High School Chemistry — Thermochemistry and Energy — Solutions

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Section A · Worked SolutionsA 部分 · 详细解答

Q1 EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1-2 Enthalpy sign & exo/endo焓变符号与放/吸热 · HS-PS3-1 [3 marks][3 分]

Which statement correctly pairs an enthalpy sign with the direction of heat flow for a reaction at constant pressure?以下哪项正确地将焓变符号与恒压反应的热流方向配对？

Answer:答案： (A) $\Delta H > 0$: heat flows from surroundings into the system (endothermic)

(a) Identify the correct sign convention判断正确的符号约定 A1·A1·A1

At constant pressure, $q_p = \Delta H$. When the system absorbs heat from the surroundings, the system gains energy, so $\Delta H > 0$ (endothermic). When the system releases heat to the surroundings, the system loses energy, so $\Delta H < 0$ (exothermic). Option (A) correctly pairs $\Delta H > 0$ with heat flowing from surroundings into the system (endothermic).恒压时 $q_p = \Delta H$。体系从环境吸热时，体系能量增加，故 $\Delta H > 0$（吸热）。体系向环境放热时，体系能量减少，故 $\Delta H < 0$（放热）。选项 (A) 正确地将 $\Delta H > 0$ 与热量从环境流入体系（吸热）配对。

(B) is wrong: $\Delta H < 0$ is exothermic, not endothermic.(B) 错误：$\Delta H < 0$ 是放热，非吸热。
(C) is wrong: $\Delta H > 0$ is endothermic; heat does not flow from system to surroundings.(C) 错误：$\Delta H > 0$ 是吸热，热量不从体系流向环境。
(D) is correct sign but wrong label; $\Delta H < 0$ is exothermic, but option (D) correctly matches sign with direction yet the question pairs exo with the wrong flow.(D) 符号与热流方向的组合自身正确，但题目的干扰项将其与"放热"而非"体系放热"描述混淆。

The sign of $\Delta H$ reflects the system's perspective: positive means the system gains energy (endothermic), negative means the system loses energy (exothermic).$\Delta H$ 的符号反映体系的视角：正值意味体系得到能量（吸热），负值意味体系损失能量（放热）。 A memory device: think of $\Delta H$ as the system's "bank balance change." Endothermic reactions deposit energy into the system (balance increases, $\Delta H > 0$); exothermic reactions withdraw energy from the system (balance decreases, $\Delta H < 0$). This perspective never changes regardless of what the surroundings feel.记忆方法：把 $\Delta H$ 想成体系"账户余额的变化"。吸热反应往体系账户存钱（余额增加，$\Delta H > 0$）；放热反应从体系账户取钱（余额减少，$\Delta H < 0$）。这一视角不随环境的感受而改变。

Q2 EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Exothermic vs endothermic放热与吸热反应 · SCH3U Unit 4 [4 marks][4 分]

$\text{NH}_4\text{NO}_3$ dissolves in water; temperature drops from $22.0\ \text{°C}$ to $14.5\ \text{°C}$.$\text{NH}_4\text{NO}_3$ 溶于水，温度从 $22.0\ \text{°C}$ 降至 $14.5\ \text{°C}$。

Answer:答案： (a) endothermic吸热 · (b) $\Delta H > 0$ (system absorbs energy from surroundings)（体系从环境吸收能量）

(a) Classify the process from temperature data从温度数据判断反应类型 M1·A1

The solution temperature drops by $22.0 - 14.5 = 7.5\ \text{°C}$. The surroundings (solution) lost heat to the dissolving solid, meaning the dissolving process absorbed heat from the surroundings. Therefore the process is endothermic.溶液温度下降 $22.0 - 14.5 = 7.5\ \text{°C}$。环境（溶液）向溶解的固体输出热量，说明溶解过程从环境吸热。因此该过程为吸热过程。

(b) State the sign of $\Delta H$ and explain energy flow说明 $\Delta H$ 的符号并解释能量流动 M1·A1

$\Delta H > 0$ (positive). The system (the dissolving process) gains energy from the surroundings. Energy flows from the surroundings (the water solution, which cools down) into the system, increasing the system's enthalpy.$\Delta H > 0$（正值）。体系（溶解过程）从环境获得能量。能量从环境（冷却的水溶液）流入体系，使体系的焓值增大。

Temperature change of the surroundings reveals the direction of heat flow for the reaction.环境温度的变化揭示了反应热流的方向。 In a coffee-cup calorimeter the solution is the surroundings and the reaction is the system. If the surroundings cool, heat flowed out of the surroundings into the system: endothermic. If the surroundings warm, heat flowed from the system to the surroundings: exothermic. $\text{NH}_4\text{NO}_3$ dissolving endothermically is why it is used in instant cold packs: dissolving it in water pulls heat from the pack's interior, making it cold.在咖啡杯量热计中，溶液是环境，反应是体系。若环境温度降低，热量从环境流入体系：吸热。若环境温度升高，热量从体系流向环境：放热。$\text{NH}_4\text{NO}_3$ 的吸热溶解正是即用型冰袋的工作原理：将其溶于水会从冰袋内部抽取热量，从而产生冷却效果。

Q3 MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Calorimetry ($q = mc\Delta T$)量热法（$q = mc\Delta T$） · Chemistry 11 [6 marks][6 分]

$150.0\ \text{g}$ water heated by candle: $22.0\ \text{°C} \to 35.5\ \text{°C}$. $c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$.$150.0\ \text{g}$ 水被蜡烛加热：$22.0\ \text{°C} \to 35.5\ \text{°C}$。$c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。

Answer:答案： (a) $\Delta T = 13.5\ \text{°C}$ · (b) $q = 8464.5\ \text{J} = 8.46\ \text{kJ}$ · (c) two stated assumptions两个假设条件

(a) Calculate $\Delta T$计算 $\Delta T$ A1

$$ \Delta T \;=\; T_{\text{final}} - T_{\text{initial}} \;=\; 35.5 - 22.0 \;=\; 13.5\ \text{°C.} $$

(b) Calculate heat absorbed, $q$, in J and kJ计算吸收的热量 $q$，以 J 和 kJ 表示 M1·A1·A1

$$ q \;=\; mc\Delta T \;=\; 150.0\ \text{g} \times 4.18\ \text{J g}^{-1}\ \text{°C}^{-1} \times 13.5\ \text{°C} \;=\; 8464.5\ \text{J} \;\approx\; 8465\ \text{J.} $$ $$ q \;=\; \frac{8464.5}{1000} \;=\; 8.46\ \text{kJ.} $$

(c) Two assumptions for the coffee-cup calorimeter咖啡杯量热计的两个假设条件 A1·A1

All heat released by the candle is absorbed by the water (no heat lost to the calorimeter walls, air, or surroundings).蜡烛释放的全部热量都被水吸收（无热量散逸至量热计壁、空气或其他环境）。
The heat capacity of the calorimeter cup itself is negligible compared to that of the water.量热计杯体本身的热容与水相比可以忽略不计。

(Also acceptable: the density of water is $1.00\ \text{g mL}^{-1}$, or the specific heat of the solution equals that of water.)（亦可接受：水的密度为 $1.00\ \text{g mL}^{-1}$，或溶液比热容等于水的比热容。）

$q = mc\Delta T$: three inputs, one output. Identify each before substituting.$q = mc\Delta T$：三个输入，一个输出。代入前逐一确认每个量。 Common errors: (1) forgetting to convert J to kJ for the final answer; (2) using $\Delta T = T_i - T_f$ (wrong order, gives negative $q$ for warming); (3) omitting units in intermediate steps. In a calorimetry problem, heat always flows from the hotter object to the cooler one; the calorimetry equation gives the magnitude and sign automatically if $\Delta T$ is computed as $T_f - T_i$.常见错误：(1) 忘记将 J 换算为 kJ；(2) 将 $\Delta T$ 写成 $T_i - T_f$（顺序相反，对升温情况给出负 $q$）；(3) 中间步骤省略单位。在量热题中，热量总是从高温物体流向低温物体；只要 $\Delta T$ 按 $T_f - T_i$ 计算，量热方程会自动给出正确的大小和符号。

Q4 MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §4 Thermochemical equations热化学方程式 · Chem 30-A1.1k [6 marks][6 分]

$\text{CH}_4(g) + 2\,\text{O}_2(g) \to \text{CO}_2(g) + 2\,\text{H}_2\text{O}(l)$, $\Delta H = -890\ \text{kJ mol}^{-1}$. Sample: $3.20\ \text{g}$ of $\text{CH}_4$.$\text{CH}_4(g) + 2\,\text{O}_2(g) \to \text{CO}_2(g) + 2\,\text{H}_2\text{O}(l)$，$\Delta H = -890\ \text{kJ mol}^{-1}$。样品：$3.20\ \text{g}$ 甲烷。

Answer:答案： (a) exothermic; $890\ \text{kJ}$ released per mol放热；每摩尔释放 $890\ \text{kJ}$ · (b) $n = 0.1995\ \text{mol}$ · (c) $q = 177.5\ \text{kJ}$

(a) Classify and interpret $\Delta H = -890\ \text{kJ mol}^{-1}$判断类型并解释 $\Delta H = -890\ \text{kJ mol}^{-1}$ A1·A1

The reaction is exothermic because $\Delta H$ is negative. The value $-890\ \text{kJ mol}^{-1}$ means that $890\ \text{kJ}$ of heat is released to the surroundings when exactly $1\ \text{mol}$ of $\text{CH}_4$ undergoes complete combustion at constant pressure.由于 $\Delta H$ 为负值，该反应为放热反应。$-890\ \text{kJ mol}^{-1}$ 表示在恒压条件下，每燃烧 $1\ \text{mol}$ 甲烷时向环境释放 $890\ \text{kJ}$ 的热量。

(b) Moles of $\text{CH}_4$ in $3.20\ \text{g}$$3.20\ \text{g}$ 甲烷的物质的量 M1·A1

$$ n(\text{CH}_4) \;=\; \frac{m}{M} \;=\; \frac{3.20\ \text{g}}{16.05\ \text{g mol}^{-1}} \;=\; 0.1995\ \text{mol.} $$

(c) Heat released for $3.20\ \text{g}$ of $\text{CH}_4$$3.20\ \text{g}$ 甲烷完全燃烧时释放的热量 M1·A1·A1

$$ |q| \;=\; n \times |\Delta H| \;=\; 0.1995\ \text{mol} \times 890\ \text{kJ mol}^{-1} \;=\; 177.5\ \text{kJ.} $$ Since the reaction is exothermic, $q_{\text{released}} = 177.5\ \text{kJ}$ (heat released to surroundings).由于反应为放热，释放热量 $q = 177.5\ \text{kJ}$（向环境放出）。

Thermochemical stoichiometry: scale $\Delta H$ by the mole ratio, just like any other stoichiometric quantity.热化学计量：像其他计量量一样，用摩尔比对 $\Delta H$ 进行等比例换算。 The three-step pattern for any thermochemical calculation: (1) convert mass to moles; (2) multiply by $\Delta H$ per mole to get the energy change for the actual amount; (3) state the sign clearly. Notice that $0.1995 \approx 0.2$ mol: a $3.20\ \text{g}$ sample is about one-fifth of a mole of methane, so the heat released is about one-fifth of $890\ \text{kJ}$, which is close to $178\ \text{kJ}$. This order-of-magnitude check is a fast exam-room sanity test.热化学计算的三步模式：(1) 将质量换算为物质的量；(2) 乘以每摩尔的 $\Delta H$ 得到实际量的能量变化；(3) 明确写出符号。注意 $0.1995 \approx 0.2$ mol：$3.20\ \text{g}$ 样品约为 $\frac{1}{5}$ mol 甲烷，释放热量约为 $890\ \text{kJ}$ 的五分之一，接近 $178\ \text{kJ}$。这种数量级估算是考场上的快速合理性检验。

Q5 MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §7 Potential-energy diagrams势能图 · HS-PS3-4 [6 marks][6 分]

PE diagram: reactants at $+40\ \text{kJ}$, peak at $+110\ \text{kJ}$, products at $-20\ \text{kJ}$.势能图：反应物位于 $+40\ \text{kJ}$，峰值位于 $+110\ \text{kJ}$，产物位于 $-20\ \text{kJ}$。

Answer:答案： (a) $E_a = 70\ \text{kJ}$ · (b) $\Delta H = -60\ \text{kJ}$ (exothermic)（放热） · (c) $E_a$ decreases; $\Delta H$ unchanged减小；$\Delta H$ 不变

(a) Activation energy for the forward reaction正反应的活化能 M1·A1

$E_a$ is the energy difference between the transition state (peak) and the reactants:$E_a$ 是过渡态（峰值）与反应物之间的能量差： $$ E_a \;=\; 110 - 40 \;=\; 70\ \text{kJ.} $$

(b) $\Delta H$ and exo/endo classification$\Delta H$ 及放热/吸热判断 M1·A1

$$ \Delta H \;=\; E_{\text{products}} - E_{\text{reactants}} \;=\; (-20) - (+40) \;=\; -60\ \text{kJ.} $$ $\Delta H = -60\ \text{kJ}$, so the reaction is exothermic.$\Delta H = -60\ \text{kJ}$，故该反应为放热反应。

(c) Effect of a catalyst on $E_a$ and $\Delta H$催化剂对 $E_a$ 和 $\Delta H$ 的影响 A1·A1

$E_a$ decreases: a catalyst provides an alternative reaction pathway with a lower energy transition state, so less energy is required to initiate the reaction.$E_a$ 减小：催化剂提供具有较低能量过渡态的替代反应路径，使反应所需的活化能降低。
$\Delta H$ stays the same: enthalpy is a state function that depends only on the initial (reactants) and final (products) energy levels, not on the pathway. A catalyst does not change the reactants or products, so $\Delta H$ is unchanged.$\Delta H$ 不变：焓是状态函数，只取决于初始（反应物）和最终（产物）的能量水平，与路径无关。催化剂不改变反应物或产物，故 $\Delta H$ 不变。

On a PE diagram: $E_a$ = peak minus reactants; $\Delta H$ = products minus reactants. Only $E_a$ changes with a catalyst.在势能图上：$E_a$ = 峰值减反应物；$\Delta H$ = 产物减反应物。催化剂只改变 $E_a$。 A common exam error is claiming a catalyst changes $\Delta H$. Remember: catalysts speed up reactions by lowering the energy barrier but do not alter the thermodynamic starting or ending points. The reaction releases the same total energy regardless of whether a catalyst is present. This is analogous to taking a mountain pass versus a tunnel to cross a mountain: the altitude difference between start and finish is the same no matter which route you take; only the maximum altitude on the route changes.常见考试错误是声称催化剂改变了 $\Delta H$。记住：催化剂通过降低能量壁垒来加速反应，但不改变热力学的起点或终点。无论是否使用催化剂，反应释放的总能量相同。这好比翻越山脉走山口还是穿隧道：起点和终点的海拔差相同，不论走哪条路；只是路上的最高点不同。

Section B · Worked SolutionsB 部分 · 详细解答

Q6 MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3-4 Calorimetry + enthalpy of neutralisation量热法 + 中和反应焓变 · SCH4U Unit 5 [8 marks][8 分]

$50.0\ \text{mL}$ of $1.00\ \text{mol L}^{-1}$ HCl + $50.0\ \text{mL}$ of $1.00\ \text{mol L}^{-1}$ NaOH; $T_i = 21.0\ \text{°C}$, $T_f = 27.5\ \text{°C}$; density $1.00\ \text{g mL}^{-1}$, $c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$.$50.0\ \text{mL}$ 的 $1.00\ \text{mol L}^{-1}$ 盐酸与 $50.0\ \text{mL}$ 的 $1.00\ \text{mol L}^{-1}$ 氢氧化钠混合；$T_i = 21.0\ \text{°C}$，$T_f = 27.5\ \text{°C}$；密度 $1.00\ \text{g mL}^{-1}$，$c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。

Answer:答案： (a) $m = 100.0\ \text{g}$, $\Delta T = 6.5\ \text{°C}$ · (b) $q_{\text{soln}} = 2717\ \text{J}$ · (c) $0.0500\ \text{mol}$ · (d) $\Delta H_{\text{neut}} = -54.3\ \text{kJ mol}^{-1}$ (exothermic)（放热）

(a) Total mass and $\Delta T$溶液总质量和 $\Delta T$ A1·A1

Total volume $= 50.0 + 50.0 = 100.0\ \text{mL}$. At density $1.00\ \text{g mL}^{-1}$, $m = 100.0\ \text{g}$.总体积 $= 50.0 + 50.0 = 100.0\ \text{mL}$。密度为 $1.00\ \text{g mL}^{-1}$，故 $m = 100.0\ \text{g}$。 $$ \Delta T \;=\; 27.5 - 21.0 \;=\; 6.5\ \text{°C.} $$

(b) Heat absorbed by the solution溶液吸收的热量 M1·A1

$$ q_{\text{soln}} \;=\; mc\Delta T \;=\; 100.0 \times 4.18 \times 6.5 \;=\; 2717\ \text{J.} $$

(c) Moles of water produced生成水的物质的量 A1

The reaction is $\text{H}^+(aq) + \text{OH}^-(aq) \to \text{H}_2\text{O}(l)$, with a 1:1 mole ratio. Moles of HCl $= 0.0500\ \text{L} \times 1.00\ \text{mol L}^{-1} = 0.0500\ \text{mol}$. Since HCl and NaOH are in equal amounts and react 1:1, moles of $\text{H}_2\text{O}$ produced $= 0.0500\ \text{mol}$.反应为 $\text{H}^+(aq) + \text{OH}^-(aq) \to \text{H}_2\text{O}(l)$，1:1 的摩尔比。盐酸物质的量 $= 0.0500\ \text{L} \times 1.00\ \text{mol L}^{-1} = 0.0500\ \text{mol}$。HCl 与 NaOH 等量且 1:1 反应，故生成 $\text{H}_2\text{O}$ 为 $0.0500\ \text{mol}$。

(d) Enthalpy of neutralisation per mole每摩尔中和焓变 M1·A1·A1

Heat evolved by the reaction equals heat absorbed by the solution: $q_{\text{rxn}} = -q_{\text{soln}} = -2717\ \text{J}$ (negative because heat is released to the solution).反应放出的热量等于溶液吸收的热量：$q_{\text{rxn}} = -q_{\text{soln}} = -2717\ \text{J}$（负号表示向溶液放热）。 $$ \Delta H_{\text{neut}} \;=\; \frac{q_{\text{rxn}}}{n(\text{H}_2\text{O})} \;=\; \frac{-2717\ \text{J}}{0.0500\ \text{mol}} \;=\; -54\,340\ \text{J mol}^{-1} \;=\; -54.3\ \text{kJ mol}^{-1.} $$ The reaction is exothermic ($\Delta H < 0$). The literature value for strong acid/strong base neutralisation is approximately $-57.3\ \text{kJ mol}^{-1}$; the experimental value is slightly lower due to heat losses to the calorimeter.该反应为放热反应（$\Delta H < 0$）。强酸强碱中和的文献值约为 $-57.3\ \text{kJ mol}^{-1}$；实验值偏低，是因为有热量散失至量热计。

The sign convention: $q_{\text{rxn}} = -q_{\text{soln}}$. The solution absorbs what the reaction releases, and vice versa.符号约定：$q_{\text{rxn}} = -q_{\text{soln}}$。溶液吸收的热量正是反应释放的热量，反之亦然。 This is the single most important sign relationship in calorimetry. When the solution warms up ($q_{\text{soln}} > 0$), the reaction released heat ($q_{\text{rxn}} < 0$, exothermic). Dividing by moles gives $\Delta H$ per mole. A common error is to divide by the total volume or mass instead of by the moles of the limiting reagent (here, the moles of water formed). Remember the experimental value for strong acid/strong base neutralisation ($\approx -57\ \text{kJ mol}^{-1}$) as a benchmark: values near this confirm a correct calculation.这是量热法中最重要的符号关系。当溶液升温（$q_{\text{soln}} > 0$），反应放热（$q_{\text{rxn}} < 0$，放热反应）。除以物质的量得每摩尔的 $\Delta H$。常见错误是除以总体积或总质量，而非限制试剂的摩尔数（此题为生成水的摩尔数）。记住强酸强碱中和的实验参考值（约 $-57\ \text{kJ mol}^{-1}$）：接近该值可确认计算正确。

Q7 MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Thermochemical equation stoichiometry热化学方程式计量 · HS-PS3-1 [7 marks][7 分]

$2\,\text{C}(s) + \text{H}_2(g) \to \text{C}_2\text{H}_2(g)$, $\Delta H_f^\circ = +226\ \text{kJ mol}^{-1}$. Sample: $10.0\ \text{g}$ of $\text{C}_2\text{H}_2$, $M = 26.04\ \text{g mol}^{-1}$.$2\,\text{C}(s) + \text{H}_2(g) \to \text{C}_2\text{H}_2(g)$，$\Delta H_f^\circ = +226\ \text{kJ mol}^{-1}$。样品：$10.0\ \text{g}$ 乙炔，$M = 26.04\ \text{g mol}^{-1}$。

Answer:答案： (a) positive sign means acetylene is less stable than its elements正值表示乙炔比其单质能量更高（不稳定） · (b) $n = 0.3840\ \text{mol}$ · (c) $\Delta H = +86.8\ \text{kJ}$ (heat absorbed)（吸热） · (d) $\text{C}_2\text{H}_2(g) \to 2\,\text{C}(s) + \text{H}_2(g)$, $\Delta H = -226\ \text{kJ mol}^{-1}$

(a) Interpret the positive $\Delta H_f^\circ$解释 $\Delta H_f^\circ$ 为正值的含义 A1·A1

A positive $\Delta H_f^\circ = +226\ \text{kJ mol}^{-1}$ means energy must be supplied to form $1\ \text{mol}$ of acetylene from its elements in their standard states. Acetylene has more enthalpy than its constituent elements (carbon and hydrogen); it is therefore thermodynamically less stable (higher energy) relative to its elements. Such compounds are called endothermic or "energy-rich" compounds.$\Delta H_f^\circ = +226\ \text{kJ mol}^{-1}$ 为正值，意味着由标准状态下的单质生成 $1\ \text{mol}$ 乙炔需要输入能量。乙炔的焓值高于其构成单质（碳和氢），因此相对于单质，乙炔在热力学上不稳定（能量更高）。这类化合物称为吸热化合物或"富能"化合物。

(b) Moles of $\text{C}_2\text{H}_2$ in $10.0\ \text{g}$$10.0\ \text{g}$ 乙炔的物质的量 M1·A1

$$ n(\text{C}_2\text{H}_2) \;=\; \frac{10.0\ \text{g}}{26.04\ \text{g mol}^{-1}} \;=\; 0.3840\ \text{mol.} $$

(c) Enthalpy change for synthesis of $10.0\ \text{g}$ of acetylene合成 $10.0\ \text{g}$ 乙炔的焓变 M1·A1

$$ \Delta H \;=\; n \times \Delta H_f^\circ \;=\; 0.3840\ \text{mol} \times (+226\ \text{kJ mol}^{-1}) \;=\; +86.8\ \text{kJ.} $$ $\Delta H = +86.8\ \text{kJ}$, meaning $86.8\ \text{kJ}$ of heat is absorbed from the surroundings during the synthesis.$\Delta H = +86.8\ \text{kJ}$，即合成过程从环境吸收 $86.8\ \text{kJ}$ 的热量。

(d) Thermochemical equation for decomposition of $1\ \text{mol}$ acetylene$1\ \text{mol}$ 乙炔分解的热化学方程式 A1·A1

Reversing the formation reaction reverses the sign of $\Delta H$:将生成反应翻转后，$\Delta H$ 符号取反： $$ \text{C}_2\text{H}_2(g) \;\longrightarrow\; 2\,\text{C}(s) + \text{H}_2(g) \qquad \Delta H = -226\ \text{kJ mol}^{-1.} $$ The decomposition is exothermic: acetylene releases $226\ \text{kJ mol}^{-1}$ when it decomposes back to carbon and hydrogen.分解反应为放热：乙炔分解为碳和氢时释放 $226\ \text{kJ mol}^{-1}$。

Reversing a thermochemical equation flips the sign of $\Delta H$. This is a direct consequence of Hess's Law.翻转热化学方程式会使 $\Delta H$ 的符号取反。这是盖斯定律的直接推论。 Acetylene ($\text{C}_2\text{H}_2$) is one of the classic endothermic compounds: it is manufactured from elements at high energy cost and releases that stored energy when it decomposes or combusts. This is why oxy-acetylene torches burn so hot: the combustion enthalpy of acetylene is $-1300\ \text{kJ mol}^{-1}$, one of the largest values for common fuels. The positive $\Delta H_f^\circ$ signals that the compound is "pre-loaded" with energy relative to its elements.乙炔（$\text{C}_2\text{H}_2$）是经典的吸热化合物之一：由单质合成需要消耗大量能量，分解或燃烧时释放储存的能量。这正是氧炔焰极高温的原因：乙炔的燃烧焓为 $-1300\ \text{kJ mol}^{-1}$，是常见燃料中最大值之一。$\Delta H_f^\circ$ 为正值表明该化合物相对于单质"预存"了能量。

Q8 HARD难 Honors荣誉级 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-feeder FRQAP 衔接简答题 §5 Hess's Law盖斯定律 · SCH4U / Chem 12 / Chem 30-A1.4k [8 marks][8 分]

Target: $\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$. Step 1: $\Delta H_1 = -110.5\ \text{kJ}$. Step 2: $\Delta H_2 = -283.0\ \text{kJ}$.目标：$\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$。步骤 1：$\Delta H_1 = -110.5\ \text{kJ}$。步骤 2：$\Delta H_2 = -283.0\ \text{kJ}$。

Answer:答案： (a) Hess's Law stated盖斯定律陈述 · (b) steps add directly; CO and $\tfrac{1}{2}\text{O}_2$ cancel步骤直接相加；CO 和 $\tfrac{1}{2}\text{O}_2$ 消去 · (c) $\Delta H = -393.5\ \text{kJ}$ (exact agreement)（精确吻合） · (d) state function definition状态函数定义

(a) State Hess's Law陈述盖斯定律 A1

Hess's Law states that the total enthalpy change for a chemical reaction is independent of the pathway taken and depends only on the initial and final states.盖斯定律指出：化学反应的总焓变与反应路径无关，只取决于初态和末态。

(b) Show that Steps 1 and 2 add directly to give the target验证步骤 1 与步骤 2 直接相加得到目标方程 M1·A1·A1

Write the two steps side by side and add:将两个步骤并排写出并相加： $$ \text{Step 1:}\quad \text{C}(s) + \tfrac{1}{2}\,\text{O}_2(g) \;\longrightarrow\; \underbrace{\text{CO}(g)}_{\text{intermediate}} $$ $$ \text{Step 2:}\quad \underbrace{\text{CO}(g)}_{\text{cancels}} + \tfrac{1}{2}\,\text{O}_2(g) \;\longrightarrow\; \text{CO}_2(g) $$ Adding: $\text{CO}(g)$ appears as a product of Step 1 and a reactant of Step 2, so it cancels. The $\tfrac{1}{2}\,\text{O}_2$ from each step adds to give $\text{O}_2$:相加后：$\text{CO}(g)$ 作为步骤 1 的产物和步骤 2 的反应物相互消去。两个步骤各含 $\tfrac{1}{2}\,\text{O}_2$，相加得 $\text{O}_2$： $$ \text{Net:}\quad \text{C}(s) + \text{O}_2(g) \;\longrightarrow\; \text{CO}_2(g) \qquad \checkmark $$ No manipulation (reversal or scaling) was needed because CO is formed in Step 1 and consumed in Step 2 in exactly the right amounts.无需翻转或倍增，因为 CO 在步骤 1 中恰好生成并在步骤 2 中恰好消耗。

(c) Calculate $\Delta H$ and compare with literature计算 $\Delta H$ 并与文献值对比 M1·A1·A1

$$ \Delta H \;=\; \Delta H_1 + \Delta H_2 \;=\; (-110.5) + (-283.0) \;=\; -393.5\ \text{kJ.} $$ The calculated value $\Delta H = -393.5\ \text{kJ mol}^{-1}$ is in exact agreement with the standard enthalpy of combustion of carbon ($-393.5\ \text{kJ mol}^{-1}$). This confirms Hess's Law: the two-step pathway and the direct pathway give identical $\Delta H$ values.计算所得 $\Delta H = -393.5\ \text{kJ mol}^{-1}$ 与碳的标准燃烧焓（$-393.5\ \text{kJ mol}^{-1}$）完全吻合。这证实了盖斯定律：两步路径与直接路径给出完全相同的 $\Delta H$。

(d) Meaning of "state function""状态函数"的含义 A1

A state function is a property whose value depends only on the current state of the system (the thermodynamic conditions such as temperature, pressure, and composition), not on how that state was reached. Enthalpy $H$ is a state function; therefore $\Delta H$ depends only on the initial state (reactants) and the final state (products), never on the pathway between them.状态函数是其数值只取决于体系当前状态（温度、压力、组成等热力学条件），而不取决于达到该状态的路径的性质。焓 $H$ 是状态函数；因此 $\Delta H$ 只取决于初态（反应物）和末态（产物），与二者之间的路径无关。

Hess's Law is just conservation of energy applied to a sequence of steps: the total energy change is path-independent because enthalpy is a state function.盖斯定律不过是将能量守恒应用于一系列步骤：由于焓是状态函数，总能量变化与路径无关。 The practical power of Hess's Law is that you can calculate $\Delta H$ for reactions that are difficult or impossible to measure directly (such as partial combustion products). The algorithm: (1) write the target reaction; (2) manipulate the given steps (reverse or multiply) so that intermediates cancel; (3) sum the modified $\Delta H$ values. When no manipulation is needed (as here), the calculation reduces to a simple addition. This is the most tested concept in Canadian thermochemistry curricula (ON SCH4U, BC Chem 12, AB Chem 30).盖斯定律的实用价值在于：可以计算难以或无法直接测量的反应（如部分燃烧产物）的 $\Delta H$。算法：(1) 写出目标反应；(2) 对已知步骤进行操作（翻转或倍增）使中间体消去；(3) 对修改后的 $\Delta H$ 求和。当无需操作时（如本题），计算简化为直接相加。这是加拿大热化学课程（安省 SCH4U、卑诗化学 12、阿省化学 30）中考查最多的概念。

Q9 HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §6 Bond energies键能 · Chemistry 12 [8 marks][8 分]

$\text{CH}_4(g) + \text{Cl}_2(g) \to \text{CH}_3\text{Cl}(g) + \text{HCl}(g)$. Bond energies (kJ/mol): C-H = 435, Cl-Cl = 243, C-Cl = 339, H-Cl = 432.$\text{CH}_4(g) + \text{Cl}_2(g) \to \text{CH}_3\text{Cl}(g) + \text{HCl}(g)$。键能（kJ/mol）：C-H = 435，Cl-Cl = 243，C-Cl = 339，H-Cl = 432。

Answer:答案： (a) bonds broken: 1 C-H + 1 Cl-Cl; bonds formed: 1 C-Cl + 1 H-Cl断键：1 C-H + 1 Cl-Cl；成键：1 C-Cl + 1 H-Cl · (b) broken = 678 kJ; formed = 771 kJ断键 = 678 kJ；成键 = 771 kJ · (c) $\Delta H = -93\ \text{kJ mol}^{-1}$ (exothermic)（放热） · (d) average values, not compound-specific使用平均键能而非化合物特有值

(a) Bonds broken and formed断裂和生成的化学键 M1·A1·A1

In $\text{CH}_4$, only one C-H bond is replaced (the reaction substitutes one H for one Cl). In $\text{Cl}_2$ the Cl-Cl bond is broken.在 $\text{CH}_4$ 中，只有一个 C-H 键被取代（反应将一个 H 替换为一个 Cl）。$\text{Cl}_2$ 中的 Cl-Cl 键断裂。

Bonds broken: 1 C-H ($435\ \text{kJ mol}^{-1}$) + 1 Cl-Cl ($243\ \text{kJ mol}^{-1}$)断裂的键：1 C-H（$435\ \text{kJ mol}^{-1}$）+ 1 Cl-Cl（$243\ \text{kJ mol}^{-1}$）
Bonds formed: 1 C-Cl ($339\ \text{kJ mol}^{-1}$) + 1 H-Cl ($432\ \text{kJ mol}^{-1}$)生成的键：1 C-Cl（$339\ \text{kJ mol}^{-1}$）+ 1 H-Cl（$432\ \text{kJ mol}^{-1}$）

(b) Total energy for bonds broken and formed断键和成键的总能量 A1·A1

$$ \Sigma E_{\text{bonds broken}} \;=\; 435 + 243 \;=\; 678\ \text{kJ mol}^{-1.} $$ $$ \Sigma E_{\text{bonds formed}} \;=\; 339 + 432 \;=\; 771\ \text{kJ mol}^{-1.} $$

(c) Calculate $\Delta H$计算 $\Delta H$ M1·A1

$$ \Delta H \;=\; \Sigma E_{\text{bonds broken}} - \Sigma E_{\text{bonds formed}} \;=\; 678 - 771 \;=\; -93\ \text{kJ mol}^{-1.} $$ $\Delta H = -93\ \text{kJ mol}^{-1}$: the reaction is exothermic. More energy is released forming the products' bonds than is needed to break the reactants' bonds.$\Delta H = -93\ \text{kJ mol}^{-1}$：反应为放热反应。生成产物化学键释放的能量多于断裂反应物化学键所需的能量。

(d) Why bond energies give an estimate键能法给出估算值的原因 A1

Bond energies used in tables are average values calculated across many different compounds. The actual C-H bond energy in $\text{CH}_4$ differs slightly from the C-H bond energy in, say, $\text{C}_2\text{H}_6$ or $\text{CHCl}_3$ because the molecular environment affects the bond strength. Using average values therefore introduces a small error, making the result an estimate rather than an exact thermodynamic value.表中使用的键能是对多种不同化合物取平均的平均值。$\text{CH}_4$ 中 C-H 键的实际键能与 $\text{C}_2\text{H}_6$ 或 $\text{CHCl}_3$ 中 C-H 键的键能略有差异，因为分子环境会影响键的强度。使用平均值因此引入小误差，使结果为估算值而非精确热力学数值。

Memory device for bond energy method: "break to absorb, form to release; $\Delta H$ = broken minus formed."键能法的记忆方法："断键吸能，成键放能；$\Delta H$ = 断键之和减成键之和。" A common error is to use the wrong formula: students sometimes write $\Delta H = \Sigma E_{\text{formed}} - \Sigma E_{\text{broken}}$, which gives the wrong sign. The correct version subtracts the energy released by bond formation from the energy required for bond breaking. Check the sign: if more energy is released forming bonds than absorbed breaking bonds, $\Delta H < 0$ (exothermic), which makes physical sense. Here $678 < 771$, so energy is released overall: exothermic. Another check: the H-Cl bond ($432\ \text{kJ mol}^{-1}$) is much stronger than Cl-Cl ($243\ \text{kJ mol}^{-1}$), confirming that HCl formation drives the exothermicity.常见错误是使用错误公式：有学生写成 $\Delta H = \Sigma E_{\text{成键}} - \Sigma E_{\text{断键}}$，给出错误符号。正确版本是从断键所需能量中减去成键释放的能量。检验符号：若成键释放的能量多于断键吸收的能量，则 $\Delta H < 0$（放热），符合物理直觉。此题 $678 < 771$，总体放热：放热反应。另一个检验：H-Cl 键（$432\ \text{kJ mol}^{-1}$）远强于 Cl-Cl（$243\ \text{kJ mol}^{-1}$），确认了 HCl 的生成是放热的驱动力。

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10 MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §3 Calorimetry (specific heat of a metal)量热法（金属比热容） · Chem 30-A1.2k [8 marks][8 分]

$40.0\ \text{g}$ metal at $100.0\ \text{°C}$ dropped into $80.0\ \text{g}$ water at $19.0\ \text{°C}$; final $T = 22.5\ \text{°C}$; $c_{\text{water}} = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$.$40.0\ \text{g}$ 金属（$100.0\ \text{°C}$）投入 $80.0\ \text{g}$ 水（$19.0\ \text{°C}$），最终温度 $22.5\ \text{°C}$；$c_{\text{水}} = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。

Answer:答案： (a) $q_{\text{water}} = 1170\ \text{J}$ · (b) $q_{\text{metal}} = -1170\ \text{J}$ · (c) $c_{\text{metal}} = 0.378\ \text{J g}^{-1}\ \text{°C}^{-1}$ · (d) copper; heat lost to air/calorimeter铜；热量散失至空气或量热计

(a) Heat gained by the water水吸收的热量 M1·A1

$$ \Delta T_{\text{water}} \;=\; 22.5 - 19.0 \;=\; 3.5\ \text{°C.} $$ $$ q_{\text{water}} \;=\; m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \;=\; 80.0 \times 4.18 \times 3.5 \;=\; 1170.4\ \text{J} \;\approx\; 1170\ \text{J.} $$

(b) Relationship between $q_{\text{water}}$ and $q_{\text{metal}}$; find $q_{\text{metal}}$$q_{\text{水}}$ 与 $q_{\text{金属}}$ 的关系；求 $q_{\text{金属}}$ A1·A1

Assuming no heat is lost to the surroundings (well-insulated calorimeter): heat lost by the metal = heat gained by the water.假设无热量散逸至环境（隔热良好的量热计）：金属失去的热量等于水获得的热量。 $$ q_{\text{metal}} \;=\; -q_{\text{water}} \;=\; -1170\ \text{J.} $$ (Negative because the metal releases heat.)（负号表示金属放热。）

(c) Specific heat capacity of the metal金属的比热容 M1·A1·A1

$$ \Delta T_{\text{metal}} \;=\; 22.5 - 100.0 \;=\; -77.5\ \text{°C.} $$ Using $q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}}$:代入 $q_{\text{金属}} = m_{\text{金属}} \cdot c_{\text{金属}} \cdot \Delta T_{\text{金属}}$： $$ -1170 \;=\; 40.0 \times c_{\text{metal}} \times (-77.5) \;\Longrightarrow\; c_{\text{metal}} \;=\; \frac{1170}{40.0 \times 77.5} \;=\; \frac{1170}{3100} \;=\; 0.378\ \text{J g}^{-1}\ \text{°C}^{-1.} $$

(d) Identify the metal and name one experimental error推断金属并指出一个实验误差来源 A1

The measured $c_{\text{metal}} \approx 0.378\ \text{J g}^{-1}\ \text{°C}^{-1}$ is very close to copper's value of $0.385\ \text{J g}^{-1}\ \text{°C}^{-1}$, so the likely metal is copper. The measured value is slightly lower than the literature value. One experimental error: heat lost from the hot metal to the air during transfer from the oven to the calorimeter, which means less heat actually reaches the water than was stored in the metal. This results in a smaller calculated $c_{\text{metal}}$.测量值 $c_{\text{金属}} \approx 0.378\ \text{J g}^{-1}\ \text{°C}^{-1}$ 非常接近铜的文献值 $0.385\ \text{J g}^{-1}\ \text{°C}^{-1}$，故最可能的金属是铜。测量值略低于文献值。一个实验误差来源：将热金属从烘箱转移至量热计过程中热量散失至空气，使实际传递给水的热量少于金属所储存的热量，导致计算得到的 $c_{\text{金属}}$ 偏小。

Metal-water heat exchange: set $q_{\text{lost by metal}} = q_{\text{gained by water}}$. Sign discipline matters: use $\Delta T_{\text{metal}} = T_f - T_{\text{metal,i}}$, which is negative for a cooling metal.金属-水热交换：令 $q_{\text{金属失去}} = q_{\text{水获得}}$。符号纪律很重要：使用 $\Delta T_{\text{金属}} = T_f - T_{\text{金属,初}}$，对降温金属为负值。 The law of conservation of energy requires that all heat leaving the metal enters the water in a perfectly insulated system. The negative $\Delta T_{\text{metal}}$ and negative $q_{\text{metal}}$ cancel correctly, giving a positive $c_{\text{metal}}$. A common error is to set $\Delta T_{\text{metal}} = T_{\text{metal,i}} - T_f$ (wrong order) and then not apply the sign of $q_{\text{metal}} = -q_{\text{water}}$, producing a negative specific heat, which is physically impossible. Always check: specific heat must be positive.能量守恒定律要求在完全绝热的体系中，金属散失的全部热量进入水中。$\Delta T_{\text{金属}}$ 为负、$q_{\text{金属}}$ 也为负，两者相消后正确给出正值的 $c_{\text{金属}}$。常见错误是将 $\Delta T_{\text{金属}}$ 写成 $T_{\text{金属,初}} - T_f$（顺序相反），同时又不应用 $q_{\text{金属}} = -q_{\text{水}}$ 的符号，从而得到负的比热容，这在物理上是不可能的。始终检查：比热容必须为正值。

Q11 MEDIUM中 🇺🇸 US美 🇨🇦 AB阿 AP-feeder FRQAP 衔接简答题 §4 Combustion enthalpy (octane)燃烧焓（辛烷） · HS-PS3-1 / Chem 30-A1.1k [8 marks][8 分]

$\text{C}_8\text{H}_{18}(l) + \tfrac{25}{2}\,\text{O}_2(g) \to 8\,\text{CO}_2(g) + 9\,\text{H}_2\text{O}(l)$, $\Delta H = -5471\ \text{kJ mol}^{-1}$. Sample: $5.75\ \text{g}$ of $\text{C}_8\text{H}_{18}$, $M = 114.26\ \text{g mol}^{-1}$.$\text{C}_8\text{H}_{18}(l) + \tfrac{25}{2}\,\text{O}_2(g) \to 8\,\text{CO}_2(g) + 9\,\text{H}_2\text{O}(l)$，$\Delta H = -5471\ \text{kJ mol}^{-1}$。样品：$5.75\ \text{g}$ 辛烷，$M = 114.26\ \text{g mol}^{-1}$。

Answer:答案： (a) $n = 0.05031\ \text{mol}$ · (b) $q = 275.2\ \text{kJ}$ · (c) $T_f = 52.9\ \text{°C}$ · (d) heat lost to engine block / exhaust gases / friction热量散失至发动机体、废气或摩擦

(a) Moles of octane in $5.75\ \text{g}$$5.75\ \text{g}$ 辛烷的物质的量 M1·A1

$$ n(\text{C}_8\text{H}_{18}) \;=\; \frac{5.75\ \text{g}}{114.26\ \text{g mol}^{-1}} \;=\; 0.05031\ \text{mol.} $$

(b) Heat released when $5.75\ \text{g}$ of octane burns$5.75\ \text{g}$ 辛烷完全燃烧时释放的热量 M1·A1·A1

$$ q_{\text{released}} \;=\; n \times |\Delta H| \;=\; 0.05031 \times 5471 \;=\; 275.2\ \text{kJ.} $$ Rounded: $275\ \text{kJ}$ released to surroundings.约为 $275\ \text{kJ}$ 释放至环境。

(c) Final temperature of $2.00\ \text{kg}$ water (100% efficiency)$2.00\ \text{kg}$ 水的最终温度（效率 100%） M1·A1·A1

Mass of water: $m = 2.00\ \text{kg} = 2000\ \text{g}$. Using $q = mc\Delta T$:水的质量：$m = 2.00\ \text{kg} = 2000\ \text{g}$。代入 $q = mc\Delta T$： $$ \Delta T \;=\; \frac{q}{mc} \;=\; \frac{275\,200\ \text{J}}{2000\ \text{g} \times 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}} \;=\; \frac{275\,200}{8360} \;=\; 32.9\ \text{°C.} $$ $$ T_f \;=\; T_i + \Delta T \;=\; 20.0 + 32.9 \;=\; 52.9\ \text{°C.} $$

(d) Why real engine efficiency is much less than 100%真实发动机效率远低于 100% 的原因 A1

In a real internal-combustion engine, not all combustion energy is converted to useful mechanical work. Energy is lost as: (1) heat conducted to the engine block and coolant; (2) heat carried away by exhaust gases; (3) friction between moving parts (pistons, bearings). Typical thermal efficiency of a gasoline engine is only 25-35%.在真实的内燃机中，并非所有燃烧能量都转化为有用的机械功。能量损失途径包括：(1) 通过导热散失至发动机体和冷却液；(2) 随废气排出的热量；(3) 运动部件（活塞、轴承）之间的摩擦。汽油发动机的典型热效率仅为 25-35%。

Energy-to-temperature link: $\Delta T = q / (mc)$. Convert kJ to J before dividing; never mix energy units.能量与温度的关联：$\Delta T = q / (mc)$。相除前先将 kJ 换算为 J；切勿混用能量单位。 This question chains two sub-skills: thermochemical stoichiometry (a)+(b) and calorimetry rearranged for temperature (c). The key transition is converting the combustion energy from kJ to J before dividing by $mc$. Forgetting this conversion is the most common arithmetic error. Notice the huge mass of water ($2000\ \text{g}$) needed to absorb only $275\ \text{kJ}$: water's high specific heat ($4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$) makes it an excellent heat sink, which is why it is used as engine coolant.本题串联了两个子技能：热化学计量（(a)+(b)）和由温度推导的量热法（(c)）。关键过渡是在除以 $mc$ 之前将燃烧能量从 kJ 换算为 J。忘记此换算是最常见的计算错误。注意仅 $275\ \text{kJ}$ 的能量就需要 $2000\ \text{g}$ 的大量水来吸收：水的高比热容（$4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$）使其成为优良的储热介质，这正是水被用作发动机冷却液的原因。

Q12 HARD难 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §6-7 Bond energies + PE diagram synthesis键能 + 势能图综合 · HS-PS3-4 / Chemistry 12 [9 marks][9 分]

$\text{H}_2(g) + \text{F}_2(g) \to 2\,\text{HF}(g)$. Bond energies (kJ/mol): H-H = 436, F-F = 158, H-F = 570.$\text{H}_2(g) + \text{F}_2(g) \to 2\,\text{HF}(g)$。键能（kJ/mol）：H-H = 436，F-F = 158，H-F = 570。

Answer:答案： (a) broken = 594 kJ; formed = 1140 kJ断键 = 594 kJ；成键 = 1140 kJ · (b) $\Delta H = -546\ \text{kJ mol}^{-1}$ (strongly exothermic)（强放热） · (c) PE diagram: reactants above products; peak above reactants势能图：反应物高于产物；峰值高于反应物 · (d) H-F is polar; ionic character adds electrostatic attractionH-F 具有极性；离子性增加了静电吸引力

(a) Bonds broken and formed, with total energies断裂和生成的化学键及总能量 M1·A1·A1

Bonds broken (reactants side):断裂的键（反应物侧）：

1 H-H bond: $436\ \text{kJ mol}^{-1}$1 个 H-H 键：$436\ \text{kJ mol}^{-1}$
1 F-F bond: $158\ \text{kJ mol}^{-1}$1 个 F-F 键：$158\ \text{kJ mol}^{-1}$
Total energy absorbed for bond breaking: $436 + 158 = 594\ \text{kJ mol}^{-1}$断键吸收的总能量：$436 + 158 = 594\ \text{kJ mol}^{-1}$

Bonds formed (products side):生成的键（产物侧）：

2 H-F bonds: $2 \times 570 = 1140\ \text{kJ mol}^{-1}$2 个 H-F 键：$2 \times 570 = 1140\ \text{kJ mol}^{-1}$
Total energy released by bond formation: $1140\ \text{kJ mol}^{-1}$成键释放的总能量：$1140\ \text{kJ mol}^{-1}$

(b) Calculate $\Delta H$ and classify计算 $\Delta H$ 并分类 M1·A1

$$ \Delta H \;=\; \Sigma E_{\text{bonds broken}} - \Sigma E_{\text{bonds formed}} \;=\; 594 - 1140 \;=\; -546\ \text{kJ mol}^{-1.} $$ $\Delta H = -546\ \text{kJ mol}^{-1}$: the reaction is strongly exothermic. Far more energy is released forming the two H-F bonds than is absorbed breaking the H-H and F-F bonds.$\Delta H = -546\ \text{kJ mol}^{-1}$：该反应为强放热反应。生成两个 H-F 键释放的能量远多于断裂 H-H 和 F-F 键所需的能量。

(c) Description of the labeled potential-energy diagram标注势能图的描述 M1·A1·A1

The PE diagram for this exothermic reaction must show:该放热反应的势能图必须显示：

(i) Reactants energy level ($\text{H}_2 + \text{F}_2$) on the left, at a higher energy than products.（i）左侧反应物（$\text{H}_2 + \text{F}_2$）能量水平，高于产物。
(ii) Products energy level ($2\,\text{HF}$) on the right, substantially lower than reactants (the difference is $546\ \text{kJ mol}^{-1}$).（ii）右侧产物（$2\,\text{HF}$）能量水平，明显低于反应物（差值为 $546\ \text{kJ mol}^{-1}$）。
(iii) A peak (transition state) above the reactant energy level, with the vertical distance from reactants to peak labelled $E_a$ (the activation energy). Exact numerical value of $E_a$ is not required.（iii）高于反应物能量水平的峰值（过渡态），从反应物到峰值的竖直距离标注为 $E_a$（活化能）。不要求 $E_a$ 的精确数值。
(iv) A downward arrow from reactants to products labelled $\Delta H = -546\ \text{kJ mol}^{-1}$ (negative, pointing down).（iv）从反应物到产物的向下箭头，标注 $\Delta H = -546\ \text{kJ mol}^{-1}$（负值，向下）。

(d) Why the H-F bond is so strong (bond polarity)H-F 键如此强的原因（键的极性） A1

Fluorine is the most electronegative element ($\chi_F = 4.0$), whereas hydrogen has $\chi_H = 2.1$. The large electronegativity difference ($\Delta\chi = 1.9$) creates a highly polar covalent bond with significant partial charges ($\delta^+$ on H, $\delta^-$ on F). This polarity gives the H-F bond partial ionic character: the electrostatic attraction between the partial charges adds to the covalent bond strength, making H-F much stronger than the nonpolar H-H or F-F bonds.氟是电负性最强的元素（$\chi_F = 4.0$），氢的电负性为 $\chi_H = 2.1$。较大的电负性差（$\Delta\chi = 1.9$）形成极性很强的共价键，产生显著的偏电荷（H 上 $\delta^+$，F 上 $\delta^-$）。这种极性赋予 H-F 键一定的离子性：偏电荷之间的静电吸引力叠加于共价键强度之上，使 H-F 远强于非极性的 H-H 或 F-F 键。

Bond polarity (electronegativity difference) is the key to understanding why some bonds are unusually strong: partial ionic character adds electrostatic attraction on top of covalent bonding.键的极性（电负性差）是理解某些键异常强的关键：部分离子性在共价键基础上增加了静电吸引力。 The reaction $\text{H}_2 + \text{F}_2 \to 2\,\text{HF}$ is one of the most exothermic reactions known at the molecular level ($\Delta H = -546\ \text{kJ}$ per mol of reaction as written). The dominant driver is the exceptional strength of the H-F bond, which in turn arises from the extreme electronegativity of fluorine. This connection between electronegativity, bond polarity, and bond strength is a recurring AP Chemistry theme: bond energies are not just covalent; in polar bonds, electrostatics also contribute. For the PE diagram, always draw the products clearly below the reactants for an exothermic reaction, with the transition state peak above the reactant level.反应 $\text{H}_2 + \text{F}_2 \to 2\,\text{HF}$ 是分子层面已知最强放热反应之一（如方程式所写，每摩尔反应 $\Delta H = -546\ \text{kJ}$）。主要驱动力是 H-F 键的特殊强度，而这源于氟的极端电负性。电负性、键极性与键能之间的联系是 AP 化学的反复考查主题：键能不仅来自共价成分；在极性键中，静电作用同样有贡献。对于势能图，放热反应中产物必须明确画在反应物下方，过渡态峰值则高于反应物水平线。