📚 A-Level Chemistry: Calculating Enthalpy Change of Hydration of Anhydrous Salts | A-Level 化学：无水盐水合焓变的计算
Many students find enthalpy of hydration confusing because it appears as a number in Born-Haber cycles and solubility calculations. In this revision guide, we break down the definition, the sign convention and the step-by-step method for calculating the hydration enthalpy of anhydrous salts.
许多同学觉得水合焓变难以理解，因为它既出现在玻恩-哈伯循环中，又出现在溶解度的计算里。本复习指南将拆解定义、符号约定以及无水盐水合焓变的逐步计算方法。
1. What is Enthalpy of Hydration? | 什么是水合焓变？
Enthalpy of hydration is the enthalpy change when one mole of gaseous ions is dissolved in an unlimited amount of water to form an infinitely dilute solution. It is usually represented as ΔH(hyd).
水合焓变是指 1 摩尔气态离子在过量水中溶解，形成无限稀溶液时的焓变，通常用 ΔH(hyd) 表示。
Because ion–dipole attractions between the ion and water molecules release energy, hydration enthalpy is always exothermic, meaning it has a negative sign for positive or negative ions.
由于离子与水分子的离子-偶极作用会释放能量，水合焓变总是放热过程，因此对阳离子或阴离子而言均为负值。
2. Lattice Enthalpy and Enthalpy of Solution | 晶格焓与溶解焓变
The lattice dissociation enthalpy is the endothermic enthalpy change for separating one mole of solid ionic compound into gaseous ions. For example, NaCl(s) → Na⁺(g) + Cl⁻(g) has a positive ΔH.
晶格离解焓是指将 1 摩尔离子化合物固体分离成气态离子时吸收的热量。例如 NaCl(s) → Na⁺(g) + Cl⁻(g) 对应的 ΔH 为正值。
The enthalpy of solution of an anhydrous salt is the enthalpy change when one mole of the solid dissolves in water to form an infinitely dilute solution. An anhydrous salt contains no water of crystallisation, such as CuSO₄ rather than CuSO₄·5H₂O.
无水盐的溶解焓变是 1 摩尔该固体溶解于水形成无限稀溶液时的焓变。无水盐不含结晶水，例如 CuSO₄，而不是 CuSO₄·5H₂O。
3. The Key Relationship | 关键关系式
For an anhydrous ionic salt, the enthalpy change of solution can be imagined in two steps: first break the ionic lattice into gaseous ions, then hydrate those gaseous ions. This is an application of Hess’s Law.
对于无水离子盐，溶解焓变可以想象为两步：先将离子晶格拆散成气态离子，再将这些气态离子水合。这是赫斯定律的应用。
ΔH(sol) = ΔH(lattice dissociation) + ΔH(hydration of cation) + ΔH(hydration of anion)
The lattice dissociation term is positive because energy is absorbed to break attractions. The hydration terms are negative because ion–water attractions release energy.
晶格离解项为正，因为断键需要吸收能量；水合项为负，因为离子与水的吸引会释放能量。
4. Worked Example: Sodium Chloride | 实例计算：氯化钠
Use the following data to calculate the enthalpy change of solution of anhydrous NaCl.
利用以下数据计算无水 NaCl 的溶解焓变。
- Lattice dissociation enthalpy of NaCl = +787 kJ mol⁻¹ | NaCl 的晶格离解焓 = +787 kJ mol⁻¹
- ΔH(hyd) of Na⁺ = -406 kJ mol⁻¹ | Na⁺ 的水合焓 = -406 kJ mol⁻¹
- ΔH(hyd) of Cl⁻ = -364 kJ mol⁻¹ | Cl⁻ 的水合焓 = -364 kJ mol⁻¹
ΔH(sol) = (+787) + (-406) + (-364) = +17 kJ mol⁻¹
The value is small and positive, so dissolving anhydrous NaCl is slightly endothermic. In practice, the entropy increase of the system makes the process favourable.
计算值较小且为正，说明无水 NaCl 溶解过程轻微吸热。实际上，体系的熵增使该过程能够自发进行。
5. Worked Example: Magnesium Chloride | 实例计算：氯化镁
For MgCl₂, remember that the formula contains two chloride ions, so the hydration enthalpy of Cl⁻ must be multiplied by 2.
对于 MgCl₂，要注意化学式中含有两个氯离子，因此 Cl⁻ 的水合焓必须乘以 2。
Data: ΔH(lattice dissociation) = +2493 kJ mol⁻¹; ΔH(hyd) of Mg²⁺ = -1920 kJ mol⁻¹; ΔH(hyd) of Cl⁻ = -364 kJ mol⁻¹.
数据：ΔH(晶格离解) = +2493 kJ mol⁻¹；Mg²⁺ 的水合焓 = -1920 kJ mol⁻¹；Cl⁻ 的水合焓 = -364 kJ mol⁻¹。
ΔH(sol) = (+2493) + (-1920) + 2(-364) = -155 kJ mol⁻¹
The overall enthalpy change is negative, so the dissolving of anhydrous MgCl₂ is exothermic.
总焓变为负，说明无水 MgCl₂ 溶解是放热过程。
6. Effect of Ionic Charge and Radius | 离子电荷与半径的影响
Hydration enthalpy becomes more exothermic as the charge density of an ion increases. Higher charge and smaller radius both strengthen the ion–dipole attractions between the ion and water molecules.
离子电荷密度越大，水合焓变越负。电荷越高、半径越小，离子与水分子之间的离子-偶极作用越强。
For example, Mg²⁺ has a higher charge and smaller radius than Na⁺, so its hydration enthalpy is far more negative. Al³⁺ is even more exothermic because of its very high charge density.
例如，Mg²⁺ 比 Na⁺ 电荷更高、半径更小，因此其水合焓变显著更负。Al³⁺ 由于电荷密度极高，水合焓变更加放热。
7. Comparing Hydration Enthalpies | 比较水合焓变
When comparing ions in exam questions, always explain using charge density, which is charge divided by ionic radius. A similar argument is used when explaining trends in lattice enthalpy.
在考试中比较离子时，务必使用电荷密度（电荷除以离子半径）来解释。这种论证方式与解释晶格焓变化趋势时一致。
| Ion | 离子 | Charge density | 电荷密度 | ΔH(hyd) / kJ mol⁻¹ |
| Na⁺ | low | 较低 | -406 |
| Mg²⁺ | medium | 中等 | -1920 |
| Al³⁺ | very high | 极高 | -4665 |
| Cl⁻ | low | 较低 | -364 |
For ions with the same charge, the smaller ion will have the more exothermic hydration enthalpy. Therefore, Li⁺ is more exothermic than K⁺ because Li⁺ has a smaller ionic radius.
对于电荷相同的离子，半径较小的离子水合焓更负。因此，Li⁺ 的水合焓比 K⁺ 更放热，因为 Li⁺ 的离子半径更小。
8. Calculating an Unknown Ion Hydration Enthalpy | 计算未知离子的水合焓
The key relationship can be rearranged to find any missing value. If the lattice dissociation enthalpy, the enthalpy of solution, and one ion’s hydration enthalpy are known, the other ion’s hydration enthalpy can be calculated.
关键关系式可以变形，从而求出任意未知量。如果已知晶格离解焓、溶解焓和其中一种离子的水合焓，就可以求出另一种离子的水合焓。
Using the NaCl example: ΔH(sol) = +17 kJ mol⁻¹, ΔH(lattice dissociation) = +787 kJ mol⁻¹, and ΔH(hyd) of Na⁺ = -406 kJ mol⁻¹.
以 NaCl 为例：ΔH(溶解) = +17 kJ mol⁻¹，ΔH(晶格离解) = +787 kJ mol⁻¹，Na⁺ 的水合焓 = -406 kJ mol⁻¹。
ΔH(hyd, Cl⁻) = (+17) – (+787) – (-406) = -364 kJ mol⁻¹
Always keep the signs carefully arranged. Subtracting a negative sign is equivalent to adding the value.
计算时务必小心整理符号。减去一个负数等价于加上该数值。
9. Sign Conventions and Born-Haber Cycles | 符号约定与玻恩-哈伯循环
In a Born-Haber cycle, lattice enthalpy may be given as lattice formation enthalpy, which is negative, or as lattice dissociation enthalpy, which is positive. You must check which convention the question uses.
在玻恩-哈伯循环中，晶格焓可能以晶格生成焓（负值）给出，也可能以晶格离解焓（正值）给出。考生必须首先判断题目采用哪种约定。
If the lattice formation enthalpy is used, substitute +ΔH(lattice dissociation) with -ΔH(lattice formation) in the solution relationship.
如果题目给出晶格生成焓，就将上述关系中的 +ΔH(晶格离解) 替换为 -ΔH(晶格生成)。
ΔH(sol) = -ΔH(lattice formation) + ΔH(hyd cation) + ΔH(hyd anion)
This mistake costs many marks in A-Level exams, so always label the energy cycle clearly.
这一符号错误在 A-Level 考试中非常常见，因此在画能量循环图时必须明确标注。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
-
Forgetting to multiply the hydration enthalpy by the number of ions in the formula. For example, MgCl₂ contains two Cl⁻ ions, so use 2 × ΔH(hyd, Cl⁻).
忘记将水合焓乘以化学式中的离子数目。例如 MgCl₂ 含两个 Cl⁻，应使用 2 × ΔH(hyd, Cl⁻)。
-
Using the wrong sign for lattice enthalpy. The lattice dissociation enthalpy is positive, while lattice formation enthalpy is negative.
晶格焓的符号使用错误。晶格离解焓为正，晶格生成焓为负。
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Confusing hydrated salts with anhydrous salts. Hydrated salts require additional steps for removing water of crystallisation.
混淆水合盐与无水盐。水合盐溶解需要额外考虑脱去结晶水的步骤。
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Omitting state symbols. Always write (s), (g) and (aq) in thermochemical equations.
漏写状态符号。热化学方程式中务必写清 (s)、(g) 和 (aq)。
11. Practice Question | 练习
The lattice dissociation enthalpy of anhydrous CaCl₂ is +2258 kJ mol⁻¹. The hydration enthalpies of Ca²⁺ and Cl⁻ are -1650 kJ mol⁻¹ and -364 kJ mol⁻¹ respectively. Calculate the enthalpy change of solution of CaCl₂, and state whether the process is exothermic or endothermic.
已知无水 CaCl₂ 的晶格离解焓为 +2258 kJ mol⁻¹，Ca²⁺ 和 Cl⁻ 的水合焓分别为 -1650 kJ mol⁻¹ 和 -364 kJ mol⁻¹。计算 CaCl₂ 的溶解焓变，并判断该过程是放热还是吸热。
ΔH(sol) = (+2258) + (-1650) + 2(-364) = -120 kJ mol⁻¹
Since the overall value is negative, dissolving anhydrous CaCl₂ is exothermic.
因为总值为负，所以无水 CaCl₂ 的溶解是放热过程。
12. Summary | 总结
For anhydrous salts, the enthalpy change of solution is the sum of the lattice dissociation enthalpy and the hydration enthalpies of the gaseous ions. Hydration enthalpy is exothermic and increases in magnitude with ionic charge density.
对于无水盐，溶解焓变等于晶格离解焓与气态离子水合焓之和。水合焓为放热项，且随离子电荷密度增大而变得更负。
Always check the sign convention, multiply by the correct stoichiometric number, and use the state symbols carefully in Born-Haber cycles.
做题时始终检查符号约定，乘以正确的化学计量数，并在玻恩-哈伯循环中仔细标注状态符号。
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