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. Assume all equilibria are in aqueous solution unless stated otherwise. 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。除非另有说明，均假设平衡在水溶液中进行。Q1-Q2 不可使用计算器；Q3-Q5 可用计算器。
For the reversible reaction $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$ in a sealed flask, which statement correctly describes the system at dynamic equilibrium?对于密封烧瓶中的可逆反应 $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$，下列哪项正确描述了动态平衡状态下的体系？
(A)The reaction has stopped; no $\text{N}_2\text{O}_4$ or $\text{NO}_2$ molecules are reacting.反应已停止；没有 $\text{N}_2\text{O}_4$ 或 $\text{NO}_2$ 分子在反应。
(B)The concentrations of $\text{N}_2\text{O}_4$ and $\text{NO}_2$ are equal.$\text{N}_2\text{O}_4$ 与 $\text{NO}_2$ 的浓度相等。
(C)The forward and reverse rates are equal, so concentrations remain constant.正向反应速率与逆向反应速率相等，因此浓度保持不变。
(D)Only the forward (decomposition) reaction continues at equilibrium.平衡时只有正向（分解）反应继续进行。
For the equilibrium $\text{A(g)} \rightleftharpoons 2\,\text{B(g)}$ at a given temperature, $K_{eq} = 4.00$. A mixture is prepared with $[\text{A}] = 0.500\ \text{mol/L}$ and $[\text{B}] = 0.600\ \text{mol/L}$.在某温度下，平衡 $\text{A(g)} \rightleftharpoons 2\,\text{B(g)}$ 的 $K_{eq} = 4.00$。准备一个混合物，其中 $[\text{A}] = 0.500\ \text{mol/L}$，$[\text{B}] = 0.600\ \text{mol/L}$。
(a)Calculate the reaction quotient $Q$.计算反应商 $Q$。[2]
(b)Compare $Q$ with $K_{eq}$ and predict the direction the reaction will shift to reach equilibrium. Justify your answer.将 $Q$ 与 $K_{eq}$ 比较，预测反应向哪个方向移动以达到平衡。请说明理由。[3]
(c)State what would be true about $Q$ if the system were already at equilibrium.若体系已处于平衡状态，$Q$ 的值有何特点？[1]
Consider the following equilibrium system in a sealed container: $2\,\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{SO}_3\text{(g)}$ $\Delta H = -198\ \text{kJ/mol}$在密封容器中考虑以下平衡体系：$2\,\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{SO}_3\text{(g)}$ $\Delta H = -198\ \text{kJ/mol}$
Using Le Chatelier's principle, predict and explain the effect of each of the following changes on the equilibrium position. State the direction of shift (left, right, or no shift) and the qualitative effect on $[\text{SO}_3]$ at the new equilibrium.运用勒沙特列原理，预测并解释以下各变化对平衡位置的影响。说明移动方向（左移、右移或不移动）以及新平衡时 $[\text{SO}_3]$ 的定性变化。
(a)$\text{SO}_2$ is added to the container at constant volume and temperature.在体积和温度不变的情况下向容器中加入 $\text{SO}_2$。[2]
(b)The total pressure is increased by decreasing the volume at constant temperature.在温度不变的情况下通过减小体积来增大总压。[2]
At a certain temperature, the following equilibrium is established in a $2.00\ \text{L}$ closed container: $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$. At equilibrium, the container holds $0.240\ \text{mol}$ $\text{H}_2$, $0.240\ \text{mol}$ $\text{I}_2$, and $1.92\ \text{mol}$ $\text{HI}$.在某温度下，以下平衡在 $2.00\ \text{L}$ 密封容器中建立：$\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$。平衡时容器中有 $0.240\ \text{mol}$ $\text{H}_2$、$0.240\ \text{mol}$ $\text{I}_2$ 和 $1.92\ \text{mol}$ $\text{HI}$。
(a)Calculate the equilibrium concentration of each species.计算每种物质的平衡浓度。[2]
(b)Calculate $K_{eq}$ for this reaction. Show the expression and your substitution.计算该反应的 $K_{eq}$。写出表达式并代入数值。[2]
(c)Interpret the magnitude of $K_{eq}$: does equilibrium strongly favour products or reactants? Explain in one sentence.解读 $K_{eq}$ 的大小：平衡是强烈偏向产物还是反应物？用一句话解释。[2]
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. Write the $K_{eq}$ or $K_{sp}$ expression before substituting values. State units where applicable. Use full ICE-table notation in Part-II questions that require it. Calculator permitted on Q6-Q9.每一步推理都要写出。在代入数值前先写出 $K_{eq}$ 或 $K_{sp}$ 表达式。在适用处写出单位。在需要 ICE 表格的题目中使用完整 ICE 表格符号。Q6-Q9 可用计算器。
Q6MEDIUM中🇺🇸 US美🇨🇦 ON安AP-feeder FRQAP 衔接简答题§3 + §4 $Q$ and Le Chatelier$Q$ 与勒沙特列 · HS-PS1-6Honors荣誉级[8 marks][8 分]
For the reaction $\text{CO(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$, $K_{eq} = 3.92$ at $800\ \text{K}$. A mixture is prepared with $[\text{CO}] = 0.200\ \text{mol/L}$, $[\text{H}_2] = 0.200\ \text{mol/L}$, $[\text{CH}_4] = 0.100\ \text{mol/L}$, and $[\text{H}_2\text{O}] = 0.100\ \text{mol/L}$.对于反应 $\text{CO(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$，$800\ \text{K}$ 时 $K_{eq} = 3.92$。准备一个混合物：$[\text{CO}] = 0.200\ \text{mol/L}$，$[\text{H}_2] = 0.200\ \text{mol/L}$，$[\text{CH}_4] = 0.100\ \text{mol/L}$，$[\text{H}_2\text{O}] = 0.100\ \text{mol/L}$。
(a)Write the $K_{eq}$ expression for this reaction.写出该反应的 $K_{eq}$ 表达式。[1]
(b)Calculate the reaction quotient $Q$ for this mixture. Show all steps.计算该混合物的反应商 $Q$。写出所有步骤。[3]
(c)Predict the direction the reaction will shift to reach equilibrium. Justify by comparing $Q$ and $K_{eq}$.预测反应将向哪个方向移动以达到平衡。通过比较 $Q$ 和 $K_{eq}$ 来说明理由。[2]
(d)If the volume of the container is suddenly halved at constant temperature, predict the direction of shift and explain using Le Chatelier's principle. Consider the moles of gas on each side.若在温度不变的情况下容器体积突然减半，预测移动方向，并用勒沙特列原理解释。考虑两边的气体物质的量。[2]
Barium sulfate, $\text{BaSO}_4$, is a sparingly soluble salt with $K_{sp} = 1.1 \times 10^{-10}$ at $25\ ^\circ\text{C}$. The dissolving equilibrium is: $\text{BaSO}_4\text{(s)} \rightleftharpoons \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}$.硫酸钡 $\text{BaSO}_4$ 是一种微溶盐，$25\ ^\circ\text{C}$ 时 $K_{sp} = 1.1 \times 10^{-10}$。溶解平衡为：$\text{BaSO}_4\text{(s)} \rightleftharpoons \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}$。
(a)Write the $K_{sp}$ expression for this equilibrium.写出该平衡的 $K_{sp}$ 表达式。[1]
(b)Calculate the molar solubility of $\text{BaSO}_4$ in pure water at $25\ ^\circ\text{C}$.计算 $\text{BaSO}_4$ 在 $25\ ^\circ\text{C}$ 纯水中的摩尔溶解度。[3]
(c)A student dissolves $\text{BaSO}_4$ in a $0.100\ \text{mol/L}\ \text{Na}_2\text{SO}_4$ solution instead of pure water. Without full calculation, predict whether the molar solubility will be greater than, less than, or equal to your answer in (b). Explain using Le Chatelier's principle.某同学将 $\text{BaSO}_4$ 溶于 $0.100\ \text{mol/L}$ $\text{Na}_2\text{SO}_4$ 溶液中而非纯水中。不必完整计算，预测摩尔溶解度是大于、小于还是等于 (b) 的答案。用勒沙特列原理解释。[3]
At a certain temperature, $K_{eq} = 0.360$ for the reaction $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$. A $1.00\ \text{L}$ flask is charged with $1.00\ \text{mol}$ of pure $\text{N}_2\text{O}_4$ and allowed to reach equilibrium.在某温度下，反应 $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$ 的 $K_{eq} = 0.360$。向 $1.00\ \text{L}$ 烧瓶中充入 $1.00\ \text{mol}$ 纯 $\text{N}_2\text{O}_4$，并使其达到平衡。
(a)Set up a complete ICE table (I = initial, C = change, E = equilibrium) using $x$ as the change in $[\text{N}_2\text{O}_4]$.用 $x$ 表示 $[\text{N}_2\text{O}_4]$ 的变化量，建立完整的 ICE 表格（I = 初始，C = 变化，E = 平衡）。[2]
(b)Write the equilibrium expression in terms of $x$ and solve the resulting quadratic equation for $x$. Show your algebra fully.用 $x$ 写出平衡表达式，并求解所得一元二次方程的 $x$ 值。完整写出代数步骤。[4]
(c)State the equilibrium concentrations of $\text{N}_2\text{O}_4$ and $\text{NO}_2$, then verify your answers by substituting back into the $K_{eq}$ expression.写出 $\text{N}_2\text{O}_4$ 和 $\text{NO}_2$ 的平衡浓度，然后代回 $K_{eq}$ 表达式进行验证。[2]
Q9HARD难Honors荣誉级🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§5 + §3 $K_{sp}$ and precipitation$K_{sp}$ 与沉淀判断 · Chem 30 Unit D[7 marks][7 分]
$K_{sp}(\text{PbCl}_2) = 1.7 \times 10^{-5}$ at $25\ ^\circ\text{C}$. The dissolving equilibrium is: $\text{PbCl}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+}\text{(aq)} + 2\,\text{Cl}^-\text{(aq)}$.$25\ ^\circ\text{C}$ 时 $K_{sp}(\text{PbCl}_2) = 1.7 \times 10^{-5}$。溶解平衡为：$\text{PbCl}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+}\text{(aq)} + 2\,\text{Cl}^-\text{(aq)}$。
(a)Write the $K_{sp}$ expression for $\text{PbCl}_2$.写出 $\text{PbCl}_2$ 的 $K_{sp}$ 表达式。[1]
(b)Calculate the molar solubility of $\text{PbCl}_2$ in pure water. Let $s$ = mol/L dissolved.计算 $\text{PbCl}_2$ 在纯水中的摩尔溶解度。设 $s$ = 溶解量（mol/L）。[3]
(c)A solution is prepared by mixing equal volumes of $0.100\ \text{mol/L}\ \text{Pb(NO}_3)_2$ and $0.060\ \text{mol/L}\ \text{NaCl}$. Calculate the reaction quotient $Q_{sp}$ and determine whether $\text{PbCl}_2$ precipitates upon mixing. Show full calculation.将等体积的 $0.100\ \text{mol/L}$ $\text{Pb(NO}_3)_2$ 与 $0.060\ \text{mol/L}$ $\text{NaCl}$ 溶液混合。计算反应商 $Q_{sp}$ 并判断混合后 $\text{PbCl}_2$ 是否会沉淀。写出完整计算过程。[3]
PART III · MODELING / APPLIED第三部分 · 建模与应用Haber process + ICE calculations · 26 marks哈伯法 + ICE 定量计算 · 共 26 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define all symbols with units at the start of each question. Write the equilibrium expression 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）。代入前先写出平衡表达式。每题以一句结合情境的完整句子作结。第三部分全程可用计算器。
Q10MEDIUM中🇺🇸 US美🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§7 Haber process哈伯法 · Chem 30 Unit D[8 marks][8 分]
The Haber process for industrial ammonia synthesis uses the reaction: $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$ $\Delta H = -92\ \text{kJ/mol}$工业合成氨的哈伯法使用以下反应：$\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$ $\Delta H = -92\ \text{kJ/mol}$
Industrial plants typically operate at $400-500\ ^\circ\text{C}$ and $150-300\ \text{atm}$, and use an iron catalyst.工业装置通常在 $400-500\ ^\circ\text{C}$ 和 $150-300\ \text{atm}$ 下运行，并使用铁催化剂。
(a)Using Le Chatelier's principle, predict the effect of increasing pressure on the yield of $\text{NH}_3$. Justify your answer by counting moles of gas on each side of the equation.利用勒沙特列原理，预测增大压强对 $\text{NH}_3$ 产率的影响。通过计算方程两侧的气体物质的量来说明理由。[3]
(b)The reaction is exothermic. A higher temperature shifts the equilibrium toward reactants and reduces the $\text{NH}_3$ yield. Yet industrial plants operate at $400-500\ ^\circ\text{C}$ rather than at room temperature. Explain this apparent contradiction in terms of both thermodynamics and kinetics.该反应是放热反应。较高温度使平衡向反应物方向移动，降低 $\text{NH}_3$ 产率。然而工业装置却在 $400-500\ ^\circ\text{C}$ 而非室温下运行。从热力学和动力学两方面解释这一表面矛盾。[3]
(c)Explain why adding an iron catalyst does not change the value of $K_{eq}$ or shift the equilibrium position, even though it speeds up the reaction.解释为什么加入铁催化剂不会改变 $K_{eq}$ 的值或使平衡位置移动，即使它加快了反应速率。[2]
Acetic acid (ethanoic acid) is a weak acid that ionizes according to: $\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CH}_3\text{COO}^-\text{(aq)}$ $K_a = 1.8 \times 10^{-5}$ at $25\ ^\circ\text{C}$. A $0.100\ \text{mol/L}$ solution of acetic acid is prepared.乙酸（醋酸）是一种弱酸，其电离方程式为：$\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CH}_3\text{COO}^-\text{(aq)}$ $25\ ^\circ\text{C}$ 时 $K_a = 1.8 \times 10^{-5}$。配制 $0.100\ \text{mol/L}$ 乙酸溶液。
(a)Set up an ICE table for the ionization. Use $x$ for the concentration of $\text{H}^+$ formed at equilibrium.为电离过程建立 ICE 表格。用 $x$ 表示平衡时生成的 $\text{H}^+$ 浓度。[2]
(b)Apply the approximation that $x \ll 0.100$ (valid when $K_a$ is small) and solve for $x = [\text{H}^+]$ at equilibrium.应用近似 $x \ll 0.100$（当 $K_a$ 较小时有效），求平衡时 $x = [\text{H}^+]$。[3]
(c)Calculate the pH of the solution. Show your calculation.计算该溶液的 pH。写出计算过程。[2]
(d)Verify the approximation used in (b) is valid by checking that $x$ is less than 5% of $0.100\ \text{mol/L}$.通过检验 $x$ 小于 $0.100\ \text{mol/L}$ 的 5% 来验证 (b) 中所用近似的合理性。[2]
Silver chloride has $K_{sp} = 1.8 \times 10^{-10}$ at $25\ ^\circ\text{C}$. The dissolving equilibrium is: $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$.氯化银在 $25\ ^\circ\text{C}$ 时 $K_{sp} = 1.8 \times 10^{-10}$。溶解平衡为：$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$。
(a)Calculate the molar solubility of AgCl in pure water. Let $s$ = mol/L dissolved.计算 AgCl 在纯水中的摩尔溶解度。设 $s$ = 溶解量（mol/L）。[2]
(b)Calculate the molar solubility of AgCl in a $0.0200\ \text{mol/L}$ AgNO$_3$ solution. Let $s'$ = mol/L of AgCl that dissolves. State the approximation you use and justify it.计算 AgCl 在 $0.0200\ \text{mol/L}$ $\text{AgNO}_3$ 溶液中的摩尔溶解度。设 $s'$ = AgCl 溶解量（mol/L）。说明所用近似并说明理由。[4]
(c)Compare your answers from (a) and (b). What phenomenon does this illustrate? Name it and state the general principle it follows.比较 (a) 和 (b) 的答案。这说明了什么现象？命名并阐述其遵循的一般性原理。[3]
🇺🇸 US NGSS美国 NGSSHS-PS1-6
🇨🇦 Ontario安大略SCH4U E3
🇨🇦 British Columbia不列颠哥伦比亚Chemistry 12: equilibrium, Le Chatelier, Ksp化学 12：平衡、勒沙特列、溶度积
🇨🇦 Alberta阿尔伯塔Chem 30 Unit D
Full Syllabus Map lives in ../Study Guides/Unit_12_Chemical_Equilibrium.html. Note: NGSS HS-PS1-6 assesses equilibrium conceptually; $K_{eq}$ calculation, $K_{sp}$, and ICE tables (Q6-Q9, Q11-Q12) are Honors-level for US and are core for ON SCH4U / BC Chem 12 / AB Chem 30.完整大纲对照表见 ../Study Guides/Unit_12_Chemical_Equilibrium.html。注：NGSS HS-PS1-6 以概念方式考查平衡；$K_{eq}$ 计算、$K_{sp}$ 与 ICE 表格（Q6-Q9、Q11-Q12）对美国为荣誉级，但为 ON SCH4U / BC 化学 12 / AB 化学 30 的核心内容。