Born-Haber Cycles | 玻恩-哈伯循环

📚 Born-Haber Cycles | 玻恩-哈伯循环

Born-Haber cycles are energy cycles used at A-Level to connect the formation enthalpy of an ionic compound with the enthalpy changes that break elements into gaseous ions and then bring those ions together into a lattice. They are an application of Hess’s law, so the enthalpy change is the same whether the compound forms directly from its elements or through a series of atomisation, ionisation, electron affinity and lattice-formation steps. This makes it possible to determine lattice enthalpy, which cannot be measured directly.

玻恩-哈伯循环是 A-Level 化学中用于把离子化合物的生成焓与各分步焓变联系起来的能量循环。这些分步包括元素原子化、电离、电子亲和以及离子结合成晶格。它是赫斯定律的应用，因此无论化合物由元素直接生成，还是经过多个步骤生成，总焓变相同。这样就可以求出无法直接测量的晶格焓。

1. What Is a Born-Haber Cycle? | 什么是玻恩-哈伯循环？

A Born-Haber cycle is a thermochemical cycle that uses Hess’s law to analyse the formation of an ionic compound from its elements in their standard states. It is usually drawn as an energy level diagram, but it can also be written as a set of equations. The key idea is that the direct formation route has the same overall enthalpy change as the indirect route made up of atomisation, ionisation, electron affinity and lattice formation.

玻恩-哈伯循环是一种热化学循环，利用赫斯定律分析离子化合物由标准状态下的元素生成的过程。它通常画成能级图，也可以写成一组方程式。其核心思想是：直接生成路线与由原子化、电离、电子亲和和晶格形成组成的间接路线具有相同的总焓变。

For an ionic solid such as NaCl, the cycle allows chemists to calculate the lattice enthalpy from other enthalpy changes that can be measured or estimated. This is essential because lattice enthalpy cannot be found by direct experiment: we cannot simply add one mole of gaseous Na⁺ ions to one mole of gaseous Cl⁻ ions in a calorimeter.

对于 NaCl 这类离子固体，该循环使化学家能够利用其他可测量或可估算的焓变来计算晶格焓。这一点至关重要，因为晶格焓无法通过直接实验测得：我们不可能在量热计中直接把 1 mol 气态 Na⁺ 离子与 1 mol 气态 Cl⁻ 离子混合。

Cambridge A-Level questions often ask you to construct the cycle, fill in the correct signs, or use given data to calculate an unknown enthalpy change such as lattice enthalpy, electron affinity or ionisation energy.

剑桥 A-Level 考试常要求你构建循环、写出正确的正负号，或利用给定数据计算未知焓变，例如晶格焓、电子亲和能或电离能。

2. Key Enthalpy Changes You Need | 需要掌握的关键焓变

Before building a Born-Haber cycle, you must be able to define the enthalpy changes involved and recall their standard sign conventions. The table below summarises the terms most commonly examined.

在构建玻恩-哈伯循环之前，你必须能够定义所涉及的焓变，并记住它们的标准正负号规则。下表总结了考试中最常见的术语。

Enthalpy change 焓变	Definition 定义	Typical sign 常见符号
Standard enthalpy of formation ΔHf°	Enthalpy change when 1 mol of a compound forms from its elements in their standard states. 1 mol 化合物在标准状态下由其元素生成时的焓变。	Usually negative for ionic compounds 通常为负
Standard enthalpy of atomisation ΔHat°	Enthalpy change when 1 mol of gaseous atoms forms from the element in its standard state. 1 mol 气态原子由标准状态元素生成时的焓变。	Always positive 总为正
Ionisation energy IE	Enthalpy change to remove 1 mol of electrons from 1 mol of gaseous atoms or cations. 从 1 mol 气态原子或阳离子中移走 1 mol 电子所需的焓变。	Always positive 总为正
Electron affinity EA	Enthalpy change when 1 mol of gaseous atoms or anions gains 1 mol of electrons. 1 mol 气态原子或阴离子获得 1 mol 电子时的焓变。	First EA usually negative, second EA positive 第一电子亲和能通常为负，第二电子亲和能为正
Lattice enthalpy ΔHlatt°	Enthalpy change when 1 mol of an ionic lattice forms from its gaseous ions. 1 mol 离子晶格由其气态离子生成时的焓变。	Negative for lattice formation 晶格生成时为负

One of the most common mistakes in Born-Haber calculations is using the wrong sign for electron affinity or lattice enthalpy. Always check the direction of the equation you write: if the process releases energy, the sign is negative; if it absorbs energy, the sign is positive.

玻恩-哈伯计算中最常见的错误之一是把电子亲和能或晶格焓的正负号写错。务必检查你写出的方程式方向：如果过程释放能量，符号为负；如果吸收能量，符号为正。

3. Standard Enthalpy of Formation | 标准生成焓

The standard enthalpy change of formation, ΔHf°, is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. Standard conditions are 298 K, 100 kPa and, for solutions, a concentration of 1 mol dm⁻³.

标准生成焓变 ΔHf° 是指在标准条件下，由标准状态下的元素生成 1 mol 化合物时的焓变。标准条件为 298 K、100 kPa；对溶液而言，浓度为 1 mol dm⁻³。

For sodium chloride, the formation equation is:

对于氯化钠，生成方程式为：

Na(s) + ½Cl₂(g) → NaCl(s) ΔHf° = -411 kJ mol⁻¹

The negative sign shows that the formation of NaCl from its elements is exothermic. In a Born-Haber cycle, this value is placed on the direct route from elements to the ionic compound.

负号表示由元素生成 NaCl 是放热过程。在玻恩-哈伯循环中，该数值位于从元素直接生成离子化合物的路径上。

4. Standard Enthalpy of Atomisation | 标准原子化焓

The standard enthalpy of atomisation, ΔHat°, is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. It is always endothermic because bonds between atoms in the standard state must be broken.

标准原子化焓 ΔHat° 是指由标准状态元素生成 1 mol 气态原子时的焓变。该过程总是吸热的，因为必须破坏标准状态下原子之间的化学键。

For a solid metal such as sodium, atomisation converts solid metal to gaseous metal atoms:

对于钠这类固体金属，原子化是将固态金属转化为气态金属原子：

Na(s) → Na(g) ΔHat°(Na) = +108 kJ mol⁻¹

For a diatomic non-metal such as chlorine, the equation is written per mole of chlorine atoms:

对于氯这类双原子非金属，方程式按每摩尔氯原子来写：

½Cl₂(g) → Cl(g) ΔHat°(Cl) = +122 kJ mol⁻¹

Be careful with the coefficient ½. The value is for one mole of Cl atoms, so if you use 1 mol of Cl₂ you would need twice the energy. Born-Haber cycles normally use values per mole of atoms, which matches the formula of NaCl containing one Na and one Cl.

要特别注意系数 ½。该数值对应 1 mol Cl 原子，因此如果使用 1 mol Cl₂，所需能量会是两倍。玻恩-哈伯循环通常使用每摩尔原子的数值，这与 NaCl 化学式中含有一个 Na 和一个 Cl 相匹配。

5. Ionisation Energy | 电离能

The first ionisation energy, IE₁, is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. It is always endothermic because energy must be supplied to overcome the attraction between the negative electron and the positive nucleus.

第一电离能 IE₁ 是指从 1 mol 气态原子中移走 1 mol 电子以形成 1 mol 气态 1+ 离子所需的能量。该过程总是吸热的，因为必须提供能量来克服负电子与正原子核之间的吸引力。

For sodium:

对于钠：

Na(g) → Na⁺(g) + e⁻ IE₁ = +496 kJ mol⁻¹

If the cation has a +2 charge, you must include the second ionisation energy as well. Removing a second electron from a positive ion requires even more energy because the electron is being pulled away from a more positively charged species.

如果阳离子带 +2 电荷，还必须包括第二电离能。从正离子中再移走一个电子需要更多能量，因为此时电子受到的吸引来自带更多正电荷的物种。

For magnesium in MgO:

Mg(g) → Mg⁺(g) + e⁻ IE₁ = +738 kJ mol⁻¹

Mg⁺(g) → Mg²⁺(g) + e⁻ IE₂ = +1451 kJ mol⁻¹

All ionisation energies are positive, and they are always written in the direction of removing electrons in a Born-Haber cycle.

所有电离能都为正值。在玻恩-哈伯循环中，电离能总是写成失去电子的方向。

6. Electron Affinity | 电子亲和能

The first electron affinity, EA₁, is the enthalpy change when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous 1− ions. For many non-metals, especially halogens, the first electron affinity is negative because energy is released when the nucleus attracts the incoming electron.

第一电子亲和能 EA₁ 是指 1 mol 气态原子获得 1 mol 电子形成 1 mol 气态 1− 离子时的焓变。对于许多非金属，尤其是卤素，第一电子亲和能为负值，因为原子核吸引进入的电子时会释放能量。

For chlorine:

对于氯：

Cl(g) + e⁻ → Cl⁻(g) EA₁ = -349 kJ mol⁻¹

When an oxide ion O²⁻ is formed, a second electron must be added to O⁻. The second electron affinity is endothermic because the incoming electron is repelled by the already negative O⁻ ion, so energy must be supplied to force it in.

形成氧离子 O²⁻ 时，必须向 O⁻ 再加入一个电子。第二电子亲和能是吸热的，因为进入的电子受到已经带负电的 O⁻ 离子的排斥，必须提供能量才能使其进入。

O(g) + e⁻ → O⁻(g) EA₁ = -142 kJ mol⁻¹

O⁻(g) + e⁻ → O²⁻(g) EA₂ = +844 kJ mol⁻¹

Students often forget that the second electron affinity is positive. This is a favourite exam point, especially in Born-Haber cycles for oxides such as MgO or CaO.

学生经常忘记第二电子亲和能为正值。这是考试中特别常见的考点，尤其是在 MgO 或 CaO 等氧化物的玻恩-哈伯循环中。

7. Lattice Enthalpy | 晶格焓

Lattice enthalpy can be defined in two ways. Lattice formation enthalpy is the enthalpy change when one mole of an ionic lattice is formed from its gaseous ions. It is always negative because ionic bonds are formed and energy is released.

晶格焓有两种定义方式。晶格生成焓是指由气态离子生成 1 mol 离子晶格时的焓变。该过程总是负值，因为离子键形成时会释放能量。

For sodium chloride, lattice formation is represented as:

对于氯化钠，晶格生成可表示为：

Na⁺(g) + Cl⁻(g) → NaCl(s) ΔHlatt° = -788 kJ mol⁻¹

The opposite process, lattice dissociation, is endothermic. You must state clearly which definition you are using because the sign will be different. Cambridge questions often use lattice formation enthalpy, so a negative value is expected.

相反的过程是晶格解离，为吸热过程。你必须说明使用的是哪种定义，因为符号会不同。剑桥考试题常使用晶格生成焓，因此预期为负值。

The magnitude of lattice enthalpy indicates the strength of the ionic bonding. Small, highly charged ions produce a more negative lattice enthalpy because the electrostatic attraction is stronger.

晶格焓的大小反映离子键的强度。半径小、电荷高的离子会产生更负的晶格焓，因为静电吸引力更强。

8. Constructing a Born-Haber Cycle for NaCl | 构建 NaCl 的玻恩-哈伯循环

To construct the cycle for NaCl, start with the elements in their standard states, Na(s) and ½Cl₂(g). The direct route gives solid NaCl. The indirect route first atomises sodium and chlorine, then ionises sodium, then adds an electron to chlorine, and finally forms the lattice.

构建 NaCl 的循环时，从标准状态下的元素 Na(s) 和 ½Cl₂(g) 开始。直接路线生成固态 NaCl。间接路线首先使钠和氯原子化，然后使钠电离，再让氯获得电子，最后形成晶格。

The five steps are:

五个步骤如下：

Na(s) → Na(g) ΔHat°(Na) = +108 kJ mol⁻¹

½Cl₂(g) → Cl(g) ΔHat°(Cl) = +122 kJ mol⁻¹

Na(g) → Na⁺(g) + e⁻ IE₁ = +496 kJ mol⁻¹

Cl(g) + e⁻ → Cl⁻(g) EA₁ = -349 kJ mol⁻¹

Na⁺(g) + Cl⁻(g) → NaCl(s) ΔHlatt° = ?

Applying Hess’s law, the formation enthalpy equals the sum of the indirect route:

应用赫斯定律，生成焓等于间接路线各焓变之和：

ΔHf° = ΔHat°(Na) + IE₁(Na) + ΔHat°(Cl) + EA₁(Cl) + ΔHlatt°

Substituting the known values:

代入已知数值：

-411 = +108 + 496 + 122 – 349 + ΔHlatt°

Sum of the four measured steps:

四个可测量步骤之和为：

108 + 496 + 122 – 349 = +377 kJ mol⁻¹

Therefore:

因此：

ΔHlatt° = -411 – 377 = -788 kJ mol⁻¹

The

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Born-Haber Cycles | 玻恩-哈伯循环