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Reactivity 1 · SolutionsReactivity 1 · 解析

Thermochemistry & Spontaneity — Solutions热化学与自发性 —— 解析

Companion to the Reactivity 1 Practice SetReactivity 1 练习题的解析配套

EASY MEDIUM HARD Paper 1 Paper 2 Paper 3 HL HL

Topics Reactivity 1.1 – 1.4考点 Reactivity 1.1 – 1.4HL

PART I  ·  PAPER 1第一部分  ·  第一卷Multiple Choice — Worked Answers选择题 —— 详细解析

Multiple Choice选择题

Q1EASYPaper 11.1 Exo/Endo Signs

Which combination correctly describes an exothermic reaction?哪一组描述正确对应放热反应？

Answer:答案： (B)
Exothermic means the system releases energy to the surroundings, so the surroundings' temperature rises. From the system's perspective it has lost energy, so $\Delta H < 0$. Losing energy also means the products sit at lower energy (more stable) than the reactants: the energy difference is what gets released. (A) and (D) wrongly pair a temperature decrease with exothermic release; (C) has the sign of $\Delta H$ backwards.放热意味着体系向环境释放能量，故环境温度升高。从体系的角度看，它损失了能量，所以 $\Delta H < 0$。损失能量也意味着产物的能量比反应物更低（更稳定）：这部分能量差正是被释放的部分。(A)、(D) 错误地将温度降低与放热配对；(C) 把 $\Delta H$ 的符号弄反了。
Insight.要点。 Always fix the sign of $\Delta H$ from the system's perspective, never the surroundings'. A quick check: surroundings warm up $\Rightarrow$ heat left the system $\Rightarrow$ exothermic $\Rightarrow$ $\Delta H < 0$. This single rule resolves every exo/endo sign question, including the trickier calorimetry ones later in this set.务必从体系而非环境的角度来判断 $\Delta H$ 的符号。快速判断法：环境升温 $\Rightarrow$ 热量离开体系 $\Rightarrow$ 放热 $\Rightarrow$ $\Delta H < 0$。这一条规则可以解决所有放热/吸热符号题，包括本套题后面更棘手的量热法题目。

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Q2MEDIUMPaper 11.1 Calorimetry Sign

Water surrounding a reaction vessel warms up. Correct sign of $\Delta H$ and reason?反应容器周围的水温升高。正确的 $\Delta H$ 符号及理由？

Answer:答案： (B)
$Q = mc\Delta T$ calculates heat gained by the water (the surroundings). If the water heats up, the reaction (the system) must have released that energy, so $\Delta H < 0$. (A) and (C) reverse the logic: they treat the water's warming as the system absorbing heat, which is backwards, since the water gained the heat that the system lost. (D) describes a scenario that didn't happen (the water cooling), which would signal an endothermic reaction instead.$Q = mc\Delta T$ 计算的是水（环境）吸收的热量。若水升温，说明反应（体系）释放了这部分能量，故 $\Delta H < 0$。(A)、(C) 把逻辑弄反了：它们把水的升温当成体系在吸热，实际相反，水获得的正是体系失去的热量。(D) 描述的是没有发生的情形（水降温），那才会对应吸热反应。
Insight.要点。 This is the single most common calorimetry mistake: forgetting that $Q = mc\Delta T$ measures the surroundings' heat gain, and that $\Delta H$ (the system's enthalpy change) is the negative of it: $\Delta H = -Q/n$. Students who skip this translation step consistently get the sign backwards on exam calorimetry questions.这是量热法题目中最常见的错误：忘记 $Q = mc\Delta T$ 测的是环境吸收的热量，而 $\Delta H$（体系的焓变）是它的负值：$\Delta H = -Q/n$。跳过这一步"符号转换"的学生，在考试的量热计算题中总会把符号弄反。

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Q3MEDIUMPaper 11.2 Bond Enthalpy Direction

Using average bond enthalpies, $\Delta H$ is calculated as?用平均键能计算 $\Delta H$ 的正确表达式？

Answer:答案： (B)
Breaking bonds always costs energy (endothermic, positive contribution); forming bonds always releases energy (exothermic, so it is subtracted). $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$. Option (A) has the direction backwards. (C) and (D) add the two sums instead of subtracting, which double-counts energy instead of finding the net change.断键总是需要吸收能量（吸热，取正贡献）；成键总是释放能量（放热，因此要减去）。$\Delta H = \sum(\text{断键}) - \sum(\text{成键})$。选项 (A) 方向弄反了。(C)、(D) 把两个和相加而非相减，这是把能量重复计入而非求净变化。
Insight.要点。 Memory device: "Break, then Build." Broken comes first (and positive) in the formula; formed comes second (and subtracted). If a reaction is exothermic, the bonds formed in the products must be stronger overall than the bonds broken in the reactants, since more energy comes out (formed) than went in (broken).记忆窍门："先断后建"。公式中断键在前（取正），成键在后（要减）。若反应放热，说明产物中形成的键整体上比反应物中断裂的键更强，因为释放的能量（成键）比消耗的能量（断键）更多。

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Q4MEDIUMPaper 11.2 Hess's Law: Combustion Data

Using standard enthalpies of combustion, $\Delta H^\ominus$ is correctly calculated as?用标准燃烧焓计算 $\Delta H^\ominus$ 的正确表达式？

Answer:答案： (C)
Combustion data is the mirror image of formation data. Enthalpies of formation build reactants and products up from elements, so the formula reads "products minus reactants." Combustion enthalpies instead describe reactants and products both burning down to a common set of products ($\mathrm{CO_2}$, $\mathrm{H_2O}$), so the natural cycle direction is reversed: "reactants minus products." (A) and (B) apply the formation-style formula to the wrong quantity.燃烧焓数据与生成焓数据互为镜像。生成焓是从单质向上构建反应物和产物，因此公式是"产物减反应物"。燃烧焓则描述反应物和产物都向下燃烧到同一组终产物（$\mathrm{CO_2}$、$\mathrm{H_2O}$），循环的自然方向因而反过来："反应物减产物"。(A)、(B) 把生成焓式的公式用错了对象。
Insight.要点。 Draw the Hess cycle before writing any formula: elements/reactants at the top, common combustion products ($\mathrm{CO_2} + \mathrm{H_2O}$) at the bottom. Formation arrows point down from elements; combustion arrows point down from both reactants and products. Reading "the direct route equals the sum of the indirect route" off the diagram removes the need to memorise which formula goes with which data type.写公式之前先画出盖斯循环：单质/反应物在上方，共同的燃烧终产物（$\mathrm{CO_2} + \mathrm{H_2O}$）在下方。生成焓的箭头从单质向下；燃烧焓的箭头则从反应物和产物都向下。从图中读出"直接路径等于间接路径之和"，就无需死记哪种数据配哪个公式。

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Q5MEDIUMPaper 11.3 Incomplete Combustion

Why are larger hydrocarbons more prone to incomplete combustion?为何较大的烃分子更容易不完全燃烧？

Answer:答案： (B)
A larger hydrocarbon molecule (more C and H atoms) needs proportionally more $\mathrm{O_2}$ per mole for complete combustion to $\mathrm{CO_2}$ and $\mathrm{H_2O}$. If the local oxygen supply is fixed (e.g. a candle flame, an engine cylinder), a bigger fuel molecule is more likely to run short of oxygen before every carbon atom is fully oxidized, so some carbon ends up as CO or soot instead. C–H bond strength (A) and boiling point (C) are not the controlling factors here: this is a stoichiometric oxygen-supply argument, not a bond-strength or volatility one.较大的烃分子（碳、氢原子更多）完全燃烧为 $\mathrm{CO_2}$ 和 $\mathrm{H_2O}$ 所需的 $\mathrm{O_2}$ 按比例更多。若局部供氧固定（如蜡烛火焰、发动机气缸），较大的燃料分子在所有碳原子被完全氧化之前更容易缺氧，于是部分碳最终生成 CO 或炭黑。C–H 键强度 (A) 和沸点 (C) 并非此处的决定因素：这是化学计量供氧的论证，不是键强或挥发性的问题。
Insight.要点。 Think in moles of $\mathrm{O_2}$ per mole of fuel, not just molecule size: octane ($\mathrm{C_8H_{18}}$) needs $12.5$ mol $\mathrm{O_2}$ per mole, methane ($\mathrm{CH_4}$) needs only $2$. In a fixed-air environment, the fuel with the larger oxygen "appetite" per molecule is the one most likely to combust incompletely.要按"每摩尔燃料需要多少摩尔 $\mathrm{O_2}$"来思考，而不只是看分子大小：辛烷（$\mathrm{C_8H_{18}}$）每摩尔需要 $12.5$ mol $\mathrm{O_2}$，甲烷（$\mathrm{CH_4}$）仅需 $2$ mol。在供氧固定的环境中，每分子"氧气胃口"更大的燃料最容易发生不完全燃烧。

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Q6MEDIUMPaper 11.3 Fuel Cells

Which statement about hydrogen fuel cells is correct?关于氢燃料电池，哪一项正确？

Answer:答案： (B)
Overall: $2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O}$. Every hydrogen and oxygen atom fed in ends up in water; there is no carbon anywhere in the cell, so no $\mathrm{CO_2}$ can form. (A) is wrong: hydrogen is oxidized (loses electrons) at the anode, not reduced. (C) contradicts the whole appeal of fuel cells: zero-carbon operation. (D) is wrong by definition: a fuel cell's entire function is converting the chemical energy of the H–H and O=O bonds into electrical energy via the external circuit.总反应：$2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O}$。输入的每个氢、氧原子最终都进入水中；电池中不含碳，因此不可能生成 $\mathrm{CO_2}$。(A) 错误：氢气在阳极被氧化（失电子），而非还原。(C) 与燃料电池的核心卖点（零碳运行）相矛盾。(D) 从定义上就错：燃料电池的全部功能正是通过外电路把 H–H 和 O=O 键中的化学能转化为电能。
Insight.要点。 Track atoms, not just words: if a device contains no carbon-based fuel, it is structurally impossible for it to emit $\mathrm{CO_2}$. This atom-tracking habit catches many plausible-sounding wrong options in fuels questions.要追踪原子，而不只是字面表述：若某装置不含碳基燃料，它在结构上就不可能排放 $\mathrm{CO_2}$。这种"追踪原子"的习惯能识破燃料类题目中许多听起来合理但实际错误的选项。

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Q7HARDPaper 1HL1.4 Entropy Prediction

For which reaction is $\Delta S^\ominus$ expected to be most positive?哪个反应的 $\Delta S^\ominus$ 预期最正？

Answer:答案： (B)
Entropy tracks the number of moles of gas most strongly. Count moles of gas, reactants $\to$ products: (A) $3 \to 2$ (decrease); (B) $0 \to 1$ (increase: a pure solid decomposes to give a solid and a gas from nothing); (C) $4 \to 2$ (decrease); (D) gas $\to$ liquid (decrease). Only (B) increases the number of gas moles, from a starting point of zero, giving the largest possible relative jump in disorder among these four.熵变最主要由气体的物质的量决定。统计气体摩尔数由反应物到产物的变化：(A) $3 \to 2$（减小）；(B) $0 \to 1$（增大：纯固体分解，凭空生成一种固体和一种气体）；(C) $4 \to 2$（减小）；(D) 气体 $\to$ 液体（减小）。四者中只有 (B) 使气体摩尔数增大，且是从零出发，是这四者中混乱度相对跃升幅度最大的一个。
Insight.要点。 For a fast Paper 1 entropy call, just count $\Delta n_{\text{gas}} = n_{\text{gas, products}} - n_{\text{gas, reactants}}$: positive $\Delta n_{\text{gas}}$ almost always means $\Delta S^\ominus > 0$. Only reach for solution/mixing entropy arguments when every option has the same $\Delta n_{\text{gas}}$.在 Paper 1 中快速判断熵变，只需数 $\Delta n_{\text{gas}} = n_{\text{气体，产物}} - n_{\text{气体，反应物}}$：$\Delta n_{\text{gas}}$ 为正几乎总意味着 $\Delta S^\ominus > 0$。只有当所有选项的 $\Delta n_{\text{gas}}$ 都相同时，才需要考虑溶解/混合熵之类的论证。

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PART II  ·  PAPER 2第二部分  ·  第二卷Structured Response — Worked Solutions结构化解答题 —— 详细解析

Structured Response结构化解答题

SR 1MEDIUMPaper 21.1 Calorimetry + Hess's Law

Ethanol calorimetry: 200.0 g water, 21.0 °C → 47.5 °C; spirit burner mass 87.42 g → 86.57 g.乙醇量热：200.0 g 水，21.0 °C → 47.5 °C；酒精灯质量 87.42 g → 86.57 g。

(a) Heat, moles, experimental $\Delta H_c^\ominus$.热量、物质的量、实验 $\Delta H_c^\ominus$。 M1·A1·M1·A1
$$Q = mc\Delta T = 200.0 \times 4.18 \times (47.5 - 21.0) = 200.0 \times 4.18 \times 26.5 = 22{,}154~\mathrm{J} = 22.2~\mathrm{kJ}$$
Mass of ethanol burned $= 87.42 - 86.57 = 0.85~\mathrm{g}$.燃烧的乙醇质量 $= 87.42 - 86.57 = 0.85~\mathrm{g}$。
$$n(\mathrm{C_2H_5OH}) = \dfrac{0.85}{46.08} = 0.01845~\mathrm{mol}$$
$$\Delta H_c^\ominus(\mathrm{exp}) = -\dfrac{Q}{n} = -\dfrac{22{,}154}{0.01845} \approx -1.20 \times 10^{6}~\mathrm{J\,mol^{-1}} = -1.20 \times 10^{3}~\mathrm{kJ\,mol^{-1}}$$
Insight.要点。 Notice the chain: $Q$ (surroundings' heat, always positive here) $\to$ $n$ (from the mass lost by the fuel, not the water) $\to$ $\Delta H = -Q/n$ (the sign flip back to the system). Mixing up which mass belongs to $Q$'s formula and which belongs to $n$'s is the most common slip in this calculation.留意这条计算链：$Q$（环境吸热，此处恒为正）$\to$ $n$（由燃料损失的质量求得，而非水的质量）$\to$ $\Delta H = -Q/n$（符号翻转回体系视角）。把 $Q$ 公式该用的质量和 $n$ 公式该用的质量搞混，是这类计算中最常见的失误。
(b) Why the experimental magnitude is smaller than the theoretical value.为何实验值的绝对值小于理论值。 A1·A1·A1
  • Heat loss to the surroundings (air, calorimeter walls) rather than all of it going into the water, because the open spirit-burner setup is very poorly insulated.部分热量散失到环境（空气、量热计壁），并非全部进入水中，因为敞口酒精灯装置隔热极差。
  • Incomplete combustion of the ethanol (some carbon forms soot/CO instead of $\mathrm{CO_2}$), which releases less energy per mole burned.乙醇不完全燃烧（部分碳生成炭黑/CO 而非 $\mathrm{CO_2}$），每摩尔燃烧释放的能量更少。
  • Evaporation of ethanol from the wick before it burns, so the mass lost overstates the ethanol that actually combusted.部分乙醇在燃烧前从灯芯挥发，使损失的质量高估了实际燃烧的乙醇量。
(c) Theoretical $\Delta H_c^\ominus$ via Hess's law and comparison.用盖斯定律求理论 $\Delta H_c^\ominus$ 并比较。 M1·A1·A1·R1
$$\Delta H = [2(-394) + 3(-286)] - [(-278) + 0] = [-788 - 858] - (-278) = -1646 + 278 = -1368~\mathrm{kJ\,mol^{-1}}$$
Percentage difference: $\dfrac{1368 - 1200}{1368} \times 100\% \approx 12\%$. The experimental value is about 12% smaller in magnitude, consistent with the heat-loss and incomplete-combustion reasons given in (b).百分比差异：$\dfrac{1368 - 1200}{1368} \times 100\% \approx 12\%$。实验值的绝对值约小 12%，与 (b) 中给出的散热与不完全燃烧原因一致。
(d) Apparatus improvement.装置改进。 M1·R1
Add a draft shield / lid around the burner and calorimeter (or use a copper calorimeter with an insulating jacket, and move the flame closer to the base). This reduces heat lost by convection and radiation before it reaches the water, so a larger fraction of the released heat is captured by $Q = mc\Delta T$, bringing the experimental value closer to the theoretical one.在酒精灯与量热计周围加挡风罩/盖子（或改用带隔热套的铜制量热计，并使火焰更靠近容器底部）。这可减少热量在到达水之前因对流和辐射散失，使 $Q = mc\Delta T$ 捕获到更大比例的释放热量，令实验值更接近理论值。

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SR 2HARDPaper 2HL1.4 Entropy + Gibbs Energy + Equilibrium

$\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$, $\Delta H^\ominus = +57.2~\mathrm{kJ\,mol^{-1}}$. $S^\ominus(\mathrm{N_2O_4}) = 304$, $S^\ominus(\mathrm{NO_2}) = 240~\mathrm{J\,K^{-1}\,mol^{-1}}$.$\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$，$\Delta H^\ominus = +57.2~\mathrm{kJ\,mol^{-1}}$。$S^\ominus(\mathrm{N_2O_4}) = 304$，$S^\ominus(\mathrm{NO_2}) = 240~\mathrm{J\,K^{-1}\,mol^{-1}}$。

(a) $\Delta S^\ominus$.$\Delta S^\ominus$。 M1·A1
$$\Delta S^\ominus = 2(240) - 304 = 480 - 304 = +176~\mathrm{J\,K^{-1}\,mol^{-1}}$$
(b) $\Delta G^\ominus$ at 298 K and spontaneity.298 K 时的 $\Delta G^\ominus$ 及自发性。 M1·A1·R1
$$\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus = 57{,}200 - 298(176) = 57{,}200 - 52{,}448 = +4752~\mathrm{J\,mol^{-1}} = +4.75~\mathrm{kJ\,mol^{-1}}$$
Since $\Delta G^\ominus > 0$ at 298 K, the forward (dissociation) reaction is not spontaneous at this temperature: $\mathrm{N_2O_4}$ is favored.由于 298 K 时 $\Delta G^\ominus > 0$，正反应（解离）在此温度不自发：$\mathrm{N_2O_4}$ 占优。
(c) Temperature above which forward reaction is spontaneous.正反应自发所需的最低温度。 M1·A1·R1
$$0 = \Delta H^\ominus - T\Delta S^\ominus \;\Rightarrow\; T = \dfrac{\Delta H^\ominus}{\Delta S^\ominus} = \dfrac{57{,}200}{176} = 325~\mathrm{K}$$
Above 325 K, $\Delta G^\ominus < 0$ and dissociation becomes spontaneous.高于 325 K 时 $\Delta G^\ominus < 0$，解离变得自发。
(d) Equilibrium constant $K$ at 298 K.298 K 下的平衡常数 $K$。 M1·A1·R1
$$\ln K = -\dfrac{\Delta G^\ominus}{RT} = -\dfrac{4752}{(8.314)(298)} = -\dfrac{4752}{2477.6} = -1.918$$
$$K = e^{-1.918} \approx 0.147$$
$K < 1$ confirms that reactants ($\mathrm{N_2O_4}$) are favored at 298 K, consistent with the positive $\Delta G^\ominus$ found in (b): a positive $\Delta G^\ominus$ always corresponds to $K < 1$.$K < 1$ 证实 298 K 时反应物（$\mathrm{N_2O_4}$）占优，与 (b) 中所得的正 $\Delta G^\ominus$ 一致：$\Delta G^\ominus$ 为正必然对应 $K < 1$。
(e) Why $\Delta S^\ominus$ is positive.为何 $\Delta S^\ominus$ 为正。 A1
One mole of gas becomes two moles of gas: more independent particles means more ways to arrange positions and momenta, so the number of accessible microstates (disorder) increases.一摩尔气体变为两摩尔气体：独立粒子数增多意味着位置与动量的排列方式增多，可及微观状态数（混乱度）随之增大。
Insight.要点。 Parts (b)–(d) are three views of the same fact and must agree: $\Delta G^\ominus > 0 \Leftrightarrow$ not spontaneous $\Leftrightarrow K < 1$. If your sign in (b) and your $K$ in (d) ever disagree (e.g. $\Delta G^\ominus > 0$ but $K > 1$), you have an arithmetic error; use this cross-check every time.(b)–(d) 是同一个事实的三种视角，必须彼此一致：$\Delta G^\ominus > 0 \Leftrightarrow$ 不自发 $\Leftrightarrow K < 1$。若 (b) 中的符号与 (d) 中的 $K$ 出现矛盾（例如 $\Delta G^\ominus > 0$ 却算出 $K > 1$），说明存在计算错误，每次都应做这一交叉检验。

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SR 3HARDPaper 21.2 Bond Enthalpies vs Formation Data

$\mathrm{C_3H_8(g)} + 5\mathrm{O_2(g)} \rightarrow 3\mathrm{CO_2(g)} + 4\mathrm{H_2O(g)}$. Bond enthalpies (kJ mol⁻¹): C–C 346, C–H 414, O=O 498, C=O 804, O–H 463.$\mathrm{C_3H_8(g)} + 5\mathrm{O_2(g)} \rightarrow 3\mathrm{CO_2(g)} + 4\mathrm{H_2O(g)}$。键能（kJ mol⁻¹）：C–C 346，C–H 414，O=O 498，C=O 804，O–H 463。

(a) $\Delta H$ via bond enthalpies.用键能求 $\Delta H$。 M1·M1·A1·A1·A1
Propane ($\mathrm{CH_3CH_2CH_3}$) has 2 C–C bonds and 8 C–H bonds.丙烷（$\mathrm{CH_3CH_2CH_3}$）含 2 个 C–C 键与 8 个 C–H 键。
$$\text{Bonds broken} = 2(346) + 8(414) + 5(498) = 692 + 3312 + 2490 = 6494~\mathrm{kJ}$$
Products: 3 $\mathrm{CO_2}$ (2 C=O each = 6) and 4 $\mathrm{H_2O}$ (2 O–H each = 8).产物：3 个 $\mathrm{CO_2}$（各 2 个 C=O，共 6 个）与 4 个 $\mathrm{H_2O}$（各 2 个 O–H，共 8 个）。
$$\text{Bonds formed} = 6(804) + 8(463) = 4824 + 3704 = 8528~\mathrm{kJ}$$
$$\Delta H = 6494 - 8528 = -2034~\mathrm{kJ\,mol^{-1}}$$
(b) $\Delta H$ via enthalpies of formation.用生成焓求 $\Delta H$。 M1·A1
$$\Delta H = [3(-394) + 4(-242)] - [(-104) + 0] = [-1182 - 968] - (-104) = -2150 + 104 = -2046~\mathrm{kJ\,mol^{-1}}$$
(c) Comparison.比较。 R1·R1
The two values ($-2034$ vs $-2046~\mathrm{kJ\,mol^{-1}}$) are close but not identical (about 12 kJ mol⁻¹ apart, <1%). Average bond enthalpies are means taken across many different molecules containing that bond type; the actual C–H and C–C bonds in propane specifically differ very slightly from those averages, so the bond-enthalpy method gives only an estimate. Formation data, by contrast, refers to the exact compounds in this reaction and gives the more reliable value.两个值（$-2034$ 与 $-2046~\mathrm{kJ\,mol^{-1}}$）接近但不完全相同（相差约 12 kJ mol⁻¹，<1%）。平均键能是对许多含该类型键的不同分子取平均得到的；丙烷中实际的 C–H 和 C–C 键与这些平均值略有差异，因此键能法只能给出估计值。相比之下，生成焓数据针对的正是本反应中的具体化合物，给出的值更可靠。
(d) Indirect determination of $\Delta H_f^\ominus(\mathrm{C_3H_8})$.间接求 $\Delta H_f^\ominus(\mathrm{C_3H_8})$。 M1·R1·A1
Build a Hess cycle with two routes from the elements ($3\mathrm{C(graphite)} + 4\mathrm{H_2(g)} + 5\mathrm{O_2(g)}$) to the same combustion products ($3\mathrm{CO_2} + 4\mathrm{H_2O}$): Route 1 burns the elements directly; Route 2 first forms $\mathrm{C_3H_8}$ ($\Delta H_f^\ominus$) and then burns it ($\Delta H_c^\ominus(\mathrm{C_3H_8})$). Equating the two routes:构建一个盖斯循环，从单质（$3\mathrm{C(graphite)} + 4\mathrm{H_2(g)} + 5\mathrm{O_2(g)}$）出发经两条路径到达同一组燃烧产物（$3\mathrm{CO_2} + 4\mathrm{H_2O}$）：路径 1 直接燃烧单质；路径 2 先生成 $\mathrm{C_3H_8}$（$\Delta H_f^\ominus$），再燃烧它（$\Delta H_c^\ominus(\mathrm{C_3H_8})$）。令两条路径相等：
$$\Delta H_f^\ominus(\mathrm{C_3H_8}) = 3\,\Delta H_c^\ominus(\mathrm{C,\,graphite}) + 4\,\Delta H_c^\ominus(\mathrm{H_2}) - \Delta H_c^\ominus(\mathrm{C_3H_8})$$
All three combustion enthalpies on the right can be measured directly by calorimetry, even though $\Delta H_f^\ominus(\mathrm{C_3H_8})$ itself cannot.右边三个燃烧焓都可以直接用量热法测得，即便 $\Delta H_f^\ominus(\mathrm{C_3H_8})$ 本身无法直接测量。
Insight.要点。 Bond enthalpies and formation/combustion data are two independent routes to the same $\Delta H$, and small disagreement between them (here about 0.6%) is expected and diagnostic: it is evidence that bond enthalpies are averages, not a sign that one method is "wrong." A large disagreement (tens of kJ) would instead suggest an arithmetic slip.键能法与生成焓/燃烧焓数据是通向同一个 $\Delta H$ 的两条独立路径，二者出现小幅差异（此处约 0.6%）是预期之中、且具有诊断意义的：这正是键能为平均值的证据，而非某种方法"出错"的标志。若差异达到几十 kJ，则更可能提示计算失误。

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PART III  ·  PAPER 3 HL第三部分  ·  第三卷 HLData-Based — Worked Solutions数据题 —— 详细解析

Data-Based Questions (HL)数据题（HL）

P3-1HARDPaper 3 HL1.3 Fuels, Energy Density & Emissions

Methanol ($M = 32.05$, $\Delta H_c^\ominus = -726~\mathrm{kJ\,mol^{-1}}$) vs octane ($M = 114.26$, $\Delta H_c^\ominus = -5470~\mathrm{kJ\,mol^{-1}}$).甲醇（$M = 32.05$，$\Delta H_c^\ominus = -726~\mathrm{kJ\,mol^{-1}}$）与辛烷（$M = 114.26$，$\Delta H_c^\ominus = -5470~\mathrm{kJ\,mol^{-1}}$）。

(a) Energy density.能量密度。 M1·A1·A1·R1
$$\text{Methanol: } \dfrac{726}{32.05} \approx 22.7~\mathrm{kJ\,g^{-1}} \qquad \text{Octane: } \dfrac{5470}{114.26} \approx 47.9~\mathrm{kJ\,g^{-1}}$$
Octane has roughly double the energy density by mass. This reflects methanol already being partially oxidized (it carries an O atom), so it releases less energy per gram than a pure hydrocarbon.辛烷的单位质量能量密度约为甲醇的两倍。这反映出甲醇本身已部分氧化（含有一个 O 原子），因此每克释放的能量比纯烃更少。
(b) $\mathrm{CO_2}$ per MJ from octane.辛烷每 MJ 产生的 $\mathrm{CO_2}$。 M1·M1·A1·A1
Per mole octane: 5470 kJ released, 8 mol $\mathrm{CO_2}$ formed. Per MJ (1000 kJ):每摩尔辛烷释放 5470 kJ，生成 8 mol $\mathrm{CO_2}$。每 MJ（1000 kJ）：
$$n(\mathrm{octane}) = \dfrac{1000}{5470} = 0.1828~\mathrm{mol} \;\Rightarrow\; n(\mathrm{CO_2}) = 8 \times 0.1828 = 1.463~\mathrm{mol}$$
$$m(\mathrm{CO_2}) = 1.463 \times 44.01 \approx 64.4~\mathrm{g}$$
(c) "Carbon neutral" claim."碳中和"的说法。 R1·R1·R1
The label is only partly justified. The $\mathrm{CO_2}$ released on combustion was indeed recently absorbed by the growing crop, so that step alone can be roughly balanced. However, "carbon neutral" ignores the fossil-fuel energy typically used to farm, harvest, transport, and distill/process the biomass into fuel, plus any land-use change (e.g. clearing forest for cropland releases stored carbon). A full life-cycle assessment, not just the combustion step, is needed before the "carbon neutral" claim can be accepted.这一说法只是部分成立。燃烧释放的 $\mathrm{CO_2}$ 确实是作物近期生长时吸收的，仅就这一步而言大致可以平衡。但"碳中和"忽略了种植、收割、运输以及将生物质蒸馏/加工成燃料通常所耗费的化石能源，以及任何土地利用变化（例如清除森林以获得耕地会释放储存的碳）。要接受"碳中和"的说法，需要完整的生命周期评估，而不仅仅是燃烧这一步。
(d) Hydrogen fuel cells: advantage and limitation.氢燃料电池：优点与局限。 R1·R1·R1·R1
Advantage: zero tailpipe emissions, since the only product is water, so there is no local $\mathrm{CO_2}$, $\mathrm{CO}$, or particulate emission at the point of use, unlike combustion.优点：尾气零排放，因为唯一产物是水，不像燃烧那样在使用端产生局部 $\mathrm{CO_2}$、CO 或颗粒物排放。
Limitation: hydrogen has a very low volumetric energy density as a gas, requiring heavy high-pressure or cryogenic storage tanks, and most hydrogen today is produced from natural gas (releasing $\mathrm{CO_2}$ upstream) rather than by clean electrolysis, so the "zero-carbon" benefit depends heavily on how the hydrogen itself was made.局限：氢气作为气体的体积能量密度很低，需要笨重的高压或低温储罐；且目前大多数氢气由天然气制取（上游仍释放 $\mathrm{CO_2}$）而非清洁电解制氢，因此"零碳"效益很大程度上取决于氢气本身的制取方式。
Insight.要点。 Whenever a question asks you to "evaluate" a fuel or claim a technology is "clean," always separate the point-of-use emissions from the whole life-cycle emissions (production, transport, storage). A fuel can be emission-free at the tailpipe and still carry a large carbon footprint upstream; examiners reward answers that explicitly make this distinction.凡遇到"评价"某种燃料或"清洁"技术主张的题目，务必把使用端排放与全生命周期排放（生产、运输、储存）区分开。某种燃料在尾气端可以零排放，但上游仍可能有很大的碳足迹，阅卷标准会奖励明确作出这一区分的答案。

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P3-2HARDPaper 3 HLHL1.2 Born–Haber Cycle

Born–Haber cycle for $\mathrm{MgO(s)}$. Data (kJ mol⁻¹): $\Delta H_{at}(\mathrm{Mg}) = +148$, $IE_1(\mathrm{Mg}) = +738$, $IE_2(\mathrm{Mg}) = +1451$, $\Delta H_{at}(\mathrm{O}) = +249$, $EA_1(\mathrm{O}) = -141$, $EA_2(\mathrm{O}) = +798$, $\Delta H_f^\ominus(\mathrm{MgO}) = -602$.$\mathrm{MgO(s)}$ 的玻恩-哈伯循环。数据（kJ mol⁻¹）：$\Delta H_{at}(\mathrm{Mg}) = +148$，$IE_1(\mathrm{Mg}) = +738$，$IE_2(\mathrm{Mg}) = +1451$，$\Delta H_{at}(\mathrm{O}) = +249$，$EA_1(\mathrm{O}) = -141$，$EA_2(\mathrm{O}) = +798$，$\Delta H_f^\ominus(\mathrm{MgO}) = -602$。

(a) Electron affinity; why $EA_2(\mathrm{O})$ is endothermic.电子亲和能；为何 $EA_2(\mathrm{O})$ 是吸热的。 A1·R1
Electron affinity is the enthalpy change when one mole of gaseous atoms (or ions) each gains one electron. $EA_2(\mathrm{O})$ adds an electron to $\mathrm{O^-(g)}$, which is already negatively charged; the incoming electron is repelled by that existing negative charge, so energy must be supplied to force it on, hence $EA_2 > 0$, unlike the usually-exothermic first electron affinity.电子亲和能是指一摩尔气态原子（或离子）各获得一个电子时的焓变。$EA_2(\mathrm{O})$ 是给已带负电的 $\mathrm{O^-(g)}$ 再加一个电子；新加入的电子会被已有的负电荷排斥，因此必须提供能量才能把它加上去，故 $EA_2 > 0$，与通常放热的第一电子亲和能不同。
(b) Lattice enthalpy of $\mathrm{MgO}$.$\mathrm{MgO}$ 的晶格焓。 M1·M1·M1·A1·A1·R1
By Hess's law, the cycle sum equals $\Delta H_f^\ominus$:由盖斯定律，循环各步之和等于 $\Delta H_f^\ominus$：
$$\Delta H_f^\ominus = \Delta H_{at}(\mathrm{Mg}) + IE_1 + IE_2 + \Delta H_{at}(\mathrm{O}) + EA_1 + EA_2 + \Delta H_{\text{latt}}^\ominus$$
$$148 + 738 + 1451 + 249 - 141 + 798 = 3243~\mathrm{kJ\,mol^{-1}}$$
$$-602 = 3243 + \Delta H_{\text{latt}}^\ominus \;\Rightarrow\; \Delta H_{\text{latt}}^\ominus = -602 - 3243 = -3845~\mathrm{kJ\,mol^{-1}}$$
(c) Why $\mathrm{CaO}$'s lattice enthalpy is less exothermic than $\mathrm{MgO}$'s.为何 $\mathrm{CaO}$ 的晶格焓放热程度小于 $\mathrm{MgO}$。 A1·R1·R1·R1
Lattice enthalpy magnitude increases with ionic charge and decreases with ionic radius (roughly $\propto q_+q_-/r$). $\mathrm{Mg^{2+}}$ and $\mathrm{Ca^{2+}}$ carry the same charge, but $\mathrm{Ca^{2+}}$ is a larger ion (one period further down the group) than $\mathrm{Mg^{2+}}$. The larger $\mathrm{Ca^{2+}}$–$\mathrm{O^{2-}}$ separation weakens the electrostatic attraction between the ions, so less energy is released on forming the lattice: $\mathrm{CaO}$'s lattice enthalpy is smaller in magnitude (less exothermic) than $\mathrm{MgO}$'s.晶格焓的大小随离子电荷增大而增大，随离子半径增大而减小（大致 $\propto q_+q_-/r$）。$\mathrm{Mg^{2+}}$ 与 $\mathrm{Ca^{2+}}$ 电荷相同，但 $\mathrm{Ca^{2+}}$（同族下一周期）比 $\mathrm{Mg^{2+}}$ 离子半径更大。更大的 $\mathrm{Ca^{2+}}$–$\mathrm{O^{2-}}$ 间距削弱了离子间的静电吸引，形成晶格时释放的能量更少，因此 $\mathrm{CaO}$ 的晶格焓绝对值比 $\mathrm{MgO}$ 的更小（放热程度更弱）。
(d) $\mathrm{AgCl}$ vs $\mathrm{MgO}$: experimental vs theoretical lattice enthalpy.$\mathrm{AgCl}$ 与 $\mathrm{MgO}$：实验与理论晶格焓。 R1·R1·R1
A purely ionic model assumes perfectly spherical, non-interacting charge clouds. $\mathrm{Ag^+}$ is a small, highly polarizing cation with a distorted (non-noble-gas) electron configuration, and $\mathrm{Cl^-}$ is a large, easily polarized anion; $\mathrm{Ag^+}$ pulls and distorts the $\mathrm{Cl^-}$ electron cloud toward itself, giving the bond significant covalent (electron-sharing) character that a purely ionic model does not capture. The Born–Haber (experimental) value, which reflects the real bonding, therefore comes out more exothermic than the theoretical ionic-model prediction. $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ are comparatively closer to the "hard sphere" ionic picture, so the two values agree much more closely for $\mathrm{MgO}$.纯离子模型假设电荷云是完全球形且互不作用的。$\mathrm{Ag^+}$ 是一个小而极化能力强的阳离子（电子构型非稀有气体型），$\mathrm{Cl^-}$ 是一个大而易被极化的阴离子；$\mathrm{Ag^+}$ 会吸引并使 $\mathrm{Cl^-}$ 的电子云向自身偏移，使该键带有明显的共价（共享电子）成分，这是纯离子模型无法体现的。因此反映真实成键情况的玻恩-哈伯（实验）值，其放热程度大于理论离子模型的预测值。而 $\mathrm{Mg^{2+}}$ 与 $\mathrm{O^{2-}}$ 相对更接近"硬球"离子图像，因此 $\mathrm{MgO}$ 的两个值吻合得多。
Insight.要点。 Set up every Born–Haber cycle the same way: write the full chain from elements $\to$ gaseous atoms $\to$ gaseous ions $\to$ solid lattice, sum all known steps, then set that sum plus the unknown equal to $\Delta H_f^\ominus$. Getting the atomization/ionization/electron-affinity signs right (all endothermic except electron affinity, which is usually but not always exothermic) is worth more marks than the arithmetic itself.每个玻恩-哈伯循环都用同一套流程处理：写出从单质 $\to$ 气态原子 $\to$ 气态离子 $\to$ 固体晶格的完整链条，把所有已知步骤相加，再令该和加上未知项等于 $\Delta H_f^\ominus$。把原子化/电离/电子亲和能的符号弄对（除电子亲和能外均为吸热，电子亲和能通常但不总是放热）比算术本身更值得拿分。

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