Enthalpy, bond energies, Hess's law, fuels, and the thermodynamic principles that determine whether a reaction will occur. Master the energy accounting behind every chemical change.焓（enthalpy）、键能（bond enthalpy）、盖斯定律（Hess's law）、燃料以及决定反应能否发生的热力学原理。掌握每一次化学变化背后的能量账本。
Chemical reactions involve energy transfer between a system and its surroundings. The system is the reaction itself; the surroundings include the solvent, container, and everything else. Total energy is always conserved, but it moves — and the direction of that transfer is what we measure.化学反应伴随着体系（system）与环境（surroundings）之间的能量转移。体系就是反应本身；环境包括溶剂、容器以及其余一切。总能量守恒，但能量会流动——我们要测量的正是这一流动的方向。
Heat vs. Temperature热量与温度的区别Heat ($Q$) is energy transferred due to a temperature difference — it's measured in joules. Temperature ($T$) is a measure of average kinetic energy of particles. Adding the same amount of heat to a small beaker of water causes a larger temperature increase than adding it to a bathtub. They are fundamentally different quantities.热量 ($Q$) 是因温度差而转移的能量，单位是焦耳。温度 ($T$) 衡量粒子的平均动能。把同样多的热量加到一小杯水里，温升远大于加到一整缸洗澡水中。两者是本质上不同的物理量。
Exothermic and Endothermic Reactions放热反应与吸热反应
An exothermic reaction releases energy from the system to the surroundings — the temperature of the surroundings increases. $\Delta H$ is negative. An endothermic reaction absorbs energy from the surroundings — the temperature decreases. $\Delta H$ is positive. The sign convention reflects the system's perspective: losing energy means $\Delta H < 0$.放热（exothermic）反应把能量从体系释放给环境——环境温度升高，$\Delta H$ 为负。吸热（endothermic）反应从环境吸收能量——环境温度降低，$\Delta H$ 为正。符号约定从体系的视角出发：体系失去能量即 $\Delta H < 0$。
Relative Stability相对稳定性In an exothermic reaction, the products are more stable (lower energy) than the reactants. The energy difference is released. In an endothermic reaction, the products are less stable (higher energy) than the reactants and energy must be absorbed for the reaction to proceed.放热反应中，产物比反应物更稳定（能量更低），多出的能量被释放出来。吸热反应中，产物比反应物更不稳定（能量更高），反应必须吸收能量才能进行。
Energy Profiles能量图（Energy Profile）
Exothermic: products below reactants on energy axis. $\Delta H < 0$.放热：产物在能量轴上低于反应物，$\Delta H < 0$。
Endothermic: products above reactants on energy axis. $\Delta H > 0$.吸热：产物在能量轴上高于反应物，$\Delta H > 0$。
Axes: reaction coordinate ($x$), potential energy ($y$).坐标轴：横轴是反应坐标 ($x$)，纵轴是势能 ($y$)。
Standard Enthalpy Change标准焓变
The standard enthalpy change $\Delta H^\ominus$ refers to the heat transferred at constant pressure under standard conditions (298 K, 100 kPa) with all substances in their standard states. It is measured using calorimetry — the temperature change of a known mass of water.标准焓变（standard enthalpy change）$\Delta H^\ominus$ 指在标准条件下（298 K、100 kPa，所有物质处于标准状态）恒压时转移的热量。通过量热法（calorimetry）测定——即测量一定质量水的温度变化。
Calorimetry Equations量热公式
$$Q = mc\Delta T$$
$Q$ = heat (J), $m$ = mass of water (g), $c$ = specific heat capacity of water (4.18 J g⁻¹ K⁻¹), $\Delta T$ = temperature change (K).$Q$ = 热量（J），$m$ = 水的质量（g），$c$ = 水的比热容（specific heat capacity，4.18 J g⁻¹ K⁻¹），$\Delta T$ = 温度变化（K）。
Enthalpy Change of Reaction反应的焓变
$$\Delta H = -\frac{Q}{n}$$
$n$ = moles of limiting reagent. The negative sign: if the surroundings gain heat ($Q > 0$), the system loses it ($\Delta H < 0$).$n$ = 限量试剂（limiting reagent）的物质的量。负号的含义：若环境吸热（$Q > 0$），则体系失热（$\Delta H < 0$）。
Why the Negative Sign?为什么要加负号？$Q = mc\Delta T$ calculates the heat gained by the surroundings (the water). If the water heats up, the reaction released energy — the system lost energy. So the enthalpy change of the reaction is negative of the heat gained by water. Students frequently forget this sign and get the wrong direction.$Q = mc\Delta T$ 算出的是环境（即水）吸收的热量。如果水升温，说明反应释放了能量——体系失去了能量。因此反应的焓变是水吸热量取负号。学生最常忘掉这个负号，结果方向写反。
Worked Example — Enthalpy of Neutralization例题 — 中和焓（enthalpy of neutralization）
50.0 cm³ of 1.00 mol dm⁻³ HCl is added to 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises from 22.0 °C to 28.8 °C. Calculate $\Delta H$ for the neutralization.
Step 1 — Calculate heat gained by water
Total volume = 100.0 cm³, so mass ≈ 100.0 g (assuming density = 1.00 g cm⁻³)
In a calorimetry experiment, the temperature of 200 g of water increases by 5.0 °C. What is the heat gained by the water? ($c$ = 4.18 J g⁻¹ K⁻¹)在一次量热实验中，200 g 水的温度升高了 5.0 °C。水吸收的热量是多少？（$c$ = 4.18 J g⁻¹ K⁻¹）
$Q = mc\Delta T = 200 \times 4.18 \times 5.0 = 4180$ J. Answer: (B).$Q = mc\Delta T = 200 \times 4.18 \times 5.0 = 4180$ J。答案：(B)。
Reactivity 1.2
Energy Cycles in Reactions反应中的能量循环
Bond Enthalpies键能（Bond Enthalpies）
Breaking bonds requires energy (endothermic). Forming bonds releases energy (exothermic). The enthalpy change of a reaction can be estimated by comparing the total energy needed to break all bonds in the reactants with the total energy released when forming all bonds in the products.断键需要吸收能量（吸热），成键则释放能量（放热）。一个反应的焓变可以这样估算：把反应物中所有键的断裂能量加起来，再减去产物中所有键的生成释放能量。这里的键能（bond enthalpy）是核心。
Enthalpy Change from Bond Enthalpies由键能求焓变
$$\Delta H = \sum (\text{bonds broken}) - \sum (\text{bonds formed})$$
Bond enthalpy values are averages and may differ from measured values for specific molecules.键能数值是平均值，对于具体分子可能与实测值有偏差。
Why "Average" Bond Enthalpies?为什么是"平均"键能？The O–H bond in water has a slightly different bond enthalpy than the O–H bond in ethanol. Published values are averages across many compounds. This is why calculations using bond enthalpies give estimates, not exact answers. Hess's law calculations with formation/combustion data give more accurate results.水中的 O–H 键能与乙醇中的 O–H 键能略有差别。数据表给出的数值是大量化合物的平均值。因此用键能算出的焓变只是估计值，并不精确。用盖斯定律配合生成焓（enthalpy of formation）/燃烧焓（enthalpy of combustion）的数据计算才更准确。
Worked Example — Bond Enthalpy Calculation例题 — 键能法计算焓变
$$\Delta H = 2652 - 3460 = -808\;\text{kJ mol}^{-1}$$
该反应是放热的——成键释放的能量比断键吸收的更多。
Hess's Law盖斯定律（Hess's Law）
Hess's law states that the enthalpy change for a reaction is independent of the pathway between initial and final states. This is a consequence of enthalpy being a state function. It allows us to calculate enthalpy changes that are difficult to measure directly.盖斯定律（Hess's law）指出：反应的焓变只取决于始态和终态，与具体路径无关。这是因为焓是一个状态函数。它让我们能算出那些难以直接测量的焓变。
Note: the order is reversed compared to formation data.注意：方向与生成焓公式相反。
Memory Trick for the Formulas记忆窍门Formation: "products minus reactants" (same as the natural direction — you go from reactants to products).
Combustion: "reactants minus products" (reversed — because combustion data gives an alternative pathway downward to common products like CO₂ and H₂O).生成焓：产物减反应物（与反应自然方向一致——反应物 → 产物）。
燃烧焓：反应物减产物（反过来——因为燃烧数据相当于走另一条向下到 CO₂、H₂O 这类共同终点的路径）。
Born–Haber Cycles (HL)玻恩-哈伯循环（HL）
A Born–Haber cycle applies Hess's law to the formation of an ionic compound. It breaks the process into steps: atomization of the metal, ionization energies, atomization of the non-metal, electron affinities, and lattice enthalpy. The cycle allows you to calculate any one unknown step from the others.玻恩-哈伯循环把盖斯定律应用到离子化合物的形成上，把过程拆成若干步：金属的原子化、电离能、非金属的原子化、电子亲和能以及晶格能。只要循环中其他几步已知，就能算出任一未知步骤。
HL Only — Born–Haber Steps仅 HL — 玻恩-哈伯各步骤The cycle includes: enthalpy of atomization (sublimation and/or bond enthalpies), ionization energies of the metal, electron affinities of the non-metal, lattice enthalpy, and the overall enthalpy of formation. The sum around the cycle must equal zero (Hess's law).循环包含：原子化焓（升华和/或键能）、金属的电离能、非金属的电子亲和能、晶格能以及总生成焓。整个循环各步之和必须等于零（这正是盖斯定律的体现）。
Reactive metals, non-metals, and organic compounds undergo combustion when heated in oxygen. Complete combustion of hydrocarbons produces CO₂ and H₂O. Incomplete combustion (limited oxygen supply) produces CO and/or C (soot) instead — this is dangerous because CO is a toxic, odorless gas.活泼金属、非金属以及有机化合物在氧气中受热都会发生燃烧。烃类完全燃烧的产物是 CO₂ 和 H₂O。不完全燃烧（氧气供应不足）则会生成 CO 和/或 C（炭黑）——这非常危险，因为 CO 是无色无味的有毒气体。
Complete vs. Incomplete Combustion完全燃烧与不完全燃烧Complete: CH₄ + 2O₂ → CO₂ + 2H₂O. Incomplete: 2CH₄ + 3O₂ → 2CO + 4H₂O, or CH₄ + O₂ → C + 2H₂O. Larger hydrocarbons are more prone to incomplete combustion because they require more oxygen per molecule.完全燃烧：CH₄ + 2O₂ → CO₂ + 2H₂O。不完全燃烧：2CH₄ + 3O₂ → 2CO + 4H₂O，或 CH₄ + O₂ → C + 2H₂O。烃分子越大越容易不完全燃烧，因为每个分子需要的氧气更多。
Fossil Fuels and the Greenhouse Effect化石燃料与温室效应
Fossil fuels (coal, crude oil, natural gas) are non-renewable resources formed from ancient organisms. Burning them releases CO₂, which is a greenhouse gas. Increasing atmospheric CO₂ enhances the greenhouse effect and contributes to climate change and ocean acidification.化石燃料（煤、原油、天然气）是由远古生物形成的不可再生资源。燃烧它们会释放 CO₂——一种温室气体。大气中 CO₂ 浓度升高会加剧温室效应，导致气候变化和海洋酸化。
Biofuels生物燃料
Biofuels are produced from biological materials through recent photosynthesis, making them renewable on shorter timescales. They are considered potentially "carbon neutral" because the CO₂ released during combustion was recently fixed from the atmosphere. However, their production can compete with food crops and may involve significant energy inputs.生物燃料来源于经过近代光合作用的生物质材料，因此在较短的时间尺度上是可再生的。它们被认为有可能实现"碳中和"，因为燃烧释放的 CO₂ 是近期才从大气中固定下来的。不过，其生产可能与粮食作物争地，且本身可能要消耗大量能量。
Fuel Cells燃料电池
A fuel cell converts chemical energy directly to electrical energy. In a hydrogen fuel cell, hydrogen is oxidized at the anode and oxygen is reduced at the cathode. The only product is water, making it an appealing clean energy technology.燃料电池把化学能直接转化为电能。在氢燃料电池中，氢气在阳极被氧化，氧气在阴极被还原。唯一的产物是水，因此是一项颇具吸引力的清洁能源技术。
Which of the following is a product of the incomplete combustion of octane (C₈H₁₈)?下列哪一项是辛烷（C₈H₁₈）不完全燃烧的产物？
CO
CO₃²⁻
O₃
H₂
Correct! Incomplete combustion produces carbon monoxide (CO) and/or carbon (soot) along with water.正确！不完全燃烧会生成一氧化碳（CO）和/或炭（炭黑），并伴随水的生成。
Incomplete combustion of hydrocarbons produces CO and/or C (soot) along with H₂O. Answer: (A).烃类不完全燃烧的产物是 CO 和/或 C（炭黑），并伴随 H₂O。答案：(A)。
Reactivity 1.4 — Additional Higher Level
Entropy and Spontaneity熵与自发性
HL Only仅 HLThis entire sub-topic is assessed at Higher Level only. SL students do not need to learn entropy, Gibbs energy, or spontaneity calculations.整个子主题仅在 HL 考察。SL 学生不需要学习熵（entropy）、吉布斯自由能（Gibbs free energy）或自发性（spontaneous）计算。
Entropy熵（Entropy）
Entropy ($S$) measures the dispersal or distribution of matter and/or energy in a system. More possible arrangements means higher entropy. Under the same conditions: $S_\text{gas} > S_\text{liquid} > S_\text{solid}$.熵（entropy，$S$）衡量体系中物质和/或能量的分散程度。可能的微观状态越多，熵越大。在相同条件下：$S_\text{gas} > S_\text{liquid} > S_\text{solid}$。
Predicting Entropy Changes如何判断熵的变化Entropy typically increases when: more moles of gas are produced, a solid or liquid becomes a gas, a solute dissolves, or temperature increases. Entropy typically decreases in the reverse scenarios.下列情况下熵通常增大：气体的物质的量增加、固体或液体变为气体、溶质溶解、温度升高。反向过程熵通常减小。
Gibbs Free Energy吉布斯自由能
The Gibbs free energy change ($\Delta G$) determines whether a reaction is spontaneous at constant temperature and pressure. A negative $\Delta G$ means the reaction is spontaneous in the forward direction.吉布斯自由能变（Gibbs free energy，$\Delta G$）决定一个反应在恒温恒压下是否自发（spontaneous）。$\Delta G$ 为负意味着反应在正方向上自发。
$T$ must be in Kelvin. Convert $\Delta S$ to kJ K⁻¹ mol⁻¹ (÷1000) if $\Delta H$ is in kJ mol⁻¹.$T$ 必须用开尔文。如果 $\Delta H$ 的单位是 kJ mol⁻¹，把 $\Delta S$ 除以 1000 换成 kJ K⁻¹ mol⁻¹ 后再代入。
Spontaneity Summary Table自发性判别速查表
$\Delta H$
$\Delta S$
$\Delta G$
Spontaneous?
− (exo)
+ (increase)
Always −
Yes, at all temperatures
+ (endo)
− (decrease)
Always +
Never spontaneous
− (exo)
− (decrease)
Depends on $T$
Yes at low $T$
+ (endo)
+ (increase)
Depends on $T$
Yes at high $T$
$\Delta H$
$\Delta S$
$\Delta G$
是否自发？
− (放热)
+ (增大)
始终为 −
所有温度下都自发
+ (吸热)
− (减小)
始终为 +
永远不自发
− (放热)
− (减小)
取决于 $T$
低温下自发
+ (吸热)
+ (增大)
取决于 $T$
高温下自发
Relationship to Equilibrium与化学平衡的关系
At equilibrium, $\Delta G = 0$. The relationship between Gibbs energy and the equilibrium constant is:达到平衡时，$\Delta G = 0$。吉布斯自由能与平衡常数的关系为：
Worked Example — Gibbs Energy Calculation例题 — 吉布斯自由能计算
For the decomposition of CaCO₃: $\Delta H^\ominus$ = +178 kJ mol⁻¹, $\Delta S^\ominus$ = +161 J K⁻¹ mol⁻¹. At what temperature does this reaction become spontaneous?
A reaction has $\Delta H$ = −30 kJ mol⁻¹ and $\Delta S$ = −100 J K⁻¹ mol⁻¹. Below what temperature is the reaction spontaneous?某反应 $\Delta H$ = −30 kJ mol⁻¹，$\Delta S$ = −100 J K⁻¹ mol⁻¹。在低于多少温度时反应是自发的？
$30$ K
$3000$ K
$300$ K
$0.3$ K
Correct! At the boundary: $T = \Delta H / \Delta S = 30000 / 100 = 300$ K. Below 300 K, $\Delta G < 0$ and the reaction is spontaneous.正确！临界温度：$T = \Delta H / \Delta S = 30000 / 100 = 300$ K。低于 300 K 时 $\Delta G < 0$，反应自发。
$T = \Delta H / \Delta S = 30000\text{ J} / 100\text{ J K}^{-1} = 300$ K. Since $\Delta H < 0$ and $\Delta S < 0$, the reaction is spontaneous below this temperature. Answer: (C).$T = \Delta H / \Delta S = 30000\text{ J} / 100\text{ J K}^{-1} = 300$ K。$\Delta H < 0$ 且 $\Delta S < 0$，所以低于该温度反应才自发。答案：(C)。
Preparation备考
Exam Strategy考试策略
Paper 1 (Multiple Choice)Paper 1（选择题）
Know the signs: exothermic = negative $\Delta H$, endothermic = positive. Be careful with the bond enthalpy formula — it's "broken minus formed," not the other way around. For Hess's law questions, draw the energy cycle before trying to calculate.记牢符号：放热（exothermic）= $\Delta H$ 为负，吸热（endothermic）= 正。键能公式小心方向——是"断键减成键"，不是反过来。盖斯定律（Hess's law）题先画能量循环再计算。
Paper 2 (Extended Response)Paper 2（简答题）
Show all working clearly in calorimetry calculations: state $Q = mc\Delta T$, then $\Delta H = -Q/n$. Don't forget units and the negative sign. For Hess's law, write out the full calculation showing each term. Energy profile sketches must label axes, show activation energy, $\Delta H$, reactants, and products.量热（calorimetry）计算要写清每一步：先 $Q = mc\Delta T$，再 $\Delta H = -Q/n$。别忘了单位和负号。盖斯定律题要把每一项都列出来。能量图（energy profile）必须标坐标轴、活化能、$\Delta H$ 以及反应物和产物。
Data Booklet数据手册
Familiarize yourself with the location of: specific heat capacity of water, average bond enthalpies, standard enthalpies of formation and combustion, standard entropy values, and the equations ($Q = mc\Delta T$, $\Delta G = \Delta H - T\Delta S$, $\Delta G^\ominus = -RT\ln K$).熟悉这些信息在数据手册中的位置：水的比热容（specific heat capacity）、平均键能、标准生成焓（standard enthalpy change）与燃烧焓、标准熵值，以及公式（$Q = mc\Delta T$、$\Delta G = \Delta H - T\Delta S$、$\Delta G^\ominus = -RT\ln K$）。
Watch Out小心陷阱
Common Mistakes常见错误
Mistake 1 — Forgetting the Negative Sign in ΔH错误 1 — ΔH 忘了写负号$Q = mc\Delta T$ gives the heat gained by the surroundings. The enthalpy change of the reaction is $\Delta H = -Q/n$. Many students write a positive value when the reaction is exothermic.$Q = mc\Delta T$ 算出的是环境（surroundings）吸收的热量。反应的焓变是 $\Delta H = -Q/n$。许多学生在放热反应上写成了正值。
Mistake 2 — Reversing the Bond Enthalpy Formula错误 2 — 键能公式方向搞反It's $\Delta H = \text{bonds broken} - \text{bonds formed}$. Breaking requires energy (positive), forming releases energy (positive too, because we subtract it). If you get a positive $\Delta H$ for combustion, you've reversed the formula.公式是 $\Delta H = \text{bonds broken} - \text{bonds formed}$。断键吸能（取正），成键放能（也取正，因为后面要减去）。如果你算出来燃烧反应是正的 $\Delta H$，那就是公式方向反了。
Mistake 3 — Unit Mismatch in Gibbs Calculations (HL)错误 3 — 吉布斯计算中单位不匹配（HL）$\Delta H$ is in kJ mol⁻¹ but $\Delta S$ is in J K⁻¹ mol⁻¹. You must convert one to match the other before using $\Delta G = \Delta H - T\Delta S$. Dividing $\Delta S$ by 1000 to get kJ K⁻¹ mol⁻¹ is the most common approach.$\Delta H$ 的单位是 kJ mol⁻¹，$\Delta S$ 的单位是 J K⁻¹ mol⁻¹。代入 $\Delta G = \Delta H - T\Delta S$ 之前必须先统一单位。常用做法是把 $\Delta S$ 除以 1000，换成 kJ K⁻¹ mol⁻¹。
Mistake 4 — Hess's Law Formula Mix-up错误 4 — 盖斯定律公式混用For formation data: products minus reactants. For combustion data: reactants minus products. These are reversed! Drawing the energy cycle diagram first helps prevent this error.生成焓数据：产物减反应物。燃烧焓数据：反应物减产物。两者方向相反！动笔前先画能量循环图能避免这个错误。
Review复习
Flashcards闪卡
Click a card to flip it.点击卡片翻面。
What is the sign of $\Delta H$ for an exothermic reaction?放热反应 $\Delta H$ 的符号？
Negative ($\Delta H < 0$). The system releases energy to the surroundings.负（$\Delta H < 0$）。体系向环境释放能量。
State Hess's Law陈述盖斯定律
The enthalpy change for a reaction is independent of the pathway between the initial and final states.反应的焓变只取决于始态与终态，与具体路径无关。
Bond breaking: endothermic or exothermic?断键是吸热还是放热？
Endothermic — energy must be supplied to break bonds.吸热——必须提供能量才能断键。
What does $\Delta G < 0$ mean?$\Delta G < 0$ 意味着什么？
The reaction is spontaneous in the forward direction at the given temperature and pressure.在给定温度和压强下，反应在正方向自发。
Formula for $\Delta H$ from bond enthalpies?由键能求 $\Delta H$ 的公式？
$\Delta H = \Sigma$(bonds broken) $-$ $\Sigma$(bonds formed)
Why are calorimetry values less than theoretical?为什么量热实验值通常小于理论值？
Heat loss to the surroundings (container, air) means less heat is captured by the water, giving a smaller temperature change.部分热量散失到环境（容器、空气），水吸收的热量减少，温升偏小。
Products of incomplete combustion?不完全燃烧的产物？
Carbon monoxide (CO) and/or carbon (soot/C), plus water.一氧化碳（CO）和/或炭（炭黑 C），并伴随水。
When is an endothermic reaction spontaneous? (HL)吸热反应何时自发？（HL）
When $\Delta S > 0$ and the temperature is high enough that $T\Delta S > \Delta H$, making $\Delta G < 0$.当 $\Delta S > 0$ 且温度足够高使 $T\Delta S > \Delta H$，从而 $\Delta G < 0$。
Assessment自测
Unit Quiz单元测验
1. A reaction absorbs heat from the surroundings. Which statement is correct?1. 一个反应从环境中吸热。下列哪一项正确？
$\Delta H < 0$, the products are lower in energy$\Delta H < 0$，产物能量更低
$\Delta H > 0$, the products are higher in energy$\Delta H > 0$，产物能量更高
$\Delta H > 0$, the products are lower in energy$\Delta H > 0$，产物能量更低
$\Delta H < 0$, the products are higher in energy$\Delta H < 0$，产物能量更高
Correct! An endothermic reaction absorbs heat, so $\Delta H > 0$ and the products are at a higher energy level than the reactants.正确！吸热反应吸热，$\Delta H > 0$，产物的能量高于反应物。
Absorbing heat = endothermic = $\Delta H > 0$. Products are higher in energy. Answer: (B).吸热 = endothermic = $\Delta H > 0$，产物能量更高。答案：(B)。
2. Which equation correctly represents Hess's law using standard enthalpies of combustion?2. 下列哪个公式正确表示用标准燃烧焓计算盖斯定律？
$\Delta H = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$
$\Delta H = \sum \Delta H_c(\text{products}) - \sum \Delta H_c(\text{reactants})$
$\Delta H = \sum \Delta H_c(\text{broken}) - \sum \Delta H_c(\text{formed})$
$\Delta H = \sum \Delta H_f(\text{reactants}) - \sum \Delta H_f(\text{products})$
Correct! For combustion data, it's reactants minus products — the reverse of the formation data formula.正确！燃烧焓数据是反应物减产物——与生成焓公式的方向相反。
For combustion data: $\Delta H = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$. Answer: (A).燃烧焓数据：$\Delta H = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$。答案：(A)。
3. (HL) For a reaction with $\Delta H = +50$ kJ mol⁻¹ and $\Delta S = +200$ J K⁻¹ mol⁻¹, the reaction becomes spontaneous above:3. （HL）某反应 $\Delta H = +50$ kJ mol⁻¹，$\Delta S = +200$ J K⁻¹ mol⁻¹。反应在多少温度以上变得自发？
$0.25$ K
$4000$ K
$100$ K
$250$ K
Correct! $T = \Delta H / \Delta S = 50000 / 200 = 250$ K. Above 250 K, $\Delta G < 0$ and the reaction is spontaneous.正确！$T = \Delta H / \Delta S = 50000 / 200 = 250$ K。高于 250 K 时 $\Delta G < 0$，反应自发。
$T = \Delta H / \Delta S = 50000\text{ J} / 200\text{ J K}^{-1} = 250$ K. Answer: (D).$T = \Delta H / \Delta S = 50000\text{ J} / 200\text{ J K}^{-1} = 250$ K。答案：(D)。