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)。
Worked Example — Enthalpy of Combustion of Ethanol (Spirit Burner)例题 — 乙醇的燃烧焓（酒精灯法）
A spirit burner of ethanol ($\text{C}_2\text{H}_5\text{OH}$, $M = 46.0\;\text{g mol}^{-1}$) heats 100.0 g of water. The temperature rises from 20.0 °C to 46.5 °C, and 0.46 g of ethanol is burned. Calculate the experimental enthalpy of combustion and compare it with the data-booklet value of $-1367\;\text{kJ mol}^{-1}$.
The experimental value ($-1110$) is less exothermic than the literature value ($-1367$) — a shortfall of about 19%. The difference comes from heat lost to the surroundings (air, calorimeter), incomplete combustion (soot on the beaker), and evaporation of ethanol and water. All of these lower the measured $\Delta T$, so the calculated $\Delta H_c$ is always numerically smaller than the true value.
用盛有乙醇（$\text{C}_2\text{H}_5\text{OH}$，$M = 46.0\;\text{g mol}^{-1}$）的酒精灯加热 100.0 g 水。温度从 20.0 °C 升到 46.5 °C，燃烧掉的乙醇质量为 0.46 g。求实验测得的燃烧焓，并与数据手册值 $-1367\;\text{kJ mol}^{-1}$ 比较。
Going Deeper — Why Calorimetry Under-reads深入一步 — 量热法为何总是偏小Simple calorimetry assumes all the heat from the reaction is transferred to the water and none is lost. In practice the flame heats the air and the container, and unburned fuel escapes. Because the assumption fails in the same direction every time, the systematic error is one-sided: combustion enthalpies measured this way are always less exothermic than the true value. Insulating the calorimeter, using a lid, and shielding from draughts all reduce (but never eliminate) the loss.简单量热法假设反应放出的热量全部传给了水、没有任何损失。实际上火焰会加热空气和容器，还有未燃烧的燃料逸散。由于这个假设每次都朝同一方向失效，系统误差是单向的：用这种方法测得的燃烧焓总是比真实值放热更少。给量热器保温、加盖、挡风都能减小（但无法消除）这种损失。
Exam Tip — Calorimetry on Paper 2考试提示 — Paper 2 的量热计算
Graders award marks for the method, not just the final number: state $Q = mc\Delta T$ with the correct mass (mass of the solution/water, not the fuel), then $\Delta H = -Q/n$ with $n$ = moles of the limiting species. Carry the negative sign and the correct unit (kJ mol⁻¹). An "evaluate the experiment" part almost always wants: heat loss to surroundings, incomplete combustion, and non-standard conditions.阅卷给分看方法，不只看最终数字：先写 $Q = mc\Delta T$，质量用溶液/水的质量（不是燃料的质量），再写 $\Delta H = -Q/n$，其中 $n$ 是限量物质的物质的量。别丢负号和正确单位（kJ mol⁻¹）。"评价实验"这一问几乎总是要答：向环境散热、不完全燃烧、非标准条件。
A student measures the enthalpy of combustion of a fuel using a spirit burner and beaker. Their value is $-880\;\text{kJ mol}^{-1}$ but the data-booklet value is $-1300\;\text{kJ mol}^{-1}$. Which is the best explanation?某学生用酒精灯和烧杯测某燃料的燃烧焓，得到 $-880\;\text{kJ mol}^{-1}$，而数据手册值为 $-1300\;\text{kJ mol}^{-1}$。下列哪项是最佳解释？
The reaction is endothermic该反应是吸热的
Too much fuel was burned燃烧的燃料太多了
Heat was lost to the surroundings and combustion was incomplete热量散失到环境且燃烧不完全
The specific heat capacity of water was too high水的比热容取得太大
Correct! Heat loss and incomplete combustion both lower $\Delta T$, so the measured enthalpy is less exothermic than the true value — exactly the one-sided error seen here.正确！散热和不完全燃烧都会降低 $\Delta T$，使实测焓比真实值放热更少——正是这里出现的单向误差。
The reaction is still exothermic; the gap is a systematic under-read from heat loss and incomplete combustion, which lower the measured temperature rise. Answer: (C).反应仍是放热的；差距源于散热和不完全燃烧造成的系统性偏小，二者都会降低实测温升。答案：(C)。
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。
Going Deeper — Checking the "Estimate" Claim深入一步 — 核实"估计值"这一说法The bond-enthalpy method above gives $\Delta H \approx -808\;\text{kJ mol}^{-1}$ for methane combustion. The value from standard enthalpies of formation (a more direct, accurate route) is $-890\;\text{kJ mol}^{-1}$ — a difference of about 80 kJ mol⁻¹, or roughly 9%. That gap is exactly the "average vs. specific" error described above: the C–H and O–H bonds in this particular molecule don't have exactly the tabulated average bond enthalpy. On an exam, if asked to explain a discrepancy between a bond-enthalpy answer and a formation-data answer, this is the reason to give — not a claim that either calculation contains an arithmetic error.上面用键能法算出甲烷燃烧的 $\Delta H \approx -808\;\text{kJ mol}^{-1}$。而用标准生成焓（更直接、更准确的方法）算出的值是 $-890\;\text{kJ mol}^{-1}$——相差约 80 kJ mol⁻¹，约 9%。这一差距正是前面所说的"平均值 vs. 具体值"误差：这个特定分子中的 C–H 键与 O–H 键并不恰好等于表中给出的平均键能。在考试中，若被要求解释键能法答案与生成焓法答案之间的差异，应给出这个原因——而不是声称其中某一个计算存在算术错误。
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）。
Worked Example — Applying Hess's Law with Formation Data例题 — 用生成焓套用盖斯定律
Calculate $\Delta H^\ominus$ for the reaction $\text{CH}_4\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow \text{CH}_3\text{Cl(g)} + \text{HCl(g)}$, given $\Delta H_f^\ominus(\text{CH}_4) = -75\;\text{kJ mol}^{-1}$, $\Delta H_f^\ominus(\text{CH}_3\text{Cl}) = -82\;\text{kJ mol}^{-1}$, $\Delta H_f^\ominus(\text{HCl}) = -92\;\text{kJ mol}^{-1}$. ($\Delta H_f^\ominus(\text{Cl}_2) = 0$, an element in its standard state.)
Don't forget elements in their standard state (here, Cl₂(g)) always contribute $0$ — a frequent source of arithmetic slips is accidentally omitting this term instead of setting it to zero.
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）。
Worked Example — Lattice Enthalpy via a Born–Haber Cycle (HL)例题 — 用玻恩-哈伯循环求晶格能（HL）
Determine the lattice enthalpy $\Delta H_\text{lat}^\ominus$ of NaCl (formation of the lattice from gaseous ions) from: $\Delta H_f^\ominus(\text{NaCl}) = -411$; atomization of Na $= +107$; first ionization energy of Na $= +496$; atomization of Cl (½ Cl–Cl) $= +122$; electron affinity of Cl $= -349$ (all kJ mol⁻¹).
Lattice formation is strongly exothermic, as expected — gaseous Na⁺ and Cl⁻ release a large amount of energy when they assemble into the ionic lattice.
Exam Tip — Hess's Law & Born–Haber考试提示 — 盖斯定律与玻恩-哈伯
Draw the cycle before substituting numbers — arrows fix the sign of every step. Reverse an arrow → reverse the sign. Multiply a species by its coefficient → multiply its enthalpy. For Born–Haber, the only step you cannot measure directly is the lattice enthalpy, so it is almost always the unknown you solve for. Watch the sign of electron affinity (usually negative for the first electron) and remember ionization energies are always positive.代数字之前先把循环画出来——箭头决定每一步的符号。反转箭头 → 符号取反。物种乘以系数 → 其焓也乘以该系数。玻恩-哈伯循环中唯一无法直接测量的就是晶格能，所以它几乎总是要求解的未知量。注意电子亲和能的符号（第一个电子通常为负），并记住电离能永远为正。
Going Deeper — Why Lattice Enthalpies Vary So Much深入一步 — 为什么晶格能差异如此之大NaCl's lattice enthalpy ($-787\;\text{kJ mol}^{-1}$, calculated above) is much smaller in magnitude than MgO's ($-3791\;\text{kJ mol}^{-1}$). Coulomb's law explains both factors driving the difference: lattice enthalpy scales with the product of the ionic charges and inversely with the sum of the ionic radii. MgO has ions of charge $+2$ and $-2$ (versus $+1$ and $-1$ for NaCl) — a factor of roughly 4 from charge alone — and $\text{Mg}^{2+}$/$\text{O}^{2-}$ are also smaller than $\text{Na}^+$/$\text{Cl}^-$, which pulls the ions closer and strengthens the attraction further. When comparing two lattice enthalpies, always check charge first (it dominates), then radius.NaCl 的晶格能（上面算出的 $-787\;\text{kJ mol}^{-1}$）在数值上远小于 MgO 的晶格能（$-3791\;\text{kJ mol}^{-1}$）。库仑定律解释了造成这一差异的两个因素：晶格能与离子电荷的乘积成正比，与离子半径之和成反比。MgO 的离子电荷为 $+2$ 与 $-2$（而 NaCl 为 $+1$ 与 $-1$）——仅电荷这一项就带来约 4 倍的差异——而且 $\text{Mg}^{2+}$/$\text{O}^{2-}$ 也比 $\text{Na}^+$/$\text{Cl}^-$ 更小，使离子靠得更近、进一步增强了吸引力。比较两个晶格能时，应先看电荷（其影响占主导），再看半径。
Using combustion enthalpies, the enthalpy of formation is found from "reactants minus products." A student instead used "products minus reactants." What happens to their answer?用燃烧焓求生成焓时，公式是"反应物减产物"。某学生却用了"产物减反应物"。他的答案会怎样？
It is too small by a factor of two会小一半
It has the correct magnitude but the wrong sign数值大小正确但符号相反
It is unaffected不受影响
It becomes zero会变成零
Correct! Swapping the order of a subtraction negates the result, so the magnitude is right but the sign flips — turning an exothermic formation into an apparent endothermic one.正确！把减法的两项交换会使结果取负，因此数值大小不变、符号相反——把放热的生成过程错看成吸热的。
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。
Fuel Cells vs. Direct Combustion燃料电池 vs. 直接燃烧Burning hydrogen directly (as in a combustion engine) releases the same overall energy as a fuel cell reaction, because both start and end at the same states — but a combustion engine wastes much of that energy as heat before it can be converted to mechanical work, capped by thermodynamic (Carnot) efficiency limits. A fuel cell converts chemical energy directly to electrical energy without an intermediate heat stage, so it typically achieves a higher energy-conversion efficiency (often 40–60%, versus roughly 20–30% for an internal combustion engine).直接燃烧氢气（如内燃机）释放的总能量与燃料电池反应相同，因为两者的始态和终态一致——但内燃机会把其中大部分能量以热的形式浪费掉，才能转化为机械功，并受热力学（卡诺）效率极限的限制。燃料电池把化学能直接转化为电能，中间没有经过热的阶段，因此通常能达到更高的能量转化效率（常为 40–60%，而内燃机约为 20–30%）。
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)。
A fuel's usefulness depends not only on how exothermic it is per mole, but on how much energy it stores per gram (or per dm³). This is why fuel choice is a trade-off between energy density, storage, and emissions.燃料的实用性不仅取决于每摩尔放热多少，还取决于每克（或每 dm³）储存多少能量。因此选燃料是能量密度、储存和排放之间的权衡。
Worked Example — Energy Density: Octane vs Hydrogen例题 — 能量密度：辛烷 vs 氢气
Compare the specific energy of octane ($\Delta H_c^\ominus = -5470\;\text{kJ mol}^{-1}$, $M = 114.0$) with hydrogen ($\Delta H_c^\ominus = -286\;\text{kJ mol}^{-1}$, $M = 2.02$).
Octane
$$\frac{5470}{114.0} = 48.0\;\text{kJ g}^{-1}$$
Hydrogen
$$\frac{286}{2.02} = 142\;\text{kJ g}^{-1}$$
Interpret
Per gram, hydrogen releases about three times as much energy as octane — which is why it is attractive for aerospace. The catch is storage: as a gas it has a very low energy density per unit volume, so it must be compressed or liquefied.
Worked Example — Mass of CO₂ from Complete Combustion例题 — 完全燃烧产生的 CO₂ 质量
Propane burns completely: $\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$. What mass of CO₂ is produced when 22.0 g of propane ($M = 44.0$) is burned?
"Evaluate biofuels / fuel cells" questions want a balanced answer: name a benefit and a drawback. Biofuels are near carbon-neutral but compete with food land; hydrogen fuel cells emit only water but need energy to produce the hydrogen and are hard to store. For "carbon neutral," state explicitly that combustion CO₂ was recently fixed by photosynthesis — a bare "it's renewable" rarely scores."评价生物燃料 / 燃料电池"这类题要给出平衡的答案：既说一个优点又说一个缺点。生物燃料接近碳中和，但与粮食用地争地；氢燃料电池只排放水，但制氢耗能且难以储存。答"碳中和"时要明确指出：燃烧释放的 CO₂ 是近期由光合作用固定的——只写"它可再生"通常不得分。
Worked Example — Mass of CO₂ Emitted per Unit of Energy例题 — 每单位能量排放的 CO₂ 质量
Octane, $\text{C}_8\text{H}_{18}$ ($M = 114$, $\Delta H_c^\ominus = -5470\;\text{kJ mol}^{-1}$), burns completely: $\text{C}_8\text{H}_{18} + \tfrac{25}{2}\text{O}_2 \rightarrow 8\text{CO}_2 + 9\text{H}_2\text{O}$. Find the mass of CO₂ released per $1000\;\text{kJ}$ of energy produced.
This is exactly the same three-step pattern as any other combustion stoichiometry problem (moles from energy instead of from mass) — it just uses $\Delta H_c$ as the bridge from "energy released" to "moles of fuel burned" before the usual mole-ratio and mass steps.
Methane has $\Delta H_c^\ominus = -890\;\text{kJ mol}^{-1}$ ($M = 16.0$). What is its specific energy, to 3 s.f.?甲烷的 $\Delta H_c^\ominus = -890\;\text{kJ mol}^{-1}$（$M = 16.0$）。它的比能量是多少（保留 3 位有效数字）？
$14.2\;\text{kJ g}^{-1}$
$55.6\;\text{kJ g}^{-1}$
$890\;\text{kJ g}^{-1}$
$5.56\;\text{kJ g}^{-1}$
Correct! Specific energy $= |\Delta H_c| / M = 890 / 16.0 = 55.6\;\text{kJ g}^{-1}$.正确！比能量 $= |\Delta H_c| / M = 890 / 16.0 = 55.6\;\text{kJ g}^{-1}$。
Divide the (magnitude of the) molar combustion enthalpy by the molar mass: $890 / 16.0 = 55.6\;\text{kJ g}^{-1}$. Answer: (B).用摩尔燃烧焓的绝对值除以摩尔质量：$890 / 16.0 = 55.6\;\text{kJ g}^{-1}$。答案：(B)。
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：$S_\text{gas} > S_\text{liquid} > S_\text{solid}$。
Going Deeper — Entropy as Counting Microstates深入一步 — 熵是对微观状态的计数The statistical definition makes "dispersal" precise: $S = k_B \ln W$, where $W$ is the number of equivalent microscopic arrangements (microstates) consistent with the system's observable (macroscopic) state, and $k_B$ is the Boltzmann constant. A gas has vastly more ways to arrange its particles' positions and momenta than a solid does, which is why $S_\text{gas} > S_\text{solid}$ rather than just an empirical rule to memorize. IB does not require calculations with this formula, but it's the reason every qualitative entropy argument you make ("more disorder," "more ways to arrange") is actually quantitative underneath.统计定义让"分散程度"变得精确：$S = k_B \ln W$，其中 $W$ 是与体系可观测（宏观）状态相符的等效微观排列（微观状态）数目，$k_B$ 是玻尔兹曼常数。气体排列其粒子位置和动量的方式远多于固体，这正是 $S_\text{gas} > S_\text{solid}$ 的原因，而不只是一条需要死记的经验规则。IB 并不要求用这个公式做计算，但你所做的每一个定性熵论证（"更无序""排列方式更多"）本质上都是这个量化关系的体现。
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。
Worked Example — Predicting and Ranking Entropy Change例题 — 预测并比较熵变
Without calculating any values, rank the following changes from most positive to most negative $\Delta S$: (i) $\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(g)}$, (ii) $\text{NaCl(s)} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$, (iii) $2\text{NO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{NO}_2\text{(g)}$.
(i) Liquid → gas
A phase change from liquid to gas is one of the largest entropy increases possible — molecules go from a fairly ordered liquid to a highly disordered, freely-moving gas. Strongly positive $\Delta S$.
(ii) Solid dissolving
A solid lattice breaking apart into freely-moving aqueous ions increases disorder, but far less dramatically than forming a gas. Moderately positive $\Delta S$.
(iii) 3 mol gas → 2 mol gas
Fewer gas particles means fewer possible arrangements. Negative $\Delta S$.
This kind of "rank without calculating" question rewards recognizing the size of a disorder change, not just its direction — a phase change to gas dominates any other effect listed here.
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)。
Worked Example — Entropy Change of the Haber Process例题 — 哈伯法的熵变
$\Delta S^\ominus$ is negative because 4 mol of gas become 2 mol of gas — fewer gas particles means fewer arrangements, so entropy falls. Predicting the sign from the change in moles of gas is a reliable first check.
Worked Example — Gibbs Energy & the Crossover Temperature例题 — 吉布斯自由能与临界温度
The Haber process has $\Delta H^\ominus = -92\;\text{kJ mol}^{-1}$ and $\Delta S^\ominus = -199\;\text{J K}^{-1}\text{mol}^{-1}$. (a) Is it spontaneous at 298 K? (b) Above what temperature does it become non-spontaneous?
Both $\Delta H$ and $\Delta S$ are negative, so the reaction is spontaneous below 462 K and non-spontaneous above it. This is why the Haber process runs at a compromise temperature — hot enough for a workable rate, but not so hot that yield collapses.
$\Delta H$ 和 $\Delta S$ 都为负，所以反应在 462 K 以下自发、以上不自发。这正是哈伯法采用折中温度的原因——高到能有可用的反应速率，但又不至于高到产率崩溃。
Going Deeper — Why ΔG Works: Total Entropy深入一步 — ΔG 为何有效：总熵The real driver of spontaneity is the total entropy change of the universe, $\Delta S_\text{total} = \Delta S_\text{sys} + \Delta S_\text{surr}$, where $\Delta S_\text{surr} = -\Delta H / T$. Multiplying $\Delta S_\text{total}$ by $-T$ gives exactly $\Delta G = \Delta H - T\Delta S$. So "$\Delta G < 0$" is just a convenient restatement of "$\Delta S_\text{total} > 0$" — the second law in disguise.自发性真正的驱动力是宇宙的总熵变 $\Delta S_\text{total} = \Delta S_\text{sys} + \Delta S_\text{surr}$，其中 $\Delta S_\text{surr} = -\Delta H / T$。把 $\Delta S_\text{total}$ 乘以 $-T$ 恰好得到 $\Delta G = \Delta H - T\Delta S$。所以"$\Delta G < 0$"不过是"$\Delta S_\text{total} > 0$"的便捷改写——本质上就是热力学第二定律。
Which reaction is expected to have the most positive $\Delta S^\ominus$?下列哪个反应的 $\Delta S^\ominus$ 预期最正（增大最多）？
Correct! (D) produces a gas from a solid, increasing the moles of gas from 0 to 1 — the largest entropy gain. (A) and (B) reduce gas moles; (C) keeps them constant.正确！(D) 由固体生成气体，气体物质的量从 0 增加到 1——熵增最大。(A) 和 (B) 使气体物质的量减少；(C) 保持不变。
Judge by the change in moles of gas: (D) goes 0 → 1 mol gas (biggest increase). (A) 3 → 0, (B) 4 → 2 both decrease; (C) 2 → 2 no change. Answer: (D).看气体物质的量的变化：(D) 从 0 → 1 mol 气体（增加最多）。(A) 3 → 0、(B) 4 → 2 都减少；(C) 2 → 2 不变。答案：(D)。
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（$Q = mc\Delta T$、$\Delta G = \Delta H - T\Delta S$、$\Delta G^\ominus = -RT\ln K$）。
A frequent question format gives $\Delta H^\ominus$ and $\Delta S^\ominus$ and asks you to (a) calculate $\Delta G^\ominus$ at a stated temperature, and (b) find the crossover temperature where the reaction changes from spontaneous to non-spontaneous (or vice versa). Practice both directions: computing $\Delta G$ at a given $T$, and solving for $T$ when $\Delta G = 0$. Always state which temperature range makes the reaction spontaneous, not just the crossover value itself — a bare number without a direction ("above" or "below") usually loses the final mark.常见题型会给出 $\Delta H^\ominus$ 和 $\Delta S^\ominus$，要求 (a) 在给定温度下计算 $\Delta G^\ominus$，(b) 求出反应由自发变为不自发（或相反）的临界温度。两个方向都要练习：在给定 $T$ 下算 $\Delta G$，以及令 $\Delta G = 0$ 反过来求 $T$。一定要说明是哪个温度区间使反应自发，而不只是给出临界值本身——只写数字、不写"高于"或"低于"通常会丢掉最后一分。
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 — Using Mass of Fuel Instead of Mass of Water in Q = mcΔT错误 — 在 Q = mcΔT 中用了燃料质量而不是水的质量In a calorimetry calculation, $m$ in $Q = mc\Delta T$ is the mass of the water (or solution) being heated — not the mass of the fuel burned or the reactant added. The fuel/reactant mass is used later, to find moles of the limiting species for the $\Delta H = -Q/n$ step. Mixing these two masses up is one of the most common ways to get a wildly wrong final answer despite correct-looking working.在量热计算中，$Q = mc\Delta T$ 里的 $m$ 是被加热的水（或溶液）的质量——而不是燃烧的燃料或加入的反应物的质量。燃料/反应物的质量要在之后 $\Delta H = -Q/n$ 那一步、用来求限量物质的物质的量时才用到。把这两个质量搞混，是明明过程看起来正确、结果却严重错误的最常见原因之一。
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。
Mistake 5 — Judging Entropy by Mass Instead of Particle Count and State错误 5 — 用质量而不是粒子数与状态来判断熵Entropy comparisons are about the number of gas particles and physical state, not about molar mass or "how big" a molecule looks. A reaction that converts 2 mol of gas into 2 mol of a different gas can still have a nonzero $\Delta S$ (more complex molecules generally have somewhat higher entropy per mole than simpler ones at the same state), but this effect is small compared to a change in the number of moles of gas or a change of state — don't let molecular size override the state/mole-count check.判断熵变要看气体粒子数与物理状态，而不是摩尔质量或分子"看起来有多大"。把 2 mol 气体转化为 2 mol 另一种气体，$\Delta S$ 仍可能不为零（在相同状态下，结构更复杂的分子摩尔熵通常略高于结构简单的分子），但这一效应远小于气体物质的量的变化或状态的变化——不要让分子大小凌驾于状态/摩尔数检验之上。
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)。
4. In a reaction, breaking all the bonds in the reactants requires 2500 kJ and forming all the bonds in the products releases 2900 kJ. What is $\Delta H$?4. 某反应中，断裂反应物所有键需要 2500 kJ，生成产物所有键释放 2900 kJ。$\Delta H$ 是多少？
$-400$ kJ mol⁻¹
$+400$ kJ mol⁻¹
$+5400$ kJ mol⁻¹
$-5400$ kJ mol⁻¹
Correct! $\Delta H = \text{broken} - \text{formed} = 2500 - 2900 = -400$ kJ mol⁻¹ — exothermic, since more energy is released forming bonds than is used breaking them.正确！$\Delta H = \text{broken} - \text{formed} = 2500 - 2900 = -400$ kJ mol⁻¹——放热，因为成键释放的能量比断键消耗的更多。
5. In a neutralization, 100.0 g of solution rises by 6.0 °C. Taking $c = 4.18$ J g⁻¹ K⁻¹, what heat is released to the solution, and what is the sign of $\Delta H$ for the reaction?5. 一次中和反应中，100.0 g 溶液温度升高 6.0 °C。取 $c = 4.18$ J g⁻¹ K⁻¹，放给溶液的热量是多少，反应的 $\Delta H$ 符号如何？
$2.51$ kJ, $\Delta H > 0$$2.51$ kJ，$\Delta H > 0$
$25.1$ kJ, $\Delta H < 0$$25.1$ kJ，$\Delta H < 0$
$2.51$ kJ, $\Delta H < 0$$2.51$ kJ，$\Delta H < 0$
$0.418$ kJ, $\Delta H < 0$$0.418$ kJ，$\Delta H < 0$
Correct! $Q = mc\Delta T = 100.0 \times 4.18 \times 6.0 = 2508$ J $= 2.51$ kJ. The solution warms up, so the reaction is exothermic: $\Delta H < 0$.正确！$Q = mc\Delta T = 100.0 \times 4.18 \times 6.0 = 2508$ J $= 2.51$ kJ。溶液升温，说明反应放热：$\Delta H < 0$。
$Q = mc\Delta T = 100.0 \times 4.18 \times 6.0 = 2508$ J $\approx 2.51$ kJ; a temperature rise means an exothermic reaction ($\Delta H < 0$). Answer: (C).$Q = mc\Delta T = 100.0 \times 4.18 \times 6.0 = 2508$ J $\approx 2.51$ kJ；温度升高说明反应放热（$\Delta H < 0$）。答案：(C)。
6. (HL) Which change increases the entropy of the system the most?6. （HL）下列哪种变化使体系的熵增加最多？
Water freezing to ice水结成冰
Solid iodine subliming to gas固态碘升华为气体
A gas being compressed气体被压缩
Two gases reacting to form one gas两种气体反应生成一种气体
Correct! Sublimation (solid → gas) is the largest jump in disorder, since $S_\text{gas} \gg S_\text{solid}$. Freezing and compression lower entropy; forming fewer gas moles also lowers it.正确！升华（固 → 气）是无序度增加最大的过程，因为 $S_\text{gas} \gg S_\text{solid}$。结冰和压缩使熵减小；生成更少的气体物质的量也使熵减小。
Entropy rises most when a solid becomes a gas: $S_\text{gas} \gg S_\text{liquid} \gg S_\text{solid}$. Answer: (B).当固体变为气体时熵增加最多：$S_\text{gas} \gg S_\text{liquid} \gg S_\text{solid}$。答案：(B)。