PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 25 marksAP 风格选择题 + 安/卑省考短答 · 共 25 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, state units in every answer. Use $c_{\text{water}} = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$ throughout unless otherwise stated. No calculator on Q1-Q2; calculator permitted on Q3-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题每道都要写出单位。除另有说明外，全卷取 $c_{\text{水}} = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。Q1-Q2 不可使用计算器；Q3-Q5 可用计算器。
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？
(A)$\Delta H > 0$: heat flows from surroundings into the system (endothermic).$\Delta H > 0$：热量从环境流入体系（吸热）。
(B)$\Delta H < 0$: heat flows from surroundings into the system (endothermic).$\Delta H < 0$：热量从环境流入体系（吸热）。
(C)$\Delta H > 0$: heat flows from the system to the surroundings (exothermic).$\Delta H > 0$：热量从体系流向环境（放热）。
(D)$\Delta H < 0$: heat flows from the system to the surroundings (exothermic).$\Delta H < 0$：热量从体系流向环境（放热）。
Q2EASY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]
A student dissolves solid ammonium nitrate ($\text{NH}_4\text{NO}_3$) in water. The temperature of the solution drops from $22.0\ \text{°C}$ to $14.5\ \text{°C}$.学生将固体硝酸铵（$\text{NH}_4\text{NO}_3$）溶于水，溶液温度从 $22.0\ \text{°C}$ 降至 $14.5\ \text{°C}$。
(a)Classify the dissolving process as exothermic or endothermic. Justify your answer using the temperature data.根据温度数据，判断溶解过程为放热还是吸热，并说明理由。[2]
(b)State the sign of $\Delta H$ for this process and explain what it means in terms of energy flow between system and surroundings.说明该过程 $\Delta H$ 的符号，并从体系与环境之间能量流动的角度加以解释。[2]
A $150.0\ \text{g}$ sample of water in a coffee-cup calorimeter is heated by a small burning candle. The water temperature rises from $22.0\ \text{°C}$ to $35.5\ \text{°C}$. Use $c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$.咖啡杯量热计中 $150.0\ \text{g}$ 水被小燃烧的蜡烛加热，温度从 $22.0\ \text{°C}$ 上升至 $35.5\ \text{°C}$。取 $c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。
(a)Calculate $\Delta T$ for the water.计算水的 $\Delta T$。[1]
(b)Calculate the heat absorbed by the water, $q$, in joules and in kilojoules. Show all working with units.计算水吸收的热量 $q$，以焦耳和千焦表示。写出完整计算过程及单位。[3]
(c)State two assumptions made when using this coffee-cup calorimeter to measure heat from combustion.写出本实验使用咖啡杯量热计测量燃烧热时的两个假设条件。[2]
(a)State whether this reaction is exothermic or endothermic and explain what the value $-890\ \text{kJ mol}^{-1}$ means in words.说明该反应是放热还是吸热，并用文字解释 $-890\ \text{kJ mol}^{-1}$ 的含义。[2]
(b)Calculate the moles of $\text{CH}_4$ in a $3.20\ \text{g}$ sample. ($M(\text{CH}_4) = 16.05\ \text{g mol}^{-1}$)计算 $3.20\ \text{g}$ 甲烷的物质的量。（$M(\text{CH}_4) = 16.05\ \text{g mol}^{-1}$）[1]
(c)Calculate the heat released when $3.20\ \text{g}$ of $\text{CH}_4$ is completely combusted. Express your answer in kJ.计算 $3.20\ \text{g}$ 甲烷完全燃烧时释放的热量，以 kJ 表示。[3]
A potential-energy diagram for a reaction shows: reactants at $+40\ \text{kJ}$, the transition state (peak) at $+110\ \text{kJ}$, and products at $-20\ \text{kJ}$ (all relative to a zero baseline).某反应的势能图显示：反应物位于 $+40\ \text{kJ}$，过渡态（峰值）位于 $+110\ \text{kJ}$，产物位于 $-20\ \text{kJ}$（均相对于零基线）。
(a)Calculate the activation energy ($E_a$) for the forward reaction.计算正反应的活化能（$E_a$）。[2]
(b)Calculate $\Delta H$ for the reaction and state whether it is exothermic or endothermic.计算该反应的 $\Delta H$，并说明是放热还是吸热。[2]
(c)A catalyst is added. State how $E_a$ and $\Delta H$ each change (increase, decrease, or stay the same) and explain why.加入催化剂后，$E_a$ 和 $\Delta H$ 各如何变化（增大、减小还是不变），并说明原因。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 31 marksAP 衔接简答题 + 荣誉级 · 共 31 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning with units. Identify the equation used before substituting values. State your assumptions where relevant. Calculator permitted throughout Part II. Use $c_{\text{water}} = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$ unless given otherwise.每一步推理都要写出，并附单位。代入数值前先注明所用方程。在适当位置写出假设条件。第二部分全程可用计算器。除另有说明外，取 $c_{\text{水}} = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。
Q6MEDIUM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]
In a neutralisation experiment, $50.0\ \text{mL}$ of $1.00\ \text{mol L}^{-1}$ HCl is mixed with $50.0\ \text{mL}$ of $1.00\ \text{mol L}^{-1}$ NaOH in a coffee-cup calorimeter. The temperature rises from $21.0\ \text{°C}$ to $27.5\ \text{°C}$. Assume the total solution has density $1.00\ \text{g mL}^{-1}$ and $c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$.在中和实验中，$50.0\ \text{mL}$ 的 $1.00\ \text{mol L}^{-1}$ 盐酸与 $50.0\ \text{mL}$ 的 $1.00\ \text{mol L}^{-1}$ 氢氧化钠在咖啡杯量热计中混合，温度从 $21.0\ \text{°C}$ 升至 $27.5\ \text{°C}$。假设混合溶液密度为 $1.00\ \text{g mL}^{-1}$，$c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。
(a)Calculate the total mass of the solution and $\Delta T$.计算溶液总质量和 $\Delta T$。[2]
(b)Calculate the heat absorbed by the solution, $q_{\text{soln}}$, in joules.计算溶液吸收的热量 $q_{\text{soln}}$，以焦耳表示。[2]
(c)Calculate the moles of water produced in the neutralisation.计算中和反应生成水的物质的量。[1]
(d)Calculate $\Delta H_{\text{neutralisation}}$ per mole of water produced, in $\text{kJ mol}^{-1}$. Include sign and units. State whether the reaction is exothermic or endothermic.计算每摩尔水生成的中和焓变 $\Delta H_{\text{中和}}$，以 $\text{kJ mol}^{-1}$ 表示。写出符号和单位，并说明反应是放热还是吸热。[3]
(a)Explain what the positive sign of $\Delta H_f^\circ$ tells you about the stability of acetylene relative to its elements.解释 $\Delta H_f^\circ$ 为正值说明了乙炔相对于其单质的稳定性如何。[2]
(b)Calculate the moles of $\text{C}_2\text{H}_2$ in $10.0\ \text{g}$ of the gas. ($M = 26.04\ \text{g mol}^{-1}$)计算 $10.0\ \text{g}$ 乙炔气体的物质的量。（$M = 26.04\ \text{g mol}^{-1}$）[1]
(c)Calculate the enthalpy change when $10.0\ \text{g}$ of acetylene is synthesised from its elements. State whether heat is absorbed or released.计算由单质合成 $10.0\ \text{g}$ 乙炔时的焓变，并说明是吸热还是放热。[2]
(d)Write the thermochemical equation for the decomposition of $1\ \text{mol}$ of acetylene back into its elements, including the numerical value of $\Delta H$.写出 $1\ \text{mol}$ 乙炔分解为其单质的热化学方程式，含 $\Delta H$ 数值。[2]
(a)State Hess's Law in one sentence.用一句话陈述盖斯定律。[1]
(b)Explain why no manipulation (reversal or scaling) of the given steps is needed. Show that Steps 1 and 2 add directly to give the target.说明为何无需对已知步骤进行翻转或倍增。验证步骤 1 与步骤 2 直接相加后恰好等于目标方程。[3]
(c)Calculate $\Delta H$ for the target reaction and compare it with the standard enthalpy of combustion of carbon ($-393.5\ \text{kJ mol}^{-1}$). Comment on the agreement.计算目标反应的 $\Delta H$，与碳的标准燃烧焓（$-393.5\ \text{kJ mol}^{-1}$）对比，评价二者的吻合程度。[3]
(d)Hess's Law relies on enthalpy being a state function. What does "state function" mean in this context?盖斯定律依赖焓是状态函数。在此情境下"状态函数"是什么意思？[1]
Q9HARD难🇨🇦 BC卑BC Provincial-style卑诗省考风格§6 Bond energies键能 · Chemistry 12[8 marks][8 分]
Estimate $\Delta H$ for the gas-phase chlorination of methane using average bond energies:用平均键能估算甲烷气相氯化反应的 $\Delta H$：
(a)Identify all bonds broken and all bonds formed. List them with their bond energies.列出该反应中断裂的所有化学键和生成的所有化学键，并写出各自的键能。[3]
(b)Calculate the total energy required to break bonds and the total energy released when bonds form.计算断键所需的总能量和成键释放的总能量。[2]
(c)Calculate $\Delta H$ using $\Delta H = \Sigma E_{\text{bonds broken}} - \Sigma E_{\text{bonds formed}}$. State whether the reaction is exothermic or endothermic.用公式 $\Delta H = \Sigma E_{\text{断键}} - \Sigma E_{\text{成键}}$ 计算 $\Delta H$，并说明是放热还是吸热。[2]
(d)Explain why the bond-energy method gives an estimate rather than an exact value for $\Delta H$.解释为何键能法给出的 $\Delta H$ 是估算值而非精确值。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 25 marks阿省毕业考 + 通用题型 · 共 25 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define all symbols with units at the start of each question. Show the equation before substituting. Conclude each question with a one-sentence contextual answer. Calculator permitted throughout Part III.每题开始时定义所有符号（含单位）。代入数值前先写出方程。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
Q10MEDIUM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]
A student heats a $40.0\ \text{g}$ sample of an unknown metal to $100.0\ \text{°C}$ and drops it into $80.0\ \text{g}$ of water initially at $19.0\ \text{°C}$ in a well-insulated calorimeter. The final equilibrium temperature is $22.5\ \text{°C}$. Use $c_{\text{water}} = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$.学生将 $40.0\ \text{g}$ 未知金属加热至 $100.0\ \text{°C}$，再投入隔热良好的量热计中盛有的 $80.0\ \text{g}$ 初温为 $19.0\ \text{°C}$ 的水中，最终平衡温度为 $22.5\ \text{°C}$。取 $c_{\text{水}} = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。
(a)Calculate the heat gained by the water, $q_{\text{water}}$, in joules.计算水吸收的热量 $q_{\text{水}}$，以焦耳表示。[2]
(b)State the relationship between $q_{\text{water}}$ and $q_{\text{metal}}$ and determine $q_{\text{metal}}$.写出 $q_{\text{水}}$ 与 $q_{\text{金属}}$ 之间的关系，并求 $q_{\text{金属}}$。[2]
(c)Calculate the specific heat capacity of the metal, $c_{\text{metal}}$, in $\text{J g}^{-1}\ \text{°C}^{-1}$. Show all working with units.计算金属的比热容 $c_{\text{金属}}$，以 $\text{J g}^{-1}\ \text{°C}^{-1}$ 表示。写出完整计算过程及单位。[3]
(d)Given that copper has $c = 0.385\ \text{J g}^{-1}\ \text{°C}^{-1}$, identify the likely metal and state one experimental error that would cause the measured $c$ to differ from the literature value.已知铜的比热容 $c = 0.385\ \text{J g}^{-1}\ \text{°C}^{-1}$，推断最可能的金属，并说明一个导致测量值与文献值偏差的实验误差来源。[1]
(a)Calculate the moles of octane in $5.75\ \text{g}$ of the liquid. ($M(\text{C}_8\text{H}_{18}) = 114.26\ \text{g mol}^{-1}$)计算 $5.75\ \text{g}$ 液态辛烷的物质的量。（$M(\text{C}_8\text{H}_{18}) = 114.26\ \text{g mol}^{-1}$）[1]
(b)Calculate the heat released when $5.75\ \text{g}$ of octane is completely burned. Express your answer in kJ.计算 $5.75\ \text{g}$ 辛烷完全燃烧时释放的热量，以 kJ 表示。[3]
(c)If this heat were used to warm $2.00\ \text{kg}$ of water from $20.0\ \text{°C}$ (assuming 100% efficiency), calculate the final temperature of the water. Use $c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$.假设该热量全部用于将 $2.00\ \text{kg}$ 的水从 $20.0\ \text{°C}$ 加热（效率 100%），计算水的最终温度。取 $c = 4.18\ \text{J g}^{-1}\ \text{°C}^{-1}$。[3]
(d)In a real engine the efficiency is far below 100%. Suggest one reason why not all combustion energy is converted to useful work.在真实发动机中效率远低于 100%。请给出一个燃烧能量不能全部转化为有用功的原因。[1]
Q12HARD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]
Consider the gas-phase reaction between hydrogen and fluorine:考虑氢气与氟气的气相反应：
(a)List all bonds broken and all bonds formed. Calculate the total energy input (bonds broken) and total energy output (bonds formed).列出所有断裂的化学键和生成的化学键。计算断键所需总能量（输入）和成键释放的总能量（输出）。[3]
(b)Calculate $\Delta H$ for the reaction. State whether it is exothermic or endothermic.计算该反应的 $\Delta H$，说明是放热还是吸热。[2]
(c)Sketch a labeled potential-energy diagram for this reaction. Mark: (i) reactants energy level, (ii) products energy level, (iii) activation energy $E_a$, and (iv) $\Delta H$. Exact values are not required.为该反应绘制并标注势能图。标出：(i) 反应物能量水平，(ii) 产物能量水平，(iii) 活化能 $E_a$，(iv) $\Delta H$。无需使用精确数值。[3]
(d)The H-F bond ($570\ \text{kJ mol}^{-1}$) is much stronger than H-H ($436\ \text{kJ mol}^{-1}$) or F-F ($158\ \text{kJ mol}^{-1}$). Using bond polarity, explain why the H-F bond is so strong.H-F 键（$570\ \text{kJ mol}^{-1}$）比 H-H（$436\ \text{kJ mol}^{-1}$）或 F-F（$158\ \text{kJ mol}^{-1}$）强得多。从键的极性角度解释 H-F 键为何如此强。[1]
🇺🇸 US NGSS美国 NGSSHS-PS3-1HS-PS3-4
🇨🇦 Ontario安大略SCH3U Unit 4 · SCH4U Unit 5
🇨🇦 British Columbia不列颠哥伦比亚Chemistry 11/12: thermochemistry, enthalpy, calorimetry化学 11/12：热化学、焓、量热法
🇨🇦 Alberta阿尔伯塔Chem 30 Unit A · 30-A1.1k · 30-A1.2k · 30-A1.4k
Full Syllabus Map lives in the companion Study Guide. Hess's Law (Q8) is Honors-flagged for NGSS; it is core for ON SCH4U / BC Chem 12 / AB Chem 30.完整大纲对照表见配套学习指南。盖斯定律（Q8）在 NGSS 层级标记为荣誉级；在安大略 SCH4U / 卑诗化学 12 / 阿省化学 30 中为核心内容。