° 为正，反应在所有温度下都可行。若 ΔH° 为正且 ΔS° 为负，则在标准条件下永不可行。若两者同号，则反应在高于或低于某一温度时变得可行。令 ΔG° = 0 可求出边界温度。
T = ΔH° ÷ ΔS°
Feasibility is not the same as rate. A reaction with a negative ΔG° may still be too slow to observe at room temperature because of a high activation energy. This is why thermodynamic predictions must always be paired with kinetic reasoning in combined Topic 109 questions.
可行并不等于速率快。即使 ΔG° 为负，若活化能很高，反应在室温下仍可能慢到无法观察。因此在综合专题 109 题目中，热力学预测必须与动力学推理结合。
The equilibrium constant is also linked to Gibbs free energy. For a reaction at temperature T, the relationship is shown below, where R is the gas constant 8.31 J mol⁻¹ K⁻¹ and T is in kelvin.
平衡常数也与吉布斯自由能相联系。对于温度 T 下的反应，关系式如下，其中 R 为气体常数 8.31 J mol⁻¹ K⁻¹，T 的单位为开尔文。
ΔG° = −RT ln K
A large positive K means a negative ΔG° and a product-favoured reaction. A very small K means a positive ΔG° and a reactant-favoured reaction. This equation is a powerful bridge between the energetics and equilibrium parts of the specification.
K 很大意味着 ΔG° 为负、反应偏向生成物；K 很小意味着 ΔG° 为正、反应偏向反应物。这个方程是连接能量学与平衡部分的有力桥梁。
5. Dynamic Equilibrium and Le Chatelier's Principle | 动态平衡与勒夏特列原理
Dynamic equilibrium is reached in a closed system when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant. It is dynamic because both reactions continue to occur; it is not static. The equilibrium position can be changed by altering temperature, pressure, concentration or by adding a catalyst.
在封闭体系中，当正反应速率等于逆反应速率，且反应物与生成物浓度保持不变时，即达到动态平衡。之所以称为动态，是因为两个反应仍在进行，并非静止。改变温度、压强、浓度或加入催化剂，都可改变平衡位置。
Le Chatelier's principle states that if a change is imposed on a system at equilibrium, the position of equilibrium shifts in the direction that tends to oppose the change. Increasing the concentration of a reactant shifts equilibrium to the right. Increasing pressure shifts equilibrium towards the side with fewer moles of gas. Increasing temperature shifts equilibrium in the endothermic direction.
勒夏特列原理指出，若对平衡体系施加改变，平衡位置将向抵消该改变的方向移动。增大反应物浓度，平衡右移；增大压强，平衡向气体摩尔数较少的一侧移动；升高温度，平衡向吸热方向移动。
A catalyst increases both forward and reverse rates equally. It does not change the position of equilibrium or the value of Kc or Kp; it only reduces the time taken to reach equilibrium. This distinction is frequently tested in multiple-choice and written questions.
催化剂同等提高正、逆反应速率。它不改变平衡位置，也不改变 Kc 或 Kp 的数值，只缩短达到平衡所需时间。这一区别在选择题和书面题中经常考查。
Only temperature changes the value of the equilibrium constant. If the forward reaction is exothermic, increasing temperature decreases Kc. If the forward reaction is endothermic, increasing temperature increases Kc. Changes in pressure or concentration change the position of equilibrium but not the value of Kc, because Kc depends only on temperature.
只有温度会改变平衡常数的数值。若正反应放热，升温使 Kc 减小；若正反应吸热，升温使 Kc 增大。改变压强或浓度会改变平衡位置，但不改变 Kc 的数值，因为 Kc 只取决于温度。
Industrially, the Haber process illustrates the compromise between thermodynamics and kinetics. The forward reaction is exothermic and involves a decrease in the number of gas moles, so low temperature and high pressure favour a high yield. However, low temperature gives too slow a rate, so a moderate temperature of about 700 K and a pressure of about 200 atm are used with an iron catalyst.
工业上，哈伯法体现了热力学与动力学之间的折中。正反应放热且气体摩尔数减少，因此低温和高压有利于提高产率。但低温使速率过慢，故实际采用约 700 K 的温度、约 200 atm 的压强以及铁催化剂。
6. Kc Calculations and ICE Tables | Kc 计算与 ICE 表格
The equilibrium constant Kc gives a quantitative measure of the position of equilibrium in terms of concentrations. For a general reaction aA + bB ⇌ cC + dD, the expression is shown below. Pure solids and pure liquids are omitted because their concentrations are effectively constant.
平衡常数 Kc 以浓度定量表示平衡位置。对于一般反应 aA + bB ⇌ cC + dD，表达式如下。纯固体和纯液体不写入表达式，因为它们的浓度实际上恒定。
Kc = [C]ᶜ[D]ᵈ ÷ [A]ᵃ[B]ᵇ
The units of Kc depend on the stoichiometry. For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), the units cancel and Kc has no units. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc has units of mol⁻² dm⁶. Always work out the units from the expression rather than memorising them.
Kc 的单位取决于化学计量关系。对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，单位相消，Kc 无单位。对于 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Kc 的单位为 mol⁻² dm⁶。应始终由表达式推导单位，而不是死记。
centrations, calculate the change using the mole ratio, then find the equilibrium values. For a reaction A ⇌ nge is 0.200 mol dm⁻³, so equilibrium [A] = 0.300 mol dm⁻³ and Kc = 0.200 ÷ 0.300 = 0.667.
ICE 表格是组织平衡计算的可靠方法。ICE 代表初始、变化和平衡。先写出初始摩尔数或浓度，再按摩尔比计算变化量，最后得到平衡值。对于反应 A ⇌ B，若 A 的初始浓度为 0.500 mol dm⁻³，平衡时 B 的浓度为 0.200 mol dm⁻³，则变化量为 0.200 mol dm⁻³，故平衡时 [A] = 0.300 mol dm⁻³，Kc = 0.200 ÷ 0.300 = 0.667。
If a question gives initial moles and volume, convert to concentrations before substituting into the Kc expression. If the volume changes, the concentrations change, and the system may shift to restore equilibrium. A useful check is to compare the reaction quotient Q with Kc: if Q < Kc, the forward reaction is favoured; if Q > Kc, the reverse reaction is favoured.
若题目给出初始摩尔数和体积，应先换算为浓度再代入 Kc 表达式。若体积改变，浓度随之改变，体系可能移动以恢复平衡。一个有用的检查是比较反应商 Q 与 Kc：若 Q < Kc，正反应占优；若 Q > Kc，逆反应占优。
Common errors include using moles instead of concentration, forgetting to multiply the change by the stoichiometric coefficient, and rounding too early. Keep extra significant figures during the calculation and round only at the end. State the units clearly, and if the answer is asked to three significant figures, give exactly that.
常见错误包括用摩尔数代替浓度、忘记将变化量乘以化学计量系数、以及过早四舍五入。计算过程中应保留多余有效数字，最后才取整。单位要写清楚；若要求保留三位有效数字，就应恰好给出三位。
7. Kp and Partial Pressures | Kp 与分压
For gaseous equilibria, it is often more convenient to use partial pressures instead of concentrations. The mole fraction of a gas is the number of moles of that gas divided by the total number of moles of all gases present. The partial pressure is the mole fraction multiplied by the total pressure.
对于气体平衡，用分压代替浓度往往更方便。某气体的摩尔分数等于该气体的摩尔数除以所有气体的总摩尔数。分压等于摩尔分数乘以总压。
partial pressure = mole fraction × total pressure
For the general reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), the equilibrium constant Kp is defined in terms of partial pressures.
对于一般反应 aA(g) + bB(g) ⇌ cC(g) + dD(g)，平衡常数 Kp 用分压定义如下。
Kp = p(C)ᶜ × p(D)ᵈ ÷ (p(A)ᵃ × p(B)ᵇ)
The units of Kp depend on the change in the number of moles of gas, Δn. The relationship between Kp and Kc is shown below, where R is 8.31 J mol⁻¹ K⁻¹ and T is in kelvin.
Kp 的单位取决于气体摩尔数的变化 Δn。Kp 与 Kc 的关系如下，其中 R 为 8.31 J mol⁻¹ K⁻¹，T 的单位为开尔文。
Kp = Kc(RT)Δⁿ
When Δn = 0, Kp = Kc and both are dimensionless. For the Haber process, N₂(g) + 3H₂(g) ⇌ d the total pressure, then asks for Kp. The safest method is to calculate the total moles, find each mole fraction, convert to partial pressures, and substitute.
当 Δn = 0 时，Kp = Kc，且两者均无单位。对于哈伯法 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Δn = −2，因此 Kp 与 Kc 的单位和数值都不同。典型考题给出各气体的平衡摩尔数和总压，要求计算 Kp。最稳妥的方法是先求总摩尔数，再求各摩尔分数，换算为分压后代入。
Changes in total pressure do not change Kp at constant temperature, but they do change the position of equilibrium if Δn ≠ 0. Increasing pressure shifts the equilibrium towards the side with fewer gas moles, which for the Haber process is the product side. Catalysts do not change Kp either; they only affect the rate of attainment.
在恒温下，改变总压不改变 Kp；但若 Δn ≠ 0，它会改变平衡位置。增大压强使平衡向气体摩尔数较少的一侧移动，对哈伯法而言即生成物一侧。催化剂也不改变 Kp，只影响达到平衡的速率。
