Bond Enthalpies and Hess’s Law | A-Level Chemistry | 键焓与赫斯定律

What Are Bond Enthalpies? | 什么是键焓？

Bond enthalpy (also called bond energy or bond dissociation energy) is the amount of energy required to break one mole of a specific covalent bond in the gaseous state, averaged over a range of different compounds.

键焓（也称为键能或键解离能）是指在气态下断裂一摩尔特定共价键所需的能量，该数值是在一系列不同化合物中取的平均值。

The key definition points you must remember for A-Level exams:

• It applies to gaseous molecules only — all species must be in the gas phase.
• It is an endothermic process — breaking bonds requires energy input (ΔH is positive).
• It is a mean (average) value — the same bond type (e.g., C–H) has slightly different strengths in different molecules (CH₄ vs. C₂H₆), so an average is used.

A-Level 考试中必须记住的定义要点：

• 仅适用于气态分子——所有物质必须处于气相。
• 这是一个吸热过程——断裂化学键需要吸收能量（ΔH 为正值）。
• 这是平均（均值）值——相同类型的键（如 C–H）在不同分子中（CH₄ vs. C₂H₆）的强度略有不同，因此使用平均值。

Key Equation | 关键公式
ΔH_reaction = Σ (bond enthalpies of bonds broken) − Σ (bond enthalpies of bonds formed)

ΔH_反应 = Σ (断裂键的键焓之和) − Σ (形成键的键焓之和)

Why Use Mean Bond Enthalpies? | 为什么使用平均键焓？

In reality, the energy required to break a particular bond depends on its chemical environment. For example, the C–H bond in methane (CH₄) has a bond enthalpy of approximately 439 kJ mol⁻¹, while the C–H bond in ethane (C₂H₆) is about 423 kJ mol⁻¹. Rather than memorising hundreds of slightly different values, chemists use mean bond enthalpies — the average across many compounds — which are tabulated in data books.

实际上，断裂特定键所需的能量取决于其化学环境。例如，甲烷 (CH₄) 中 C–H 键的键焓约为 439 kJ mol⁻¹，而乙烷 (C₂H₆) 中 C–H 键的键焓约为 423 kJ mol⁻¹。化学家不记忆数百个略有不同的值，而是使用平均键焓——跨多种化合物的平均值——这些数值在数据手册中以表格形式列出。

This means calculations using mean bond enthalpies are less accurate than those using experimental enthalpy changes of formation or combustion. However, they are very useful when experimental data is not available, and they give a good estimate of the enthalpy change for a reaction.

这意味着使用平均键焓的计算不如使用实验生成焓或燃烧焓的计算准确。然而，当实验数据不可用时，它们非常有用，并且能够给出反应焓变的良好估计值。

Bond	Mean Bond Enthalpy (kJ mol⁻¹)	键
C–H	413	C–H
C–C	347	C–C
C=C	612	C=C
C≡C	838	C≡C
C–O	358	C–O
C=O	805	C=O
O–H	464	O–H
O=O	498	O=O
H–H	436	H–H
H–Cl	431	H–Cl
Cl–Cl	243	Cl–Cl
C–Cl	346	C–Cl
N≡N	945	N≡N

Calculating ΔH Using Bond Enthalpies | 使用键焓计算 ΔH

The method is straightforward and always follows the same three-step process:

Step 1: Draw Out All Bonds | 第一步：画出所有化学键

Write the balanced equation and draw the displayed formula (structural formula) for every reactant and product. Identify every single covalent bond present in each molecule.

写出配平方程式，并为每个反应物和产物画出结构式（展示式）。识别每个分子中存在的每一个共价键。

Step 2: Count Bonds Broken and Formed | 第二步：统计断裂和形成的键

List every bond that is broken in the reactants and every bond that is formed in the products. Remember: bonds in reactants are broken (endothermic, positive), bonds in products are formed (exothermic, negative).

列出反应物中断裂的每一个键，以及产物中形成的每一个键。记住：反应物中的键被断裂（吸热，正值），产物中的键被形成（放热，负值）。

Step 3: Apply the Formula | 第三步：应用公式

ΔH = Σ (bond enthalpies of bonds broken) − Σ (bond enthalpies of bonds formed)

Worked Example: Combustion of Methane | 例题：甲烷的燃烧

Reaction | 反应: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Bonds Broken (Reactants) | 断裂的键（反应物）

In CH₄: four C–H bonds → 4 × 413 = 1652 kJ mol⁻¹
In 2O₂: two O=O bonds → 2 × 498 = 996 kJ mol⁻¹
Total bonds broken = 1652 + 996 = 2648 kJ mol⁻¹

Bonds Formed (Products) | 形成的键（产物）

In CO₂: two C=O bonds → 2 × 805 = 1610 kJ mol⁻¹
In 2H₂O: four O–H bonds → 4 × 464 = 1856 kJ mol⁻¹
Total bonds formed = 1610 + 1856 = 3466 kJ mol⁻¹

Calculation | 计算

ΔH = 2648 − 3466 = −818 kJ mol⁻¹

The negative sign tells us the reaction is exothermic — more energy is released forming bonds than is absorbed breaking them. The experimental value for the combustion of methane is −890 kJ mol⁻¹. Our calculated value (−818 kJ mol⁻¹) is close but not exact, which illustrates the limitation of using mean bond enthalpies — they are averages, not exact values for this specific reaction.

负号告诉我们该反应是放热的——形成键释放的能量大于断裂键吸收的能量。甲烷燃烧的实验值为 −890 kJ mol⁻¹。我们的计算值（−818 kJ mol⁻¹）接近但不精确，这说明了使用平均键焓的局限性——它们是平均值，不是该特定反应的精确值。

Hess’s Law: The Principle of Energy Conservation | 赫斯定律：能量守恒原理

Hess’s Law states that the total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same. In simpler terms: if you can go from reactants to products by two different pathways, the total enthalpy change is the same for both routes.

赫斯定律指出，化学反应的总焓变与所采取的路径无关，只要初始条件和最终条件相同。简单来说：如果你可以通过两条不同的路径从反应物到产物，两条路径的总焓变是相同的。

Hess’s Law | 赫斯定律:
ΔH_{route 1} = ΔH_{route 2} = ΔH_{route 3}

This is a direct consequence of the First Law of Thermodynamics (conservation of energy). Enthalpy is a state function — it depends only on the current state of the system, not on how the system got there.

这是热力学第一定律（能量守恒）的直接结果。焓是一个状态函数——它只取决于系统的当前状态，而不取决于系统如何到达该状态。

Constructing Hess’s Law Cycles | 构建赫斯定律循环

There are three main types of enthalpy cycles you need to master for A-Level Chemistry:

1. Enthalpy of Formation Cycles | 生成焓循环

Uses the standard enthalpies of formation (ΔH_f°) of compounds to calculate an unknown enthalpy change. The cycle goes from elements → reactants → products, or elements → reactants and elements → products as two alternative routes.

使用化合物的标准生成焓 (ΔH_f°) 来计算未知的焓变。循环路径为：元素 → 反应物 → 产物，或者元素 → 反应物和元素 → 产物作为两条替代路径。

Formula | 公式:
ΔH_reaction = Σ ΔH_f°(products) − Σ ΔH_f°(reactants)

2. Enthalpy of Combustion Cycles | 燃烧焓循环

Uses the standard enthalpies of combustion (ΔH_c°) to calculate an unknown enthalpy change. The cycle goes from reactants → combustion products and products → combustion products. This is particularly useful for organic compounds.

使用标准燃烧焓 (ΔH_c°) 来计算未知的焓变。循环路径为：反应物 → 燃烧产物和产物 → 燃烧产物。这对有机化合物特别有用。

Formula | 公式:
ΔH_reaction = Σ ΔH_c°(reactants) − Σ ΔH_c°(products)

Note the order: reactants MINUS products for combustion cycles — this is the opposite of formation cycles!

注意顺序：燃烧循环是反应物减去产物——这与生成循环相反！

3. Bond Enthalpy Cycles | 键焓循环

Uses mean bond enthalpies, as covered in detail above. The cycle goes from reactants → gaseous atoms → products.

使用平均键焓，如上文所述。循环路径为：反应物 → 气态原子 → 产物。

Worked Example: Hess’s Law with Enthalpy of Formation | 例题：使用生成焓的赫斯定律

Problem | 问题: Calculate ΔH for the reaction: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

Data | 数据:

• ΔH_f° of Fe₂O₃(s) = −824 kJ mol⁻¹
• ΔH_f° of CO(g) = −111 kJ mol⁻¹
• ΔH_f° of CO₂(g) = −394 kJ mol⁻¹
• ΔH_f° of Fe(s) = 0 kJ mol⁻¹ (element in standard state)

Solution | 解答

ΔH = Σ ΔH_f°(products) − Σ ΔH_f°(reactants)

Products:
2Fe(s): 2 × 0 = 0 kJ mol⁻¹
3CO₂(g): 3 × (−394) = −1182 kJ mol⁻¹
Total products = −1182 kJ mol⁻¹

Reactants:
Fe₂O₃(s): 1 × (−824) = −824 kJ mol⁻¹
3CO(g): 3 × (−111) = −333 kJ mol⁻¹
Total reactants = −1157 kJ mol⁻¹

ΔH = (−1182) − (−1157) = −25 kJ mol⁻¹

The reaction is slightly exothermic. This is the reaction that occurs in a blast furnace during the extraction of iron — the negative ΔH means the reaction is thermodynamically favourable and helps maintain the high temperatures needed in the furnace.

该反应略微放热。这是高炉炼铁过程中发生的反应——负的 ΔH 意味着该反应在热力学上是有利的，有助于维持高炉中所需的高温。

Common Exam Mistakes to Avoid | 常见考试错误

Mistake 1: Forgetting State Symbols | 错误 1：忘记状态符号

Bond enthalpy calculations only work for gases. If a reactant or product is a liquid or solid, you must account for the enthalpy of vaporisation or sublimation. Many A-Level questions specify “(g)” for all species to avoid this complication.

键焓计算仅适用于气体。如果反应物或产物是液体或固体，你必须考虑汽化焓或升华焓。许多 A-Level 题目将所有物质标记为 “(g)” 以避免这种复杂情况。

Mistake 2: Confusing Breaking vs. Forming | 错误 2：混淆断裂与形成

Bonds broken = ENDOTHERMIC (+), bonds formed = EXOTHERMIC (−). The formula is broken MINUS formed. Many students get the signs wrong. Double-check: if more energy is released forming bonds than is needed to break bonds, the reaction must be exothermic (negative ΔH).

断裂键 = 吸热 (+)，形成键 = 放热 (−)。公式是断裂减去形成。许多学生把符号搞错了。再次确认：如果形成键释放的能量大于断裂键所需的能量，反应一定是放热的（负 ΔH）。

Mistake 3: Forgetting the Coefficient | 错误 3：忘记化学计量系数

When counting bonds, multiply by the coefficient from the balanced equation. For 3O₂, that is three O=O bonds, not one. This is the single most common error in bond enthalpy calculations.

计算键的数量时，要乘以配平方程式中的系数。对于 3O₂，那是三个 O=O 键，而不是一个。这是键焓计算中最常见的错误。

Mistake 4: Reversing the Combustion Formula | 错误 4：燃烧公式顺序颠倒

ΔH = Σ ΔH_c°(reactants) − Σ ΔH_c°(products). This is reactants minus products, not products minus reactants. Many students blindly apply the formation formula and lose marks.

ΔH = Σ ΔH_c°(反应物) − Σ ΔH_c°(产物)。这是反应物减去产物，而不是产物减去反应物。许多学生盲目套用生成公式而丢分。

Mistake 5: Ignoring the Direction of the Arrow | 错误 5：忽略箭头方向

In a Hess cycle diagram, if you go against the direction of an arrow, you must reverse the sign of the enthalpy change. This is a classic exam trap — always trace your route carefully through the cycle.

在赫斯循环图中，如果你逆着箭头方向走，必须反转焓变的符号。这是典型的考试陷阱——始终仔细地追踪你在循环中的路径。

Practice Questions | 练习题

Question 1 | 第 1 题

Calculate ΔH for the hydrogenation of ethene using mean bond enthalpies:
C₂H₄(g) + H₂(g) → C₂H₆(g)

使用平均键焓计算乙烯加氢反应的 ΔH：C₂H₄(g) + H₂(g) → C₂H₆(g)

Hint: In C₂H₄, the C=C double bond counts as ONE bond (not two single bonds). In C₂H₆, there is a C–C single bond.

Question 2 | 第 2 题

Use Hess’s Law and the following enthalpy of combustion data to calculate the enthalpy of formation of propane (C₃H₈):

• ΔH_c° of C(s) = −394 kJ mol⁻¹
• ΔH_c° of H₂(g) = −286 kJ mol⁻¹
• ΔH_c° of C₃H₈(g) = −2220 kJ mol⁻¹

使用赫斯定律和以下燃烧焓数据计算丙烷 (C₃H₈) 的生成焓。

Question 3 | 第 3 题

Explain why the bond enthalpy method gives only an approximate value for the enthalpy change of a reaction. Use the combustion of methane as an example to support your answer. (6 marks — typical A-Level exam question)

解释为什么键焓法只能给出反应焓变的近似值。以甲烷的燃烧为例来支持你的答案。（6 分——典型的 A-Level 考题）

Summary | 总结

• Bond enthalpy is the energy to break a bond in the gaseous state, averaged across compounds.
• ΔH = Σ (bonds broken) − Σ (bonds formed).
• Hess’s Law: the enthalpy change is path-independent — use cycles to find unknown ΔH values.
• Formation cycles: ΔH = Σ ΔH_f°(products) − Σ ΔH_f°(reactants).
• Combustion cycles: ΔH = Σ ΔH_c°(reactants) − Σ ΔH_c°(products). Note the reversal!
• Always draw out the structures, count bonds carefully, and double-check your signs.

• 键焓是在气态中断裂化学键所需的能量，取跨化合物的平均值。
• ΔH = Σ (断裂的键) − Σ (形成的键)。
• 赫斯定律：焓变与路径无关——使用循环来求解未知的 ΔH 值。
• 生成循环：ΔH = Σ ΔH_f°(产物) − Σ ΔH_f°(反应物)。
• 燃烧循环：ΔH = Σ ΔH_c°(反应物) − Σ ΔH_c°(产物)。注意顺序相反！
• 始终画出结构式，仔细计算键的数量，并再次检查符号。

Mastering these concepts is essential not just for A-Level Chemistry exams, but also for university-level thermodynamics and physical chemistry. The principles of energy conservation and bond energetics underpin everything from designing fuels to understanding biochemical pathways.

掌握这些概念不仅对 A-Level 化学考试至关重要，对大学级别的热力学和物理化学也同样重要。能量守恒和键能学的原理是从设计燃料到理解生化途径的一切的基础。