Energetics and Bond Enthalpy | 能量学与键焓

📚 Energetics and Bond Enthalpy | 能量学与键焓

Energetics is one of the most calculation-heavy topics in Edexcel A-Level Chemistry. It connects laboratory calorimetry, Hess’s law cycles and bond-breaking models, and it regularly appears in both AS and A2 exam papers. This revision guide covers the definitions, equations and common errors you need to handle enthalpy questions confidently.

能量学是 Edexcel A-Level 化学中计算量最大的主题之一。它把实验室量热法、赫斯定律循环和断键模型联系起来，并且在 AS 和 A2 考试中经常出现。这份复习指南涵盖你需要掌握的焓变定义、计算方程和常见错误，帮助你从容应对焓变题目。

1. Enthalpy Change | 焓变

Enthalpy, H, is the heat content of a system at constant pressure. The enthalpy change, ΔH, is the heat transferred in a reaction at constant pressure. It is defined as ΔH = H(products) − H(reactants).

焓 H 是体系在恒压下的热含量。焓变 ΔH 是反应在恒压下传递的热量，定义为 ΔH = H(生成物) − H(反应物)。

Standard conditions are a pressure of 100 kPa, a temperature of 298 K, and solutions with a concentration of 1 mol dm⁻³. An enthalpy change measured under these conditions is shown with the standard symbol, for example ΔH°.

标准条件是压强 100 kPa、温度 298 K、溶液浓度为 1 mol dm⁻³。在这些条件下测得的焓变用标准符号表示，例如 ΔH°。

The units of enthalpy change are usually kJ mol⁻¹. This refers to the energy change per mole of the reaction as written in the balanced equation.

焓变的单位通常是 kJ mol⁻¹。它表示按照配平方程式所写的每摩尔反应对应的能量变化。

2. Exothermic and Endothermic Reactions | 放热反应与吸热反应

In an exothermic reaction, heat is released to the surroundings, so the surrounding temperature rises. The products have less enthalpy than the reactants, and ΔH is negative. Combustion, neutralisation and respiration are common examples.

在放热反应中，热量释放到周围环境，因此环境温度升高。生成物的焓比反应物低，ΔH 为负值。燃烧、中和和呼吸作用都是常见例子。

In an endothermic reaction, heat is absorbed from the surroundings, so the surrounding temperature falls. The products have more enthalpy than the reactants, and ΔH is positive. Thermal decomposition and photosynthesis are examples.

在吸热反应中，热量从周围环境吸收，因此环境温度下降。生成物的焓比反应物高，ΔH 为正值。热分解和光合作用都是例子。

Feature	Exothermic	Endothermic
Heat transfer	Released to surroundings	Absorbed from surroundings
Temperature change	Increases	Decreases
Sign of ΔH	Negative	Positive

When drawing reaction profile diagrams, an exothermic profile has products lower than reactants, while an endothermic profile has products higher than reactants. The activation energy is always shown as the energy barrier from reactants to the transition state.

在绘制反应能量图时，放热反应的产物能量低于反应物，而吸热反应的产物能量高于反应物。活化能始终表示为从反应物到过渡态的能量障碍。

3. Standard Enthalpy Changes | 标准焓变

Standard enthalpy change of reaction, ΔH°r, is the enthalpy change when a reaction occurs in the molar quantities shown in the balanced equation under standard conditions.

标准反应焓变 ΔH°r 是指在标准条件下，按配平方程式所示摩尔量进行反应时的焓变。

Standard enthalpy change of formation, ΔH°f, is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. For an element in its standard state, ΔH°f is defined as zero.

标准生成焓变 ΔH°f 是指在标准条件下，由标准状态下的元素生成一摩尔化合物时的焓变。标准状态下元素的 ΔH°f 被定义为零。

Standard enthalpy change of combustion, ΔH°c, is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions. Combustion reactions are always exothermic, so ΔH°c values are negative.

标准燃烧焓变 ΔH°c 是指在标准条件下，一摩尔物质在氧气中完全燃烧时的焓变。燃烧反应总是放热的，因此 ΔH°c 为负值。

Standard enthalpy change of neutralisation, ΔH°neut, is the enthalpy change when one mole of water is formed from the reaction of an acid and an alkali under standard conditions. For strong acids with strong alkalis it is usually about −57 kJ mol⁻¹.

标准中和焓变 ΔH°neut 是指在标准条件下，酸与碱反应生成一摩尔水时的焓变。强酸与强碱反应的中和焓变通常约为 −57 kJ mol⁻¹。

4. Measuring Enthalpy Changes by Calorimetry | 量热法测量焓变

Calorimetry measures the heat transferred by observing the temperature change of a known mass of water or solution. The heat absorbed or released is calculated using the equation:

量热法通过观察已知质量的水或溶液的温度变化来测量传递的热量。吸收或放出的热量用以下方程计算：

q = m × c × ΔT

Here q is the heat transferred in joules, m is the mass of the solution or water in grams, c is the specific heat capacity in J g⁻¹ K⁻¹, and ΔT is the temperature change in Kelvin or °C.

其中 q 是传递的热量，单位为焦耳；m 是溶液或水的质量，单位为克；c 是比热容，单位为 J g⁻¹ K⁻¹；ΔT 是温度变化，单位为 K 或 °C。

For aqueous solutions, the specific heat capacity is often taken as 4.18 J g⁻¹ K⁻¹ and the density is taken as 1 g cm⁻³, so 1 cm³ of solution has a mass of 1 g.

对于水溶液，比热容通常取 4.18 J g⁻¹ K⁻¹，密度取 1 g cm⁻³，因此 1 cm³ 溶液的质量为 1 g。

The enthalpy change per mole is then found by dividing the heat transferred by the number of moles of the limiting reactant:

然后每摩尔的焓变通过将传递的热量除以限制反应物的摩尔数得到：

ΔH = −q ÷ n

The negative sign is included because q is measured as the heat gained by the water. If the water gains heat, the reaction must have lost heat, and so ΔH is negative.

公式中包含负号，因为 q 测量的是水获得的热量。如果水获得热量，反应必然失去热量，因此 ΔH 为负值。

Example: 50 cm³ of 1 mol dm⁻³ HCl is mixed with 50 cm³ of 1 mol dm⁻³ NaOH. The temperature rises by 6.8 °C. Calculate the enthalpy change of neutralisation.
示例：将 50 cm³ 的 1 mol dm⁻³ HCl 与 50 cm³ 的 1 mol dm⁻³ NaOH 混合，温度升高 6.8 °C。计算中和焓变。

Total volume = 100 cm³, mass = 100 g. q = 100 × 4.18 × 6.8 = 2842 J. Moles of water formed = 0.050 mol. ΔH = −2842 J ÷ 0.050 mol = −56840 J mol⁻¹ = −56.8 kJ mol⁻¹.

总体积 = 100 cm³，质量 = 100 g。q = 100 × 4.18 × 6.8 = 2842 J。生成水的摩尔数 = 0.050 mol。ΔH = −2842 J ÷ 0.050 mol = −56840 J mol⁻¹ = −56.8 kJ mol⁻¹。

5. Hess’s Law | 赫斯定律

Hess’s law states that the total enthalpy change for a reaction depends only on the initial and final states, not on the route taken. This allows enthalpy changes to be calculated indirectly when direct measurement is difficult or impossible.

赫斯定律指出，反应的总焓变只取决于始态和终态，与反应途径无关。因此，当直接测量困难或不可能时，可以通过间接方法计算焓变。

A Hess cycle is drawn by linking reactants and products through alternative routes involving elements or other known reactions. The principle is:

赫斯循环通过元素或其他已知反应将反应物和产物连接起来。其原理为：

ΔH direct = ΔH route 1 + ΔH route 2

When using standard enthalpy changes of formation, the cycle goes from elements to both reactants and products. The enthalpy change of reaction is calculated from:

使用标准生成焓变时，循环从元素分别指向反应物和产物。反应焓变由下式计算：

ΔH°r = ΣΔH°f(products) − ΣΔH°f(reactants)

When using standard enthalpy changes of combustion, the cycle burns both reactants and products to common combustion products. The enthalpy change of reaction is:

使用标准燃烧焓变时，循环将反应物和产物都燃烧为共同的燃烧产物。反应焓变为：

ΔH°r = ΣΔH°c(reactants) − ΣΔH°c(products)

Be careful with the direction of subtraction. Formation cycles subtract reactant terms, while combustion cycles subtract product terms.

注意相减的方向。生成焓循环减去反应物项，而燃烧焓循环减去产物项。

6. Enthalpy Calculations Using Formation and Combustion | 用生成焓与燃烧焓计算

Example 1: Calculate ΔH°r for the reaction 2SO₂(g) + O₂(g) → 2SO₃(g) given the standard enthalpies of formation: ΔH°f[SO₂] = −297 kJ mol⁻¹, ΔH°f[SO₃] = −395 kJ mol⁻¹, and ΔH°f[O₂] = 0.

示例 1：已知标准生成焓 ΔH°f[SO₂] = −297 kJ mol⁻¹，ΔH°f[SO₃] = −395 kJ mol⁻¹，ΔH°f[O₂] = 0，计算反应 2SO₂(g) + O₂(g) → 2SO₃(g) 的 ΔH°r。

ΣΔH°f(products) = 2 × (−395) = −790 kJ mol⁻¹. ΣΔH°f(reactants) = 2 × (−297) + 0 = −594 kJ mol⁻¹. ΔH°r = −790 − (−594) = −196 kJ mol⁻¹.

ΣΔH°f(生成物) = 2 × (−395) = −790 kJ mol⁻¹。ΣΔH°f(反应物) = 2 × (−297) + 0 = −594 kJ mol⁻¹。ΔH°r = −790 − (−594) = −196 kJ mol⁻¹。

Example 2: Calculate ΔH°r for the reaction CH₃COOH(l) + C₂H₅OH(l) → CH₃COOC₂H₅(l) + H₂O(l) using combustion data: ΔH°c[CH₃COOH] = −875 kJ mol⁻¹, ΔH°c[C₂H₅OH] = −1367 kJ mol⁻¹, ΔH°c[CH₃COOC₂H₅] = −2231 kJ mol⁻¹, ΔH°c[H₂O] = 0.

示例 2：已知燃烧焓数据 ΔH°c[CH₃COOH] = −875 kJ mol⁻¹、ΔH°c[C₂H₅OH] = −1367 kJ mol⁻¹、ΔH°c[CH₃COOC₂H₅] = −2231 kJ mol⁻¹、ΔH°c[H₂O] = 0，计算反应 CH₃COOH(l) + C₂H₅OH(l) → CH₃COOC₂H₅(l) + H₂O(l) 的 ΔH°r。

ΣΔH°c(reactants) = −875 + (−1367) = −2242 kJ mol⁻¹. ΣΔH°c(products) = −2231 + 0 = −2231 kJ mol⁻¹. ΔH°r = −2242 − (−2231) = −11 kJ mol⁻¹.

ΣΔH°c(反应物) = −875 + (−1367) = −2242 kJ mol⁻¹。ΣΔH°c(生成物) = −2231 + 0 = −2231 kJ mol⁻¹。ΔH°r = −2242 − (−2231) = −11 kJ mol⁻¹。

7. Bond Enthalpy and Mean Bond Enthalpy | 键焓与平均键焓

Bond enthalpy is the energy required to break one mole of a given covalent bond in the gaseous state. The higher the bond enthalpy, the stronger the bond.

键焓是在气态下断裂一摩尔某种共价键所需的能量。键焓越大，键越强。

Mean bond enthalpy is the average energy required to break one mole of a particular type of bond in a range of different gaseous compounds. It is an average value because the same bond, such as C–H, has slightly different strengths in different molecules.

平均键焓是在一系列不同气态化合物中断裂一摩尔某种特定类型键所需的平均能量。它是平均值，因为同一种键（如 C–H）在不同分子中的强度略有不同。

Bond breaking is always endothermic, so energy must be supplied to break bonds. Bond formation is always exothermic, so energy is released when new bonds form.

断键总是吸热的，因此需要提供能量来断裂键。成键总是放热的，因此形成新键时释放能量。

Bond enthalpy calculations are best used for reactions where all reactants and products are gases. If liquids or solids appear, additional enthalpy changes such as vaporisation may be ignored, which introduces error.

键焓计算最适合所有反应物和产物都是气体的反应。如果出现液体或固体，额外的焓变（如汽化）可能被忽略，这会带来误差。

8. Calculating Enthalpy Changes from Bond Enthalpies | 由键焓计算焓变

The enthalpy change of a reaction can be estimated using the equation:

反应的焓变可以用以下方程估算：

ΔH = ΣΔH(bonds broken) − ΣΔH(bonds formed)

Bonds broken are the bonds present in the reactants. Bonds formed are the bonds present in the products. Add up all the bond enthalpies for bonds broken, then subtract all the bond enthalpies for bonds formed.

断裂的键是反应物中存在的键。形成的键是产物中存在的键。将所有断裂键的键焓相加，再减去所有形成键的键焓。

Example: Estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) using mean bond enthalpies: C–H = 413, O=O = 498, C=O = 799, O–H = 463 kJ mol⁻¹.

示例：已知平均键焓 C–H = 413、O=O = 498、C=O = 799、O–H = 463 kJ mol⁻¹，估算 CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) 的 ΔH。

Bonds broken: 4 × C–H = 4 × 413 = 1652; 2 × O=O = 2 × 498 = 996. Total broken = 2648 kJ mol⁻¹.

断裂的键：4 × C–H = 4 × 413 = 1652；2 × O=O = 2 × 498 = 996。断裂总计 = 2648 kJ mol⁻¹。

Bonds formed: 2 × C=O = 2 × 799 = 1598; 4 × O–H = 4 × 463 = 1852. Total formed = 3450 kJ mol⁻¹.

形成的键：2 × C=O = 2 × 799 = 1598；4 × O–H = 4 × 463 = 1852。形成总计 = 3450 kJ mol⁻¹。

ΔH = 2648 − 3450 = −802 kJ mol⁻¹. The reaction is exothermic because more energy is released by bond formation than is absorbed by bond breaking.

ΔH = 2648 − 3450 = −802 kJ mol⁻¹。该反应为放热反应，因为成键释放的能量大于断键吸收的能量。

9. Limitations of Bond Enthalpy Calculations | 键焓计算的局限性

Bond enthalpy calculations give only approximate values. Mean bond enthalpies do not account for the specific molecular environment of each bond.

键焓计算只能给出近似值。平均键焓没有考虑每个键所处的具体分子环境。

If water vapour is formed in a combustion reaction, the calculated value assumes gaseous water. The experimental enthalpy change for liquid water will be more exothermic because condensation of water releases additional energy.

如果燃烧反应生成水蒸气，计算值假设水为气态。生成液态水的实验焓变会放热更多，因为水蒸气冷凝会释放额外能量。

Bond enthalpies apply only to covalent bonds in the gaseous state. Intermolecular forces, such as hydrogen bonds, are not included, which can make calculated values differ from experimental values in molecules with strong intermolecular interactions.

键焓仅适用于气态共价键。分子间作用力（如氢键）未被包含在内，因此对于具有强分子间作用的分子，计算值可能与实验值不同。

Despite these limitations, bond enthalpy calculations are useful for estimating enthalpy changes when experimental data are unavailable and for explaining why reactions are exothermic or endothermic.

尽管存在这些局限，键焓计算在缺乏实验数据时仍可用于估算焓变，并解释反应为什么是放热或吸热。

10. Exam Technique and Common Errors | 考试技巧与常见错误

Always write the sign of ΔH. A missing negative sign loses marks in exothermic reactions. Link the sign to the direction of heat transfer.

始终写出 ΔH 的符号。放热反应中缺少负号会扣分。将符号与热量传递的方向联系起来。

Check the units. Calorimetry gives joules first, but final answers are usually required in kJ mol⁻¹. Divide by 1000 when converting from J to kJ.

检查单位。量热法首先得到焦耳，但最终答案通常要求以 kJ mol⁻¹ 表示。从 J 转换为 kJ 时要除以 1000。

Balance the chemical equation before calculating bond enthalpies. Count the exact number of each bond broken and formed, including all O=O, C=O and O–H groups.

在使用键焓计算之前配平化学方程式。准确统计每种断裂和形成的键的数量，包括所有 O=O、C=O 和 O–H 基团。

For Hess’s law cycles, draw arrows carefully and use the correct subtraction order. If the cycle uses formation, the equation is products minus reactants. If it uses combustion, the equation is reactants minus products.

对于赫斯定律循环，仔细画出箭头并使用正确的相减顺序。如果循环使用生成焓，方程为生成物减反应物。如果使用燃烧焓，方程则为反应物减生成物。

In calorimetry, state the assumptions: no heat loss to the surroundings, the solution has the same specific heat capacity as water, and complete mixing occurs. Exam questions often ask why an experimental value is less exothermic than a data book value.

在量热法中，说明假设条件：没有热量散失到周围环境、溶液的比热容与水相同、混合完全。考题常问为什么实验值不如数据手册值放热多。

The usual answer is that heat is lost to the surroundings, or that incomplete combustion occurred, or that the experiment was not carried out under standard conditions. Use precise wording to gain full marks.

通常的答案是热量散失到周围环境、发生不完全燃烧，或实验不是在标准条件下进行。使用准确的表述才能得到满分。

Published by TutorHao | A-Level Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Energetics and Bond Enthalpy | 能量学与键焓