📚 Enthalpy Change of Reaction from Enthalpy Changes of Formation | 由生成焓变计算反应焓变
In A-Level Chemistry, many important reactions cannot be measured directly in a calorimeter. We therefore use tabulated standard enthalpy changes of formation and apply Hess’s law to calculate an unknown reaction enthalpy. This article develops the master equation ΔH°r = Σ ΔH°f(products) − Σ ΔH°f(reactants) and shows worked examples relevant to the Cambridge specification.
在 A-Level 化学中，许多重要反应无法用量热计直接测量。因此我们使用表列的标准生成焓变，并应用赫斯定律来计算未知的反应焓变。本文推导主公式 ΔH°r = Σ ΔH°f(生成物) − Σ ΔH°f(反应物)，并展示与剑桥考试大纲相关的例题。
1. Introduction to Enthalpy Changes | 焓变简介
The enthalpy change of a reaction, ΔH°r, is the heat absorbed or released when a reaction occurs at constant pressure. A negative value means the reaction is exothermic; a positive value means it is endothermic. Direct measurement is not always possible, so chemists calculate ΔH°r from standard enthalpy changes of formation.
反应的焓变 ΔH°r 是恒压条件下反应发生时吸收或释放的热量。负值表示反应放热，正值表示反应吸热。由于直接测量并不总是可行，化学家常通过标准生成焓变来计算 ΔH°r。
2. Standard Enthalpy of Formation Defined | 标准生成焓变的定义
The standard enthalpy change of formation, ΔH°f, is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions: 298 K, 1 atm or 100 kPa, and 1 mol dm⁻³ for solutions. For example, the formation equation for liquid water is H₂(g) + ½O₂(g) → H₂O(l).
标准摩尔生成焓变 ΔH°f 是指在标准条件下（298 K、1 atm 或 100 kPa、溶液浓度 1 mol dm⁻³），由最稳定单质生成 1 mol 化合物时的焓变。例如液态水的生成方程式为 H₂(g) + ½O₂(g) → H₂O(l)。
3. Key Conventions and Standard States | 关键规定与标准状态
Standard state symbols must be memorised. Carbon is graphite, oxygen is O₂(g), hydrogen is H₂(g), nitrogen is N₂(g), and bromine is Br₂(l). A correct formation equation always produces exactly one mole of the target compound, even if fractional coefficients are used.
必须记住标准状态符号。碳为石墨，氧为 O₂(g)，氢为 H₂(g)，氮为 N₂(g)，溴为 Br₂(l)。正确的生成方程式总是恰好生成 1 mol 目标化合物，即使使用分数计量数。
The table below shows common elements and their standard states with ΔH°f = 0:
下表列出常见元素及其 ΔH°f = 0 的标准状态：
| Element | Standard state | ΔH°f / kJ mol⁻¹ |
|---|---|---|
| Carbon | C(s, graphite) | 0 |
| Hydrogen | H₂(g) | 0 |
| Oxygen | O₂(g) | 0 |
| Bromine | Br₂(l) | 0 |
| Iodine | I₂(s) | 0 |
4. Why Elements Have Zero Formation Enthalpy | 为什么单质的生成焓为零
By definition, an element in its standard state has ΔH°f = 0 because forming an element from itself involves no chemical change. The zero is a reference point, not an absence of energy. For example, ΔH°f[O₂(g)] = 0, but ΔH°f[O₃(g)] is not zero because ozone is not the standard state of oxygen.
根据定义，标准状态下的单质 ΔH°f = 0，因为单质由自身生成不涉及化学变化。零值是参考点，不代表没有能量。例如 ΔH°f[O₂(g)] = 0，但 ΔH°f[O₃(g)] 不为零，因为臭氧不是氧的标准状态。
5. 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 reaction is independent of the route taken. We can imagine breaking all reactants into their elements in standard states, then recombining those elements to form products. The sum of these two hypothetical steps equals the direct reaction enthalpy.
赫斯定律指出，反应的总焓变与所采取的路径无关。我们可以设想将所有反应物分解为标准状态下的单质，再将这些单质重新组合成产物。这两个假设步骤的焓变之和等于直接反应的焓变。
6. Deriving the Master Equation | 推导主公式
For a general reaction aA + bB → cC + dD, the standard enthalpy change is calculated using:
对于一般反应 aA + bB → cC + dD，标准焓变通过下式计算：
ΔH°r = Σ ΔH°f(products) − Σ ΔH°f(reactants)
More explicitly, ΔH°r = [c ΔH°f(C) + d ΔH°f(D)] − [a ΔH°f(A) + b ΔH°f(B)]. Each formation enthalpy must be multiplied by its stoichiometric coefficient. If a coefficient is fractional, such as ½, use the fraction directly.
更明确地，ΔH°r = [c ΔH°f(C) + d ΔH°f(D)] − [a ΔH°f(A) + b ΔH°f(B)]。每个生成焓必须乘以其化学计量数。如果计量数为分数（如 ½），直接使用该分数。
7. Worked Example 1: Combustion of Methane | 例题 1：甲烷的燃烧
Calculate ΔH°r for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Given: ΔH°f[CH₄(g)] = −74.8 kJ mol⁻¹, ΔH°f[CO₂(g)] = −393.5 kJ mol⁻¹, ΔH°f[H₂O(l)] = −285.8 kJ mol⁻¹, ΔH°f[O₂(g)] = 0.
计算 CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) 的 ΔH°r。已知：ΔH°f[CH₄(g)] = −74.8 kJ mol⁻¹，ΔH°f[CO₂(g)] = −393.5 kJ mol⁻¹，ΔH°f[H₂O(l)] = −285.8 kJ mol⁻¹，ΔH°f[O₂(g)] = 0。
Products: [−393.5 + 2 × (−285.8)] = −965.1 kJ. Reactants: [−74.8 + 2 × 0] = −74.8 kJ. Therefore ΔH°r = −965.1 − (−74.8) = −890.3 kJ mol⁻¹. The negative sign confirms that methane combustion is strongly exothermic.
生成物：[−393.5 + 2 × (−285.8)] = −965.1 kJ。反应物：[−74.8 + 2 × 0] = −74.8 kJ。因此 ΔH°r = −965.1 − (−74.8) = −890.3 kJ mol⁻¹。负号证实甲烷燃烧是强放热反应。
8. Worked Example 2: Thermal Decomposition of Limestone | 例题 2：石灰石的热分解
For CaCO₃(s) → CaO(s) + CO₂(g), use ΔH°f[CaCO₃(s)] = −1206.9 kJ mol⁻¹, ΔH°f[CaO(s)] = −635.1 kJ mol⁻¹, ΔH°f[CO₂(g)] = −393.5 kJ mol⁻¹.
对于 CaCO₃(s) → CaO(s) + CO₂(g)，使用 ΔH°f[CaCO₃(s)] = −1206.9 kJ mol⁻¹、ΔH°f[CaO(s)] = −635.1 kJ mol⁻¹、ΔH°f[CO₂(g)] = −393.5 kJ mol⁻¹。
Products: [−635.1 + (−393.5)] = −1028.6 kJ. Reactants: [−1206.9] = −1206.9 kJ. Thus ΔH°r = −1028.6 − (−1206.9) = +178.3 kJ mol⁻¹. The positive value shows the decomposition is endothermic and requires sustained heating.
生成物：[−635.1 + (−393.5)] = −1028.6 kJ。反应物：[−1206.9] = −1206.9 kJ。因此 ΔH°r = −1028.6 − (−1206.9) = +178.3 kJ mol⁻¹。正值表明分解是吸热反应，需要持续加热。
9. Common Errors and Exam Tips | 常见错误与应试技巧
Students often forget to multiply each ΔH°f by its stoichiometric coefficient, or they incorrectly assign a non-zero ΔH°f to an element in a compound
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