PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 24 marksAP 风格选择题 + 安/卑省考短答 · 共 24 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. No calculator on Q1-Q3; calculator permitted on Q4-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。Q1-Q3 不可使用计算器；Q4-Q5 可用计算器。
For the reaction $\text{A} \to 2\text{B}$, the concentration of A decreases from $1.20\ \text{mol L}^{-1}$ to $0.60\ \text{mol L}^{-1}$ in $30\ \text{s}$. What is the average rate of appearance of B?对于反应 $\text{A} \to 2\text{B}$，A 的浓度在 $30\ \text{s}$ 内从 $1.20\ \text{mol L}^{-1}$ 降至 $0.60\ \text{mol L}^{-1}$。B 的平均生成速率是多少？
According to collision theory, which two conditions must be satisfied simultaneously for a molecular collision to result in a chemical reaction?根据碰撞理论，分子碰撞导致化学反应必须同时满足哪两个条件？
(A)Equal masses and equal speeds of the colliding particles碰撞粒子质量相等和速度相等
(B)Sufficient collision energy and correct molecular orientation足够的碰撞能量和正确的分子方向
(C)High temperature and the presence of a catalyst高温和催化剂的存在
(D)High concentration and high pressure only仅高浓度和高压
A reaction has a forward activation energy $E_{a,\text{fwd}} = 80\ \text{kJ mol}^{-1}$ and an enthalpy change $\Delta H = -50\ \text{kJ mol}^{-1}$.某反应的正向活化能 $E_{a,\text{fwd}} = 80\ \text{kJ mol}^{-1}$，焓变 $\Delta H = -50\ \text{kJ mol}^{-1}$。
(a)State whether the reaction is endothermic or exothermic and justify your answer.说明该反应是吸热还是放热，并说明理由。[2]
(b)Calculate the reverse activation energy $E_{a,\text{rev}}$.计算逆向活化能 $E_{a,\text{rev}}$。[2]
(c)Sketch the potential energy diagram. Label reactants, products, activated complex, $E_{a,\text{fwd}}$, $E_{a,\text{rev}}$, and $\Delta H$.勾画势能图，标注反应物、生成物、活化络合物、$E_{a,\text{fwd}}$、$E_{a,\text{rev}}$ 和 $\Delta H$。[2]
A student investigates the reaction between marble chips (CaCO$_3$) and dilute hydrochloric acid (HCl) under three conditions: (I) large marble chips at 20 °C; (II) powdered marble at 20 °C; (III) large marble chips at 40 °C. The acid concentration and total mass of marble are identical in all three experiments.学生在三种条件下研究大理石块（CaCO$_3$）与稀盐酸（HCl）的反应：（I）大块大理石，20 °C；（II）大理石粉末，20 °C；（III）大块大理石，40 °C。三组实验的酸浓度和大理石总质量完全相同。
(a)Rank conditions I, II, and III from slowest to fastest initial reaction rate and justify your ranking using collision theory.将条件 I、II、III 从最慢到最快排列初始反应速率，并用碰撞理论说明排序依据。[3]
(b)State one variable that must be kept constant to make conditions I and II a valid comparison, and explain why it must be controlled.写出使条件 I 和 II 成为有效比较所必须保持不变的一个变量，并解释为何必须控制该变量。[2]
(c)Will the total volume of CO$_2$ produced at the end of the reaction be the same or different across conditions I, II, and III? Explain.条件 I、II、III 反应结束时产生 CO$_2$ 的总体积是否相同？解释原因。[1]
Hydrogen peroxide decomposes slowly at room temperature: $2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$. Adding manganese dioxide (MnO$_2$) dramatically speeds up the reaction, but MnO$_2$ can be recovered unchanged after the reaction is complete.过氧化氢在室温下缓慢分解：$2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$。加入二氧化锰（MnO$_2$）可显著加速反应，但反应结束后 MnO$_2$ 可以不变地回收。
(a)Identify MnO$_2$ as a homogeneous or heterogeneous catalyst and justify your answer.判断 MnO$_2$ 是均相还是非均相催化剂，并说明理由。[2]
(b)Explain, using activation energy, how MnO$_2$ increases the decomposition rate without changing the overall enthalpy change $\Delta H$.用活化能解释 MnO$_2$ 如何在不改变总焓变 $\Delta H$ 的情况下提高分解速率。[3]
(c)Name one biological catalyst and state the type of reaction it catalyses.写出一种生物催化剂，并说明其催化的反应类型。[1]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. State the formula used before substituting values. State units in every final answer. Calculator permitted on Q6-Q9.每一步推理都要写出。代入数值前先写出所用公式。每个最终答案都要写单位。Q6-Q9 可用计算器。
The decomposition of dinitrogen pentoxide proceeds as: $2\text{N}_2\text{O}_5(g) \to 4\text{NO}_2(g) + \text{O}_2(g)$. The concentration of N$_2$O$_5$ decreases from $0.80\ \text{mol L}^{-1}$ to $0.20\ \text{mol L}^{-1}$ over $90\ \text{s}$.五氧化二氮的分解反应为：$2\text{N}_2\text{O}_5(g) \to 4\text{NO}_2(g) + \text{O}_2(g)$。N$_2$O$_5$ 的浓度在 $90\ \text{s}$ 内从 $0.80\ \text{mol L}^{-1}$ 降至 $0.20\ \text{mol L}^{-1}$。
(a)Calculate the average rate of disappearance of N$_2$O$_5$. Include units.计算 N$_2$O$_5$ 消失的平均速率，写出单位。[2]
(b)Calculate the average rate of appearance of NO$_2$.计算 NO$_2$ 生成的平均速率。[2]
(c)Calculate the average rate of appearance of O$_2$.计算 O$_2$ 生成的平均速率。[2]
(d)State one reason why the instantaneous rate at $t = 0$ is higher than the average rate you calculated in (a).说明为何 $t = 0$ 时的瞬时速率高于 (a) 中计算的平均速率。[1]
Q7MEDIUM中🇨🇦 ON安🇨🇦 BC卑ON Provincial-style安大略省考风格§2 + §3 Collision theory + PE diagrams (integrated)碰撞理论 + 势能图（综合） · SCH4U D3.5-D3.6[8 marks][8 分]
Two reactions, X and Y, are carried out at the same temperature. Reaction X has an activation energy of $40\ \text{kJ mol}^{-1}$ and reaction Y has an activation energy of $120\ \text{kJ mol}^{-1}$. Both reactions have the same negative enthalpy change $\Delta H$.反应 X 和 Y 在相同温度下进行。反应 X 的活化能为 $40\ \text{kJ mol}^{-1}$，反应 Y 的活化能为 $120\ \text{kJ mol}^{-1}$。两个反应具有相同的负焓变 $\Delta H$。
(a)State which reaction proceeds faster and explain why in terms of the fraction of molecules with sufficient energy to react.说明哪个反应进行更快，并用能量足以反应的分子比例进行解释。[3]
(b)Sketch PE diagrams for both reactions X and Y on the same axes. Label $E_a$, $\Delta H$, reactants, and products for each curve.在同一坐标轴上绘制反应 X 和 Y 的势能图，分别为每条曲线标注 $E_a$、$\Delta H$、反应物和生成物。[3]
(c)A catalyst is added to reaction Y, reducing its activation energy to $60\ \text{kJ mol}^{-1}$. Describe two changes to the PE diagram of reaction Y when the catalyst is added.向反应 Y 中加入催化剂后，其活化能降至 $60\ \text{kJ mol}^{-1}$。描述催化剂加入后反应 Y 势能图发生的两个变化。[2]
Q8HARD难🇨🇦 BC卑🇨🇦 AB阿BC Provincial-style卑诗省考风格§4 + §5 Factors and catalysis: Haber process影响因素与催化剂：哈伯法 · BC Chem 12 / Chem 30 GO3[8 marks][8 分]
The Haber process produces ammonia: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g),\ \Delta H = -92\ \text{kJ mol}^{-1}$. Industrial plants use an iron catalyst, high pressure (150-300 atm), and a temperature of about 450 °C.哈伯法合成氨：$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g),\ \Delta H = -92\ \text{kJ mol}^{-1}$。工业装置使用铁催化剂、高压（150-300 atm）以及约 450 °C 的温度。
(a)Explain how increasing the pressure increases the reaction rate, using collision theory.用碰撞理论解释增大压强如何提高反应速率。[2]
(b)The iron catalyst is a solid while N$_2$ and H$_2$ are gases. Identify this as homogeneous or heterogeneous catalysis and describe where the reaction occurs on the catalyst.铁催化剂为固体而 N$_2$ 和 H$_2$ 为气体。判断这是均相还是非均相催化，并描述反应在催化剂上发生的位置。[2]
(c)Although a higher temperature would further increase the reaction rate, industrial plants operate at only 450 °C. Use your knowledge of both kinetics and Le Chatelier's principle to explain why this temperature is a compromise.虽然更高温度能进一步提高反应速率，工业装置仍只在 450 °C 运行。利用动力学和勒夏特列原理的知识解释为何这一温度是折中方案。[2]
(d)The iron catalyst becomes "poisoned" by small amounts of sulfur. Suggest what "catalyst poisoning" means at the molecular level and why it reduces the reaction rate.铁催化剂会被少量硫"中毒"。在分子层面解释"催化剂中毒"的含义，以及为何它会降低反应速率。[2]
The following initial-rate data were collected for the reaction $\text{A} + \text{B} \to \text{products}$ at constant temperature.在恒定温度下，收集了反应 $\text{A} + \text{B} \to \text{products}$ 的以下初始速率数据。
(a)Determine the order of reaction with respect to A and with respect to B. Show your reasoning explicitly.确定反应对 A 和对 B 的反应级数，明确写出推导过程。[3]
(b)Write the overall rate law and calculate the rate constant $k$ with units, using data from Experiment 1.写出总速率定律，并使用实验 1 的数据计算速率常数 $k$（含单位）。[3]
(c)Predict the initial rate when $[\text{A}]_0 = 0.30\ \text{mol L}^{-1}$ and $[\text{B}]_0 = 0.20\ \text{mol L}^{-1}$.预测 $[\text{A}]_0 = 0.30\ \text{mol L}^{-1}$，$[\text{B}]_0 = 0.20\ \text{mol L}^{-1}$ 时的初始速率。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 28 marks阿省毕业考 + 通用题型 · 共 28 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define symbols (with units) at the start of each question. State the formula used before substituting. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。代入数值前先写出所用公式。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Honors荣誉级🇨🇦 ON安🇨🇦 BC卑AP-feeder FRQAP 衔接简答题§7 Reaction mechanisms and rate-determining step反应机理与速率决定步骤 · SCH4U / BC Chem 12[9 marks][9 分]
A proposed two-step mechanism for the reaction $2\text{NO}(g) + \text{Cl}_2(g) \to 2\text{NOCl}(g)$ is:反应 $2\text{NO}(g) + \text{Cl}_2(g) \to 2\text{NOCl}(g)$ 的拟议两步机理为：
(a)Identify the rate-determining step and explain your choice.确定速率决定步骤并说明理由。[2]
(b)Identify the reaction intermediate. Explain how a reaction intermediate differs from a catalyst.确定反应中间体，并解释反应中间体与催化剂的区别。[2]
(c)Write the rate law predicted by this mechanism based on the rate-determining step. Then substitute the expression for $[\text{NOCl}_2]$ from the fast equilibrium in Step 1 to eliminate the intermediate and obtain the overall rate law in terms of only $[\text{NO}]$ and $[\text{Cl}_2]$.根据速率决定步骤写出该机理预测的速率定律。再利用步骤 1 快速平衡中 $[\text{NOCl}_2]$ 的表达式消去中间体，得到仅含 $[\text{NO}]$ 和 $[\text{Cl}_2]$ 的总速率定律。[4]
(d)State whether this mechanism is consistent with the experimentally observed rate law $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$.说明该机理是否与实验观察到的速率定律 $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ 一致。[1]
A chemist studies the reaction $\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \to \text{ZnSO}_4(aq) + \text{H}_2(g)$. In Experiment A, 2.0 g of zinc granules reacts with $50\ \text{mL}$ of $1.0\ \text{mol L}^{-1}$ H$_2$SO$_4$ at 25 °C. After $60\ \text{s}$, the concentration of H$_2$SO$_4$ has fallen to $0.40\ \text{mol L}^{-1}$.化学家研究反应 $\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \to \text{ZnSO}_4(aq) + \text{H}_2(g)$。实验 A 中，2.0 g 锌粒与 $50\ \text{mL}$ 的 $1.0\ \text{mol L}^{-1}$ H$_2$SO$_4$ 在 25 °C 反应，$60\ \text{s}$ 后 H$_2$SO$_4$ 浓度降至 $0.40\ \text{mol L}^{-1}$。
(a)Calculate the average rate of disappearance of H$_2$SO$_4$ in Experiment A. Include units.计算实验 A 中 H$_2$SO$_4$ 消失的平均速率，写出单位。[2]
(b)Experiment B is identical to A except that zinc granules are replaced by zinc powder. Predict whether the rate in Experiment B is greater than, less than, or equal to the rate in Experiment A. Explain using collision theory.实验 B 与 A 完全相同，只是将锌粒换成锌粉。预测实验 B 的速率是大于、小于还是等于实验 A，用碰撞理论解释。[3]
(c)In Experiment C, the acid concentration is doubled to $2.0\ \text{mol L}^{-1}$ while all other conditions match Experiment A. Explain how this change affects the initial reaction rate and why.实验 C 将酸浓度加倍至 $2.0\ \text{mol L}^{-1}$，其他条件与实验 A 相同。解释此变化如何影响初始反应速率以及原因。[2]
(d)Will the total mass of H$_2$ gas produced at the end of the reaction be the same or different in Experiments A, B, and C? Identify which experiment(s), if any, will produce a different total, and explain.实验 A、B、C 反应结束时产生 H$_2$ 气体的总质量是否相同？说明哪组（如有）总量不同，并解释。[2]
A student measures the decomposition of hydrogen peroxide: $2\text{H}_2\text{O}_2(aq) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$. The table shows $[\text{H}_2\text{O}_2]$ over time at 25 °C without any catalyst.学生测量过氧化氢的分解：$2\text{H}_2\text{O}_2(aq) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$。下表显示 25 °C 下无催化剂时 $[\text{H}_2\text{O}_2]$ 随时间的变化。
Time / s时间 / s
$[\text{H}_2\text{O}_2]$ / mol L$^{-1}$
0
0.900
60
0.720
120
0.576
180
0.461
(a)Calculate the average rate of disappearance of H$_2$O$_2$ over $t = 0$ to $60\ \text{s}$, and over $t = 120\ \text{s}$ to $180\ \text{s}$.计算 $t = 0$ 至 $60\ \text{s}$ 以及 $t = 120\ \text{s}$ 至 $180\ \text{s}$ 两段的 H$_2$O$_2$ 消失平均速率。[3]
(b)The rate decreases over time. Using collision theory, explain why the rate slows as the reaction proceeds.速率随时间降低。用碰撞理论解释速率随反应进行而减慢的原因。[2]
(c)Calculate the ratio $[\text{H}_2\text{O}_2]_{60} / [\text{H}_2\text{O}_2]_0$ and the ratio $[\text{H}_2\text{O}_2]_{120} / [\text{H}_2\text{O}_2]_{60}$. What does a constant ratio over equal time intervals imply about the order of the reaction?计算 $[\text{H}_2\text{O}_2]_{60} / [\text{H}_2\text{O}_2]_0$ 和 $[\text{H}_2\text{O}_2]_{120} / [\text{H}_2\text{O}_2]_{60}$ 的比值。等时间间隔内比值恒定对反应级数意味着什么？[3]
(d)A biology student adds catalase enzyme to the same initial concentration of H$_2$O$_2$. Describe two observable differences between the catalysed and uncatalysed reactions and explain each using activation energy.一名生物学学生向相同初始浓度的 H$_2$O$_2$ 中加入过氧化氢酶。描述催化与非催化反应之间的两个可观察到的差异，并分别用活化能解释。[2]
🇺🇸 US NGSS美国 NGSSHS-PS1-5
🇨🇦 Ontario安大略SCH4U D2.1 · D3.5 · D3.6
🇨🇦 British Columbia不列颠哥伦比亚Chemistry 12: reaction rates, collision theory, PE diagrams, rate laws化学 12：反应速率、碰撞理论、势能图、速率定律
🇨🇦 Alberta阿尔伯塔Chem 30 GO2 · GO3 · GO4
Full Syllabus Map lives in ../Study Guides/Unit_11_Reaction_Rates_and_Kinetics.html. Sections 6-7 (rate laws, mechanisms) are Honors level for US NGSS; questions Q9-Q10 are honors-flagged for this content.完整大纲对照表见 ../Study Guides/Unit_11_Reaction_Rates_and_Kinetics.html。第 6-7 节（速率定律、反应机理）在 US NGSS 属荣誉级；Q9-Q10 为荣誉级题目。