📚 Born–Haber Cycles: Complete A-Level Guide | 玻恩-哈伯循环：A-Level 完整指南
Born–Haber cycles are a fundamental tool in thermodynamics used to determine the lattice enthalpy of an ionic compound. They combine several experimentally measurable enthalpy changes by applying Hess's law, allowing chemists to calculate a value that cannot be measured directly.
玻恩-哈伯循环是热力学中用于确定离子化合物晶格焓的基本工具。它通过应用赫斯定律，将几个实验可测的焓变组合起来，从而计算出无法直接测量的数值。
This guide covers the full A-Level treatment: the definitions you must know, how to build a cycle step by step, two fully worked calculations, the difference between theoretical and experimental lattice enthalpy, and the exam traps that cost the most marks.
本指南涵盖 A-Level 的完整内容：必须掌握的各个定义、如何一步步构建循环、两个完整的计算实例、理论晶格焓与实验晶格焓的差异，以及最容易失分的考试陷阱。
1. Introduction to Born–Haber Cycles | 玻恩-哈伯循环简介
The Born–Haber cycle is named after the German scientists Max Born and Fritz Haber, who developed it in 1919. It is essentially an application of Hess's law to the formation of an ionic solid from its constituent elements in their standard states.
玻恩-哈伯循环以德国科学家马克斯·玻恩和弗里茨·哈伯命名，他们于 1919 年提出了这一方法。它本质上是将赫斯定律应用于从其组成元素的标态生成离子固体的过程。
The cycle breaks down the overall formation reaction into a series of steps, such as atomisation, ionisation, electron gain and lattice formation. This allows the calculation of the lattice enthalpy, which is a measure of the strength of the ionic bonding.
这种循环将总生成反应分解为一系列步骤，如原子化、电离、电子获得和晶格形成。这样就可以计算晶格焓，它是离子键强度的量度。
In A-Level Chemistry, constructing and interpreting Born–Haber cycles is an essential skill for understanding the energetics of ionic compounds and the factors that stabilise ionic lattices. It also underpins later topics such as enthalpy of solution, hydration enthalpy and the polarising power of cations.
在 A-Level 化学中，构建和解读玻恩-哈伯循环是理解离子化合物能量学和稳定离子晶格因素的关键技能。它也为后续主题打下基础，例如溶解焓、水合焓以及阳离子的极化能力。
2. Hess's Law and Energy Cycles | 赫斯定律与能量循环
Hess's law states that the total enthalpy change for a chemical reaction is the same, regardless of the route by which the reaction takes place, provided the initial and final conditions are identical.
赫斯定律指出，只要初始和最终条件相同，一个化学反应的总焓变与反应所采取的路径无关。
In a Born–Haber cycle, two routes connect the same starting point (elements in their standard states) and the same finishing point (the solid ionic lattice). Route one is the direct formation of the solid. Route two is the indirect journey: atomise the elements, ionise the metal, add electrons to the non-metal, then bring the gaseous ions together.
在玻恩-哈伯循环中，有两条路径连接同一个起点（标态下的单质）和同一个终点（固态离子晶格）。第一条路径是直接生成固体；第二条是间接路径：先将单质原子化，再使金属电离、非金属获得电子，最后让气态离子结合成晶格。
Because both routes start and end at the same energy levels, the sum of the enthalpy changes along route two must equal the enthalpy change along route one. This gives the master equation that every Born–Haber calculation is built on.
由于两条路径的起点与终点的能量状态相同，路径二上各焓变之和必然等于路径一的焓变。由此得到所有玻恩-哈伯计算所依据的主方程。
ΔHf = ΔHat(metal) + IE + ΔHat(non-metal) + EA + ΔHlatt
It is best practice to draw the cycle as an energy level diagram, with the elements at a defined baseline, gaseous ions above them, and the solid lattice drawn at its true (usually lowest) energy level.
最好把循环画成能级图：把单质放在基准线上，气态离子画在其上方，固态晶格画在它真实（通常最低）的能级位置。
3. Key Enthalpy Definitions You Must Know | 必须掌握的关键焓变定义
Mark schemes award definitions as precisely as calculations, so learn each one word for word, including the standard conditions of 298 K, 100 kPa and one mole of substance.
评分标准对定义的严格程度不亚于计算，因此请逐字记忆每一个定义，包括 298 K、100 kPa 以及一摩尔物质的标态条件。
| Quantity | 量 | Symbol | 符号 | Definition | 定义 | Typical sign | 常见符号 |
|---|---|---|---|
| Standard enthalpy of formation | 标准生成焓 | ΔHf | Enthalpy change when 1 mol of a compound forms from its elements in their standard states | 由标态单质生成 1 mol 化合物时的焓变 | Negative for most ionic solids | 大多数离子固体为负 |
| Enthalpy of atomisation | 原子化焓 | ΔHat | Enthalpy change when 1 mol of gaseous atoms forms from the element in its standard state | 由标态单质生成 1 mol 气态原子时的焓变 | Always positive | 恒为正 |
| First ionisation energy | 第一电离能 | IE₁ | Enthalpy change when 1 mol of gaseous atoms loses 1 mol of electrons | 1 mol 气态原子失去 1 mol 电子时的焓变 | Always positive | 恒为正 |
| First electron affinity | 第一电子亲和能 | EA₁ | Enthalpy change when 1 mol of gaseous atoms gains 1 mol of electrons | 1 mol 气态原子获得 1 mol 电子时的焓变 | Usually negative | 通常为负 |
| Lattice enthalpy (formation) | 晶格焓（生成方向） | ΔHlatt | Enthalpy change when 1 mol of a solid ionic lattice forms from its gaseous ions | 由气态离子生成 1 mol 固态离子晶格时的焓变 | Always negative | 恒为负 |
| Enthalpy of hydration | 水合焓 | ΔHhyd | Enthalpy change when 1 mol of gaseous ions becomes hydrated in water | 1 mol 气态离子在水中变为水合离子时的焓变 | Always negative | 恒为负 |
| Enthalpy of solution | 溶解焓 | ΔHsol | Enthalpy change when 1 mol of an ionic solid dissolves in water to form a dilute solution | 1 mol 离子固体溶于足量水形成稀溶液时的焓变 | Positive or negative | 可正可负 |
Note that some specifications and textbooks define lattice enthalpy in the opposite (dissociation) direction. Always read the definition given in the question before you assign a sign.
注意，某些考纲和教材把晶格焓定义为相反方向（解离方向）。在确定符号之前，务必先看清题目所给的定义。
4. Lattice Enthalpy: Meaning and Sign Conventions | 晶格焓的含义与符号约定
Lattice enthalpy is a direct measure of the strength of the electrostatic attraction in an ionic crystal. The more exothermic the value, the stronger the ionic bonding and the higher the melting point.
晶格焓是离子晶体中静电引力强度的直接量度。数值越放热，离子键越强，熔点越高。
Using the formation definition, the process is exothermic because oppositely charged gaseous ions are attracted together and fall to a lower energy state.
采用生成方向的定义时，该过程是放热的，因为带相反电荷的气态离子相互吸引并落到更低的能量状态。
Na⁺(g) + Cl⁻(g) → NaCl(s) ΔHlatt = -788 kJ mol⁻¹
Using the dissociation definition, the same quantity is written as a positive value, because energy must be supplied to pull the ions apart against the electrostatic attraction.
采用解离方向的定义时，同一个量写成正值，因为必须提供能量才能使离子克服静电引力而彼此分开。
NaCl(s) → Na⁺(g) + Cl⁻(g) ΔHlatt = +788 kJ mol⁻¹
The two numbers are identical in magnitude and opposite in sign. Both are acceptable if the definition used is stated and applied consistently through the whole calculation.
这两个数值大小相同、符号相反。只要所用定义明确并在整个计算中保持一致，两者都可以接受。
5. Constructing the Cycle for NaCl | 构建 NaCl 循环
Take sodium chloride as the simplest case, because sodium forms only a 1+ ion and chlorine forms only a 1- ion in its common ionic compounds.
以氯化钠为例最简单，因为钠在常见离子化合物中只形成 1+ 离子，氯只形成 1- 离子。
The indirect route from the elements to NaCl(s) has five steps. Write them in the order below and keep each arrow pointing in the direction of the enthalpy change shown.
从单质到 NaCl(s) 的间接路径共五步。按下面的顺序书写，并让每个箭头的方向与所标焓变方向一致。
- Na(s) → Na(g): enthalpy of atomisation of sodium, ΔHat(Na) = +108 kJ mol⁻¹ | 钠的原子化焓，ΔHat(Na) = +108 kJ mol⁻¹
- Na(g) → Na⁺(g) + e⁻: first ionisation energy of sodium, IE₁(Na) = +496 kJ mol⁻¹ | 钠的第一电离能，IE₁(Na) = +496 kJ mol⁻¹
- of chlorine, which is half the Cl-Cl bond dissociation enthalpy, ΔHat(Cl) = +122 kJ mol⁻¹ | 氯的原子化焓，即 Cl-Cl 键解离焓的一半，ΔHat(Cl) = +122 kJ mol⁻¹
- Cl(g) + e⁻ → Cl⁻(g): first electron affinity of chlorine, EA₁(Cl) = -349 kJ mol⁻¹ | 氯的第一电子亲和能，EA₁(Cl) = -349 kJ mol⁻¹
- Na⁺(g) + Cl⁻(g) → NaCl(s): lattice enthalpy of formation, the unknown | 钠离子与氯离子生成 NaCl(s)：晶格焓（生成方向），即未知量
Notice that chlorine is introduced as ½Cl₂ because the standard state of chlorine is the diatomic molecule, but the cycle requires exactly one mole of chlorine atoms.
注意氯是以 ½Cl₂ 引入的，因为氯的标态是双原子分子，而循环中恰好需要一摩尔氯原子。
6. Worked Calculation: Lattice Enthalpy of NaCl | 计算实例：NaCl 的晶格焓
Apply Hess's law by equating the direct route with the sum of the indirect route. Substituting the standard values gives the working below.
应用赫斯定律，令直接路径等于间接路径各项之和。代入标准数值后得到如下计算过程。
ΔHf(NaCl) = ΔHat(Na) + IE₁(Na) + ½ΔHdiss(Cl₂) + EA₁(Cl) + ΔHlatt
-411 = (+108) + (+496) + (+122) + (-349) + ΔHlatt
-411 = +377 + ΔHlatt
ΔHlatt = -788 kJ mol⁻¹
The answer is negative, which confirms that forming the NaCl lattice from gaseous ions releases energy. If the question defines lattice enthalpy as dissociation, simply reverse the sign and answer +788 kJ mol⁻¹.
结果为负，说明由气态离子形成 NaCl 晶格会放出能量。若题目把晶格焓定义为解离方向，则直接把符号反过来，答案写 +788 kJ mol⁻¹。
Two arithmetic checks are worth doing: make sure the four known terms sum to +377, and make sure the final sign matches the definition requested.
有两个值得做的算术检查：确认四个已知项之和为 +377，以及确认最终符号与题目要求的定义相符。
7. A 2+ Ion Example: Lattice Enthalpy of MgO | 二价离子实例：MgO 的晶格焓
Magnesium oxide requires extra care because magnesium forms a 2+ ion and oxygen forms a 2- ion. That means a second ionisation energy and a second electron affinity must both appear in the cycle.
氧化镁需要格外小心，因为镁形成 2+ 离子，氧形成 2- 离子。这意味着循环中必须同时出现第二电离能和第二电子亲和能。
- ΔHat(Mg) = +148 kJ mol⁻¹ | 镁的原子化焓 = +148 kJ mol⁻¹
- IE₁(Mg) = +738 kJ mol⁻¹; IE₂(Mg) = +1451 kJ mol⁻¹ | 第一电离能 = +738；第二电离能 = +1451 kJ mol⁻¹
- ΔHat(O) = +249 kJ mol⁻¹, which is half the O=O bond dissociation enthalpy | 氧的原子化焓 = +249 kJ mol⁻¹，即 O=O 键解离焓的一半
- EA₁(O) = -141 kJ mol⁻¹; EA₂(O) = +798 kJ mol⁻¹ | 第一电子亲和能 = -141；第二电子亲和能 = +798 kJ mol⁻¹
- ΔHf(MgO) = -602 kJ mol⁻¹ | 氧化镁的标准生成焓 = -602 kJ mol⁻¹
ΔHf(MgO) = ΔHat(Mg) + IE₁ + IE₂ + ΔHat(O) + EA₁ + EA₂ + ΔHlatt
-602 = 148 + 738 + 1451 + 249 + (-141) + 798 + ΔHlatt
-602 = +3243 + ΔHlatt
ΔHlatt = -3845 kJ mol⁻¹
Compare this with sodium chloride. The MgO value is roughly five times larger in magnitude, which explains why magnesium oxide melts at about 2852 °C while sodium chloride melts at 801 °C.
将其与氯化钠相比。MgO 的数值在数量级上大约是 NaCl 的五倍，这解释了为什么氧化镁熔点约为 2852 °C，而氯化钠只有 801 °C。
8. First and Second Electron Affinity | 第一与第二电子亲和能
The first electron affinity is the enthalpy change when one mole of gaseous atoms each gains one electron. It is usually exothermic because the incoming electron is attracted to the positive nuclear charge of the neutral atom.
第一电子亲和能是一摩尔气态原子各自获得一个电子时的焓变。它通常是放热的，因为进入的电子受到中性原子核正电荷的吸引。
The second electron affinity is the enthalpy change when one mole of gaseous 1- ions each gains a further electron. It is always endothermic.
第二电子亲和能是一摩尔气态 1- 离子各自再获得一个电子时的焓变。它总是吸热的。
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