📚 Enthalpy change of formation from enthalpy changes of combustion | 利用燃烧焓变计算生成焓变
In A-Level Chemistry, thermodynamic data tables often provide standard enthalpy changes of combustion (ΔcH°) for a wide range of substances. By combining these values with Hess's Law, we can calculate the standard enthalpy change of formation (ΔfH°) for a compound without performing a direct synthesis experiment. This technique is especially powerful because combustion reactions are experimentally straightforward and heavily tabulated.
在 A-Level 化学中，热力学数据表通常会提供大量物质的标准摩尔燃烧焓变 (ΔcH°)。结合这些数值与赫斯定律，我们可以计算一种化合物的标准摩尔生成焓变 (ΔfH°)，而不必进行直接的合成实验。这种方法特别有用，因为燃烧反应实验操作简单且已有大量表格数据。
1. Standard enthalpy of formation (ΔfH°) | 标准生成焓 (ΔfH°)
tes under standard conditions (298 K, 100 kPa). For example, the formation of liquid water is H₂(g) + ½O₂(g) → H₂O(l), with ΔfH° often quoted as −286 kJ mol⁻¹.
标准生成焓变，记为 ΔfH°，是指在标准条件 (298 K, 100 kPa) 下，由处于标准状态的元素生成一摩尔化合物时的焓变。例如，液态水的生成反应为 H₂(g) + ½O₂(g) → H₂O(l)，其 ΔfH° 通常为 −286 kJ mol⁻¹。
By definition, the ΔfH° of any element in its standard state is zero. This includes O₂(g), H₂(g), C(graphite), and S(s, rhombic). These zeros are essential reference points in the energy cycle.
根据定义，任何处于标准状态的元素其 ΔfH° 为零。这包括 O₂(g)、H₂(g)、C(石墨) 和 S(s, 斜方硫)。这些零值是能量循环中必不可少的参考点。
2. Standard enthalpy of combustion (ΔcH°) | 标准燃烧焓 (ΔcH°)
andard conditions. For carbon, the combustion is C(s) + O₂(g) → CO₂(g), and its ΔcH° is −394 kJ mol⁻¹. Notice that for many elements, the combustion product is an oxide, so the ΔcH° of an element often equals the ΔfH° of its most common oxide.
标准燃烧焓变，记为 ΔcH°，是指在标准条件下，一摩尔物质在过量氧气中完全燃烧时的焓变。对于碳，燃烧反应为 C(s) + O₂(g) → CO₂(g)，其 ΔcH° 是 −394 kJ mol⁻¹。注意，对于很多元素，燃烧产物是氧化物，因此元素的 ΔcH° 常等于其最常见氧化物的 ΔfH°。
ΔcH° values are always negative (exothermic) for fuels. Data books list ΔcH° for many organic compounds, alcohols, hydrocarbons and even elements. Oxygen itself does not have a meaningful combustion enthalpy, so its ΔcH° is taken as zero in calculations.
对于燃料来说，ΔcH° 总是负值 (放热)。数据手册中列出了许多有机化合物、醇、烃甚至元素的 ΔcH°。氧气本身没有有意义的燃烧焓，因此在计算中其 ΔcH° 按零处理。
3. Hess's Law and energy cycles | 赫斯定律与能量循环
Hess's Law states that the total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. We can design an alternative pathway: instead of forming the compound directly from its elements, we can burn the elements to a common set of combustion products, and then imagine the compound burning back to those same products. The enthalpy change around a closed cycle must sum to zero.
赫斯定律指出，只要始态和终态相同，反应的总焓变与途径无关。我们可以设计一条替代途径：不直接由元素生成化合物，而是将元素燃烧成一组共同的燃烧产物，再想象该化合物燃烧回到同样的产物。在一个闭合循环中，焓变之和必定为零。
The cycle is usually drawn with the elements at the top, the compound at the bottom left, and the common combustion products (CO₂, H₂O, etc.) at the bottom right. The direct route is ΔfH°, and the indirect route goes via the sum of the ΔcH° values of the elements, then subtracts the ΔcH° of the compound.
通常绘制循环时，元素放在最上方，化合物放在左下方，共同的燃烧产物 (CO₂, H₂O 等) 放在右下方。直接途径是 ΔfH°，间接途径则先经过各元素 ΔcH° 之和，再减去化合物的 ΔcH°。
4. Key relationship: ΔfH° from ΔcH° | 关键关系：由燃烧焓求生成焓
From the energy cycle we obtain the fundamental formula: for the formation reaction elements → compound,
ΔfH°(compound) = Σ ΔcH°(elements) – ΔcH°(compound)
由能量循环我们得出基本公式：对于生成反应 元素 → 化合物，
ΔfH°(化合物) = Σ ΔcH°(元素) – ΔcH°(化合物)
The summation Σ runs over all the elements in their standard states, multiplied by the appropriate stoichiometric coefficients as they appear in the balanced formation equation. If any element is oxygen, its ΔcH° term is zero.
求和符号 Σ 涵盖所有处于标准状态的元素，并按配平的生成方程中的化学计量数相乘。如果某一元素是氧气，其 ΔcH° 项为零。
y a rearrangement of the more general expression for any reaction: ΔH° = Σ ΔcH°(reactants) – ial case where the reactants are the elements and the product is the compound of interest.
这个关系式其实是对任意反应都成立的通用表达式 ΔH° = Σ ΔcH°(反应物) – Σ ΔcH°(产物) 的重新排列，只是将其应用于反应物为元素、产物为目标化合物这一特殊情形。
5. Constructing the energy cycle step by step | 逐步构建能量循环
Write the target formation equation, e.g. 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l).
写出目标生成方程，如 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)。
Draw a cycle: place the elements (with their coefficients) at the top. Below left, place the compound. Both the elements and the compound can be burnt to exactly the same set of stable oxides: CO₂(g) and H₂O(l).
绘制循环图：将元素 (连同化学计量数) 放在顶部。左下方放置化合物。元素和化合物都可以燃烧成完全相同的一组稳定氧化物：CO₂(g) 和 H₂O(l)。
m of the combustion enthalpies of the elements going down the right-hand path. This gives ΔfH° + ΔcH°(compound) = Σ ΔcH°(elements), which rearranges to the formula.
沿左侧路径下降的焓变 (ΔfH°) 加上化合物的燃烧焓变 (ΔcH°(化合物)) 必须等于沿着右侧路径下降的各元素燃烧焓变之和。由此可得 ΔfH° + ΔcH°(化合物) = Σ ΔcH°(元素)，重新整理即得公式。
6. Step-by-step calculation method | 逐步计算方法
Step 1: Write the balanced formation equation for one mole of the compound, with all elements in their standard states.
第一步：写出一摩尔化合物配平的生成方程式，所有元素均处于标准状态。
Step 2: bustion enthalpy for each element (ΔcH°) and for the compound. Remember that ΔcH°(O₂) = 0.
第二步：找出并列出每种元素和化合物的标准燃烧焓 (ΔcH°)。记住 ΔcH°(O₂) = 0。
Step 3: Multiply each element's ΔcH° by its stoichiometric coefficient in the formation equation, then sum them: Σ ΔcH°(elements).
第三步：将每种元素的 ΔcH° 乘以生成方程中该元素的化学计量数，然后求和：Σ ΔcH°(元素)。
Step 4: Subtract the ΔcH° of the compound from the sum: ΔfH°(compound) = Σ ΔcH°(elements) – mbustion enthalpies are negative, and subtracting a negative number adds a positive value.
第四步：用求得的和减去化合物的 ΔcH°： ΔfH°(化合物) = Σ ΔcH°(元素) – ΔcH°(化合物)。务必注意符号；燃烧焓都是负值，减去一个负数相当于加上一个正值。
7. Worked Example 1: Formation of ethanol | 实例一：乙醇的生成
Formation equation: 2C(graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)
生成方程：2C(石墨) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)
Given data:
| Substance | ΔcH° / kJ mol⁻¹ |
|---|---|
| C(graphite) | −394 |
| H₂(g) | −286 |
| C₂H₅OH(l) | −1367 |
已知数据：
| 物质 | ΔcH° / kJ mol⁻¹ |
|---|---|
| C(石墨) | −394 |
| H₂(g) | −286 |
| C₂H₅OH(l) | −1367 |
Σ ΔcH°(elements) = [2 × (−394)] + [3 × (−286)] + [½ × 0] = −788 + (−858) = −1646 kJ mol⁻¹
Σ ΔcH°(元素) = [2 × (−394)] + [3 × (−286)] + [½ × 0] = −788 + (−858) = −1646 kJ mol⁻¹
ΔfH°(C₂H₅OH) = −1646 − (−1367) = −1646 + 1367 = −279 kJ mol⁻¹
The negative value indicates an exothermic formation from its elements, consistent with the fact that ethanol is a stable compound.
负值表明由元素生成乙醇是放热的，这与乙醇是稳定化合物的事实相符。
8. Worked Example 2: Formation of butanoic acid | 实例二：丁酸的生成
O₂. Its formation equation is: 4C(s) + 4H₂(g) + O₂(g) → C₃H₇COOH(l) (equivalently C₄H₈O₂).
丁酸的分子式为 C₄H₈O₂。其生成方程为：4C(s) + 4H₂(g) + O₂(g) → C₃H₇COOH(l) (等同于 C₄H₈O₂)。
) = −394 kJ mol⁻¹, H₂(g) = −286 kJ mol⁻¹, C₄H₈O₂(l) = −2184 kJ mol⁻¹. (Again ΔcH°(O₂) = 0)
给定 ΔcH° 值：C(s) = −394 kJ mol⁻¹，H₂(g) = −286 kJ mol⁻¹，C₄H₈O₂(l) = −2184 kJ mol⁻¹。(同样 ΔcH°(O₂) = 0)
Σ ΔcH°(elements) = [4 × (−394)] + [4 × (−286)] = −1576 − 1144 = −2720 kJ mol⁻¹
ΔfH°(butanoic acid) = −2720 − (−2184) = −2720 + 2184 = −536 kJ mol⁻¹
The large negative ΔfH° reflects the stability of the carboxylic acid relative to its elements.
绝对值较大的负 ΔfH° 反映了该羧酸相对于其元素的稳定性。
9. Handling different physical states | 处理不同物态
Combustion enthalpies depend on the physical states of both reactants and products. For instance, the ΔcH° of H₂(g) is −286 kJ mol⁻¹ when the product is H₂O(l), but it would be different if the product were H₂O(g) because the enthalpy of vaporisation would be involved. Always ensure the combustion products in your cycle match those used in the given data (usually CO₂(g) and H₂O(l)).
燃烧焓依赖于反应物和产物的物态。例如，当产物为 H₂O(l) 时，H₂(g) 的 ΔcH° 是 −286 kJ mol⁻¹，但如果产物是 H₂O(g) 则数值不同，因为涉及蒸发热。一定要确保循环中使用的燃烧产物与给定数据中使用的产物一致 (通常为 CO₂(g) 和 H₂O(l))。
If an element exists in allotropes, the standard state must be used, e.g. C(graphite) not diamond. The ΔcH° of graphite is −394 kJ mol⁻¹, whereas diamond would give a slightly different value.
如果某元素有同素异形体，必须使用标准状态，例如 C(石墨) 而不是金刚石。石墨的 ΔcH° 是 −394 kJ mol⁻¹，而金刚石会略有不同。
10. Common pitfalls and sign management | 常见误区与符号处理
Many students forget that ΔcH° values are negative. When plugging into the formula, always insert the sign. Subtracting a negative ΔcH°(compound) is equivalent to adding its magnitude. Double-check that your final ΔfH° has a reasonable sign and magnitude – most stable compounds have negative formation enthalpies.
很多学生忘记了 ΔcH° 是负值。在代入公式时，一定要带入符号。减去一个负的 ΔcH°(化合物) 等于加上其绝对值。务必再次检查最终 ΔfH° 的符号和数量级是否合理——大多数稳定化合物的生成焓为负值。
Never use ΔcH° for oxygen; its value is taken as zero. Also, ensure the formation equation is balanced for exactly one mole of the product. If your compound contains nitrogen, sulfur or halogens, the combustion products are different (N₂, SO₂, HX) and the cycle must be adapted accordingly.
绝不要使用氧气的 ΔcH°；其值按零计算。同时要确保生成方程针对恰好一摩尔产物配平。如果化合物含有氮、硫或卤素，燃烧产物会有所不同 (N₂、SO₂、HX)，能量循环也需相应调整。
11. Summary and the master formula | 总结与主公式
The relationship between ΔfH° and ΔcH° is one of the most practical applications of Hess's Law in thermochemistry. The master equation can be confidently applied to any C–H–O compound with known combustion data.
ΔfH° 与 ΔcH° 之间的关系是赫斯定律在热化学中最实用的应用之一。对于任何已知燃烧数据的 C–H–O 化合物，都可以放心地应用主公式。
ΔfH°(compound) = Σ ΔcH°(elements in standard states) – ΔcH°(compound)
Memorising this equation, and understanding the energy cycle behind it, will allow you to tackle a wide range of exam questions on indirect enthalpy determinations with confidence.
记下这个方程并理解其背后的能量循环，将使你能够自信地应对各类关于间接测定焓变的考试题目。
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