High School Chemistry

Thermochemistry and Energy热化学与能量

Every chemical reaction either releases or absorbs energy — thermochemistry is the study of that energy flow. This guide builds the full picture: from the distinction between heat, temperature, and enthalpy, through exothermic and endothermic reactions, to the calorimetry equation $q = mc\Delta T$ and how it is used to measure enthalpy changes. It then covers the enthalpy of reaction ($\Delta H$), Hess's law for combining reaction steps (honors-flagged for NGSS), bond-energy calculations, and the interpretation of potential-energy diagrams. Worked examples and KaTeX formulas are used throughout.每一个化学反应都会释放或吸收能量——热化学（thermochemistry，热化学）就是研究这种能量流动的学科。本指南构建完整图景：从热量（heat，热量）、温度与焓（enthalpy，焓）的区别出发，经放热（exothermic，放热）与吸热（endothermic，吸热）反应，到量热法（calorimetry，量热法）方程 $q = mc\Delta T$ 及其测量焓变的应用，进而介绍反应焓变（ΔH，反应焓变 $\Delta H$）、用于合并反应步骤的盖斯定律（Hess's law，盖斯定律，荣誉标记），以及键能（bond energy，键能）计算和能量图（potential-energy diagram，势能图）的解读。全程使用例题与公式。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Honors block: Hess's law quantitative (ON SCH4U / AB Chem 30)荣誉级：盖斯定律定量计算（ON SCH4U / AB Chem 30）
Read me first先读这里

How to use this guide如何使用本指南

Thermochemistry is the study of heat flow in chemical reactions. All four curricula agree on a core qualitative scope: the distinction between exothermic and endothermic reactions, the concept of enthalpy change, and potential-energy diagrams. They diverge on quantitative depth. US NGSS (HS-PS1-4) keeps bond-energy and enthalpy qualitative — no calculation of total bond-energy changes is assessed. Ontario SCH4U (D2.3, D2.5) and Alberta Chemistry 30 (Unit A GO1) require quantitative calorimetry ($Q = mc\Delta T$) and Hess's law calculations. BC Chemistry 12 covers PE/KE/enthalpy and PE diagrams but has no standalone Hess's law or bond-energy calculation bullet. The table below tells you which sections are core for you; each row cites the curriculum document it was checked against.热化学是研究化学反应中热量流动的学科。四套大纲在核心定性范围上一致：区分放热与吸热反应、焓变的概念以及势能图。它们在定量深度上存在分歧。US NGSS（HS-PS1-4）保持键能和焓的定性——不评估总键能变化的计算。安大略 SCH4U（D2.3、D2.5）和阿尔伯塔 Chemistry 30（A 单元 GO1）要求定量量热法（$Q = mc\Delta T$）和盖斯定律计算。BC Chemistry 12 涵盖 PE/KE/焓与 PE 图，但没有独立的盖斯定律或键能计算条目。下表告诉你哪些节是你的核心；每行均注明所依据的课纲文件。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Chemistry美国 NGSS 高中化学 §1–§2 and §7 (energy/heat concepts, exo/endo, PE diagrams) — the qualitative core under HS-PS1-4. NGSS does not assess calculating total bond-energy changes from tabulated bond energies.§1–§2 与 §7（能量/热量概念、放热/吸热、PE 图）—— HS-PS1-4 下的定性核心。NGSS 不评估从表格键能计算总键能变化。 §3–§6 ($q=mc\Delta T$ calculation, $\Delta H$, Hess's law, bond-energy arithmetic): important skill-building but above the NGSS assessed qualitative floor.§3–§6（$q=mc\Delta T$ 计算、$\Delta H$、盖斯定律、键能算术）：重要技能训练，但高于 NGSS 的定性评估下限。 NGSS HS-PS1 (Chemistry) — HS-PS1-4 PE + Clarification + Assessment Boundary— HS-PS1-4 表现期望 + 澄清 + 评估边界
🇨🇦 ON SCH4U安大略 SCH4U All seven sections. D2.1 (terminology), D2.2 (thermochemical equations), D2.3 ($Q = mc\Delta T$ problems), D2.5 (Hess's law calculations), D3.2 (exo/endo bond breaking/forming), D3.4 (Hess's law statement and application), D3.6 (PE diagrams with activation energy) are all assessed.全部 7 节。D2.1（术语）、D2.2（热化学方程式）、D2.3（$Q = mc\Delta T$ 习题）、D2.5（盖斯定律计算）、D3.2（放热/吸热键断裂/形成）、D3.4（盖斯定律的陈述与应用）、D3.6（含活化能的 PE 图）均被评估。 Nothing — the full quantitative thermochemistry suite is in SCH4U Strand D.无 — 完整的定量热化学套件在 SCH4U D 单元。 Ontario SCH3U/4U Chemistry — SCH4U Strand D D2.1–D2.7, D3.2–D3.6— SCH4U D 单元 D2.1–D2.7，D3.2–D3.6
🇨🇦 BC Chemistry 12BC Chemistry 12 §1–§4 and §7 (energy/heat/enthalpy, exo/endo, $Q = mc\Delta T$ as part of the kinetics big idea, PE diagrams). The "energy change during a chemical reaction" content bullet covers these.§1–§4 与 §7（能量/热/焓、放热/吸热、作为动力学大概念一部分的 $Q = mc\Delta T$、PE 图）。"化学反应中的能量变化"内容条目涵盖这些。 §5 (Hess's law quantitative) and §6 (bond-energy calculation): BC has no standalone content bullet for these. Treat as enrichment.§5（盖斯定律定量）和 §6（键能计算）：BC 没有这些的独立内容条目。视为拓展内容。 BC Chemistry 11/12 — Chemistry 12 Content "energy change during a chemical reaction"; Elaboration "PE, KE, enthalpy (H), and catalysis"— Chemistry 12 内容"化学反应中的能量变化"；细化"PE、KE、焓（H）与催化"
🇨🇦 AB Chemistry 30阿尔伯塔 Chemistry 30 All seven sections. Chemistry 30 Unit A GO1 requires: $Q = mc\Delta t$, enthalpy definition, $\Delta H$ notation and calculation, Hess's law, calorimetry, and classifying exothermic/endothermic. GO2 requires: bond-breaking/forming energy changes, PE diagram analysis and labelling.全部 7 节。Chemistry 30 A 单元 GO1 要求：$Q = mc\Delta t$、焓的定义、$\Delta H$ 符号与计算、盖斯定律、量热法以及放热/吸热分类。GO2 要求：键断裂/形成的能量变化、PE 图的分析与标注。 Nothing — the quantitative thermochemistry suite is fully assessed in Chemistry 30 Unit A.无 — 定量热化学套件在 Chemistry 30 A 单元中全面评估。 Alberta Chemistry 20/30 — Chemistry 30 Unit A GO1/GO2, knowledge outcome text— Chemistry 30 A 单元 GO1/GO2，知识结果文本
🇺🇸 IB / AP feeder trackIB / AP 衔接轨道 All seven sections plus every going-deeper derivation. IB Chemistry HL Reactivity 1 and AP Chemistry assume fluent calorimetry, $\Delta H$, Hess's law, and bond-energy reasoning from the first thermochemistry problem set.全部 7 节，并完成每个"深入"推导。IB Chemistry HL Reactivity 1 与 AP Chemistry 从第一套热化学习题起就默认你熟练量热法、$\Delta H$、盖斯定律和键能推理。 Nothing — this is the conceptual and quantitative foundation for all energy, equilibrium, and kinetics units that follow.无 — 这是后续所有能量、平衡与动力学单元的概念和定量基础。 NGSS HS-PS1 (Chemistry) — see the IB Chemistry HL Reactivity 1 feeder link in "What This Feeds Into"— 见"本单元的去向"中的 IB Chemistry HL Reactivity 1 衔接链接

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

!
If you are cramming the night before如果你在临阵磨枪

Memorise five things: exothermic reactions release energy ($\Delta H < 0$); endothermic reactions absorb energy ($\Delta H > 0$); calorimetry uses $q = mc\Delta T$; Hess's law says enthalpy is additive; bond breaking absorbs energy, bond forming releases energy. Read every cram-cheat box. Skip the going-deeper derivations if time is short.背熟五件事：放热反应释放能量（$\Delta H < 0$）；吸热反应吸收能量（$\Delta H > 0$）；量热法使用 $q = mc\Delta T$；盖斯定律说明焓是可加的；断键吸收能量，成键释放能量。读每个速记框。若时间紧，可跳过深入推导。

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If you are going for the top mark如果你目标顶分

Be precise about sign conventions ($q_\text{system} = -q_\text{surroundings}$); know how to apply Hess's law by flipping and scaling equations; calculate $\Delta H$ from bond energies as $\Sigma E_\text{broken} - \Sigma E_\text{formed}$; label a PE diagram with reactants, products, $\Delta H$, $E_a$ (forward and reverse); and explain why catalysts lower $E_a$ without changing $\Delta H$. ON SCH4U D2.5 and AB Chemistry 30 GO1 expect you to carry out multi-step Hess's law calculations.精确掌握符号约定（$q_\text{system} = -q_\text{surroundings}$）；知道如何通过翻转和缩放方程来应用盖斯定律；将 $\Delta H$ 计算为键能之差 $\Sigma E_\text{broken} - \Sigma E_\text{formed}$；在 PE 图上标注反应物、产物、$\Delta H$、$E_a$（正向和逆向）；解释催化剂为何降低 $E_a$ 而不改变 $\Delta H$。ON SCH4U D2.5 和 AB Chemistry 30 GO1 要求你完成多步盖斯定律计算。

Honors flag.荣誉级标记。 Section 5 (Hess's law, quantitative multi-step enthalpy calculation) carries the Honors chip for US NGSS (HS-PS1-4 Assessment Boundary: "Assessment does not include calculating the total bond energy changes during a chemical reaction from the bond energies of reactants and products"). It is core, not honors, in Ontario SCH4U (D2.5) and Alberta Chemistry 30 (Unit A GO1). BC Chemistry 12 has no standalone Hess's law bullet (see the BC syllabus note above). If your row sends you to §5, treat it as required content.§5（盖斯定律，定量多步焓计算）在 US NGSS 中标 Honors（HS-PS1-4 评估边界："不包括从反应物和产物的键能计算化学反应中总键能变化"）。在安大略 SCH4U（D2.5）和阿尔伯塔 Chemistry 30（A 单元 GO1）中，它是核心而非荣誉内容。BC Chemistry 12 没有独立的盖斯定律条目（见上方 BC 大纲提示）。如果你的行指向 §5，就把它视为必学内容。
Section 1 · Energy, heat and enthalpy第 1 节 · 能量、热与焓

Energy, Heat, and Enthalpy能量、热与焓

Three terms, one concept: energy moving as heat at constant pressure.三个术语，一个概念：在恒压下以热量形式流动的能量。
  • Energy能量 — the capacity to do work or transfer heat. Measured in joules (J) or kilojoules (kJ). Conserved: it cannot be created or destroyed, only converted. NGSS HS-PS3-1 grounds this in conservation of energy in a system.— 做功或传递热量的能力。单位为焦耳（J）或千焦（kJ）。守恒：不可被创造或消灭，只能转化。NGSS HS-PS3-1 以系统的能量守恒为基础。
  • Heat $q$热量 $q$ — energy transferred between objects at different temperatures. Sign convention: $q > 0$ when the system gains heat (absorbs); $q < 0$ when the system loses heat (releases). $q_\text{system} + q_\text{surroundings} = 0$.— 在不同温度物体之间转移的能量。符号约定：当系统吸收热量时 $q > 0$；当系统释放热量时 $q < 0$。$q_\text{system} + q_\text{surroundings} = 0$。
  • Enthalpy $H$焓 $H$ — a thermodynamic quantity that equals the heat flow in a reaction carried out at constant pressure. We cannot measure absolute $H$; we measure the enthalpy change $\Delta H = H_\text{products} - H_\text{reactants}$. SCH4U D2.1 lists "enthalpy" as required terminology; AB Chemistry 30 GO1 requires "define enthalpy and molar enthalpy."— 一个热力学量，等于在恒压下进行的反应中的热量流动。我们无法测量绝对的 $H$；我们测量焓变 $\Delta H = H_\text{products} - H_\text{reactants}$。SCH4U D2.1 将"焓"列为必需术语；AB Chemistry 30 GO1 要求"定义焓和摩尔焓"。

Temperature vs heat — a critical distinction:温度与热量的关键区别：

Temperature is a measure of the average kinetic energy of particles; heat is the total energy transferred. Two objects can have the same temperature but transfer very different amounts of heat (e.g. a spark vs a bonfire — both at $\sim 1000\,^\circ\text{C}$ but the bonfire has far more thermal energy).温度是粒子平均动能的量度；热量是转移的总能量。两个物体可以具有相同的温度但传递非常不同的热量（例如一颗火花与一堆篝火——两者都在约 $1000\,^\circ\text{C}$，但篝火有多得多的热能）。
Worked Example 1 · Sign of $\Delta H$例题 1 · $\Delta H$ 的符号

The combustion of methane releases $890\ \mathrm{kJ}$ of heat per mole at constant pressure. Write the thermochemical equation and state the sign of $\Delta H$. Identify the system and surroundings, and state what happens to each.甲烷的燃烧在恒压下每摩尔释放 $890\ \mathrm{kJ}$ 的热量。写出热化学方程式并说明 $\Delta H$ 的符号。确定系统和环境，并说明各自发生了什么。

Thermochemical equation.热化学方程式。

$$ \text{CH}_4(g) + 2\,\text{O}_2(g) \to \text{CO}_2(g) + 2\,\text{H}_2\text{O}(l) \qquad \Delta H = -890\ \mathrm{kJ/mol} $$

Sign.符号。 Heat is released by the reaction, so $q_\text{system} < 0$, hence $\Delta H = -890\ \mathrm{kJ/mol}$ (negative). This is an exothermic reaction.热量被反应释放，所以 $q_\text{system} < 0$，因此 $\Delta H = -890\ \mathrm{kJ/mol}$（负值）。这是一个放热反应。

System and surroundings.系统与环境。 The reacting gases and products are the system; everything else (the container, the air, the calorimeter water) is the surroundings. The system loses $890\ \mathrm{kJ}$; the surroundings gain $890\ \mathrm{kJ}$. Conservation: $q_\text{sys} + q_\text{surr} = 0$. ✓反应气体与产物是系统；其他一切（容器、空气、量热器中的水）是环境。系统失去 $890\ \mathrm{kJ}$；环境获得 $890\ \mathrm{kJ}$。守恒：$q_\text{sys} + q_\text{surr} = 0$。✓

A reaction has $\Delta H = -285\ \mathrm{kJ/mol}$. What does the negative sign indicate?某反应的 $\Delta H = -285\ \mathrm{kJ/mol}$。负号表示什么？
§1 · Q1
Heat is released by the system to the surroundings (exothermic)系统向环境释放热量（放热）
Heat is absorbed by the system from the surroundings (endothermic)系统从环境吸收热量（吸热）
The products have more enthalpy than the reactants产物的焓比反应物多
The temperature of the surroundings decreases环境温度降低
$\Delta H < 0$ means $H_\text{products} < H_\text{reactants}$: the system has lost enthalpy, releasing it as heat to the surroundings. The surroundings warm up — this is an exothermic process. SCH4U D2.1 lists "exothermic" as required terminology.$\Delta H < 0$ 意味着 $H_\text{products} < H_\text{reactants}$：系统失去了焓，以热量形式释放到环境中。环境温度升高——这是一个放热过程。SCH4U D2.1 将"放热"列为必需术语。
Negative $\Delta H$ = products lower in enthalpy than reactants = system released energy = exothermic = surroundings warm up. Positive $\Delta H$ = endothermic = surroundings cool down.$\Delta H$ 为负 = 产物焓低于反应物 = 系统释放了能量 = 放热 = 环境温度升高。$\Delta H$ 为正 = 吸热 = 环境温度降低。
Which statement correctly distinguishes temperature from heat?下列哪句话正确区分了温度与热量？
§1 · Q2
Temperature and heat are the same quantity measured in different units温度和热量是用不同单位测量的同一物理量
Heat is always greater than temperature for the same substance对于相同物质，热量总是大于温度
Temperature measures the total kinetic energy of all particles温度测量所有粒子的总动能
Temperature measures average particle kinetic energy; heat is the total energy transferred between objects温度测量粒子的平均动能；热量是在物体之间转移的总能量
Temperature = average kinetic energy per particle (an intensive property; does not depend on how much material is present). Heat = total energy transferred (an extensive property; a larger sample transfers more heat at the same temperature difference). NGSS HS-PS3-4 probes heat transfer in closed systems.温度 = 每个粒子的平均动能（强度性质；不取决于存在多少材料）。热量 = 转移的总能量（广延性质；在相同温差下，更大的样品转移更多热量）。NGSS HS-PS3-4 探究封闭系统中的热传导。
Temperature is an intensive (per-particle average) property; heat is an extensive (total-transfer) property. They have different units: temperature in K or °C, heat in J or kJ.温度是强度（每粒子平均）性质；热量是广延（总转移）性质。它们有不同的单位：温度用 K 或 °C，热量用 J 或 kJ。
Going deeper — enthalpy and the first law of thermodynamics深入 — 焓与热力学第一定律

The first law of thermodynamics states that energy is conserved: $\Delta U = q + w$, where $\Delta U$ is the change in internal energy of the system, $q$ is heat absorbed by the system, and $w$ is work done on the system. In most chemistry lab reactions, the only work is pressure-volume work: $w = -P\Delta V$ (the negative sign because the system does work on surroundings when it expands). At constant pressure: $\Delta H = \Delta U + P\Delta V = q_P$. That is, the enthalpy change equals the heat flow at constant pressure — which is why enthalpy is the relevant energy function for most chemical reactions (run in open flasks at atmospheric pressure). Reactions with large $\Delta n_\text{gas}$ (moles of gaseous products minus gaseous reactants) will have $\Delta H \ne \Delta U$; for reactions with no change in moles of gas, $\Delta H \approx \Delta U$.热力学第一定律指出能量守恒：$\Delta U = q + w$，其中 $\Delta U$ 是系统内能的变化，$q$ 是系统吸收的热量，$w$ 是对系统做的功。在大多数化学实验室反应中，唯一的功是压力-体积功：$w = -P\Delta V$（负号因为系统膨胀时对环境做功）。在恒压下：$\Delta H = \Delta U + P\Delta V = q_P$。即，在恒压下，焓变等于热量流动——这就是为什么焓是大多数化学反应（在大气压下开放烧瓶中进行）的相关能量函数。气体摩尔数有较大变化 $\Delta n_\text{gas}$（气态产物摩尔数减去气态反应物摩尔数）的反应会有 $\Delta H \ne \Delta U$；对于气体摩尔数无变化的反应，$\Delta H \approx \Delta U$。

Section 2 · Exothermic vs endothermic reactions第 2 节 · 放热与吸热反应

Exothermic and Endothermic Reactions放热与吸热反应

One question decides exo or endo: do products sit lower or higher in energy than reactants?一个问题决定放热还是吸热：产物在能量上低于还是高于反应物？
  • Exothermic放热 — products have lower enthalpy than reactants. $\Delta H < 0$. The excess energy is released as heat to the surroundings, which warm up. Examples: combustion, neutralisation, respiration, condensation.— 产物的焓低于反应物。$\Delta H < 0$。多余的能量以热量形式释放到环境中，环境温度升高。示例：燃烧、中和、呼吸作用、冷凝。
  • Endothermic吸热 — products have higher enthalpy than reactants. $\Delta H > 0$. The reaction absorbs heat from the surroundings, which cool down. Examples: photosynthesis, dissolving ammonium nitrate in water, baking a cake.— 产物的焓高于反应物。$\Delta H > 0$。反应从环境中吸收热量，环境温度降低。示例：光合作用、将硝酸铵溶于水、烘焙蛋糕。

Bond energy link:键能联系：

Exothermic if energy released forming bonds in products exceeds energy absorbed breaking bonds in reactants. NGSS HS-PS1-4 grounds this in "changes in total bond energy." SCH4U D3.2 requires comparing "energy change from a reaction in which bonds are formed to one in which bonds are broken." AB Chemistry 30 GO2 requires "energy changes referring to bonds breaking and forming."如果形成产物化学键释放的能量超过断裂反应物化学键吸收的能量，则反应为放热反应。NGSS HS-PS1-4 将此建立在"总键能的变化"上。SCH4U D3.2 要求比较"成键反应与断键反应的能量变化"。AB Chemistry 30 GO2 要求"提及键断裂和形成的能量变化"。
Worked Example 2 · Classifying reactions例题 2 · 反应分类

Classify each reaction as exothermic or endothermic and explain: (a) $\text{H}_2\text{O}(g) \to \text{H}_2\text{O}(l)$, $\Delta H = -44\ \mathrm{kJ/mol}$; (b) $\text{NH}_4\text{NO}_3(s) \to \text{NH}_4^+(aq) + \text{NO}_3^-(aq)$, $\Delta H = +25\ \mathrm{kJ/mol}$.将每个反应分类为放热或吸热并解释：(a) $\text{H}_2\text{O}(g) \to \text{H}_2\text{O}(l)$，$\Delta H = -44\ \mathrm{kJ/mol}$；(b) $\text{NH}_4\text{NO}_3(s) \to \text{NH}_4^+(aq) + \text{NO}_3^-(aq)$，$\Delta H = +25\ \mathrm{kJ/mol}$。

(a) Condensation of water.(a) 水的冷凝。 $\Delta H = -44\ \mathrm{kJ/mol} < 0$ → exothermic. Products (liquid water) are at lower enthalpy than reactants (steam). The surroundings gain heat — you feel the warmth when steam touches your skin.$\Delta H = -44\ \mathrm{kJ/mol} < 0$ → 放热。产物（液态水）的焓低于反应物（蒸汽）。环境获得热量——当蒸汽触碰皮肤时你会感到温热。

(b) Dissolving ammonium nitrate.(b) 硝酸铵溶解。 $\Delta H = +25\ \mathrm{kJ/mol} > 0$ → endothermic. Products (ions in solution) are at higher enthalpy than the solid reactant. The solution cools as it draws heat from the surroundings — this is the principle behind instant cold packs.$\Delta H = +25\ \mathrm{kJ/mol} > 0$ → 吸热。产物（溶液中的离子）的焓高于固态反应物。溶液从环境中吸热而冷却——这是即时冷敷包的原理。

During an endothermic reaction, what happens to the temperature of the surroundings?在吸热反应过程中，环境温度会发生什么变化？
§2 · Q1
It increases, because the reaction releases energy升高，因为反应释放了能量
It decreases, because the reaction absorbs energy from the surroundings降低，因为反应从环境中吸收了能量
It stays the same, because energy is conserved保持不变，因为能量守恒
It increases, then decreases as equilibrium is reached先升高，然后在达到平衡时降低
Endothermic: $\Delta H > 0$, products higher in enthalpy. The system absorbs heat from the surroundings ($q_\text{surr} < 0$), so the surroundings lose energy and their temperature drops. The reaction mixture feels cold. Conservation requires $q_\text{sys} = -q_\text{surr}$.吸热：$\Delta H > 0$，产物焓更高。系统从环境中吸收热量（$q_\text{surr} < 0$），所以环境失去能量，温度降低。反应混合物感觉很冷。守恒要求 $q_\text{sys} = -q_\text{surr}$。
Endothermic = heat flows into the system = surroundings cool down. Exothermic = heat flows out of the system = surroundings warm up. Energy is still conserved in both cases — it just changes location.吸热 = 热量流入系统 = 环境冷却。放热 = 热量从系统流出 = 环境升温。在两种情况下能量仍然守恒——只是改变了位置。
Combustion of propane: $\text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}$, $\Delta H = -2220\ \mathrm{kJ/mol}$. Which statement is correct?丙烷燃烧：$\text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}$，$\Delta H = -2220\ \mathrm{kJ/mol}$。下列哪个陈述是正确的？
§2 · Q2
This is endothermic; the products are higher in energy than the reactants这是吸热反应；产物的能量高于反应物
This is exothermic; the temperature of the reaction mixture decreases这是放热反应；反应混合物的温度降低
This is exothermic; the products are lower in energy than the reactants这是放热反应；产物的能量低于反应物
This is endothermic; energy is absorbed from the flame这是吸热反应；能量从火焰中被吸收
$\Delta H = -2220\ \mathrm{kJ/mol} < 0$ → exothermic. Products ($\text{CO}_2$, $\text{H}_2\text{O}$) are at lower enthalpy than reactants ($\text{C}_3\text{H}_8$, $\text{O}_2$). The 2220 kJ is released to the surroundings (the flame heats your BBQ), not absorbed from them. AB Chemistry 30 GO1 requires classifying combustion as exothermic.$\Delta H = -2220\ \mathrm{kJ/mol} < 0$ → 放热。产物（$\text{CO}_2$、$\text{H}_2\text{O}$）的焓低于反应物（$\text{C}_3\text{H}_8$、$\text{O}_2$）。2220 kJ 释放到环境中（火焰加热你的烧烤架），而不是从环境中吸收。AB Chemistry 30 GO1 要求将燃烧分类为放热。
Negative $\Delta H$ = exothermic = energy released to surroundings = surroundings heat up. In combustion, the surroundings (air, cookware) get hot — the reaction mixture does not cool down.$\Delta H$ 为负 = 放热 = 能量释放到环境 = 环境升温。在燃烧中，环境（空气、炊具）变热——反应混合物不会降温。
Section 3 · Calorimetry ($q = mc\Delta T$)第 3 节 · 量热法（$q = mc\Delta T$）

Calorimetry: Measuring Heat with $q = mc\Delta T$量热法：用 $q = mc\Delta T$ 测量热量

The calorimetry equation — the engine of all heat calculations.量热法方程式——所有热量计算的核心。 $$ q = mc\Delta T $$
  • $q$ = heat gained or lost by the substance (J or kJ). $q > 0$ if the substance warms; $q < 0$ if it cools.= 物质获得或失去的热量（J 或 kJ）。若物质升温则 $q > 0$；若降温则 $q < 0$。
  • $m$ = mass of the substance (g or kg). Use grams when $c$ is in $\mathrm{J\,g^{-1}\,^\circ C^{-1}}$.= 物质的质量（g 或 kg）。当 $c$ 的单位为 $\mathrm{J\,g^{-1}\,^\circ C^{-1}}$ 时使用克。
  • $c$ = specific heat capacity — heat required to raise $1\ \mathrm{g}$ of the substance by $1\,^\circ\text{C}$ (or $1\ \mathrm{K}$). For water: $c = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$ (must memorise).= 比热容——将 $1\ \mathrm{g}$ 物质升温 $1\,^\circ\text{C}$（或 $1\ \mathrm{K}$）所需的热量。水的比热容：$c = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$（必须记住）。
  • $\Delta T$ $= T_\text{final} - T_\text{initial}$. Use the same unit as $c$ ($^\circ\text{C}$ or $\text{K}$; a change of $1\,^\circ\text{C} = 1\ \mathrm{K}$ in magnitude).$= T_\text{final} - T_\text{initial}$。与 $c$ 使用相同单位（$^\circ\text{C}$ 或 $\text{K}$；$1\,^\circ\text{C}$ 的变化量 $= 1\ \mathrm{K}$）。
Conservation link: $q_\text{reaction} = -q_\text{water}$ (the heat the reaction releases is absorbed by the calorimeter water). SCH4U D2.3 requires solving heat-transfer problems using $Q = mc\Delta T$; AB Chemistry 30 GO1 requires "recall the application of $Q = mc\Delta t$ to the analysis of heat transfer."守恒关联：$q_\text{reaction} = -q_\text{water}$（反应释放的热量被量热器中的水吸收）。SCH4U D2.3 要求使用 $Q = mc\Delta T$ 求解热传导问题；AB Chemistry 30 GO1 要求"回顾 $Q = mc\Delta t$ 在热传导分析中的应用"。
Worked Example 3 · Calorimetry calculation例题 3 · 量热法计算

A student dissolves $5.00\ \mathrm{g}$ of NaOH in $100.0\ \mathrm{g}$ of water in a polystyrene cup calorimeter. The temperature rises from $21.4\,^\circ\text{C}$ to $34.8\,^\circ\text{C}$. Assume the specific heat of the solution is $4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$ and the calorimeter absorbs negligible heat. Find (a) the heat released by the reaction, and (b) state whether the dissolution is exothermic or endothermic.一名学生将 $5.00\ \mathrm{g}$ NaOH 溶解于聚苯乙烯杯量热器中的 $100.0\ \mathrm{g}$ 水中。温度从 $21.4\,^\circ\text{C}$ 升高到 $34.8\,^\circ\text{C}$。假设溶液的比热容为 $4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$，量热器吸收的热量可忽略不计。求 (a) 反应释放的热量，(b) 判断溶解过程是放热还是吸热。

Identify variables.确定变量。 $m = 100.0 + 5.00 = 105.0\ \mathrm{g}$ (solution); $c = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$; $\Delta T = 34.8 - 21.4 = 13.4\,^\circ\text{C}$.$m = 100.0 + 5.00 = 105.0\ \mathrm{g}$（溶液）；$c = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$；$\Delta T = 34.8 - 21.4 = 13.4\,^\circ\text{C}$。

(a) Heat absorbed by solution.(a) 溶液吸收的热量。

$$ q_\text{solution} = mc\Delta T = (105.0)(4.18)(13.4) = 5881\ \mathrm{J} = 5.88\ \mathrm{kJ}. $$

Heat released by reaction.反应释放的热量。 $q_\text{reaction} = -q_\text{solution} = -5.88\ \mathrm{kJ}$ (negative because the reaction is the source of heat).$q_\text{reaction} = -q_\text{solution} = -5.88\ \mathrm{kJ}$（负号因为反应是热量的来源）。

(b) Classification.(b) 分类。 The temperature rose ($\Delta T > 0$) → the surroundings (water) gained heat → the reaction released heat → exothermic. ✓温度升高（$\Delta T > 0$）→ 环境（水）获得热量 → 反应释放热量 → 放热。✓

Using $q = mc\Delta T$, how much heat (in kJ) is required to raise $200\ \mathrm{g}$ of water from $20.0\,^\circ\text{C}$ to $80.0\,^\circ\text{C}$? ($c_\text{water} = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$)使用 $q = mc\Delta T$，将 $200\ \mathrm{g}$ 水从 $20.0\,^\circ\text{C}$ 加热到 $80.0\,^\circ\text{C}$ 需要多少热量（kJ）？（$c_\text{water} = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$）
§3 · Q1
$16.7\ \mathrm{kJ}$
$33.4\ \mathrm{kJ}$
$50.2\ \mathrm{kJ}$
$66.9\ \mathrm{kJ}$
$q = mc\Delta T = (200)(4.18)(80.0 - 20.0) = (200)(4.18)(60.0) = 50\,160\ \mathrm{J} = 50.2\ \mathrm{kJ}$. Always compute $\Delta T = T_\text{final} - T_\text{initial}$ first, then multiply by $m$ and $c$.$q = mc\Delta T = (200)(4.18)(80.0 - 20.0) = (200)(4.18)(60.0) = 50\,160\ \mathrm{J} = 50.2\ \mathrm{kJ}$。始终先计算 $\Delta T = T_\text{final} - T_\text{initial}$，再乘以 $m$ 和 $c$。
$q = mc\Delta T$: $m = 200\ \mathrm{g}$, $c = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$, $\Delta T = 60\,^\circ\text{C}$. Product: $200 \times 4.18 \times 60 = 50\,160\ \mathrm{J} = 50.2\ \mathrm{kJ}$.$q = mc\Delta T$：$m = 200\ \mathrm{g}$，$c = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$，$\Delta T = 60\,^\circ\text{C}$。乘积：$200 \times 4.18 \times 60 = 50\,160\ \mathrm{J} = 50.2\ \mathrm{kJ}$。
A reaction releases $2.50\ \mathrm{kJ}$ of heat into $150\ \mathrm{g}$ of water. By how many degrees Celsius does the water temperature rise? ($c_\text{water} = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$)某反应向 $150\ \mathrm{g}$ 水中释放了 $2.50\ \mathrm{kJ}$ 的热量。水的温度升高了多少摄氏度？（$c_\text{water} = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$）
§3 · Q2
$3.98\,^\circ\text{C}$
$1.99\,^\circ\text{C}$
$7.97\,^\circ\text{C}$
$16.7\,^\circ\text{C}$
Rearrange: $\Delta T = q / (mc) = 2500\ \mathrm{J} / (150 \times 4.18) = 2500 / 627 = 3.98\,^\circ\text{C}$. Note: convert kJ to J before dividing ($2.50\ \mathrm{kJ} = 2500\ \mathrm{J}$).整理：$\Delta T = q / (mc) = 2500\ \mathrm{J} / (150 \times 4.18) = 2500 / 627 = 3.98\,^\circ\text{C}$。注意：除法前将 kJ 转换为 J（$2.50\ \mathrm{kJ} = 2500\ \mathrm{J}$）。
Rearrange $q = mc\Delta T$ to get $\Delta T = q/(mc)$. Use $q = 2500\ \mathrm{J}$ (not kJ): $\Delta T = 2500/(150 \times 4.18) = 2500/627 \approx 3.98\,^\circ\text{C}$.整理 $q = mc\Delta T$ 得 $\Delta T = q/(mc)$。使用 $q = 2500\ \mathrm{J}$（不是 kJ）：$\Delta T = 2500/(150 \times 4.18) = 2500/627 \approx 3.98\,^\circ\text{C}$。
Section 4 · Enthalpy of reaction ($\Delta H$)第 4 节 · 反应焓变（$\Delta H$）

Enthalpy of Reaction and Thermochemical Equations反应焓变与热化学方程式

$\Delta H$ is a state function — it depends only on start and finish, not the path.$\Delta H$ 是状态函数——仅取决于起点和终点，而非路径。
  • Standard enthalpy of reaction $\Delta H^\circ_\text{rxn}$标准反应焓变 $\Delta H^\circ_\text{rxn}$ = enthalpy change when a reaction proceeds as written (at $25\,^\circ\text{C}$, $100\ \mathrm{kPa}$). The superscript ° denotes standard conditions.= 反应按所写方式进行时的焓变（在 $25\,^\circ\text{C}$、$100\ \mathrm{kPa}$ 下）。上标 ° 表示标准条件。
  • Scaling rule:缩放规则： $\Delta H$ scales with the amount of substance. If you double all coefficients, $\Delta H$ doubles. Example: $\text{H}_2 + \frac{1}{2}\text{O}_2 \to \text{H}_2\text{O}$, $\Delta H = -286\ \mathrm{kJ/mol}$; doubling gives $\Delta H = -572\ \mathrm{kJ}$ for 2 mol $\text{H}_2\text{O}$.$\Delta H$ 随物质的量成比例变化。若所有系数加倍，$\Delta H$ 也加倍。示例：$\text{H}_2 + \frac{1}{2}\text{O}_2 \to \text{H}_2\text{O}$，$\Delta H = -286\ \mathrm{kJ/mol}$；加倍得到 2 mol $\text{H}_2\text{O}$ 时 $\Delta H = -572\ \mathrm{kJ}$。
  • Reversal rule:逆转规则： Reversing a reaction changes the sign of $\Delta H$. If forward: $\Delta H = -286\ \mathrm{kJ/mol}$, then reverse: $\Delta H = +286\ \mathrm{kJ/mol}$.逆转反应改变 $\Delta H$ 的符号。若正向：$\Delta H = -286\ \mathrm{kJ/mol}$，则逆向：$\Delta H = +286\ \mathrm{kJ/mol}$。
  • Thermochemical equation:热化学方程式： a balanced equation with $\Delta H$ stated and state symbols (s/l/g/aq) written for all species. SCH4U D2.2 requires writing thermochemical equations "expressing the energy change as a $\Delta H$ value or as a heat term in the equation."一个带有 $\Delta H$ 值和所有物种状态符号（s/l/g/aq）的配平方程式。SCH4U D2.2 要求写出"将能量变化表示为 $\Delta H$ 值或方程式中热量项"的热化学方程式。
Worked Example 4 · Scaling thermochemical equations例题 4 · 缩放热化学方程式

Given: $\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$, $\Delta H = -393.5\ \mathrm{kJ/mol}$. Calculate the enthalpy change when (a) $3\ \mathrm{mol}$ of C is combusted completely, and (b) $44\ \mathrm{g}$ of $\text{CO}_2$ is formed from its elements. (Molar mass $\text{CO}_2 = 44\ \mathrm{g/mol}$.)已知：$\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$，$\Delta H = -393.5\ \mathrm{kJ/mol}$。计算以下情况的焓变：(a) $3\ \mathrm{mol}$ 碳完全燃烧，(b) 从其元素生成 $44\ \mathrm{g}$ $\text{CO}_2$。（$\text{CO}_2$ 摩尔质量 $= 44\ \mathrm{g/mol}$。）

(a) Scale by 3.(a) 乘以 3。 $\Delta H = 3 \times (-393.5) = -1180.5\ \mathrm{kJ}$ for $3\ \mathrm{mol}$ C.$\Delta H = 3 \times (-393.5) = -1180.5\ \mathrm{kJ}$，对应 $3\ \mathrm{mol}$ C。

(b) Convert mass to moles.(b) 将质量转换为摩尔数。 $n(\text{CO}_2) = 44/44 = 1\ \mathrm{mol}$. The equation produces $1\ \mathrm{mol}$ $\text{CO}_2$ per mole of C, so $\Delta H = 1 \times (-393.5) = -393.5\ \mathrm{kJ}$. This is the standard enthalpy of formation of $\text{CO}_2$ ($\Delta H^\circ_f$).$n(\text{CO}_2) = 44/44 = 1\ \mathrm{mol}$。该方程式每摩尔 C 产生 $1\ \mathrm{mol}$ $\text{CO}_2$，所以 $\Delta H = 1 \times (-393.5) = -393.5\ \mathrm{kJ}$。这是 $\text{CO}_2$ 的标准生成焓（$\Delta H^\circ_f$）。

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$, $\Delta H = -92\ \mathrm{kJ}$. What is $\Delta H$ for the reverse reaction $2\text{NH}_3(g) \to \text{N}_2(g) + 3\text{H}_2(g)$?对于反应 $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$，$\Delta H = -92\ \mathrm{kJ}$。逆反应 $2\text{NH}_3(g) \to \text{N}_2(g) + 3\text{H}_2(g)$ 的 $\Delta H$ 是多少？
§4 · Q1
$-92\ \mathrm{kJ}$
$+92\ \mathrm{kJ}$
$-46\ \mathrm{kJ}$
$+46\ \mathrm{kJ}$
Reversing a reaction changes the sign of $\Delta H$: forward $\Delta H = -92\ \mathrm{kJ}$ (exothermic); reverse $\Delta H = +92\ \mathrm{kJ}$ (endothermic). This makes physical sense: decomposing $\text{NH}_3$ requires energy input (the reverse of an energy-releasing synthesis).逆转反应改变 $\Delta H$ 的符号：正向 $\Delta H = -92\ \mathrm{kJ}$（放热）；逆向 $\Delta H = +92\ \mathrm{kJ}$（吸热）。这在物理上是合理的：分解 $\text{NH}_3$ 需要能量输入（与释放能量的合成相反）。
Reversal rule: reverse the reaction → negate $\Delta H$. Scaling rule: change coefficients by factor $k$ → multiply $\Delta H$ by $k$. Here, only reversal (no scaling): $\Delta H = -(-92) = +92\ \mathrm{kJ}$.逆转规则：逆转反应 → 取反 $\Delta H$。缩放规则：将系数改变因子 $k$ → 将 $\Delta H$ 乘以 $k$。此处仅逆转（无缩放）：$\Delta H = -(-92) = +92\ \mathrm{kJ}$。
$\Delta H = -286\ \mathrm{kJ}$ for $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{H}_2\text{O}(l)$. What is $\Delta H$ for producing $3\ \mathrm{mol}$ $\text{H}_2\text{O}(l)$ from its elements?$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{H}_2\text{O}(l)$ 的 $\Delta H = -286\ \mathrm{kJ}$。从元素生成 $3\ \mathrm{mol}$ $\text{H}_2\text{O}(l)$ 的 $\Delta H$ 是多少？
§4 · Q2
$-286\ \mathrm{kJ}$
$+858\ \mathrm{kJ}$
$-143\ \mathrm{kJ}$
$-858\ \mathrm{kJ}$
Scale by 3: $\Delta H = 3 \times (-286) = -858\ \mathrm{kJ}$. Producing more moles of product releases proportionally more heat. Sign stays negative (reaction is still exothermic).乘以 3：$\Delta H = 3 \times (-286) = -858\ \mathrm{kJ}$。生成更多摩尔的产物释放成比例更多的热量。符号保持为负（反应仍然是放热的）。
Scaling: multiply $\Delta H$ by the same factor applied to the stoichiometry. For $3\ \mathrm{mol}$ $\text{H}_2\text{O}$: factor $= 3$; $\Delta H = 3 \times (-286) = -858\ \mathrm{kJ}$. The sign does not change (still exothermic).缩放：将 $\Delta H$ 乘以应用于化学计量数的相同因子。对于 $3\ \mathrm{mol}$ $\text{H}_2\text{O}$：因子 $= 3$；$\Delta H = 3 \times (-286) = -858\ \mathrm{kJ}$。符号不变（仍然是放热的）。
Section 5 · Hess's law Honors — US NGSS第 5 节 · 盖斯定律 荣誉 — US NGSS

Hess's Law: Adding Enthalpy Steps盖斯定律：叠加焓步骤

Curriculum note.课纲提示。 This section is core in Ontario SCH4U (D2.5, D3.4) and Alberta Chemistry 30 (Unit A GO1). It carries the Honors chip for US NGSS students (HS-PS1-4 Assessment Boundary explicitly excludes "calculating the total bond energy changes during a chemical reaction from the bond energies of reactants and products"). BC Chemistry 12 has no standalone Hess's law content bullet; treat this as enrichment for a BC student.本节在安大略 SCH4U（D2.5、D3.4）和阿尔伯塔 Chemistry 30（A 单元 GO1）中是核心内容。对 US NGSS 学生标 Honors（HS-PS1-4 评估边界明确排除"从反应物和产物的键能计算化学反应中总键能变化"）。BC Chemistry 12 没有独立的盖斯定律内容条目；对 BC 学生视为拓展内容。
Hess's law: enthalpy is a state function — the route does not matter.盖斯定律：焓是状态函数——路径无关紧要。

Statement:陈述：

The total enthalpy change for a reaction is the sum of the enthalpy changes of any series of steps that connect the same reactants to the same products.一个反应的总焓变等于将相同反应物连接到相同产物的任何一系列步骤的焓变之和。 $$ \Delta H_\text{net} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots $$

Two rules for manipulating steps:操纵步骤的两条规则：

  • Reversal:逆转： flip the equation → change sign of $\Delta H$.翻转方程式 → 改变 $\Delta H$ 的符号。
  • Scaling:缩放： multiply all coefficients by factor $k$ → multiply $\Delta H$ by $k$.将所有系数乘以因子 $k$ → 将 $\Delta H$ 乘以 $k$。
SCH4U D3.4 states: "state Hess's law, and explain, using examples, how it is applied to find the enthalpy changes of a reaction." AB Chemistry 30 GO1: "explain and use Hess' law to calculate energy changes for a net reaction from a series of reactions."SCH4U D3.4 规定："陈述盖斯定律，并通过示例解释如何应用它求反应的焓变。"AB Chemistry 30 GO1："解释和使用盖斯定律从一系列反应计算净反应的能量变化。"
Worked Example 5 · Applying Hess's law例题 5 · 应用盖斯定律

Calculate $\Delta H$ for the reaction $\text{C}(s) + \frac{1}{2}\text{O}_2(g) \to \text{CO}(g)$ using the following data:使用以下数据计算反应 $\text{C}(s) + \frac{1}{2}\text{O}_2(g) \to \text{CO}(g)$ 的 $\Delta H$：

Reaction 1: $\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$, $\Delta H_1 = -393.5\ \mathrm{kJ}$反应 1：$\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$，$\Delta H_1 = -393.5\ \mathrm{kJ}$

Reaction 2: $\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \to \text{CO}_2(g)$, $\Delta H_2 = -283.0\ \mathrm{kJ}$反应 2：$\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \to \text{CO}_2(g)$，$\Delta H_2 = -283.0\ \mathrm{kJ}$

Target analysis.目标分析。 We need $\text{C} + \frac{1}{2}\text{O}_2 \to \text{CO}$. C appears as reactant in Reaction 1 — keep as is. CO must be a product, but it is a reactant in Reaction 2 — reverse it. $\text{CO}_2$ must cancel (it is not in the target).我们需要 $\text{C} + \frac{1}{2}\text{O}_2 \to \text{CO}$。C 作为反应物出现在反应 1 中——保持不变。CO 必须是产物，但它在反应 2 中是反应物——逆转它。$\text{CO}_2$ 必须消去（它不在目标中）。

Step 1 (unchanged):步骤 1（不变）： $\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$, $\Delta H = -393.5\ \mathrm{kJ}$.$\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$，$\Delta H = -393.5\ \mathrm{kJ}$。

Step 2 (reversed):步骤 2（逆转）： $\text{CO}_2(g) \to \text{CO}(g) + \frac{1}{2}\text{O}_2(g)$, $\Delta H = +283.0\ \mathrm{kJ}$.$\text{CO}_2(g) \to \text{CO}(g) + \frac{1}{2}\text{O}_2(g)$，$\Delta H = +283.0\ \mathrm{kJ}$。

Sum (cancel $\text{CO}_2$ and $\frac{1}{2}\text{O}_2$):求和（消去 $\text{CO}_2$ 和 $\frac{1}{2}\text{O}_2$）：

$$ \text{C}(s) + \tfrac{1}{2}\text{O}_2(g) \to \text{CO}(g) $$ $$ \Delta H = -393.5 + 283.0 = -110.5\ \mathrm{kJ}. $$
Hess's law works because enthalpy is:盖斯定律有效是因为焓是：
§5 · Q1
A path function — it depends on how the reaction proceeds路径函数——取决于反应如何进行
Always negative for chemical reactions对化学反应总是负值
A state function — it depends only on the initial and final states状态函数——仅取决于初态和末态
Conserved in all reactions without loss在所有反应中无损耗地守恒
A state function depends only on the current state (reactants and products), not the route taken. Because $\Delta H$ is a state function, you can choose any set of steps that connects the same start and finish — the total $\Delta H$ is always the same. This is Hess's law. SCH4U D3.4 and AB Chemistry 30 GO1 both cite this principle.状态函数只取决于当前状态（反应物和产物），而不取决于所走的路径。因为 $\Delta H$ 是状态函数，您可以选择连接相同起点和终点的任何一组步骤——总 $\Delta H$ 始终相同。这就是盖斯定律。SCH4U D3.4 和 AB Chemistry 30 GO1 都引用了这一原理。
Path functions (like heat $q$ or work $w$) depend on the route. State functions (like $\Delta H$, $\Delta U$, $\Delta G$) depend only on start and finish. Hess's law exploits this: any path gives the same $\Delta H$.路径函数（如热量 $q$ 或功 $w$）取决于路径。状态函数（如 $\Delta H$、$\Delta U$、$\Delta G$）只取决于起点和终点。盖斯定律利用了这一点：任何路径给出相同的 $\Delta H$。
Given: (i) $\text{S}(s) + \text{O}_2(g) \to \text{SO}_2(g)$, $\Delta H = -297\ \mathrm{kJ}$; (ii) $2\text{SO}_3(g) \to 2\text{SO}_2(g) + \text{O}_2(g)$, $\Delta H = +198\ \mathrm{kJ}$. What is $\Delta H$ for $2\text{S}(s) + 3\text{O}_2(g) \to 2\text{SO}_3(g)$?已知：(i) $\text{S}(s) + \text{O}_2(g) \to \text{SO}_2(g)$，$\Delta H = -297\ \mathrm{kJ}$；(ii) $2\text{SO}_3(g) \to 2\text{SO}_2(g) + \text{O}_2(g)$，$\Delta H = +198\ \mathrm{kJ}$。求 $2\text{S}(s) + 3\text{O}_2(g) \to 2\text{SO}_3(g)$ 的 $\Delta H$。
§5 · Q2
$-792\ \mathrm{kJ}$
$+792\ \mathrm{kJ}$
$-396\ \mathrm{kJ}$
$-99\ \mathrm{kJ}$
Step (i) × 2: $2\text{S} + 2\text{O}_2 \to 2\text{SO}_2$, $\Delta H = -594\ \mathrm{kJ}$. Step (ii) reversed: $2\text{SO}_2 + \text{O}_2 \to 2\text{SO}_3$, $\Delta H = -198\ \mathrm{kJ}$. Sum: $\Delta H = -594 + (-198) = -792\ \mathrm{kJ}$. $2\text{SO}_2$ cancels; net: $2\text{S} + 3\text{O}_2 \to 2\text{SO}_3$. ✓步骤 (i) × 2：$2\text{S} + 2\text{O}_2 \to 2\text{SO}_2$，$\Delta H = -594\ \mathrm{kJ}$。步骤 (ii) 逆转：$2\text{SO}_2 + \text{O}_2 \to 2\text{SO}_3$，$\Delta H = -198\ \mathrm{kJ}$。求和：$\Delta H = -594 + (-198) = -792\ \mathrm{kJ}$。$2\text{SO}_2$ 消去；净：$2\text{S} + 3\text{O}_2 \to 2\text{SO}_3$。✓
Target has $2\text{S}$: scale (i) by 2 ($\Delta H = -594$). Target has $\text{SO}_3$ as product, but (ii) has it as reactant: reverse (ii), so $\Delta H = -198$. Add: $-594 + (-198) = -792\ \mathrm{kJ}$.目标有 $2\text{S}$：将 (i) 乘以 2（$\Delta H = -594$）。目标有 $\text{SO}_3$ 作为产物，但 (ii) 中它是反应物：逆转 (ii)，所以 $\Delta H = -198$。相加：$-594 + (-198) = -792\ \mathrm{kJ}$。
Going deeper — standard enthalpies of formation and Hess's law深入 — 标准生成焓与盖斯定律

The standard enthalpy of formation $\Delta H^\circ_f$ is the enthalpy change when $1\ \mathrm{mol}$ of a compound is formed from its constituent elements in their standard states at $25\,^\circ\text{C}$ and $100\ \mathrm{kPa}$. By convention, $\Delta H^\circ_f = 0$ for elements in their standard state (e.g. $\text{H}_2(g)$, $\text{C}(\text{graphite})$). Hess's law gives a shortcut for any reaction:标准生成焓 $\Delta H^\circ_f$ 是在 $25\,^\circ\text{C}$ 和 $100\ \mathrm{kPa}$ 下，由标准状态的组成元素形成 $1\ \mathrm{mol}$ 化合物时的焓变。按惯例，对于标准状态的元素（例如 $\text{H}_2(g)$、$\text{C}(\text{graphite})$），$\Delta H^\circ_f = 0$。盖斯定律为任何反应提供了一个捷径：

$$ \Delta H^\circ_\text{rxn} = \sum n_\text{prod}\,\Delta H^\circ_f(\text{products}) - \sum n_\text{react}\,\Delta H^\circ_f(\text{reactants}) $$

SCH4U D2.7 requires: "calculate the heat of reaction for a formation reaction, using a table of standard enthalpies of formation and applying Hess's law." AB Chemistry 30 GO1 requires: "predict the enthalpy change for chemical equations using standard enthalpies of formation." This formula is the most efficient way to calculate $\Delta H$ for any reaction once a formation-enthalpy table is available.SCH4U D2.7 要求："使用标准生成焓表并应用盖斯定律，计算生成反应的反应热。"AB Chemistry 30 GO1 要求："使用标准生成焓预测化学方程式的焓变。"一旦生成焓表可用，这个公式是计算任何反应 $\Delta H$ 的最有效方式。

Section 6 · Bond energies第 6 节 · 键能

Bond Energies and Enthalpy Estimation键能与焓的估算

Breaking bonds costs energy; making bonds releases energy.断键耗能；成键放能。
  • Bond energy键能 = energy required to break $1\ \mathrm{mol}$ of a specific bond in the gas phase ($\mathrm{kJ/mol}$). Always positive (breaking requires input).= 在气相中断裂 $1\ \mathrm{mol}$ 特定键所需的能量（$\mathrm{kJ/mol}$）。总是正值（断键需要输入）。
  • Bond formation:成键： releases the same magnitude of energy. Breaking $\text{H--H}$ costs $436\ \mathrm{kJ/mol}$; forming $\text{H--H}$ releases $436\ \mathrm{kJ/mol}$.释放相同量的能量。断裂 $\text{H--H}$ 耗费 $436\ \mathrm{kJ/mol}$；形成 $\text{H--H}$ 释放 $436\ \mathrm{kJ/mol}$。
  • Estimation formula:估算公式：
$$ \Delta H \approx \Sigma\,E(\text{bonds broken}) - \Sigma\,E(\text{bonds formed}) $$ This gives an approximate $\Delta H$ (bond energies are averages over many molecules). A positive result ($\Sigma E_\text{broken} > \Sigma E_\text{formed}$) means endothermic; negative means exothermic. NGSS HS-PS1-4 grounds this in "changes in total bond energy." SCH4U D3.2 and AB Chemistry 30 GO2 both reference bond-breaking and bond-forming energy changes explicitly.这给出近似的 $\Delta H$（键能是许多分子的平均值）。正结果（$\Sigma E_\text{broken} > \Sigma E_\text{formed}$）意味着吸热；负结果意味着放热。NGSS HS-PS1-4 将此建立在"总键能的变化"上。SCH4U D3.2 和 AB Chemistry 30 GO2 都明确提到键断裂和键形成的能量变化。
Worked Example 6 · Bond-energy calculation例题 6 · 键能计算

Use bond energies to estimate $\Delta H$ for $\text{H}_2(g) + \text{Cl}_2(g) \to 2\,\text{HCl}(g)$. Bond energies: $\text{H--H} = 436\ \mathrm{kJ/mol}$; $\text{Cl--Cl} = 243\ \mathrm{kJ/mol}$; $\text{H--Cl} = 431\ \mathrm{kJ/mol}$.用键能估算 $\text{H}_2(g) + \text{Cl}_2(g) \to 2\,\text{HCl}(g)$ 的 $\Delta H$。键能：$\text{H--H} = 436\ \mathrm{kJ/mol}$；$\text{Cl--Cl} = 243\ \mathrm{kJ/mol}$；$\text{H--Cl} = 431\ \mathrm{kJ/mol}$。

Bonds broken (reactants).断裂的键（反应物）。 1 mol $\text{H--H}$ + 1 mol $\text{Cl--Cl}$ = $436 + 243 = 679\ \mathrm{kJ}$.1 mol $\text{H--H}$ + 1 mol $\text{Cl--Cl}$ = $436 + 243 = 679\ \mathrm{kJ}$。

Bonds formed (products).形成的键（产物）。 2 mol $\text{H--Cl}$ = $2 \times 431 = 862\ \mathrm{kJ}$.2 mol $\text{H--Cl}$ = $2 \times 431 = 862\ \mathrm{kJ}$。

Apply formula.应用公式。

$$ \Delta H \approx 679 - 862 = -183\ \mathrm{kJ}. $$

Interpretation.解读。 Negative: exothermic. More energy is released forming $\text{H--Cl}$ bonds than is consumed breaking $\text{H--H}$ and $\text{Cl--Cl}$. (Actual: $\Delta H = -184.6\ \mathrm{kJ}$ — the bond-energy estimate is very close here.) ✓负值：放热。形成 $\text{H--Cl}$ 键释放的能量多于断裂 $\text{H--H}$ 和 $\text{Cl--Cl}$ 消耗的能量。（实际值：$\Delta H = -184.6\ \mathrm{kJ}$——键能估算在这里非常接近。）✓

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$, bond energies are: $\text{N}\!\equiv\!\text{N} = 945$, $\text{H--H} = 436$, $\text{N--H} = 391\ \mathrm{kJ/mol}$. Estimate $\Delta H$.对于反应 $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$，键能为：$\text{N}\!\equiv\!\text{N} = 945$，$\text{H--H} = 436$，$\text{N--H} = 391\ \mathrm{kJ/mol}$。估算 $\Delta H$。
§6 · Q1
$+99\ \mathrm{kJ}$
$-99\ \mathrm{kJ}$
$-189\ \mathrm{kJ}$
$+189\ \mathrm{kJ}$
Bonds broken: $1\times945 + 3\times436 = 945 + 1308 = 2253\ \mathrm{kJ}$. Bonds formed (2 NH₃, each with 3 N–H): $2\times3\times391 = 6\times391 = 2346\ \mathrm{kJ}$. $\Delta H \approx 2253 - 2346 = -93\ \mathrm{kJ}$. The closest answer is $-99\ \mathrm{kJ}$ (small rounding differences are acceptable in bond-energy estimation).断裂的键：$1\times945 + 3\times436 = 945 + 1308 = 2253\ \mathrm{kJ}$。形成的键（2个NH₃，每个含3个N-H）：$2\times3\times391 = 6\times391 = 2346\ \mathrm{kJ}$。$\Delta H \approx 2253 - 2346 = -93\ \mathrm{kJ}$。最接近的答案是 $-99\ \mathrm{kJ}$（键能估算中小的舍入误差是可以接受的）。
Broken: 1 N≡N + 3 H–H = 945 + 3×436 = 2253 kJ. Formed: 6 N–H bonds in 2 NH₃ = 6×391 = 2346 kJ. $\Delta H = 2253 - 2346 = -93\ \mathrm{kJ}$ (closest: $-99$ due to rounding in the table). Sign is negative: exothermic.断裂：1 N≡N + 3 H–H = 945 + 3×436 = 2253 kJ。形成：2个NH₃中6个N–H键 = 6×391 = 2346 kJ。$\Delta H = 2253 - 2346 = -93\ \mathrm{kJ}$（最接近 $-99$，因表中的舍入）。符号为负：放热。
A reaction where $\Sigma E_\text{bonds broken} > \Sigma E_\text{bonds formed}$ is:$\Sigma E_\text{bonds broken} > \Sigma E_\text{bonds formed}$ 的反应是：
§6 · Q2
Exothermic, with $\Delta H < 0$放热反应，$\Delta H < 0$
Exothermic, with $\Delta H > 0$放热反应，$\Delta H > 0$
Endothermic, with $\Delta H < 0$吸热反应，$\Delta H < 0$
Endothermic, with $\Delta H > 0$吸热反应，$\Delta H > 0$
$\Delta H \approx \Sigma E_\text{broken} - \Sigma E_\text{formed}$. If broken > formed, then $\Delta H > 0$ → endothermic (more energy was put into breaking bonds than was released forming new ones — the net result is energy absorbed from surroundings). NGSS HS-PS1-4 grounds exo/endo in "changes in total bond energy."$\Delta H \approx \Sigma E_\text{broken} - \Sigma E_\text{formed}$。若断 > 成，则 $\Delta H > 0$ → 吸热（断键输入的能量多于成键释放的能量——净结果是从环境中吸收能量）。NGSS HS-PS1-4 将放热/吸热建立在"总键能的变化"上。
Formula: $\Delta H = \Sigma E_\text{broken} - \Sigma E_\text{formed}$. If broken > formed → result is positive → $\Delta H > 0$ → endothermic. Exothermic requires $\Sigma E_\text{formed} > \Sigma E_\text{broken}$ so that $\Delta H < 0$.公式：$\Delta H = \Sigma E_\text{broken} - \Sigma E_\text{formed}$。若断 > 成 → 结果为正 → $\Delta H > 0$ → 吸热。放热要求 $\Sigma E_\text{formed} > \Sigma E_\text{broken}$，使得 $\Delta H < 0$。
Section 7 · Potential-energy diagrams第 7 节 · 能量图

Potential-Energy Diagrams势能图

A PE diagram shows the energy profile of a reaction from reactants to products via the transition state.PE 图展示了反应从反应物经过过渡态到产物的能量轮廓。
  • Reactants and products反应物与产物 — drawn as horizontal plateaus at their respective enthalpy levels. $\Delta H = H_\text{products} - H_\text{reactants}$ is the vertical distance (negative for exothermic, positive for endothermic).——以各自的焓水平绘制为水平台阶。$\Delta H = H_\text{products} - H_\text{reactants}$ 是垂直距离（放热为负，吸热为正）。
  • Transition state (activated complex)过渡态（活化复合物） — the energy maximum along the reaction coordinate. Reactants must surmount this barrier to react.——沿反应坐标的能量最大值。反应物必须克服这一能垒才能反应。
  • Activation energy $E_a$活化能 $E_a$ — the energy difference between reactants and the transition state (always positive). $E_{a,\text{forward}}$ is the barrier from left; $E_{a,\text{reverse}}$ is the barrier from right. Note: $E_{a,\text{forward}} - E_{a,\text{reverse}} = \Delta H$.——反应物与过渡态之间的能量差（总是正值）。$E_{a,\text{forward}}$ 是从左侧的能垒；$E_{a,\text{reverse}}$ 是从右侧的能垒。注意：$E_{a,\text{forward}} - E_{a,\text{reverse}} = \Delta H$。
  • Catalyst effect催化剂效果 — lowers $E_a$ (both forward and reverse by the same amount), providing an alternate pathway. Does not change $\Delta H$ (reactant and product energies are unchanged). SCH4U D3.6 and AB Chemistry 30 GO2 both require labelling energy diagrams with $\Delta H$, $E_a$, and the effect of catalysts.——降低 $E_a$（正向和逆向降低相同数量），提供替代途径。不改变 $\Delta H$（反应物和产物能量不变）。SCH4U D3.6 和 AB Chemistry 30 GO2 都要求在能量图上标注 $\Delta H$、$E_a$ 和催化剂的效果。
Worked Example 7 · Reading a potential-energy diagram例题 7 · 解读势能图

A PE diagram shows: reactants at $40\ \mathrm{kJ}$, transition state at $150\ \mathrm{kJ}$, and products at $10\ \mathrm{kJ}$ (all relative to a reference zero). Find: (a) $\Delta H$, (b) $E_{a,\text{forward}}$, (c) $E_{a,\text{reverse}}$, and (d) what happens to these values when a catalyst is added.PE 图显示：反应物在 $40\ \mathrm{kJ}$，过渡态在 $150\ \mathrm{kJ}$，产物在 $10\ \mathrm{kJ}$（均相对于参考零点）。求：(a) $\Delta H$，(b) $E_{a,\text{forward}}$，(c) $E_{a,\text{reverse}}$，(d) 加入催化剂后这些值的变化。

(a) $\Delta H = H_\text{products} - H_\text{reactants} = 10 - 40 = -30\ \mathrm{kJ}$.(a) $\Delta H = H_\text{products} - H_\text{reactants} = 10 - 40 = -30\ \mathrm{kJ}$。 Negative → exothermic.负值 → 放热。

(b) $E_{a,\text{forward}} = H_\text{TS} - H_\text{reactants} = 150 - 40 = 110\ \mathrm{kJ}$.(b) $E_{a,\text{forward}} = H_\text{TS} - H_\text{reactants} = 150 - 40 = 110\ \mathrm{kJ}$。

(c) $E_{a,\text{reverse}} = H_\text{TS} - H_\text{products} = 150 - 10 = 140\ \mathrm{kJ}$.(c) $E_{a,\text{reverse}} = H_\text{TS} - H_\text{products} = 150 - 10 = 140\ \mathrm{kJ}$。 Check: $E_{a,\text{forward}} - E_{a,\text{reverse}} = 110 - 140 = -30 = \Delta H$. ✓核验：$E_{a,\text{forward}} - E_{a,\text{reverse}} = 110 - 140 = -30 = \Delta H$。✓

(d) Catalyst.(d) 催化剂。 Lowers both $E_{a,\text{forward}}$ and $E_{a,\text{reverse}}$ by the same amount. $\Delta H$ stays at $-30\ \mathrm{kJ}$ (reactant and product energy levels are unchanged). Reaction is faster but not different in energy terms.将 $E_{a,\text{forward}}$ 和 $E_{a,\text{reverse}}$ 降低相同数量。$\Delta H$ 保持为 $-30\ \mathrm{kJ}$（反应物和产物能量水平不变）。反应更快，但在能量方面没有差异。

On a PE diagram for an exothermic reaction, where is the transition state located relative to the reactants and products?在放热反应的 PE 图上，过渡态相对于反应物和产物位于哪里？
§7 · Q1
Below both reactants and products低于反应物和产物两者
At the same level as the products与产物在同一水平
At the same level as the reactants与反应物在同一水平
Above both reactants and products高于反应物和产物两者
The transition state is always the energy maximum — it sits above both reactants and products for any reaction (exo or endo). For an exothermic reaction, products are lower than reactants, but both are lower than the transition state. SCH4U D3.6 and AB Chemistry 30 GO2 require analysing and labelling such diagrams.过渡态总是能量最大值——对于任何反应（放热或吸热），它都位于反应物和产物的上方。对于放热反应，产物低于反应物，但两者都低于过渡态。SCH4U D3.6 和 AB Chemistry 30 GO2 要求分析和标注此类图表。
The transition state (activated complex) is always a maximum — reactants need to climb to it before falling to products. It is always above both, regardless of whether the reaction is exo or endo.过渡态（活化复合物）总是一个最大值——反应物需要攀升到它，然后才能落到产物。无论反应是放热还是吸热，它总是高于两者。
Adding a catalyst to a reaction changes which of the following?向反应中加入催化剂会改变以下哪项？
§7 · Q2
Activation energy ($E_a$) only — not $\Delta H$仅活化能（$E_a$）——不改变 $\Delta H$
$\Delta H$ only — not $E_a$仅 $\Delta H$——不改变 $E_a$
Both $E_a$ and $\Delta H$ equally$E_a$ 和 $\Delta H$ 都均等地改变
Neither $E_a$ nor $\Delta H$$E_a$ 和 $\Delta H$ 都不改变
A catalyst provides an alternate, lower-energy pathway (lowers the activation energy $E_a$, both forward and reverse). The energy of reactants and products is unchanged, so $\Delta H$ stays the same. The reaction reaches equilibrium faster, but equilibrium itself is not shifted. AB Chemistry 30 GO2 and BC Chemistry 12 elaboration both cite this principle.催化剂提供了一条替代的、能量更低的反应途径（降低活化能 $E_a$，正向和逆向均降低）。反应物和产物的能量不变，所以 $\Delta H$ 保持不变。反应更快达到平衡，但平衡本身不移动。AB Chemistry 30 GO2 和 BC Chemistry 12 细化都引用了这一原理。
Catalysts lower the energy barrier ($E_a$) but do not change the energy of reactants or products — so $\Delta H$ is unaffected. Think of a catalyst as finding a lower mountain pass between reactant valley and product valley: the pass is lower, but both valleys are at the same altitude.催化剂降低能垒（$E_a$），但不改变反应物或产物的能量——所以 $\Delta H$ 不受影响。想象催化剂找到了反应物谷和产物谷之间较低的山口：山口更低，但两个谷仍在相同的海拔。
Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Sign-convention discipline for every thermochemistry question每道热化学题的符号约定纪律
  • State the system and surroundings before you write any signs.在写任何符号之前先说明系统和环境。 $q_\text{system} + q_\text{surroundings} = 0$. If the calorimeter water warms up ($\Delta T > 0$), the reaction released heat ($q_\text{rxn} < 0$, exothermic). If the water cools ($\Delta T < 0$), the reaction absorbed heat ($q_\text{rxn} > 0$, endothermic).$q_\text{system} + q_\text{surroundings} = 0$。如果量热器中的水升温（$\Delta T > 0$），反应释放了热量（$q_\text{rxn} < 0$，放热）。如果水冷却（$\Delta T < 0$），反应吸收了热量（$q_\text{rxn} > 0$，吸热）。
  • In $q = mc\Delta T$: use grams, not kilograms, when $c$ is in $\mathrm{J\,g^{-1}\,^\circ C^{-1}}$.在 $q = mc\Delta T$ 中：当 $c$ 单位为 $\mathrm{J\,g^{-1}\,^\circ C^{-1}}$ 时使用克，不用千克。 Convert kJ to J (multiply by 1000) before dividing to find $\Delta T$ or $m$. The most common error is mixing units — always check.在除法求 $\Delta T$ 或 $m$ 之前将 kJ 转换为 J（乘以 1000）。最常见的错误是混淆单位——始终检查。
  • For Hess's law: cancel species systematically.对于盖斯定律：系统地消去物种。 Write target on one side. Check each species: if it is a reactant in target, it must appear as a reactant in the steps you keep (possibly after scaling/reversal); if it should cancel, it must appear on opposite sides in the steps and in equal amounts.将目标写在一边。检查每个物种：如果它是目标中的反应物，它必须在你保留的步骤中作为反应物出现（可能在缩放/逆转之后）；如果它应该消去，它必须在步骤的两侧出现，且数量相等。
Calorimetry (§3) and enthalpy (§4)量热法（§3）与焓（§4）
  • Mass of solution includes the dissolved solute.溶液的质量包括溶解的溶质。 If $5\ \mathrm{g}$ NaOH dissolves in $100\ \mathrm{g}$ water, use $m = 105\ \mathrm{g}$ for $q = mc\Delta T$.如果 $5\ \mathrm{g}$ NaOH 溶解在 $100\ \mathrm{g}$ 水中，使用 $m = 105\ \mathrm{g}$ 代入 $q = mc\Delta T$。
  • $\Delta H$ scales; $q$ does not.$\Delta H$ 可缩放；$q$ 不能。 $\Delta H$ is intensive when reported per mole; the $q$ from a calorimeter experiment is the total heat for the specific amount of reactant used. Convert: $\Delta H = q / n$ (kJ per mole of reactant).$\Delta H$ 以每摩尔为单位报告时是强度量；量热计实验中的 $q$ 是用于特定量反应物的总热量。转换：$\Delta H = q / n$（每摩尔反应物的 kJ）。
Hess's law (§5) Honors — US NGSS and bond energies (§6)盖斯定律（§5）荣誉 — US NGSS 与键能（§6）
  • Two rules only: reverse sign, or scale $\Delta H$.只有两条规则：反号，或缩放 $\Delta H$。 If you flip an equation, negate $\Delta H$. If you multiply coefficients by $k$, multiply $\Delta H$ by $k$. Do both in sequence if needed.如果翻转方程式，则取反 $\Delta H$。如果将系数乘以 $k$，则将 $\Delta H$ 乘以 $k$。如果需要，依次执行两者。
  • Bond-energy estimates are always approximate.键能估算总是近似的。 Bond energies in tables are averages across many compounds; your calculated $\Delta H$ will differ slightly from the true value. Always state "estimated" or "approximate" when using this method.表格中的键能是许多化合物的平均值；你计算的 $\Delta H$ 会与真实值略有不同。使用此方法时始终说明"估计"或"近似"。
  • Count bonds carefully in structural formulas.仔细计算结构式中的键。 For each reactant and product, draw the Lewis structure or count bonds from the molecular formula. $\text{CH}_4$ has 4 C–H bonds; $\text{CO}_2$ has 2 C=O double bonds (each worth the C=O double-bond energy).对于每种反应物和产物，画出路易斯结构或从分子式计算键数。$\text{CH}_4$ 有 4 个 C-H 键；$\text{CO}_2$ 有 2 个 C=O 双键（每个价值 C=O 双键能量）。
PE diagrams (§7) and answer hygienePE 图（§7）与作答规范
  • Label all five features of a PE diagram.标注 PE 图的所有五个特征。 Reactants, products, transition state, $\Delta H$ (with sign and arrow direction), $E_a$ forward (with arrow from reactants to TS), $E_a$ reverse (with arrow from products to TS). Missing a label loses marks.反应物、产物、过渡态、$\Delta H$（带符号和箭头方向）、正向 $E_a$（从反应物到过渡态的箭头）、逆向 $E_a$（从产物到过渡态的箭头）。漏掉标注会失分。
  • Significant figures in $q = mc\Delta T$.$q = mc\Delta T$ 中的有效数字。 The answer can only have as many significant figures as the least precise measurement. Typically $\Delta T$ limits precision (thermometers read to $\pm 0.1\,^\circ\text{C}$).答案的有效数字不能多于精度最低的测量值。通常 $\Delta T$ 限制了精度（温度计读数精确到 $\pm 0.1\,^\circ\text{C}$）。
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Exothermic reaction ($\Delta H$)?放热反应（$\Delta H$）？
$\Delta H < 0$. Products lower in enthalpy than reactants. Heat released to surroundings — surroundings warm up.$\Delta H < 0$。产物焓低于反应物。热量释放到环境——环境升温。
Endothermic reaction ($\Delta H$)?吸热反应（$\Delta H$）？
$\Delta H > 0$. Products higher in enthalpy than reactants. Heat absorbed from surroundings — surroundings cool down.$\Delta H > 0$。产物焓高于反应物。从环境吸收热量——环境降温。
Calorimetry formula?量热法公式？
$$q = mc\Delta T$$ $m$ = mass (g), $c$ = specific heat (J g⁻¹ °C⁻¹), $\Delta T = T_\text{f} - T_\text{i}$. Water: $c = 4.18$ J g⁻¹ °C⁻¹.$m$ = 质量（g），$c$ = 比热容（J g⁻¹ °C⁻¹），$\Delta T = T_\text{f} - T_\text{i}$。水：$c = 4.18$ J g⁻¹ °C⁻¹。
Conservation in a calorimetry experiment?量热实验中的守恒？
$q_\text{reaction} = -q_\text{water}$. Heat lost by reaction = heat gained by water (or vice versa). Total system energy is conserved.$q_\text{reaction} = -q_\text{water}$。反应失去的热量 = 水获得的热量（反之亦然）。系统总能量守恒。
Enthalpy change: scaling rule?焓变：缩放规则？
Multiply all coefficients by $k$ → multiply $\Delta H$ by $k$. Reverse the equation → change sign of $\Delta H$.将所有系数乘以 $k$ → 将 $\Delta H$ 乘以 $k$。逆转方程式 → 改变 $\Delta H$ 的符号。
Hess's law?盖斯定律？
$\Delta H_\text{net} = \Delta H_1 + \Delta H_2 + \cdots$. Enthalpy is a state function: total $\Delta H$ is the same regardless of the route taken.$\Delta H_\text{net} = \Delta H_1 + \Delta H_2 + \cdots$。焓是状态函数：总 $\Delta H$ 与所走路径无关，始终相同。
Standard enthalpy of formation ($\Delta H^\circ_f$)?标准生成焓（$\Delta H^\circ_f$）？
Enthalpy change to form 1 mol of a compound from its elements in standard states at 25°C. $\Delta H^\circ_f = 0$ for elements in standard state.在 25°C 标准状态下从元素形成 1 mol 化合物的焓变。标准状态的元素 $\Delta H^\circ_f = 0$。
Bond energy: breaking vs forming?键能：断裂与形成？
Breaking bonds absorbs energy (endothermic step; +). Forming bonds releases energy (exothermic step; −). Bond energy is always given as a positive value (for breaking).断键吸收能量（吸热步骤；+）。成键释放能量（放热步骤；−）。键能总以正值给出（断裂时）。
$\Delta H$ from bond energies?由键能计算 $\Delta H$？
$$\Delta H \approx \Sigma E(\text{broken}) - \Sigma E(\text{formed})$$ Positive result → endothermic; negative → exothermic.结果为正 → 吸热；结果为负 → 放热。
Activation energy $E_a$?活化能 $E_a$？
Energy needed to reach the transition state from reactants. Always positive. Forward $E_a$ − reverse $E_a$ = $\Delta H$.从反应物到达过渡态所需的能量。总是正值。正向 $E_a$ − 逆向 $E_a$ = $\Delta H$。
Catalyst effect on PE diagram?催化剂对 PE 图的影响？
Lowers $E_a$ (both forward and reverse by the same amount). Does NOT change $\Delta H$ — reactant and product energy levels are unchanged.降低 $E_a$（正向和逆向降低相同数量）。不改变 $\Delta H$——反应物和产物能量水平不变。
Enthalpy vs temperature: key distinction?焓与温度的关键区别？
Temperature = average KE per particle (intensive). Enthalpy = total heat flow at constant pressure (extensive). Same temperature ≠ same heat transfer.温度 = 每个粒子的平均动能（强度量）。焓 = 恒压下的总热量流动（广延量）。温度相同 ≠ 热传导量相同。
Sign of $q_\text{reaction}$ when calorimeter water warms?当量热器中的水升温时 $q_\text{reaction}$ 的符号？
Water warms → $q_\text{water} > 0$ → $q_\text{reaction} = -q_\text{water} < 0$ → reaction is exothermic ($\Delta H < 0$).水升温 → $q_\text{water} > 0$ → $q_\text{reaction} = -q_\text{water} < 0$ → 反应为放热（$\Delta H < 0$）。
Mixed Practice综合练习

Practice Quiz综合测验

A hand warmer uses the oxidation of iron: $4\text{Fe}(s) + 3\text{O}_2(g) \to 2\text{Fe}_2\text{O}_3(s)$, $\Delta H = -1648\ \mathrm{kJ}$. This reaction is: 🇺🇸 NGSS HS-PS1-4 / 🇨🇦 SCH4U D3.2暖手宝使用铁的氧化：$4\text{Fe}(s) + 3\text{O}_2(g) \to 2\text{Fe}_2\text{O}_3(s)$，$\Delta H = -1648\ \mathrm{kJ}$。此反应是：🇺🇸 NGSS HS-PS1-4 / 🇨🇦 SCH4U D3.2
Q1
Endothermic; the surroundings cool down吸热；环境温度降低
Endothermic; the surroundings warm up吸热；环境温度升高
Exothermic; the surroundings warm up放热；环境温度升高
Exothermic; the surroundings cool down放热；环境温度降低
$\Delta H = -1648\ \mathrm{kJ} < 0$ → exothermic. Products ($\text{Fe}_2\text{O}_3$) are at lower enthalpy than reactants. Heat flows to the surroundings (your hand), which warm up — the whole point of a hand warmer.$\Delta H = -1648\ \mathrm{kJ} < 0$ → 放热。产物（$\text{Fe}_2\text{O}_3$）的焓低于反应物。热量流向环境（你的手），使其升温——这正是暖手宝的作用。
Negative $\Delta H$ = exothermic = surroundings gain heat = surroundings warm up. Endothermic would have positive $\Delta H$ and cool surroundings (instant cold pack).$\Delta H$ 为负 = 放热 = 环境获得热量 = 环境升温。吸热会有正 $\Delta H$ 并使环境降温（即时冷敷包）。
$250\ \mathrm{g}$ of water absorbs $12\,500\ \mathrm{J}$ of heat. What temperature rise does it experience? ($c = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$) 🇨🇦 SCH4U D2.3 / AB Chem 30 GO1$250\ \mathrm{g}$ 水吸收了 $12\,500\ \mathrm{J}$ 的热量。它的温升是多少？（$c = 4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}$）🇨🇦 SCH4U D2.3 / AB Chem 30 GO1
Q2
$6.0\,^\circ\text{C}$
$11.96\,^\circ\text{C}$
$50.0\,^\circ\text{C}$
$23.9\,^\circ\text{C}$
$\Delta T = q/(mc) = 12\,500 / (250 \times 4.18) = 12\,500 / 1045 = 11.96\,^\circ\text{C} \approx 12.0\,^\circ\text{C}$. Rearrange $q = mc\Delta T$ to solve for $\Delta T$.$\Delta T = q/(mc) = 12\,500 / (250 \times 4.18) = 12\,500 / 1045 = 11.96\,^\circ\text{C} \approx 12.0\,^\circ\text{C}$。整理 $q = mc\Delta T$ 求解 $\Delta T$。
Rearrange $q = mc\Delta T$: $\Delta T = q/(mc) = 12500/(250 \times 4.18)$. Denominator $= 1045$; $\Delta T = 12500/1045 \approx 12\,^\circ\text{C}$.整理 $q = mc\Delta T$：$\Delta T = q/(mc) = 12500/(250 \times 4.18)$。分母 $= 1045$；$\Delta T = 12500/1045 \approx 12\,^\circ\text{C}$。
For the reaction $\text{A} \to \text{B}$, $\Delta H = -80\ \mathrm{kJ}$. What is $\Delta H$ for $3\text{B} \to 3\text{A}$? 🇨🇦 SCH4U D2.2 / AB Chem 30 GO1对于反应 $\text{A} \to \text{B}$，$\Delta H = -80\ \mathrm{kJ}$。$3\text{B} \to 3\text{A}$ 的 $\Delta H$ 是多少？🇨🇦 SCH4U D2.2 / AB Chem 30 GO1
Q3
$-80\ \mathrm{kJ}$
$-240\ \mathrm{kJ}$
$+80\ \mathrm{kJ}$
$+240\ \mathrm{kJ}$
Two operations: (1) reverse → $\text{B} \to \text{A}$, $\Delta H = +80\ \mathrm{kJ}$; (2) scale by 3 → $3\text{B} \to 3\text{A}$, $\Delta H = 3 \times (+80) = +240\ \mathrm{kJ}$. Both reversal (negate) and scaling (multiply) applied.两步操作：(1) 逆转 → $\text{B} \to \text{A}$，$\Delta H = +80\ \mathrm{kJ}$；(2) 乘以 3 → $3\text{B} \to 3\text{A}$，$\Delta H = 3 \times (+80) = +240\ \mathrm{kJ}$。同时应用了逆转（取反）和缩放（乘法）。
Reverse: negate $\Delta H$ → $+80$. Scale by 3: multiply by 3 → $+240\ \mathrm{kJ}$. The sign is positive because we reversed the direction (endothermic after reversal), then tripled the magnitude.逆转：取反 $\Delta H$ → $+80$。乘以 3：乘以 3 → $+240\ \mathrm{kJ}$。符号为正，因为我们逆转了方向（逆转后为吸热），然后三倍化了数值。
Using bond energies: $\text{H}_2(g) + \text{F}_2(g) \to 2\,\text{HF}(g)$. Bond energies: H–H = 436, F–F = 155, H–F = 565 kJ/mol. Estimated $\Delta H$? 🇺🇸 NGSS HS-PS1-4 / 🇨🇦 AB Chem 30 GO2使用键能：$\text{H}_2(g) + \text{F}_2(g) \to 2\,\text{HF}(g)$。键能：H–H = 436，F–F = 155，H–F = 565 kJ/mol。估算 $\Delta H$？🇺🇸 NGSS HS-PS1-4 / 🇨🇦 AB Chem 30 GO2
Q4
$-539\ \mathrm{kJ}$
$+539\ \mathrm{kJ}$
$-279\ \mathrm{kJ}$
$+279\ \mathrm{kJ}$
Broken: 1 H–H + 1 F–F = 436 + 155 = 591 kJ. Formed: 2 H–F = 2 × 565 = 1130 kJ. $\Delta H = 591 - 1130 = -539\ \mathrm{kJ}$. Highly exothermic (fluorine–hydrogen bond formation releases a large amount of energy).断裂：1 H–H + 1 F–F = 436 + 155 = 591 kJ。形成：2 H–F = 2 × 565 = 1130 kJ。$\Delta H = 591 - 1130 = -539\ \mathrm{kJ}$。高度放热（氟氢键的形成释放大量能量）。
$\Delta H = \Sigma E_\text{broken} - \Sigma E_\text{formed}$. Broken: $436 + 155 = 591$. Formed: $2 \times 565 = 1130$. $\Delta H = 591 - 1130 = -539\ \mathrm{kJ}$. Negative = exothermic.$\Delta H = \Sigma E_\text{broken} - \Sigma E_\text{formed}$。断裂：$436 + 155 = 591$。形成：$2 \times 565 = 1130$。$\Delta H = 591 - 1130 = -539\ \mathrm{kJ}$。负值 = 放热。
A PE diagram shows reactants at 60 kJ, transition state at 200 kJ, products at 20 kJ. What is $E_{a,\text{reverse}}$? 🇨🇦 SCH4U D3.6 / AB Chem 30 GO2PE 图显示反应物在 60 kJ，过渡态在 200 kJ，产物在 20 kJ。$E_{a,\text{reverse}}$ 是多少？🇨🇦 SCH4U D3.6 / AB Chem 30 GO2
Q5
$140\ \mathrm{kJ}$
$200\ \mathrm{kJ}$
$180\ \mathrm{kJ}$
$40\ \mathrm{kJ}$
$E_{a,\text{reverse}} = H_\text{TS} - H_\text{products} = 200 - 20 = 180\ \mathrm{kJ}$. The reverse activation energy is the energy barrier from the products' perspective (how high they must climb to reach the transition state). Check: $\Delta H = E_{a,\text{fwd}} - E_{a,\text{rev}} = (200-60) - 180 = 140 - 180 = -40\ \mathrm{kJ}$. ✓$E_{a,\text{reverse}} = H_\text{TS} - H_\text{products} = 200 - 20 = 180\ \mathrm{kJ}$。逆向活化能是从产物角度看的能垒（它们必须攀升多高才能到达过渡态）。核验：$\Delta H = E_{a,\text{fwd}} - E_{a,\text{rev}} = (200-60) - 180 = 140 - 180 = -40\ \mathrm{kJ}$。✓
$E_{a,\text{reverse}}$ = height of transition state above the products = $200 - 20 = 180\ \mathrm{kJ}$. $E_{a,\text{forward}}$ = height above reactants = $200 - 60 = 140\ \mathrm{kJ}$. Difference = $\Delta H = -40\ \mathrm{kJ}$ (exothermic). ✓$E_{a,\text{reverse}}$ = 过渡态高于产物的高度 = $200 - 20 = 180\ \mathrm{kJ}$。$E_{a,\text{forward}}$ = 高于反应物的高度 = $200 - 60 = 140\ \mathrm{kJ}$。差值 = $\Delta H = -40\ \mathrm{kJ}$（放热）。✓
Use Hess's law to find $\Delta H$ for $\text{S}(s) + \text{O}_2(g) \to \text{SO}_2(g)$ given: (i) $2\text{SO}_2(g) + \text{O}_2(g) \to 2\text{SO}_3(g)$, $\Delta H = -196\ \mathrm{kJ}$; (ii) $2\text{S}(s) + 3\text{O}_2(g) \to 2\text{SO}_3(g)$, $\Delta H = -790\ \mathrm{kJ}$. 🇨🇦 SCH4U D2.5 / AB Chem 30 GO1使用盖斯定律求 $\text{S}(s) + \text{O}_2(g) \to \text{SO}_2(g)$ 的 $\Delta H$，已知：(i) $2\text{SO}_2(g) + \text{O}_2(g) \to 2\text{SO}_3(g)$，$\Delta H = -196\ \mathrm{kJ}$；(ii) $2\text{S}(s) + 3\text{O}_2(g) \to 2\text{SO}_3(g)$，$\Delta H = -790\ \mathrm{kJ}$。🇨🇦 SCH4U D2.5 / AB Chem 30 GO1
Q6
$-790\ \mathrm{kJ}$
$-297\ \mathrm{kJ}$
$+297\ \mathrm{kJ}$
$-593\ \mathrm{kJ}$
Target: 1 mol S → 1 mol SO₂. Use (ii) ÷ 2: $\text{S} + \frac{3}{2}\text{O}_2 \to \text{SO}_3$, $\Delta H = -395\ \mathrm{kJ}$. Reverse (i) ÷ 2: $\text{SO}_3 \to \text{SO}_2 + \frac{1}{2}\text{O}_2$, $\Delta H = +98\ \mathrm{kJ}$. Sum: $\text{S} + \text{O}_2 \to \text{SO}_2$, $\Delta H = -395 + 98 = -297\ \mathrm{kJ}$. SO₃ cancels. ✓目标：1 mol S → 1 mol SO₂。用 (ii) ÷ 2：$\text{S} + \frac{3}{2}\text{O}_2 \to \text{SO}_3$，$\Delta H = -395\ \mathrm{kJ}$。逆转 (i) ÷ 2：$\text{SO}_3 \to \text{SO}_2 + \frac{1}{2}\text{O}_2$，$\Delta H = +98\ \mathrm{kJ}$。求和：$\text{S} + \text{O}_2 \to \text{SO}_2$，$\Delta H = -395 + 98 = -297\ \mathrm{kJ}$。SO₃ 消去。✓
Scale (ii) by ½ → $\text{S}+\frac{3}{2}\text{O}_2 \to \text{SO}_3$, $\Delta H = -395$. Scale and reverse (i) by ½ → $\text{SO}_3 \to \text{SO}_2+\frac{1}{2}\text{O}_2$, $\Delta H = +98$. Add: $-395+98 = -297\ \mathrm{kJ}$.将 (ii) 乘以 ½ → $\text{S}+\frac{3}{2}\text{O}_2 \to \text{SO}_3$，$\Delta H = -395$。将 (i) 乘以 ½ 并逆转 → $\text{SO}_3 \to \text{SO}_2+\frac{1}{2}\text{O}_2$，$\Delta H = +98$。相加：$-395+98 = -297\ \mathrm{kJ}$。
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Next Steps后续单元

What This Feeds Into本单元的去向

Thermochemistry is the quantitative backbone of reaction chemistry. The enthalpy concepts and calculation skills you built here recur directly in reaction rates (activation energy on PE diagrams), chemical equilibrium (Le Chatelier shifts depend on whether $\Delta H$ is positive or negative), electrochemistry (cell potentials connect to $\Delta G$), and every applied chemistry context from combustion engineering to biochemistry. The cross-references below point at the college-credit feeder and the next High School Chemistry unit.热化学是反应化学的定量骨干。你在这里建立的焓概念和计算技能直接在反应速率（PE 图上的活化能）、化学平衡（勒夏特列位移取决于 $\Delta H$ 的正负）、电化学（电池电位与 $\Delta G$ 相关）以及从燃烧工程到生物化学的每个应用化学背景中出现。以下链接指向大学学分衔接课程和下一个高中化学单元。

Within High School Chemistry.在 HS Chemistry 内部。

Reaction Rates (Unit 11) extends §7 of this guide: collision theory adds molecular-level explanation to activation energy, and the Arrhenius equation quantifies how $E_a$ and temperature determine rate. Chemical Equilibrium (Unit 12) uses $\Delta H$ to predict Le Chatelier shifts — if $\Delta H > 0$, increasing temperature shifts the equilibrium toward products; if $\Delta H < 0$, toward reactants. Electrochemistry (Unit 13) connects $\Delta H$, $\Delta G$, and cell potential through thermodynamic relationships.反应速率（第 11 单元）延伸了本指南的 §7：碰撞理论为活化能添加了分子级解释，阿伦尼乌斯方程量化了 $E_a$ 和温度如何决定速率。化学平衡（第 12 单元）使用 $\Delta H$ 来预测勒夏特列位移——如果 $\Delta H > 0$，升温使平衡向产物方向移动；如果 $\Delta H < 0$，则向反应物方向移动。电化学（第 13 单元）通过热力学关系将 $\Delta H$、$\Delta G$ 和电池电位联系起来。

College-credit feeder and cross-subject links.大学学分衔接与跨学科链接。

IB Chemistry HL · Reactivity 1: What Drives Chemical Reactions? (the college-credit feeder for enthalpy, Hess's law, bond energies, and the thermodynamic basis of spontaneity at IB depth)IB Chemistry HL · Reactivity 1：是什么驱动化学反应？（焓、盖斯定律、键能和 IB 深度下自发性热力学基础的大学学分衔接）

If you are aiming for IB Chemistry HL or AP Chemistry, the calorimetry, $\Delta H$, Hess's law, and bond-energy skills here are assumed from the first thermochemistry problem set. IB Chemistry HL Reactivity 1 extends this with entropy ($\Delta S$), Gibbs free energy ($\Delta G = \Delta H - T\Delta S$), and spontaneity. AP Chemistry Unit 6 (Thermodynamics) adds Born-Haber cycles, lattice energy, and the connection between cell potential and $\Delta G^\circ = -nFE^\circ$. The PE diagram skills you built in §7 appear in every kinetics and equilibrium treatment at the IB/AP level.如果你目标是 IB Chemistry HL 或 AP Chemistry，这里的量热法、$\Delta H$、盖斯定律和键能技能从第一套热化学习题起就被默认掌握。IB Chemistry HL Reactivity 1 通过熵（$\Delta S$）、吉布斯自由能（$\Delta G = \Delta H - T\Delta S$）和自发性来延伸这部分内容。AP Chemistry Unit 6（热力学）增加了玻恩-哈伯循环、晶格能以及电池电位与 $\Delta G^\circ = -nFE^\circ$ 之间的联系。你在 §7 中建立的 PE 图技能在 IB/AP 水平的每个动力学和平衡处理中都会出现。

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High School Chemistry · Thermochemistry and Energy · v1高中化学 · 热化学与能量 · v1