High School Chemistry

Chemical Equilibrium化学平衡

When a reaction does not go to completion it reaches a state of dynamic equilibrium — forward and reverse reactions occur at equal rates, and concentrations no longer change. This guide builds the full picture: what reversible reactions are and why they reach equilibrium, how the equilibrium constant $K_{eq}$ encodes the position of equilibrium, how the reaction quotient $Q$ tells you which way a system will shift, how Le Chatelier's principle predicts the effect of concentration, pressure, and temperature changes, how solubility equilibria ($K_{sp}$) extend the same ideas to sparingly soluble salts, how ICE tables (honors) allow quantitative equilibrium calculations, and how the Haber process applies all these principles to industrial ammonia synthesis.当反应未能进行到底时，它达到动态平衡（动态平衡）的状态——正向与逆向反应速率相等，浓度不再变化。本指南系统构建完整图景：可逆反应（可逆反应）的本质与达到平衡的原因、平衡常数（平衡常数）$K_{eq}$ 如何编码平衡位置、反应商（反应商）$Q$ 如何预测体系移动方向、勒沙特列原理（勒沙特列原理）如何预测浓度、压强与温度变化的影响、溶度积（溶度积）$K_{sp}$ 如何将平衡思想推广至微溶盐、ICE 表格（荣誉级）如何实现定量计算，以及哈伯法如何将这些原理应用于工业合成氨。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Honors block: Keq / Ksp calculation and ICE tables (ON SCH4U / BC Chem 12 / AB Chem 30)荣誉级：Keq / Ksp 计算与 ICE 表格（ON SCH4U / BC Chem 12 / AB Chem 30）
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How to use this guide如何使用本指南

Chemical equilibrium is a Grade-12 topic in every curriculum we map to, and the depth split is stark. US NGSS (HS-PS1-6) is qualitative Le Chatelier only — shift direction when one variable changes, one at a time — and explicitly excludes equilibrium-constant calculation. Ontario SCH4U Strand E, BC Chemistry 12, and Alberta Chemistry 30 Unit D all go fully quantitative: writing $K_{eq}$ and $K_{sp}$ expressions, calculating equilibrium concentrations, ICE-table problems. The table below tells you which sections are core for you; the honors chip flags every section that NGSS students can treat as enrichment.化学平衡是我们所对照的每套大纲中的 12 年级主题，深度差异显著。US NGSS（HS-PS1-6）仅进行定性勒沙特列处理——每次改变一个变量时预测移动方向——明确排除平衡常数计算。安大略 SCH4U E 单元、BC Chemistry 12 以及阿尔伯塔 Chemistry 30 D 单元均进入完整定量处理：书写 $K_{eq}$ 和 $K_{sp}$ 表达式、计算平衡浓度、ICE 表格问题。下表告诉你哪些节是你的核心；荣誉级标记标注了 NGSS 学生可视为拓展内容的每一节。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Chemistry美国 NGSS 高中化学 §1 (dynamic equilibrium) and §4 (Le Chatelier's principle) — the core under HS-PS1-6. NGSS keeps equilibrium qualitative: shift direction only, one variable at a time; no $K_{eq}$ calculation.§1（动态平衡）和 §4（勒沙特列原理）—— HS-PS1-6 下的核心。NGSS 保持定性：仅判断移动方向，每次一个变量；无 $K_{eq}$ 计算。 §2, §3, §5, §6 ($K_{eq}$, $Q$, $K_{sp}$, ICE tables): explicitly outside the NGSS Assessment Boundary for HS-PS1-6. Valuable enrichment, not assessed.§2、§3、§5、§6（$K_{eq}$、$Q$、$K_{sp}$、ICE 表格）：明确在 NGSS HS-PS1-6 评估边界之外。有价值的拓展，不被评估。 NGSS HS-PS1 (Chemistry) — HS-PS1-6 PE + Clarification + Assessment Boundary— HS-PS1-6 表现期望 + 澄清 + 评估边界
🇨🇦 ON SCH4U安大略 SCH4U §1–§7 in full. SCH4U Strand E (E2.1–E2.4, E3.1–E3.4) requires $K_{eq}$, $K_{sp}$, Le Châtelier with $Q$, dynamic equilibrium, and quantitative ICE-table calculations.§1–§7 完整学习。SCH4U E 单元（E2.1–E2.4、E3.1–E3.4）要求 $K_{eq}$、$K_{sp}$、带 $Q$ 的勒沙特列、动态平衡，以及定量 ICE 表格计算。 Nothing — Ontario SCH4U treats equilibrium fully quantitatively at Grade 12.无 — 安大略 SCH4U 在 12 年级完整定量处理平衡。 Ontario SCH3U/4U Chemistry — SCH4U Strand E, E2.1–E2.4, E3.1–E3.4— SCH4U E 单元，E2.1–E2.4，E3.1–E3.4
🇨🇦 BC Chemistry 12BC Chemistry 12 §1–§7 in full. BC Chemistry 12 Big Idea: "Dynamic equilibrium can be shifted by changes to the surrounding conditions." Core Content: "dynamic nature of chemical equilibrium"; "Le Châtelier's principle and equilibrium shift"; "equilibrium constant (Keq)"; "saturated solutions and solubility product (Ksp)." Quantitative relationships (initial concentrations, equilibrium concentrations) are assessed.§1–§7 完整学习。BC Chemistry 12 大概念："动态平衡可因周围条件变化而发生移动。"核心内容："化学平衡的动态特性"；"勒沙特列原理与平衡移动"；"平衡常数（$K_{eq}$）"；"饱和溶液与溶度积（$K_{sp}$）"。定量关系（初始浓度、平衡浓度）被考核。 Nothing — BC Chemistry 12 assesses the full quantitative scope.无 — BC Chemistry 12 考核完整定量范围。 BC Chemistry 11/12 — Chemistry 12 Big Idea "Dynamic equilibrium…"; Content "equilibrium constant (Keq)", "solubility product (Ksp)"; elaborations for $K_{eq}$, $K_{sp}$, Le Châtelier— Chemistry 12 大概念"动态平衡……"；内容"平衡常数（$K_{eq}$）"、"溶度积（$K_{sp}$）"；$K_{eq}$、$K_{sp}$、勒沙特列的细化说明
🇨🇦 AB Chemistry 30阿尔伯塔 Chemistry 30 §1–§7 in full. Chemistry 30 Unit D GO1/GO2: Le Chatelier qualitative predictions + $K_c$ expression writing + quantitative calculations (equilibrium constants and concentrations when two of the three — equilibrium concentrations, initial concentrations, equilibrium constant — are known). Note: quadratic-equation problems are excluded from required calculations.§1–§7 完整学习。Chemistry 30 D 单元 GO1/GO2：勒沙特列定性预测 + 书写 $K_c$ 表达式 + 定量计算（已知以下三项中的两项时计算平衡常数和浓度：平衡浓度、初始浓度、平衡常数）。注：需要应用二次方程的问题不在必考范围。 Quadratic-equation ICE problems: excluded from required calculations in AB Chemistry 30 (though allowed in open-ended responses).需要二次方程的 ICE 问题：在 AB Chemistry 30 中不属于必考计算（但在开放式答题中允许使用）。 Alberta Chemistry 20/30 — Chemistry 30 Unit D GO1/GO2, Key Concepts, knowledge outcome text— Chemistry 30 D 单元 GO1/GO2，关键概念，知识结果文本
🇺🇸 IB / AP feeder trackIB / AP 衔接轨道 All seven sections plus every going-deeper derivation. IB Chemistry HL Reactivity 2 and AP Chemistry Unit 7 both assume fluent $K_{eq}$, $K_{sp}$, ICE tables, and Le Chatelier reasoning from day one.全部 7 节，并完成每个"深入"推导。IB Chemistry HL Reactivity 2 与 AP Chemistry Unit 7 第一天就默认你熟练 $K_{eq}$、$K_{sp}$、ICE 表格与勒沙特列推理。 Nothing — equilibrium is the conceptual backbone of IB Reactivity 2 and AP Chemistry thermodynamics/acids.无 — 平衡是 IB Reactivity 2 与 AP Chemistry 热力学/酸碱内容的概念骨干。 NGSS HS-PS1 (Chemistry) — see the IB Chemistry HL Reactivity 2 feeder link in "What This Feeds Into"— 见"本单元的去向"中的 IB Chemistry HL Reactivity 2 链接

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise four things: at equilibrium, forward rate = reverse rate; $K_{eq} = \frac{[\text{products}]^{\text{coeff}}}{[\text{reactants}]^{\text{coeff}}}$ (omit pure solids/liquids); Le Chatelier predicts the shift direction when a stress is applied; and the Haber process runs at high pressure and moderate temperature. Read every cram-cheat box. Skip the ICE-table derivations if time is short.背熟四件事：平衡时正向速率 = 逆向速率；$K_{eq} = \frac{[\text{products}]^{\text{coeff}}}{[\text{reactants}]^{\text{coeff}}}$（纯固体/纯液体不计入）；勒沙特列预测施加压力后的移动方向；哈伯法在高压和适中温度下运行。读每个速记框。若时间紧，可跳过 ICE 表格推导。

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If you are going for the top mark如果你目标顶分

Know why $K_{eq}$ is temperature-dependent but pressure- and concentration-independent. Use $Q$ vs $K_{eq}$ comparison to predict shift direction quantitatively. Complete ICE tables from scratch when given initial concentrations and $K_{eq}$. Derive $K_{sp}$ from molar solubility and vice versa. Explain every Le Chatelier prediction using the forward/reverse rate argument, not just the shift direction. ON SCH4U E2.4 and BC Chemistry 12 expect you to calculate equilibrium concentrations, not just quote Le Chatelier.理解为何 $K_{eq}$ 取决于温度，但不受压强和浓度影响。用 $Q$ 与 $K_{eq}$ 的比较定量预测移动方向。在已知初始浓度和 $K_{eq}$ 的情况下从头完成 ICE 表格。从摩尔溶解度推导 $K_{sp}$，反之亦然。用正逆反应速率论证而非仅仅说明移动方向来解释每个勒沙特列预测。ON SCH4U E2.4 与 BC Chemistry 12 要求你计算平衡浓度，而不仅是引用勒沙特列原理。

Honors flag.荣誉级标记。 Sections 2, 3, 5, and 6 ($K_{eq}$ expressions, the reaction quotient $Q$, $K_{sp}$, ICE tables) carry the Honors chip for US NGSS, where HS-PS1-6 Assessment Boundary explicitly excludes "calculating equilibrium constants and concentrations." They are core, not honors, in ON SCH4U (E2.3, E2.4, E3.4), BC Chemistry 12 ("equilibrium constant (Keq)," "solubility product (Ksp)," "quantitative relationships"), and AB Chemistry 30 Unit D GO2. Section 7 (Haber process) is core in every curriculum as an application of Le Chatelier. If your row sends you to the quantitative sections, treat them as required content.§2、§3、§5 和 §6（$K_{eq}$ 表达式、反应商 $Q$、$K_{sp}$、ICE 表格）在 US NGSS 上标 Honors，因为 HS-PS1-6 评估边界明确排除"平衡常数和浓度的计算"。在 ON SCH4U（E2.3、E2.4、E3.4）、BC Chemistry 12（"平衡常数（$K_{eq}$）"、"溶度积（$K_{sp}$）"、"定量关系"）和 AB Chemistry 30 D 单元 GO2 中，它们是核心而非荣誉内容。§7（哈伯法）在每套大纲中都是核心内容，作为勒沙特列的应用。如果你的行指向定量部分，就把它们视为必学内容。
Section 1 · Reversible reactions and dynamic equilibrium第 1 节 · 可逆反应与动态平衡

Reversible Reactions and Dynamic Equilibrium可逆反应与动态平衡

Equilibrium is dynamic: both reactions run simultaneously at equal rates.平衡是动态的：正逆反应同时以相等速率进行。
  • Reversible reaction可逆反应 — a reaction that can proceed in both directions; written with a double arrow: $\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}$. The left-to-right direction is the forward reaction; right-to-left is the reverse.— 可沿两个方向进行的反应；用双箭头表示：$\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}$。从左到右是正向反应；从右到左是逆向反应。
  • Dynamic equilibrium动态平衡 — reached when the forward rate equals the reverse rate. Concentrations of all species are constant, but both reactions continue at the molecular level. The system must be closed (no mass exchanged with surroundings).— 当正向速率等于逆向速率时达到动态平衡。所有物种的浓度保持不变，但在分子层面正逆反应仍在继续进行。体系必须是封闭的（与外界无质量交换）。
  • Key sign of equilibrium平衡的关键标志 : macroscopic properties (concentration, colour, pressure) are constant, yet the system is not static — molecules continue to react in both directions. NGSS HS-PS1-6 Clarification Statement explicitly requires the "connection between changes at the macroscopic level and what happens at the molecular level."：宏观性质（浓度、颜色、压强）恒定不变，但体系并非静止——分子仍在双向持续反应。NGSS HS-PS1-6 澄清说明明确要求"宏观层面变化与分子层面发生情况之间的联系"。
SCH4U E3.1 requires "explain the concept of dynamic equilibrium and how it applies to concentration of reactants and products." BC Chemistry 12 Big Idea: "Dynamic equilibrium can be shifted by changes to the surrounding conditions." AB Chemistry 30 Unit D GO1 defines equilibrium as "closed system, constancy of properties, equal rates of forward and reverse reactions."SCH4U E3.1 要求"解释动态平衡的概念及其对反应物和产物浓度的适用"。BC Chemistry 12 大概念："动态平衡可因周围条件变化而发生移动。"AB Chemistry 30 D 单元 GO1 将平衡定义为"封闭体系、性质不变、正逆反应速率相等"。
Worked Example 1 · Identifying equilibrium例题 1 · 识别平衡状态

Dinitrogen tetroxide decomposes reversibly: $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$. A sealed flask initially contains only $\text{N}_2\text{O}_4$. Describe what happens over time and identify two observable signs that equilibrium has been reached.四氧化二氮可逆分解：$\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$。密封烧瓶初始只含 $\text{N}_2\text{O}_4$。描述随时间的变化过程，并给出平衡已达到的两个可观察标志。

Initially.初始阶段。 Only $\text{N}_2\text{O}_4$ (colourless) is present. The forward rate is high; the reverse rate is zero (no $\text{NO}_2$ yet). The brown gas $\text{NO}_2$ begins to form and accumulates.仅有 $\text{N}_2\text{O}_4$（无色）存在。正向速率高；逆向速率为零（尚无 $\text{NO}_2$）。棕色气体 $\text{NO}_2$ 开始生成并积累。

Over time.随时间推移。 As $[\text{NO}_2]$ rises, the reverse rate (recombination of $\text{NO}_2$) increases. As $[\text{N}_2\text{O}_4]$ falls, the forward rate decreases. Eventually the rates become equal.随着 $[\text{NO}_2]$ 升高，逆向速率（$\text{NO}_2$ 的重新结合）增大。随着 $[\text{N}_2\text{O}_4]$ 下降，正向速率降低。最终两者速率相等。

Observable signs of equilibrium.平衡的可观察标志。

(1) The brown colour of the flask becomes constant — neither deepening nor fading further, because $[\text{NO}_2]$ is no longer changing. (2) The total gas pressure in the flask becomes constant, because the total number of moles of gas (and thus pressure at fixed volume and temperature) stops changing.（1）烧瓶的棕色不再加深也不再变浅，趋于恒定——因为 $[\text{NO}_2]$ 不再变化。（2）烧瓶内总气压趋于恒定——因为气体的总物质的量（以及在固定体积和温度下的压强）停止变化。

Which statement best describes a system at dynamic equilibrium?下列哪句话最准确地描述了处于动态平衡的体系？
§1 · Q1
The reaction has stopped completely反应已完全停止
Only the forward reaction continues只有正向反应在继续
Forward and reverse reactions occur at equal rates, so concentrations remain constant正向和逆向反应以相等速率进行，因此浓度保持恒定
The concentrations of reactants and products are equal反应物与产物的浓度相等
Dynamic equilibrium means both the forward and reverse reactions are occurring simultaneously at the same rate. Concentrations are constant — not necessarily equal — because the rates of production and consumption of each species balance exactly.动态平衡意味着正向和逆向反应以相同速率同时进行。浓度恒定——不一定相等——因为每种物种的生成速率和消耗速率完全平衡。
Equilibrium is not static — both reactions continue at the molecular level. Equal rates does not mean equal concentrations; $K_{eq}$ can be very large or very small. The system must also be closed for equilibrium to be maintained.平衡不是静止的——在分子层面两个方向的反应仍在继续。速率相等并不意味着浓度相等；$K_{eq}$ 可以很大也可以很小。体系还必须是封闭的才能维持平衡。
For the reaction $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$, which condition must hold at equilibrium?对于反应 $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$，平衡时必须满足哪个条件？
§1 · Q2
Rate of HI formation = rate of HI decompositionHI 的生成速率 = HI 的分解速率
$[\text{HI}] = [\text{H}_2] + [\text{I}_2]$$[\text{HI}] = [\text{H}_2] + [\text{I}_2]$
All concentrations are equal所有物种浓度相等
The reaction has produced the maximum possible amount of HI反应已生成最大可能量的 HI
At equilibrium, the forward rate (producing HI from H$_2$ and I$_2$) equals the reverse rate (decomposing HI back to H$_2$ and I$_2$). This equality of rates — not equality of concentrations — is the defining condition of dynamic equilibrium.平衡时，正向速率（由 H$_2$ 和 I$_2$ 生成 HI）等于逆向速率（HI 分解为 H$_2$ 和 I$_2$）。速率相等——而非浓度相等——是动态平衡的定义条件。
Equal concentrations of all species is not required — the actual concentrations depend on $K_{eq}$. The key is that rates are equal, so concentrations stop changing (but may be very different from one another).不要求所有物种浓度相等——实际浓度取决于 $K_{eq}$。关键是速率相等，因此浓度停止变化（但各物种之间的浓度可能差异很大）。
Going deeper — why equilibrium is reached: rate argument深入 — 为何会达到平衡：速率论证

Consider a closed vessel initially containing only reactants A and B. At $t = 0$, the forward rate $r_f$ is high (high [A] and [B]) while the reverse rate $r_r = 0$ (no products yet). As the reaction proceeds, [A] and [B] decrease so $r_f$ drops; [C] and [D] build up so $r_r$ rises. At some time $t_\text{eq}$, $r_f = r_r$. After that, any momentary fluctuation that increases [C] also increases $r_r$, which tends to restore the balance — the system is self-regulating. This argument shows equilibrium is a kinetic phenomenon: it is the steady state at which production and consumption of every species are exactly balanced. The position of that steady state (i.e. the ratio of concentrations at equilibrium) is encoded in $K_{eq}$, derived in §2.设想一个密闭容器，初始只含反应物 A 和 B。在 $t = 0$ 时，正向速率 $r_f$ 很高（[A] 和 [B] 高），而逆向速率 $r_r = 0$（尚无产物）。随着反应进行，[A] 和 [B] 降低故 $r_f$ 下降；[C] 和 [D] 积累故 $r_r$ 升高。在某时刻 $t_\text{eq}$，$r_f = r_r$。此后，任何使 [C] 暂时增加的涨落也会增大 $r_r$，从而趋向恢复平衡——体系具有自我调节能力。这一论证表明平衡是一种动力学现象：它是每种物种的生成速率与消耗速率精确平衡的稳态。该稳态的位置（即平衡时浓度之比）由 $K_{eq}$ 编码，在 §2 中推导。

Section 2 · The equilibrium constant $K_{eq}$ Honors — US NGSS第 2 节 · 平衡常数 $K_{eq}$ 荣誉 — US NGSS

The Equilibrium Constant $K_{eq}$平衡常数 $K_{eq}$

Curriculum note.课纲提示。 Calculating $K_{eq}$ is core in ON SCH4U (E2.3, E2.4, E3.4), BC Chemistry 12 (Content: "equilibrium constant (Keq)"), and AB Chemistry 30 Unit D GO1/GO2. It carries the Honors chip for US NGSS, whose Assessment Boundary for HS-PS1-6 explicitly excludes "calculating equilibrium constants and concentrations."计算 $K_{eq}$ 在 ON SCH4U（E2.3、E2.4、E3.4）、BC Chemistry 12（内容："平衡常数（$K_{eq}$）"）和 AB Chemistry 30 D 单元 GO1/GO2 中是核心内容。对 US NGSS 标 Honors，因为其 HS-PS1-6 评估边界明确排除"平衡常数和浓度的计算"。
$K_{eq}$ is products over reactants, each raised to its stoichiometric coefficient.$K_{eq}$ 是产物除以反应物，各自取化学计量数次幂。
  • Equilibrium constant expression平衡常数表达式 for $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$:对于 $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$：
$$ K_{eq} = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} $$
  • Pure solids and pure liquids纯固体和纯液体 are omitted from the $K_{eq}$ expression (their "concentration" is constant and folded into $K_{eq}$). Example: for $\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$, $K_{eq} = [\text{CO}_2]$.不计入 $K_{eq}$ 表达式（其"浓度"为常数，已并入 $K_{eq}$）。示例：对于 $\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$，$K_{eq} = [\text{CO}_2]$。
  • Magnitude of $K_{eq}$$K_{eq}$ 的大小: $K_{eq} \gg 1$ means products are favoured (reaction lies to the right); $K_{eq} \ll 1$ means reactants are favoured (reaction lies to the left); $K_{eq} \approx 1$ means neither strongly favoured.：$K_{eq} \gg 1$ 意味着产物占优（平衡偏向右方）；$K_{eq} \ll 1$ 意味着反应物占优（平衡偏向左方）；$K_{eq} \approx 1$ 意味着两者都不强烈占优。
  • $K_{eq}$ depends only on temperature$K_{eq}$ 仅取决于温度. Changing concentration or pressure shifts the equilibrium but does not change $K_{eq}$ (at constant temperature).。改变浓度或压强会移动平衡，但不会改变 $K_{eq}$（在温度恒定时）。
Worked Example 2 · Writing and evaluating $K_{eq}$例题 2 · 书写与计算 $K_{eq}$

For the reaction $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$, write the $K_{eq}$ expression. At a certain temperature, equilibrium concentrations are $[\text{N}_2] = 0.50\ \mathrm{mol/L}$, $[\text{H}_2] = 0.60\ \mathrm{mol/L}$, and $[\text{NH}_3] = 0.80\ \mathrm{mol/L}$. Calculate $K_{eq}$.对于反应 $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$，书写 $K_{eq}$ 表达式。在某温度下，平衡浓度为 $[\text{N}_2] = 0.50\ \mathrm{mol/L}$，$[\text{H}_2] = 0.60\ \mathrm{mol/L}$，$[\text{NH}_3] = 0.80\ \mathrm{mol/L}$。计算 $K_{eq}$。

Write the expression.书写表达式。 Products over reactants, stoichiometric coefficients as exponents:产物除以反应物，化学计量数作为指数：

$$ K_{eq} = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} $$

Substitute values.代入数值。

$$ K_{eq} = \frac{(0.80)^2}{(0.50)(0.60)^3} = \frac{0.64}{(0.50)(0.216)} = \frac{0.64}{0.108} \approx 5.9 $$

Interpret.解读。 $K_{eq} \approx 5.9$ — products are moderately favoured at this temperature. (The Haber process operates at much higher pressure and carefully chosen temperature to maximise yield despite a modest $K_{eq}$; see §7.)$K_{eq} \approx 5.9$——在此温度下产物适度占优。（哈伯法在更高压强和精心选择的温度下运行，以在 $K_{eq}$ 适中的情况下最大化产率；见 §7。）

Which expression correctly represents $K_{eq}$ for $2\,\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{SO}_3\text{(g)}$?下列哪个表达式正确表示 $2\,\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{SO}_3\text{(g)}$ 的 $K_{eq}$？
§2 · Q1
$\dfrac{[\text{SO}_2]^2[\text{O}_2]}{[\text{SO}_3]^2}$
$\dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$
$\dfrac{[\text{SO}_3]}{[\text{SO}_2][\text{O}_2]}$
$\dfrac{2[\text{SO}_3]}{2[\text{SO}_2]+[\text{O}_2]}$
$K_{eq} = \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$. Products in the numerator, reactants in the denominator, each raised to its stoichiometric coefficient from the balanced equation. Never add concentrations — multiply, and raise to the power of the coefficient.$K_{eq} = \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$。产物在分子，反应物在分母，各取配平方程中化学计量数为指数。不要相加浓度——应相乘，并取化学计量数次幂。
$K_{eq}$ puts products over reactants. Each species is raised to its coefficient (from the balanced equation), not multiplied by it. The coefficient becomes an exponent.$K_{eq}$ 是产物除以反应物。每种物种取其系数（来自配平方程式）为指数，而非乘以该系数。
A reaction has $K_{eq} = 1.2 \times 10^{-8}$ at 25 °C. What does this tell you about the equilibrium position?某反应在 25°C 时 $K_{eq} = 1.2 \times 10^{-8}$。这说明平衡位置怎样？
§2 · Q2
Products are strongly favoured产物强烈占优
Neither reactants nor products are favoured反应物和产物均不占优
The reaction does not reach equilibrium反应不能达到平衡
Reactants are strongly favoured; very little product forms反应物强烈占优；只有极少量产物生成
$K_{eq} = 1.2 \times 10^{-8} \ll 1$: the equilibrium lies far to the left. At equilibrium, reactant concentrations are much greater than product concentrations. The reaction does reach equilibrium — it just produces almost no products.$K_{eq} = 1.2 \times 10^{-8} \ll 1$：平衡强烈偏向左方。平衡时反应物浓度远大于产物浓度。反应确实达到平衡——只是几乎不产生产物。
$K_{eq} \ll 1$ means reactants dominate. $K_{eq} \gg 1$ means products dominate. $K_{eq} \approx 1$ means comparable concentrations. All systems that can react in a closed vessel do reach equilibrium.$K_{eq} \ll 1$ 意味着反应物主导。$K_{eq} \gg 1$ 意味着产物主导。$K_{eq} \approx 1$ 意味着浓度相当。所有能在封闭容器中反应的体系都会达到平衡。
Section 3 · The reaction quotient $Q$ Honors — US NGSS第 3 节 · 反应商 $Q$ 荣誉 — US NGSS

The Reaction Quotient $Q$反应商 $Q$

$Q$ has the same form as $K_{eq}$, but uses any concentrations — not just equilibrium ones.$Q$ 与 $K_{eq}$ 形式相同，但使用任意浓度——不仅限于平衡浓度。
  • Reaction quotient $Q$反应商 $Q$ — calculated from the current concentrations (which need not be equilibrium concentrations). Comparing $Q$ to $K_{eq}$ tells you which way the system will shift to reach equilibrium:— 由当前浓度（不需要是平衡浓度）计算得出。将 $Q$ 与 $K_{eq}$ 比较可判断体系将向哪个方向移动以达到平衡：
Condition条件Meaning含义System shifts…体系移动方向…
$Q < K_{eq}$Too few products产物不足Forward (right) →正向（右移）→
$Q = K_{eq}$At equilibrium已处于平衡No shift不移动
$Q > K_{eq}$Too many products产物过多Reverse (left) ←逆向（左移）←
ON SCH4U E2.2 requires predicting shifts using "Le Châtelier's principle or the reaction quotient." BC Chemistry 12 elaboration on $K_{eq}$ includes the effect of concentration changes. AB Chemistry 30 Unit D GO1 uses $K_c$ to "predict the extent of the reaction."ON SCH4U E2.2 要求使用"勒沙特列原理或反应商"来预测移动。BC Chemistry 12 对 $K_{eq}$ 的细化说明包括浓度变化的影响。AB Chemistry 30 D 单元 GO1 使用 $K_c$ "预测反应进行的程度"。
Worked Example 3 · Using $Q$ to predict shift direction例题 3 · 用 $Q$ 预测移动方向

For $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$, the equilibrium constant at a certain temperature is $K_{eq} = 5.9$. A mixture at that temperature has $[\text{N}_2] = 1.0\ \mathrm{mol/L}$, $[\text{H}_2] = 1.0\ \mathrm{mol/L}$, $[\text{NH}_3] = 1.0\ \mathrm{mol/L}$. Will the reaction proceed forward or reverse?对于 $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$，在某温度下平衡常数 $K_{eq} = 5.9$。该温度下某混合物为 $[\text{N}_2] = 1.0\ \mathrm{mol/L}$，$[\text{H}_2] = 1.0\ \mathrm{mol/L}$，$[\text{NH}_3] = 1.0\ \mathrm{mol/L}$。反应将正向还是逆向进行？

Calculate $Q$.计算 $Q$。

$$ Q = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(1.0)^2}{(1.0)(1.0)^3} = \frac{1.0}{1.0} = 1.0 $$

Compare $Q$ to $K_{eq}$.将 $Q$ 与 $K_{eq}$ 比较。 $Q = 1.0 < K_{eq} = 5.9$. The numerator (products) is too small relative to the denominator (reactants) — the system needs to make more $\text{NH}_3$.$Q = 1.0 < K_{eq} = 5.9$。分子（产物）相对于分母（反应物）太小——体系需要生成更多 $\text{NH}_3$。

Conclusion.结论。 $Q < K_{eq}$, so the reaction proceeds in the forward direction (left to right), producing more $\text{NH}_3$ until $Q = K_{eq}$. ✓$Q < K_{eq}$，故反应向正方向（从左到右）进行，生成更多 $\text{NH}_3$ 直到 $Q = K_{eq}$。✓

For a reaction with $K_{eq} = 50$, the reaction quotient is calculated as $Q = 200$. Which statement is correct?某反应 $K_{eq} = 50$，计算得反应商 $Q = 200$。下列哪个说法正确？
§3 · Q1
The reaction will shift in the reverse direction to reach equilibrium反应将向逆方向移动以达到平衡
The reaction will shift in the forward direction to reach equilibrium反应将向正方向移动以达到平衡
The system is already at equilibrium体系已处于平衡状态
$K_{eq}$ will increase to match $Q$$K_{eq}$ 将增大以匹配 $Q$
$Q = 200 > K_{eq} = 50$: there are too many products relative to equilibrium. The system shifts in the reverse direction (products → reactants) to reduce the numerator and increase the denominator until $Q = K_{eq} = 50$.$Q = 200 > K_{eq} = 50$：相对于平衡状态，产物过多。体系向逆方向移动（产物 → 反应物），减小分子并增大分母，直到 $Q = K_{eq} = 50$。
$Q > K_{eq}$ means too many products — reverse shift. $Q < K_{eq}$ means too few products — forward shift. $K_{eq}$ itself never changes with concentration; it is only temperature-dependent.$Q > K_{eq}$ 意味着产物过多——向逆方向移动。$Q < K_{eq}$ 意味着产物不足——向正方向移动。$K_{eq}$ 本身不随浓度变化；它仅取决于温度。
At equilibrium, what is the relationship between $Q$ and $K_{eq}$?在平衡状态时，$Q$ 与 $K_{eq}$ 的关系是什么？
§3 · Q2
$Q > K_{eq}$
$Q < K_{eq}$
$Q = K_{eq}$
$Q$ cannot be calculated at equilibrium平衡时无法计算 $Q$
By definition, at equilibrium the concentrations have reached their equilibrium values. Substituting equilibrium concentrations into the $Q$ expression gives exactly $K_{eq}$. So $Q = K_{eq}$ is the criterion for equilibrium.根据定义，平衡时浓度已达到平衡值。将平衡浓度代入 $Q$ 表达式得到恰好等于 $K_{eq}$ 的结果。因此 $Q = K_{eq}$ 是平衡的判定标准。
$Q = K_{eq}$ defines the equilibrium state. $Q$ can always be calculated — it simply equals $K_{eq}$ when the system is at equilibrium.$Q = K_{eq}$ 定义了平衡状态。$Q$ 始终可以计算——当体系处于平衡时，它恰好等于 $K_{eq}$。
Section 4 · Le Chatelier's principle第 4 节 · 勒沙特列原理

Le Chatelier's Principle勒沙特列原理

When a stress is applied to a system at equilibrium, the system shifts to partially relieve that stress.当对平衡体系施加压力时，体系移动以部分消除该压力。
  • Adding a reactant加入反应物 → system shifts forward (right) to consume the added reactant and produce more products.→ 体系向正方向（右移）移动，消耗加入的反应物并生成更多产物。
  • Removing a product移走产物 → system shifts forward (right) to replace the removed product.→ 体系向正方向（右移）移动，补充移走的产物。
  • Increasing pressure (for gas-phase reactions)增大压强（气相反应） → system shifts toward the side with fewer moles of gas.→ 体系向气体物质的量较少的一侧移动。
  • Increasing temperature升高温度 → system shifts in the endothermic direction (favouring the reaction that absorbs heat). This is the one stress that changes $K_{eq}$ itself.→ 体系向吸热方向移动（有利于吸收热量的反应）。这是唯一会改变 $K_{eq}$ 本身的压力。
  • Catalyst催化剂 → speeds up both forward and reverse reactions equally; no shift in equilibrium position; $K_{eq}$ unchanged. Equilibrium is reached faster, not at a different position.→ 同等程度地加快正逆反应；平衡位置不移动；$K_{eq}$ 不变。平衡更快达到，位置不变。
NGSS HS-PS1-6 Clarification Statement: "Emphasis is on the application of Le Chatelier's Principle … descriptions of the connection between changes at the macroscopic level and what happens at the molecular level. Examples of designs could include different ways to increase product formation including adding reactants or removing products." Assessment Boundary: "limited to specifying the change in only one variable at a time." SCH4U E3.3 and AB Chemistry 30 Unit D GO1 both require predicting Le Chatelier shifts for concentration, temperature, pressure, and catalyst.NGSS HS-PS1-6 澄清说明："重点在于勒沙特列原理的应用……描述在宏观层面所做改变与分子层面发生情况之间的联系。设计示例可包括通过添加反应物或移除产物等不同方式增加产物生成量。"评估边界："仅限于每次指定一个变量的变化。"SCH4U E3.3 与 AB Chemistry 30 D 单元 GO1 均要求预测浓度、温度、压强和催化剂的勒沙特列移动方向。
Worked Example 4 · Applying Le Chatelier's principle例题 4 · 应用勒沙特列原理

Consider the equilibrium: $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)} + \text{heat}$. For each change below, predict the direction of the equilibrium shift and give the molecular-level reason.考虑平衡：$\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)} + \text{heat}$。对下列每种变化，预测平衡移动方向并给出分子层面的原因。

(a) Add $\text{N}_2$.(a) 加入 $\text{N}_2$。 Shift forward. More $\text{N}_2$ increases the collision frequency between $\text{N}_2$ and $\text{H}_2$, raising the forward rate. The system consumes excess $\text{N}_2$ by making more $\text{NH}_3$.移向正方向。更多 $\text{N}_2$ 增大了 $\text{N}_2$ 与 $\text{H}_2$ 之间的碰撞频率，提高了正向速率。体系通过生成更多 $\text{NH}_3$ 来消耗过量的 $\text{N}_2$。

(b) Increase pressure (decrease volume).(b) 增大压强（减小体积）。 Shift forward. Left side has $1 + 3 = 4\ \text{mol gas}$; right side has $2\ \text{mol gas}$. Shifting forward reduces the total moles of gas, partially relieving the pressure increase.移向正方向。左侧有 $1 + 3 = 4\ \text{mol}$ 气体；右侧有 $2\ \text{mol}$ 气体。向正方向移动减少了气体总物质的量，部分缓解了压强增大。

(c) Increase temperature.(c) 升高温度。 Shift reverse. The forward reaction is exothermic (releases heat). Adding heat (raising temperature) favours the endothermic (reverse) direction, decomposing $\text{NH}_3$ back to $\text{N}_2$ and $\text{H}_2$. $K_{eq}$ decreases.移向逆方向。正向反应是放热的（释放热量）。加热（升高温度）有利于吸热（逆向）方向，将 $\text{NH}_3$ 分解回 $\text{N}_2$ 和 $\text{H}_2$。$K_{eq}$ 减小。

(d) Add a catalyst.(d) 加入催化剂。 No shift. A catalyst lowers the activation energy of both reactions equally. Both the forward and reverse rates increase by the same factor, so the equilibrium position (and $K_{eq}$) is unchanged. Equilibrium is reached more quickly.不移动。催化剂同等程度地降低正逆反应的活化能。正逆反应速率以相同倍数增大，因此平衡位置（以及 $K_{eq}$）不变。平衡更快达到。

For the reaction $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$, what happens to the equilibrium when the pressure is increased by decreasing the volume?对于反应 $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$，通过减小体积增大压强时，平衡如何变化？
§4 · Q1
Shifts forward (more PCl3 and Cl2 produced)向正方向移动（生成更多 PCl3 和 Cl2）
Shifts reverse (more PCl5 produced)向逆方向移动（生成更多 PCl5）
No shift; pressure does not affect this equilibrium不移动；压强对此平衡无影响
$K_{eq}$ increases$K_{eq}$ 增大
Left side: 1 mol gas. Right side: $1 + 1 = 2$ mol gas. Increasing pressure favours the side with fewer moles of gas — the left side. The system shifts reverse, producing more $\text{PCl}_5$ to reduce the gas pressure. $K_{eq}$ does not change (temperature is constant).左侧：1 mol 气体。右侧：$1 + 1 = 2$ mol 气体。增大压强有利于气体物质的量较少的一侧——左侧。体系向逆方向移动，生成更多 $\text{PCl}_5$ 以降低气体压强。$K_{eq}$ 不变（温度恒定）。
Count gas moles: left has 1, right has 2. Increasing pressure always favours fewer gas moles — left. A catalyst or concentration change cannot shift $K_{eq}$; only temperature can.计算气体物质的量：左侧 1，右侧 2。增大压强总是有利于气体物质的量较少的一侧——即左侧。催化剂或浓度变化不能改变 $K_{eq}$；只有温度能做到。
For the endothermic reaction $\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{NO(g)}$, increasing temperature will:对于吸热反应 $\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{NO(g)}$，升高温度将：
§4 · Q2
Shift the equilibrium reverse and decrease $K_{eq}$使平衡向逆方向移动并降低 $K_{eq}$
Have no effect on the position of equilibrium对平衡位置无影响
Shift forward but leave $K_{eq}$ unchanged使平衡向正方向移动，但 $K_{eq}$ 不变
Shift the equilibrium forward and increase $K_{eq}$使平衡向正方向移动并增大 $K_{eq}$
The reaction is endothermic (absorbs heat). Adding heat by raising temperature favours the forward (endothermic) direction, producing more NO. This is the one case where $K_{eq}$ itself changes: more products at equilibrium means $K_{eq}$ increases.该反应是吸热的（吸收热量）。升高温度加热有利于正向（吸热）方向，生成更多 NO。这是 $K_{eq}$ 本身发生变化的情况：平衡时产物增多意味着 $K_{eq}$ 增大。
Temperature favours the endothermic direction. For an endothermic reaction, this is the forward direction. Only temperature changes $K_{eq}$; concentration and pressure changes shift position without changing $K_{eq}$.温度有利于吸热方向。对于吸热反应，这是正向方向。只有温度变化才能改变 $K_{eq}$；浓度和压强变化在不改变 $K_{eq}$ 的情况下移动平衡位置。
Going deeper — Le Chatelier's principle and the molecular explanation深入 — 勒沙特列原理的分子层面解释

Le Chatelier's principle is a heuristic — it predicts the direction of shift without requiring a full calculation. The molecular explanation is always a rate argument: adding a reactant increases the forward rate (more collisions) while the reverse rate is initially unchanged; the system moves forward until the rates re-equalize. Similarly, increasing pressure in a gas-phase reaction increases collision frequency of all gas molecules; the system shifts toward fewer gas moles because that reduces the total pressure more effectively. For temperature: it raises the rate of both reactions, but raises the rate of the endothermic reaction more (endothermic activation energy is higher, so it benefits more from extra thermal energy). This net increase in the endothermic direction's rate drives the equilibrium shift. NGSS HS-PS1-6 Clarification Statement explicitly requires "descriptions of the connection between changes made at the macroscopic level and what happens at the molecular level" — this going-deeper box is the molecular explanation for that requirement.勒沙特列原理是一种启发式规则——它预测移动方向而不需要完整的计算。分子层面的解释始终是速率论证：加入反应物增大了正向速率（更多碰撞），而逆向速率初始不变；体系向正方向移动直到速率重新相等。类似地，在气相反应中增大压强提高了所有气体分子的碰撞频率；体系向气体物质的量较少的方向移动，因为这样能更有效地降低总压强。对于温度：它同时提高正逆两个方向的反应速率，但对吸热反应速率的提高更多（吸热反应的活化能更高，因此从额外热能中受益更多）。这种净增大的吸热方向速率驱动了平衡移动。NGSS HS-PS1-6 澄清说明明确要求"描述在宏观层面所做改变与分子层面发生情况之间的联系"——这个深入框正是针对该要求的分子层面解释。

Section 5 · Solubility equilibria ($K_{sp}$) Honors — US NGSS第 5 节 · 溶解平衡（$K_{sp}$，溶度积）荣誉 — US NGSS

Solubility Equilibria: The Solubility Product $K_{sp}$溶解平衡：溶度积 $K_{sp}$

$K_{sp}$ is a specialised $K_{eq}$ for the dissolution of a sparingly soluble ionic solid.$K_{sp}$ 是针对微溶离子固体溶解的专门化 $K_{eq}$。
  • Dissolution equilibrium溶解平衡 for a sparingly soluble salt $\text{M}_x\text{A}_y\text{(s)} \rightleftharpoons x\,\text{M}^{n+}\text{(aq)} + y\,\text{A}^{m-}\text{(aq)}$:对于微溶盐 $\text{M}_x\text{A}_y\text{(s)} \rightleftharpoons x\,\text{M}^{n+}\text{(aq)} + y\,\text{A}^{m-}\text{(aq)}$：
$$ K_{sp} = [\text{M}^{n+}]^x [\text{A}^{m-}]^y $$
  • The pure solid $\text{M}_x\text{A}_y\text{(s)}$ does not appear in $K_{sp}$ (constant "concentration").纯固体 $\text{M}_x\text{A}_y\text{(s)}$ 不出现在 $K_{sp}$ 中（"浓度"为常数）。
  • Molar solubility $s$摩尔溶解度 $s$ = moles of salt that dissolve per litre of solution at equilibrium. For $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$: $K_{sp} = s \cdot s = s^2$. For $\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\,\text{OH}^-\text{(aq)}$: $K_{sp} = s \cdot (2s)^2 = 4s^3$.= 平衡时每升溶液中溶解的盐的物质的量。对于 $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$：$K_{sp} = s \cdot s = s^2$。对于 $\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\,\text{OH}^-\text{(aq)}$：$K_{sp} = s \cdot (2s)^2 = 4s^3$。
  • Common-ion effect同离子效应: adding a common ion (one already present in the equilibrium) shifts dissolution equilibrium in the reverse direction, reducing solubility. Predicted by Le Chatelier.：加入同离子（已在平衡中存在的离子）使溶解平衡向逆方向移动，降低溶解度。由勒沙特列原理预测。
ON SCH4U E2.3/E2.4 require determining $K_{sp}$ and solving $K_{sp}$ problems; E3.4 requires writing the $K_{sp}$ expression. BC Chemistry 12 Content: "saturated solutions and solubility product (Ksp)" — Elaboration: "Ksp as a specialized Keq expression." AB Chemistry 30 Unit D GO2 requires calculating equilibrium constants including $K_{sp}$-type problems.ON SCH4U E2.3/E2.4 要求确定 $K_{sp}$ 并解决 $K_{sp}$ 问题；E3.4 要求书写 $K_{sp}$ 表达式。BC Chemistry 12 内容："饱和溶液与溶度积（$K_{sp}$）"——细化说明："$K_{sp}$ 作为专门化的 $K_{eq}$ 表达式"。AB Chemistry 30 D 单元 GO2 要求计算包括 $K_{sp}$ 类型问题在内的平衡常数。
Worked Example 5 · $K_{sp}$ and molar solubility例题 5 · $K_{sp}$ 与摩尔溶解度

Silver chloride has $K_{sp} = 1.8 \times 10^{-10}$ at 25 °C. (a) Write the dissolution equilibrium and the $K_{sp}$ expression. (b) Calculate the molar solubility of AgCl in pure water. (c) Will AgCl precipitate if $[\text{Ag}^+] = 0.010\ \mathrm{mol/L}$ and $[\text{Cl}^-] = 2.0 \times 10^{-8}\ \mathrm{mol/L}$?氯化银在 25°C 时 $K_{sp} = 1.8 \times 10^{-10}$。(a) 写出溶解平衡和 $K_{sp}$ 表达式。(b) 计算 AgCl 在纯水中的摩尔溶解度。(c) 若 $[\text{Ag}^+] = 0.010\ \mathrm{mol/L}$，$[\text{Cl}^-] = 2.0 \times 10^{-8}\ \mathrm{mol/L}$，AgCl 是否会沉淀？

(a) Equilibrium and expression.(a) 平衡与表达式。

$$ \text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \qquad K_{sp} = [\text{Ag}^+][\text{Cl}^-] $$

(b) Molar solubility.(b) 摩尔溶解度。 Let solubility $= s$. Then $[\text{Ag}^+] = s$ and $[\text{Cl}^-] = s$.设溶解度 $= s$，则 $[\text{Ag}^+] = s$，$[\text{Cl}^-] = s$。

$$ K_{sp} = s^2 = 1.8 \times 10^{-10} \implies s = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5}\ \mathrm{mol/L} $$

(c) Precipitation check.(c) 沉淀核验。 Calculate $Q = [\text{Ag}^+][\text{Cl}^-] = (0.010)(2.0 \times 10^{-8}) = 2.0 \times 10^{-10}$. Since $Q = 2.0 \times 10^{-10} > K_{sp} = 1.8 \times 10^{-10}$, the solution is supersaturated and AgCl will precipitate. ✓计算 $Q = [\text{Ag}^+][\text{Cl}^-] = (0.010)(2.0 \times 10^{-8}) = 2.0 \times 10^{-10}$。由于 $Q = 2.0 \times 10^{-10} > K_{sp} = 1.8 \times 10^{-10}$，溶液处于过饱和状态，AgCl 将会沉淀。✓

For the dissolution of $\text{Ca}_3(\text{PO}_4)_2\text{(s)} \rightleftharpoons 3\,\text{Ca}^{2+}\text{(aq)} + 2\,\text{PO}_4^{3-}\text{(aq)}$, which expression correctly gives $K_{sp}$?对于 $\text{Ca}_3(\text{PO}_4)_2\text{(s)} \rightleftharpoons 3\,\text{Ca}^{2+}\text{(aq)} + 2\,\text{PO}_4^{3-}\text{(aq)}$ 的溶解，哪个表达式正确给出 $K_{sp}$？
§5 · Q1
$K_{sp} = [\text{Ca}^{2+}][\text{PO}_4^{3-}]$
$K_{sp} = \dfrac{[\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2}{[\text{Ca}_3(\text{PO}_4)_2]}$
$K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2$
$K_{sp} = 3[\text{Ca}^{2+}] + 2[\text{PO}_4^{3-}]$
$K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2$. The pure solid $\text{Ca}_3(\text{PO}_4)_2$ is omitted (constant). The stoichiometric coefficients (3 and 2) become exponents. Never add concentrations — multiply and raise to the stoichiometric power.$K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2$。纯固体 $\text{Ca}_3(\text{PO}_4)_2$ 被省略（常数）。化学计量数（3 和 2）变为指数。不要相加浓度——应相乘并取化学计量数次幂。
$K_{sp}$ expressions omit pure solids. Use stoichiometric coefficients as exponents. The solid does not appear in the denominator — it is simply absent.$K_{sp}$ 表达式省略纯固体。化学计量数用作指数。固体不出现在分母中——它根本不出现。
The $K_{sp}$ of $\text{BaSO}_4$ is $1.1 \times 10^{-10}$. What is its molar solubility $s$ in pure water?$\text{BaSO}_4$ 的 $K_{sp} = 1.1 \times 10^{-10}$。它在纯水中的摩尔溶解度 $s$ 是多少？
§5 · Q2
$1.0 \times 10^{-5}\ \mathrm{mol/L}$
$1.1 \times 10^{-10}\ \mathrm{mol/L}$
$5.5 \times 10^{-11}\ \mathrm{mol/L}$
$1.1 \times 10^{-5}\ \mathrm{mol/L}$
$\text{BaSO}_4 \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}$. Both ions have concentration $s$ at equilibrium. $K_{sp} = s^2 = 1.1 \times 10^{-10}$. So $s = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5} \approx 1.0 \times 10^{-5}\ \mathrm{mol/L}$.$\text{BaSO}_4 \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}$。平衡时两种离子浓度均为 $s$。$K_{sp} = s^2 = 1.1 \times 10^{-10}$。故 $s = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5} \approx 1.0 \times 10^{-5}\ \mathrm{mol/L}$。
For a 1:1 salt like BaSO$_4$, $K_{sp} = s^2$, so $s = \sqrt{K_{sp}}$. The molar solubility is not $K_{sp}$ itself — take the square root. $\sqrt{1.1 \times 10^{-10}} \approx 1.0 \times 10^{-5}$.对于 BaSO$_4$ 这样的 1:1 盐，$K_{sp} = s^2$，故 $s = \sqrt{K_{sp}}$。摩尔溶解度不是 $K_{sp}$ 本身——取平方根。$\sqrt{1.1 \times 10^{-10}} \approx 1.0 \times 10^{-5}$。
Section 6 · ICE tables Honors — US NGSS第 6 节 · ICE 表格（变化量法）荣誉 — US NGSS

ICE Tables: Quantitative Equilibrium CalculationsICE 表格：定量平衡计算

ICE = Initial, Change, Equilibrium. A systematic table to find equilibrium concentrations from initial conditions.ICE = 初始（Initial）、变化（Change）、平衡（Equilibrium）。从初始条件求平衡浓度的系统表格。
  • Row I (Initial)I 行（初始）: write the initial concentrations of all species.：写出所有物种的初始浓度。
  • Row C (Change)C 行（变化）: let the change in concentration of one reactant be $-x$ (it decreases); express all other changes in terms of $x$ using stoichiometric ratios (products increase $+x$; reactants decrease $-x$, scaled by molar ratios).：设某反应物浓度变化为 $-x$（它减小）；用化学计量比将所有其他变化用 $x$ 表示（产物增加 $+x$；反应物减小 $-x$，按物质的量比缩放）。
  • Row E (Equilibrium)E 行（平衡）: add I and C for each species. Substitute into $K_{eq}$ expression and solve for $x$.：每种物种的 I 与 C 相加。代入 $K_{eq}$ 表达式并求解 $x$。
AB Chemistry 30 Unit D GO2: "calculate equilibrium constants and concentrations for homogeneous systems … when initial concentrations and one equilibrium concentration are known; the equilibrium constant and one equilibrium concentration are known." Note: "Examples that require the application of the quadratic equation are excluded" from AB required calculations. ON SCH4U E2.4 and BC Chemistry 12 "quantitative relationships" both include ICE-table calculations.AB Chemistry 30 D 单元 GO2："计算均相体系的平衡常数和浓度……当已知初始浓度和一个平衡浓度时；当已知平衡常数和一个平衡浓度时。"注："需要应用二次方程的示例"在 AB 必考计算范围之外。ON SCH4U E2.4 与 BC Chemistry 12"定量关系"均包含 ICE 表格计算。
Worked Example 6 · Full ICE table calculation例题 6 · 完整 ICE 表格计算

For the reaction $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$, $K_{eq} = 49.0$ at 458 °C. Initially $[\text{H}_2]_0 = [\text{I}_2]_0 = 0.100\ \mathrm{mol/L}$ and $[\text{HI}]_0 = 0$. Find the equilibrium concentrations.对于反应 $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$，在 458°C 时 $K_{eq} = 49.0$。初始 $[\text{H}_2]_0 = [\text{I}_2]_0 = 0.100\ \mathrm{mol/L}$，$[\text{HI}]_0 = 0$。求平衡浓度。

Set up the ICE table.建立 ICE 表格。

H2I22 HI
I0.1000.1000
C$-x$$-x$$+2x$
E$0.100 - x$$0.100 - x$$2x$

Substitute into $K_{eq}$.代入 $K_{eq}$。

$$ K_{eq} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(2x)^2}{(0.100 - x)^2} = 49.0 $$

Solve.求解。 Take the square root of both sides (valid since the left side is a perfect square):两边取平方根（因为左边是完全平方式）：

$$ \frac{2x}{0.100 - x} = 7.00 \implies 2x = 0.700 - 7x \implies 9x = 0.700 \implies x = 0.0778\ \mathrm{mol/L} $$

Equilibrium concentrations.平衡浓度。 $[\text{H}_2] = [\text{I}_2] = 0.100 - 0.0778 = 0.0222\ \mathrm{mol/L}$; $[\text{HI}] = 2(0.0778) = 0.156\ \mathrm{mol/L}$. Check: $K_{eq} = (0.156)^2 / (0.0222)^2 = 49.3 \approx 49.0$. ✓$[\text{H}_2] = [\text{I}_2] = 0.100 - 0.0778 = 0.0222\ \mathrm{mol/L}$；$[\text{HI}] = 2(0.0778) = 0.156\ \mathrm{mol/L}$。核验：$K_{eq} = (0.156)^2 / (0.0222)^2 = 49.3 \approx 49.0$。✓

For $\text{A} \rightleftharpoons 2\text{B}$, initial $[\text{A}] = 0.50\ \mathrm{mol/L}$ and $[\text{B}] = 0$. If the change in [A] at equilibrium is $-x$, what is the equilibrium expression for [B]?对于 $\text{A} \rightleftharpoons 2\text{B}$，初始 $[\text{A}] = 0.50\ \mathrm{mol/L}$，$[\text{B}] = 0$。若平衡时 [A] 的变化量为 $-x$，[B] 的平衡表达式是什么？
§6 · Q1
$[\text{B}]_{eq} = x$
$[\text{B}]_{eq} = 0.50 - x$
$[\text{B}]_{eq} = 0.50 + 2x$
$[\text{B}]_{eq} = 2x$
For each mole of A consumed ($-x$), 2 moles of B are produced ($+2x$). Starting from $[\text{B}]_0 = 0$, the equilibrium concentration is $0 + 2x = 2x$. The stoichiometric coefficient (2) goes into the change row, not squared.每消耗 1 mol A（$-x$），生成 2 mol B（$+2x$）。从 $[\text{B}]_0 = 0$ 开始，平衡浓度为 $0 + 2x = 2x$。化学计量数（2）用于变化行，而不是平方。
ICE: Change in [B] = stoichiometric coefficient $\times$ change variable. Since 2 mol B are produced per mol A consumed, Change in [B] = $+2x$. Equilibrium [B] = Initial + Change = $0 + 2x = 2x$.ICE：[B] 的变化量 = 化学计量数 $\times$ 变化量变量。由于每消耗 1 mol A 生成 2 mol B，[B] 的变化量 = $+2x$。平衡 [B] = 初始 + 变化 = $0 + 2x = 2x$。
For $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$, $K_{eq} = 49.0$. If $[\text{H}_2]_{eq} = [\text{I}_2]_{eq} = 0.020\ \mathrm{mol/L}$, what is $[\text{HI}]_{eq}$?对于 $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$，$K_{eq} = 49.0$。若 $[\text{H}_2]_{eq} = [\text{I}_2]_{eq} = 0.020\ \mathrm{mol/L}$，则 $[\text{HI}]_{eq}$ 是多少？
§6 · Q2
$0.020\ \mathrm{mol/L}$
$0.14\ \mathrm{mol/L}$
$0.098\ \mathrm{mol/L}$
$0.28\ \mathrm{mol/L}$
$K_{eq} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = 49.0$. $[\text{HI}]^2 = 49.0 \times (0.020)(0.020) = 49.0 \times 4.0 \times 10^{-4} = 0.01960$. $[\text{HI}] = \sqrt{0.01960} = 0.140\ \mathrm{mol/L} \approx 0.14\ \mathrm{mol/L}$.$K_{eq} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = 49.0$。$[\text{HI}]^2 = 49.0 \times (0.020)(0.020) = 49.0 \times 4.0 \times 10^{-4} = 0.01960$。$[\text{HI}] = \sqrt{0.01960} = 0.140\ \mathrm{mol/L} \approx 0.14\ \mathrm{mol/L}$。
Rearrange $K_{eq}$ to find [HI]: $[\text{HI}]^2 = K_{eq} \times [\text{H}_2][\text{I}_2] = 49.0 \times (0.020)^2 = 0.0196$. Take the square root: $[\text{HI}] = \sqrt{0.0196} = 0.14\ \mathrm{mol/L}$.变形 $K_{eq}$ 求 [HI]：$[\text{HI}]^2 = K_{eq} \times [\text{H}_2][\text{I}_2] = 49.0 \times (0.020)^2 = 0.0196$。取平方根：$[\text{HI}] = \sqrt{0.0196} = 0.14\ \mathrm{mol/L}$。
Section 7 · Applications: the Haber process第 7 节 · 应用：哈伯法

Applications of Equilibrium: The Haber Process平衡的应用：哈伯法

The Haber process is the industrial application of Le Chatelier's principle: optimise conditions to maximise ammonia yield.哈伯法是勒沙特列原理的工业应用：优化条件以最大化氨的产率。
  • Reaction反应: $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)} + 92\ \mathrm{kJ/mol}$ (exothermic).：$\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)} + 92\ \mathrm{kJ/mol}$（放热）。
  • High pressure (150–300 atm)高压（150–300 atm）: 4 mol gas → 2 mol gas, so high pressure shifts equilibrium right, increasing $[\text{NH}_3]$ at equilibrium. Le Chatelier: higher pressure favours fewer gas moles.：4 mol 气体 → 2 mol 气体，故高压使平衡右移，增大平衡时的 $[\text{NH}_3]$。勒沙特列：高压有利于气体物质的量较少的一侧。
  • Moderate temperature (400–500 °C)适中温度（400–500°C）: the reaction is exothermic, so lower temperature would favour products (larger $K_{eq}$). However, too low a temperature slows the reaction rate unacceptably. The chosen temperature is a compromise: fast enough kinetically, acceptable $K_{eq}$.：该反应为放热反应，故较低温度有利于产物（$K_{eq}$ 更大）。然而温度过低会使反应速率过慢。所选温度是一个折中：动力学上足够快，同时 $K_{eq}$ 可接受。
  • Iron catalyst铁催化剂: speeds up both forward and reverse reactions equally. Does not shift the equilibrium position or change $K_{eq}$. Allows equilibrium to be reached at a lower (and more acceptable) temperature.：同等程度地加快正逆反应。不移动平衡位置也不改变 $K_{eq}$。使平衡在更低（且更可接受）的温度下达到。
  • Continuous removal of $\text{NH}_3$持续移走 $\text{NH}_3$: the product is condensed and removed, shifting equilibrium right to produce more $\text{NH}_3$. Le Chatelier: removing a product drives the forward reaction.：产物被冷凝并移走，使平衡右移以生成更多 $\text{NH}_3$。勒沙特列：移走产物推动正向反应。
NGSS HS-PS1-6 Clarification Statement: "Examples of designs could include different ways to increase product formation including adding reactants or removing products." BC Chemistry 12 Elaboration: "Le Châtelier's principle and equilibrium shift — applications (e.g., Haber process, hemoglobin and oxygen in the blood)." AB Chemistry 30 Unit D GO1 requires applying Le Chatelier to "temperature, pressure, volume, concentration or addition of a catalyst."NGSS HS-PS1-6 澄清说明："设计示例可包括通过添加反应物或移除产物等不同方式增加产物生成量。"BC Chemistry 12 细化说明："勒沙特列原理与平衡移动——应用（如哈伯法、血液中血红蛋白与氧的结合）。"AB Chemistry 30 D 单元 GO1 要求将勒沙特列原理应用于"温度、压强、体积、浓度或催化剂的加入"。
Worked Example 7 · Haber process Le Chatelier analysis例题 7 · 哈伯法勒沙特列分析

For $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)} + \text{heat}$, identify whether each of the following changes increases, decreases, or has no effect on the equilibrium concentration of $\text{NH}_3$, and whether $K_{eq}$ changes.对于 $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)} + \text{heat}$，判断下列每种变化是增大、减小还是不影响 $\text{NH}_3$ 的平衡浓度，以及 $K_{eq}$ 是否变化。

Change变化 Shift direction移动方向 $[\text{NH}_3]_{eq}$ $K_{eq}$
Add $\text{N}_2$加入 $\text{N}_2$ Forward →正向 → Increases增大 No change不变
Increase pressure增大压强 Forward → (fewer gas mol)正向 →（气体物质的量更少） Increases增大 No change不变
Increase temperature升高温度 Reverse ← (endothermic dir.)逆向 ←（吸热方向） Decreases减小 Decreases减小
Add Fe catalyst加入 Fe 催化剂 No shift不移动 No change不变 No change不变
Remove $\text{NH}_3$移走 $\text{NH}_3$ Forward →正向 → Increases (new eq.)增大（新平衡） No change不变

The industrial compromise: high pressure increases yield; the Fe catalyst achieves acceptable rate at ~450 °C; continuous $\text{NH}_3$ removal keeps $Q < K_{eq}$, driving further conversion. Only the temperature change alters $K_{eq}$.工业折中方案：高压提高产率；Fe 催化剂在约 450°C 实现可接受的反应速率；持续移走 $\text{NH}_3$ 保持 $Q < K_{eq}$，驱动进一步转化。只有温度变化才改变 $K_{eq}$。

Why does the Haber process use a moderate temperature (~450 °C) rather than a very low temperature, even though the reaction is exothermic and lower temperature would give a higher $K_{eq}$?为什么哈伯法使用适中的温度（约 450°C）而非很低的温度，即使该反应是放热的且较低温度会给出更大的 $K_{eq}$？
§7 · Q1
Higher temperature increases $K_{eq}$ for this reaction较高温度增大了该反应的 $K_{eq}$
Lower temperature would decompose the iron catalyst较低温度会分解铁催化剂
Lower temperature gives too slow a reaction rate; the moderate temperature is a kinetics–equilibrium compromise较低温度使反应速率过慢；适中温度是动力学与平衡的折中
Temperature has no effect on this reaction because a catalyst is used由于使用了催化剂，温度对该反应没有影响
The Haber process is an exothermic reaction: lower temperature means higher $K_{eq}$ (more product at equilibrium) but also a much slower reaction rate. An acceptable production rate requires ~450 °C. The operating conditions are always a kinetics–thermodynamics compromise: the catalyst reduces activation energy to make the rate acceptable at a temperature where $K_{eq}$ is still reasonable.哈伯法是一个放热反应：较低温度意味着更大的 $K_{eq}$（平衡时更多产物），但反应速率也更慢。可接受的生产速率需要约 450°C。操作条件始终是动力学与热力学的折中：催化剂降低活化能，使反应速率在 $K_{eq}$ 仍然合理的温度下变得可接受。
For an exothermic reaction, lower temperature increases $K_{eq}$. The reason for using moderate temperature is purely kinetic — reaction rate is too slow at low temperature, not because of catalyst stability or $K_{eq}$ direction.对于放热反应，较低温度增大 $K_{eq}$。使用适中温度的原因纯粹是动力学方面的——低温下反应速率太慢，而不是因为催化剂稳定性或 $K_{eq}$ 方向。
In the Haber process, ammonia is continuously removed from the reaction vessel by condensation. What is the effect on the equilibrium?在哈伯法中，氨通过冷凝不断从反应容器中移走。这对平衡有什么影响？
§7 · Q2
The equilibrium shifts forward, producing more NH3; $K_{eq}$ is unchanged平衡向正方向移动，生成更多 NH3；$K_{eq}$ 不变
The equilibrium shifts in reverse; $K_{eq}$ decreases平衡向逆方向移动；$K_{eq}$ 减小
No effect; removing a product does not affect equilibrium无影响；移走产物不影响平衡
The equilibrium shifts forward and $K_{eq}$ increases平衡向正方向移动且 $K_{eq}$ 增大
Removing a product lowers $[\text{NH}_3]$, making $Q < K_{eq}$. The system shifts forward (Le Chatelier) to produce more $\text{NH}_3$ and restore $Q = K_{eq}$. Temperature is constant, so $K_{eq}$ does not change. Continuous removal keeps $Q < K_{eq}$, sustaining the forward reaction and greatly increasing overall ammonia yield.移走产物降低了 $[\text{NH}_3]$，使 $Q < K_{eq}$。体系向正方向移动（勒沙特列）以生成更多 $\text{NH}_3$ 并恢复 $Q = K_{eq}$。温度恒定，故 $K_{eq}$ 不变。持续移走保持 $Q < K_{eq}$，维持正向反应进行，大幅提高总体氨产率。
Removing a product reduces its concentration, so $Q < K_{eq}$ — the system responds by shifting forward. $K_{eq}$ only changes with temperature. This is one of the key design features of industrial equilibrium processes.移走产物降低其浓度，使 $Q < K_{eq}$——体系响应向正方向移动。$K_{eq}$ 仅随温度变化。这是工业平衡过程的关键设计特征之一。
Going deeper — other real-world equilibrium applications深入 — 其他现实世界的平衡应用

Hemoglobin and oxygen. The binding of oxygen by hemoglobin in blood is a reversible equilibrium: $\text{Hb} + \text{O}_2 \rightleftharpoons \text{HbO}_2$. In the lungs, where $[\text{O}_2]$ is high, the equilibrium lies far right — hemoglobin loads oxygen. In tissues, where $[\text{O}_2]$ is low (consumed by cellular respiration), the equilibrium shifts left — hemoglobin releases oxygen. Le Chatelier operates here in exactly the same way as in a chemical engineering context. BC Chemistry 12 explicitly lists hemoglobin/oxygen as an application of Le Chatelier (Elaboration to "Le Châtelier's principle and equilibrium shift").血红蛋白与氧气。血红蛋白在血液中与氧气的结合是一个可逆平衡：$\text{Hb} + \text{O}_2 \rightleftharpoons \text{HbO}_2$。在肺部，$[\text{O}_2]$ 较高，平衡强烈偏向右方——血红蛋白装载氧气。在组织中，$[\text{O}_2]$ 较低（被细胞呼吸消耗），平衡左移——血红蛋白释放氧气。勒沙特列原理在这里的作用方式与化学工程中完全相同。BC Chemistry 12 明确将血红蛋白/氧气列为勒沙特列原理的应用（"勒沙特列原理与平衡移动"的细化说明）。

Contact process (sulfuric acid production). $2\,\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{SO}_3\text{(g)}$ (exothermic). Same equilibrium trade-offs as the Haber process: high pressure and moderate temperature (~450 °C with a vanadium pentoxide catalyst). The SO₃ product is absorbed in concentrated H₂SO₄ rather than water (to avoid hazardous acid mist), continuously removing product and driving the reaction forward. Both the Haber process and the Contact process illustrate that industrial chemistry is fundamentally applied equilibrium and kinetics.接触法（硫酸生产）。$2\,\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{SO}_3\text{(g)}$（放热）。与哈伯法相同的平衡权衡：高压和适中温度（约 450°C，使用五氧化二钒催化剂）。SO₃ 产物被浓 H₂SO₄ 吸收而非直接溶于水（以避免危险的酸雾），持续移走产物并推动反应向正方向进行。哈伯法和接触法都表明，工业化学从根本上是应用的平衡与动力学。

Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline for every equilibrium question每道平衡题的解题纪律
  • Write the balanced equation first.先写配平方程式。 The stoichiometric coefficients become the exponents in $K_{eq}$ and the stoichiometric multipliers in ICE-table change rows. A wrong coefficient propagates through the entire calculation.化学计量数成为 $K_{eq}$ 中的指数和 ICE 表格变化行中的化学计量乘数。系数错误会贯穿整个计算。
  • Omit pure solids and pure liquids from $K_{eq}$ and $K_{sp}$.$K_{eq}$ 和 $K_{sp}$ 中省略纯固体和纯液体。 Water in aqueous reactions is a pure liquid — omit it. Solid reactants or products in heterogeneous equilibria — omit them. Forgetting this is one of the most common errors.水溶液反应中的水是纯液体——省略。非均相平衡中的固态反应物或产物——省略。忘记这一点是最常见的错误之一。
  • Always compare $Q$ to $K_{eq}$ before predicting shifts.预测移动方向之前始终将 $Q$ 与 $K_{eq}$ 比较。 $Q < K_{eq}$ → forward; $Q > K_{eq}$ → reverse; $Q = K_{eq}$ → at equilibrium. This is faster and more precise than Le Chatelier for multi-variable questions.$Q < K_{eq}$ → 正向；$Q > K_{eq}$ → 逆向；$Q = K_{eq}$ → 处于平衡。对于多变量问题，这比勒沙特列更快且更精确。
Le Chatelier's principle (§4, §7)勒沙特列原理（§4、§7）
  • Temperature is the only variable that changes $K_{eq}$.温度是唯一能改变 $K_{eq}$ 的变量。 Adding reactants, removing products, changing pressure: all shift the equilibrium position but leave $K_{eq}$ unchanged (at constant temperature). A question asking whether $K_{eq}$ changes after a concentration or pressure change — the answer is always no.加入反应物、移走产物、改变压强：这些都移动平衡位置但不改变 $K_{eq}$（温度恒定时）。询问浓度或压强变化后 $K_{eq}$ 是否改变的问题——答案始终是否定的。
  • Pressure effects apply only to gases.压强效应仅适用于气体。 If all species are in solution or the reaction has no net change in moles of gas, pressure has no effect. Count gas moles on each side first.如果所有物种都在溶液中，或反应的气体物质的量无净变化，则压强无影响。先计算两侧的气体物质的量。
  • State the molecular-level reason, not just the direction.说明分子层面的原因，而不仅仅是方向。 Exam boards require "connection between macroscopic changes and molecular-level events" (NGSS HS-PS1-6 Clarification). Rate arguments (adding X increases forward rate) score full marks; direction-only answers typically score partial marks.考试机构要求"宏观变化与分子层面事件之间的联系"（NGSS HS-PS1-6 澄清说明）。速率论证（加入 X 增大正向速率）可得满分；仅说明方向的答案通常只得部分分。
$K_{eq}$, $K_{sp}$, and ICE tables (§2, §3, §5, §6) Honors — US NGSS$K_{eq}$、$K_{sp}$ 与 ICE 表格（§2、§3、§5、§6）荣誉 — US NGSS
  • Exponents come from coefficients, not multiplied concentrations.指数来自系数，而非相乘的浓度。 For $2\text{A} + \text{B} \rightleftharpoons 3\text{C}$: $K_{eq} = [\text{C}]^3 / ([\text{A}]^2[\text{B}])$, not $2[\text{A}][\text{B}]$ in the denominator.对于 $2\text{A} + \text{B} \rightleftharpoons 3\text{C}$：$K_{eq} = [\text{C}]^3 / ([\text{A}]^2[\text{B}])$，分母不是 $2[\text{A}][\text{B}]$。
  • ICE table change row: use stoichiometric ratios.ICE 表格变化行：使用化学计量比。 If reactant A decreases by $x$, and the equation has coefficient 2 for B and 3 for product C, then B decreases by $(2/1)x$ and C increases by $(3/1)x$ (relative to A's coefficient of 1). Never just write $\pm x$ for every species without checking the ratio.若反应物 A 减小 $x$，方程式中 B 的系数为 2，产物 C 的系数为 3，则 B 减小 $(2/1)x$，C 增大 $(3/1)x$（相对于 A 的系数 1）。不要在未检查比例的情况下对每种物种都写 $\pm x$。
  • Molar solubility: track ion stoichiometry carefully.摩尔溶解度：仔细跟踪离子的化学计量比。 For $\text{M}_x\text{A}_y$, dissolution gives $x\,\text{M}^{n+}$ and $y\,\text{A}^{m-}$. If solubility $= s$, then $[\text{M}^{n+}] = xs$ and $[\text{A}^{m-}] = ys$. $K_{sp} = (xs)^x(ys)^y$. Forgetting the $x$ or $y$ coefficients on the ion concentrations is the most common $K_{sp}$ error.对于 $\text{M}_x\text{A}_y$，溶解给出 $x\,\text{M}^{n+}$ 和 $y\,\text{A}^{m-}$。若溶解度 $= s$，则 $[\text{M}^{n+}] = xs$，$[\text{A}^{m-}] = ys$。$K_{sp} = (xs)^x(ys)^y$。忘记离子浓度中的 $x$ 或 $y$ 系数是最常见的 $K_{sp}$ 错误。
Haber process and answer hygiene (§7)哈伯法与作答规范（§7）
  • The catalyst question: always "no shift, faster equilibrium."关于催化剂的问题：始终是"不移动，平衡更快达到"。 A catalyst never changes $K_{eq}$ or the equilibrium position — it only lowers the activation energy of both reactions equally so equilibrium is reached faster. A two-part answer (shift: none; $K_{eq}$: unchanged) is always needed.催化剂从不改变 $K_{eq}$ 或平衡位置——它只是同等程度地降低正逆反应的活化能，使平衡更快达到。始终需要两部分答案（移动：无；$K_{eq}$：不变）。
  • Significant figures in equilibrium calculations.平衡计算的有效数字。 Report $K_{eq}$ to the same number of significant figures as the given concentrations (typically 2–3 sig figs). Do not round intermediate values; carry extra digits and round only at the final step.$K_{eq}$ 的有效数字与给定浓度的有效数字位数相同（通常 2–3 位）。不要对中间值四舍五入；保留额外数字，仅在最终步骤四舍五入。
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Dynamic equilibrium?动态平衡？
Forward rate = reverse rate; concentrations constant; both reactions still occur at molecular level; requires closed system.正向速率 = 逆向速率；浓度恒定；在分子层面两方向反应仍在进行；需要封闭体系。
$K_{eq}$ expression for $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$?$a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$ 的 $K_{eq}$ 表达式？
$$K_{eq} = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$$ Omit pure solids and pure liquids.省略纯固体和纯液体。
$K_{eq} \gg 1$ means… $K_{eq} \ll 1$ means…$K_{eq} \gg 1$ 意味着……$K_{eq} \ll 1$ 意味着……
$K_{eq} \gg 1$: products favoured (right). $K_{eq} \ll 1$: reactants favoured (left). $K_{eq}$ only changes with temperature.$K_{eq} \gg 1$：产物占优（偏右）。$K_{eq} \ll 1$：反应物占优（偏左）。$K_{eq}$ 仅随温度变化。
Reaction quotient $Q$ vs $K_{eq}$?反应商 $Q$ 与 $K_{eq}$ 的比较？
$Q < K_{eq}$: shift forward. $Q > K_{eq}$: shift reverse. $Q = K_{eq}$: at equilibrium.$Q < K_{eq}$：向正方向移动。$Q > K_{eq}$：向逆方向移动。$Q = K_{eq}$：处于平衡。
Le Chatelier: add reactant?勒沙特列：加入反应物？
Shift forward. More collisions between reactants raise forward rate. System makes more product until $Q = K_{eq}$ again.向正方向移动。反应物间碰撞增多，提高正向速率。体系生成更多产物直到 $Q = K_{eq}$。
Le Chatelier: increase pressure (gas reaction)?勒沙特列：增大压强（气相反应）？
Shift toward fewer moles of gas. No effect if equal gas moles on both sides. $K_{eq}$ unchanged.向气体物质的量较少的一侧移动。若两侧气体物质的量相等则无影响。$K_{eq}$ 不变。
Le Chatelier: raise temperature (exothermic reaction)?勒沙特列：升高温度（放热反应）？
Shift reverse (endothermic direction). $K_{eq}$ decreases. Temperature is the only variable that changes $K_{eq}$.向逆方向移动（吸热方向）。$K_{eq}$ 减小。温度是唯一改变 $K_{eq}$ 的变量。
Le Chatelier: add a catalyst?勒沙特列：加入催化剂？
No shift. Both rates increase equally. $K_{eq}$ unchanged. Equilibrium reached faster, not at a different position.不移动。正逆速率等量增大。$K_{eq}$ 不变。平衡更快达到，位置不变。
$K_{sp}$ for $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$?$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ 的 $K_{sp}$？
$$K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s^2$$ Solid omitted. Molar solubility $s = \sqrt{K_{sp}}$.固体省略。摩尔溶解度 $s = \sqrt{K_{sp}}$。
Common-ion effect?同离子效应？
Adding an ion already in equilibrium shifts dissolution reverse, reducing solubility. Le Chatelier: stress on product side pushes reverse.加入已在平衡中的离子使溶解平衡向逆方向移动，降低溶解度。勒沙特列：产物侧的压力推动逆向反应。
ICE table acronym?ICE 表格缩写？
I = Initial concentrations. C = Change (using stoichiometric ratios, let one reactant change by $-x$). E = Equilibrium = I + C. Substitute E row into $K_{eq}$ and solve for $x$.I = 初始浓度。C = 变化（用化学计量比，设某反应物变化 $-x$）。E = 平衡 = I + C。将 E 行代入 $K_{eq}$ 并求解 $x$。
Haber process reaction and conditions?哈伯法反应与条件？
$$\text{N}_2 + 3\,\text{H}_2 \rightleftharpoons 2\,\text{NH}_3 + \text{heat}$$ High pressure (150–300 atm), moderate temperature (450 °C), Fe catalyst, continuous NH$_3$ removal.高压（150–300 atm）、适中温度（450°C）、Fe 催化剂、持续移走 NH$_3$。
Why moderate (not low) temperature in Haber process?哈伯法为什么用适中而非低温？
Low temp gives higher $K_{eq}$ but too slow a rate. Moderate temp is a kinetics–thermodynamics compromise: acceptable rate + reasonable $K_{eq}$.低温使 $K_{eq}$ 更大，但速率过慢。适中温度是动力学与热力学的折中：可接受的速率 + 合理的 $K_{eq}$。
Why does removing product drive reaction forward?为什么移走产物会推动反应正向进行？
Removing product lowers $[\text{product}]$, so $Q < K_{eq}$. System shifts forward to restore $Q = K_{eq}$. $K_{eq}$ is unchanged (temperature constant).移走产物降低 $[\text{product}]$，使 $Q < K_{eq}$。体系向正方向移动以恢复 $Q = K_{eq}$。$K_{eq}$ 不变（温度恒定）。
Mixed Practice综合练习

Practice Quiz综合测验

What is the defining condition for dynamic equilibrium?动态平衡的定义条件是什么？
Q1
The reaction has stopped completely反应已完全停止
Concentrations of all species are equal所有物种浓度相等
Forward rate = reverse rate; concentrations are constant正向速率 = 逆向速率；浓度恒定
Only the forward reaction continues只有正向反应在继续
Dynamic equilibrium: both reactions run at equal rates; concentrations are constant (not necessarily equal); the system must be closed. "Dynamic" emphasises that reactions still occur at the molecular level.动态平衡：两个方向的反应以相等速率运行；浓度恒定（不一定相等）；体系必须封闭。"动态"强调在分子层面反应仍在进行。
Equilibrium is dynamic, not static — both reactions continue. Concentrations are constant, not equal. The defining criterion is rate equality, not concentration equality.平衡是动态的，而非静止的——两个方向的反应仍在继续。浓度恒定，而非相等。定义标准是速率相等，而非浓度相等。
Write the $K_{eq}$ expression for $\text{CO(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$. Which option is correct?写出 $\text{CO(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$ 的 $K_{eq}$ 表达式。哪个选项正确？
Q2
$\dfrac{[\text{CO}][\text{H}_2]^3}{[\text{CH}_4][\text{H}_2\text{O}]}$
$\dfrac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]}$
$\dfrac{[\text{CH}_4][\text{H}_2\text{O}]}{3[\text{CO}][\text{H}_2]}$
$\dfrac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}$
Products over reactants; stoichiometric coefficients as exponents. H$_2$ has coefficient 3, so it appears as $[\text{H}_2]^3$ in the denominator. All other species have coefficient 1 (exponent = 1, omitted). All are gases, so none are omitted on solubility grounds.产物除以反应物；化学计量数作为指数。H$_2$ 的系数为 3，故分母中出现 $[\text{H}_2]^3$。所有其他物种系数为 1（指数 = 1，省略）。所有物种均为气体，故不因溶解性而省略任何物种。
Coefficients become exponents, not multipliers. H$_2$ has coefficient 3 → $[\text{H}_2]^3$, not $3[\text{H}_2]$. Products in the numerator, reactants in the denominator.系数变成指数，而非乘数。H$_2$ 系数为 3 → $[\text{H}_2]^3$，而非 $3[\text{H}_2]$。产物在分子，反应物在分母。
For a reaction at 25 °C, $K_{eq} = 3.2 \times 10^{-4}$ and $Q = 1.5 \times 10^{-2}$. What will happen?某反应在 25°C 时，$K_{eq} = 3.2 \times 10^{-4}$，$Q = 1.5 \times 10^{-2}$。将发生什么？
Q3
Reaction shifts in reverse until $Q = K_{eq}$反应向逆方向移动，直到 $Q = K_{eq}$
Reaction shifts forward until $Q = K_{eq}$反应向正方向移动，直到 $Q = K_{eq}$
System is at equilibrium; no shift体系处于平衡，不移动
$K_{eq}$ increases to match $Q$$K_{eq}$ 增大以匹配 $Q$
$Q = 1.5 \times 10^{-2} > K_{eq} = 3.2 \times 10^{-4}$. There are too many products relative to equilibrium. The system converts products back to reactants (reverse direction) until $Q$ decreases to $K_{eq}$.$Q = 1.5 \times 10^{-2} > K_{eq} = 3.2 \times 10^{-4}$。相对于平衡状态，产物过多。体系将产物转化回反应物（逆方向），直到 $Q$ 降至 $K_{eq}$。
$Q > K_{eq}$: reverse shift. $Q < K_{eq}$: forward shift. $K_{eq}$ never adjusts to match $Q$ — it is fixed by temperature.$Q > K_{eq}$：逆向移动。$Q < K_{eq}$：正向移动。$K_{eq}$ 不会调整以匹配 $Q$——它由温度固定。
For the endothermic reaction $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$, which change produces more $\text{NO}_2$ at equilibrium? 🇺🇸 NGSS HS-PS1-6 / 🇨🇦 SCH4U E3.3对于吸热反应 $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$，哪种变化会在平衡时产生更多 $\text{NO}_2$？🇺🇸 NGSS HS-PS1-6 / 🇨🇦 SCH4U E3.3
Q4
Increase pressure (decrease volume)增大压强（减小体积）
Increase temperature升高温度
Add a catalyst加入催化剂
Remove some N$_2$O$_4$移走一些 N$_2$O$_4$
The reaction is endothermic: heat is absorbed as $\text{N}_2\text{O}_4$ decomposes. Raising temperature favours the endothermic (forward) direction, producing more $\text{NO}_2$ and increasing $K_{eq}$. Increasing pressure favours fewer gas moles (left), reducing $[\text{NO}_2]$. A catalyst has no effect on equilibrium position. Removing $\text{N}_2\text{O}_4$ shifts forward but net [NO$_2$] at new equilibrium may not increase.该反应为吸热反应：$\text{N}_2\text{O}_4$ 分解时吸收热量。升高温度有利于吸热（正向）方向，生成更多 $\text{NO}_2$ 并增大 $K_{eq}$。增大压强有利于气体物质的量较少的一侧（左），降低 $[\text{NO}_2]$。催化剂对平衡位置无影响。移走 $\text{N}_2\text{O}_4$ 使平衡右移，但新平衡时 [NO$_2$] 的净增大不确定。
For an endothermic reaction, temperature increase shifts forward (producing more products) and increases $K_{eq}$. Pressure increase for 1 mol gas → 2 mol gas favours the reactant side. Catalyst = no shift. Only temperature reliably increases $[\text{NO}_2]$ at equilibrium.对于吸热反应，温度升高使平衡正向移动（生成更多产物）并增大 $K_{eq}$。对于 1 mol 气体 → 2 mol 气体，压强增大有利于反应物侧。催化剂 = 不移动。只有升温才能可靠地增大平衡时的 $[\text{NO}_2]$。
$K_{sp}$ of $\text{PbSO}_4 = 1.6 \times 10^{-8}$. What is the molar solubility of $\text{PbSO}_4$ in pure water? 🇨🇦 SCH4U E2.4 / BC Chemistry 12$\text{PbSO}_4$ 的 $K_{sp} = 1.6 \times 10^{-8}$。$\text{PbSO}_4$ 在纯水中的摩尔溶解度是多少？🇨🇦 SCH4U E2.4 / BC Chemistry 12
Q5
$1.6 \times 10^{-8}\ \mathrm{mol/L}$
$8.0 \times 10^{-9}\ \mathrm{mol/L}$
$1.3 \times 10^{-4}\ \mathrm{mol/L}$
$4.0 \times 10^{-4}\ \mathrm{mol/L}$
$\text{PbSO}_4 \rightleftharpoons \text{Pb}^{2+} + \text{SO}_4^{2-}$. $K_{sp} = s^2 = 1.6 \times 10^{-8}$. $s = \sqrt{1.6 \times 10^{-8}} = 1.26 \times 10^{-4} \approx 1.3 \times 10^{-4}\ \mathrm{mol/L}$.$\text{PbSO}_4 \rightleftharpoons \text{Pb}^{2+} + \text{SO}_4^{2-}$。$K_{sp} = s^2 = 1.6 \times 10^{-8}$。$s = \sqrt{1.6 \times 10^{-8}} = 1.26 \times 10^{-4} \approx 1.3 \times 10^{-4}\ \mathrm{mol/L}$。
PbSO$_4$ is a 1:1 salt: $K_{sp} = s^2$. Take the square root to get molar solubility. $\sqrt{1.6 \times 10^{-8}} \approx 1.3 \times 10^{-4}\ \mathrm{mol/L}$.PbSO$_4$ 是 1:1 盐：$K_{sp} = s^2$。取平方根得摩尔溶解度。$\sqrt{1.6 \times 10^{-8}} \approx 1.3 \times 10^{-4}\ \mathrm{mol/L}$。
For $\text{A(g)} + \text{B(g)} \rightleftharpoons 2\,\text{C(g)}$, $K_{eq} = 4.0$. Initial concentrations: $[\text{A}]_0 = [\text{B}]_0 = 1.0\ \mathrm{mol/L}$, $[\text{C}]_0 = 0$. After setting up the ICE table with change $-x$ for A, what equation do you solve? 🇨🇦 SCH4U E2.4 / AB Chem 30 D-GO2对于 $\text{A(g)} + \text{B(g)} \rightleftharpoons 2\,\text{C(g)}$，$K_{eq} = 4.0$。初始浓度：$[\text{A}]_0 = [\text{B}]_0 = 1.0\ \mathrm{mol/L}$，$[\text{C}]_0 = 0$。建立 ICE 表格，设 A 的变化量为 $-x$，你需要求解哪个方程？🇨🇦 SCH4U E2.4 / AB Chem 30 D-GO2
Q6
$\dfrac{x^2}{(1.0)^2} = 4.0$
$\dfrac{(1.0-x)^2}{(2x)^2} = 4.0$
$\dfrac{2x}{(1.0-x)^2} = 4.0$
$\dfrac{(2x)^2}{(1.0-x)^2} = 4.0$
ICE: A goes from $1.0$ to $(1.0 - x)$; B from $1.0$ to $(1.0 - x)$; C from $0$ to $2x$. $K_{eq} = \frac{[\text{C}]^2}{[\text{A}][\text{B}]} = \frac{(2x)^2}{(1.0-x)(1.0-x)} = \frac{(2x)^2}{(1.0-x)^2} = 4.0$. Take square root: $\frac{2x}{1.0-x} = 2.0$, so $x = 0.50\ \mathrm{mol/L}$.ICE：A 从 $1.0$ 变为 $(1.0 - x)$；B 从 $1.0$ 变为 $(1.0 - x)$；C 从 $0$ 变为 $2x$。$K_{eq} = \frac{[\text{C}]^2}{[\text{A}][\text{B}]} = \frac{(2x)^2}{(1.0-x)(1.0-x)} = \frac{(2x)^2}{(1.0-x)^2} = 4.0$。取平方根：$\frac{2x}{1.0-x} = 2.0$，故 $x = 0.50\ \mathrm{mol/L}$。
ICE: A decreases by $x$, B decreases by $x$ (1:1 ratio with A), C increases by $2x$ (coefficient 2). Equilibrium concentrations: A = $1.0 - x$, B = $1.0 - x$, C = $2x$. Substitute: $K_{eq} = (2x)^2 / (1.0 - x)^2 = 4.0$.ICE：A 减小 $x$，B 减小 $x$（与 A 比例为 1:1），C 增大 $2x$（系数为 2）。平衡浓度：A = $1.0 - x$，B = $1.0 - x$，C = $2x$。代入：$K_{eq} = (2x)^2 / (1.0 - x)^2 = 4.0$。
Which combination of conditions does the Haber process use to achieve acceptable ammonia yield in an acceptable time? 🇺🇸 NGSS HS-PS1-6 / 🇨🇦 BC Chemistry 12哈伯法使用哪种条件组合，以在可接受时间内实现可接受的氨产率？🇺🇸 NGSS HS-PS1-6 / 🇨🇦 BC Chemistry 12
Q7
High temperature, low pressure, no catalyst高温、低压、无催化剂
High pressure, moderate temperature, iron catalyst高压、适中温度、铁催化剂
Low pressure, high temperature, iron catalyst低压、高温、铁催化剂
Low pressure, low temperature, no catalyst低压、低温、无催化剂
High pressure: 4 mol gas → 2 mol gas — shifts right (Le Chatelier). Moderate temperature (~450 °C): exothermic reaction so low temp favours products, but rate is too slow at low temp — compromise. Iron catalyst: reduces activation energy, allows acceptable rate at 450 °C without shifting equilibrium position.高压：4 mol 气体 → 2 mol 气体——向右移动（勒沙特列）。适中温度（约 450°C）：放热反应，低温有利于产物，但低温时速率过慢——折中。铁催化剂：降低活化能，使 450°C 时速率可接受，且不移动平衡位置。
Haber process: high pressure (favours fewer gas moles = right shift), moderate temperature (kinetics-thermodynamics compromise), and Fe catalyst (faster rate without position shift). High temperature would decrease $K_{eq}$ for this exothermic reaction.哈伯法：高压（有利于气体物质的量较少的一侧 = 右移）、适中温度（动力学-热力学折中）、Fe 催化剂（更快速率且不移动位置）。高温会降低这个放热反应的 $K_{eq}$。
Which statement about $K_{eq}$ is always true? 🇨🇦 SCH4U E3.2 / AB Chem 30 D-GO1关于 $K_{eq}$，哪个说法始终正确？🇨🇦 SCH4U E3.2 / AB Chem 30 D-GO1
Q8
$K_{eq}$ changes only when temperature changes$K_{eq}$ 仅当温度变化时才改变
$K_{eq}$ changes when a reactant is added加入反应物时 $K_{eq}$ 改变
$K_{eq}$ equals 1 at equilibrium平衡时 $K_{eq}$ 等于 1
$K_{eq}$ changes when pressure changes压强变化时 $K_{eq}$ 改变
$K_{eq}$ is a function of temperature only. Changes in concentration, pressure, or addition of a catalyst shift the equilibrium position (change $Q$ momentarily) but do not change $K_{eq}$. Only a temperature change changes the value of $K_{eq}$.$K_{eq}$ 仅是温度的函数。浓度、压强的变化或催化剂的加入会移动平衡位置（暂时改变 $Q$），但不会改变 $K_{eq}$。只有温度变化才会改变 $K_{eq}$ 的值。
$K_{eq}$ is fixed at a given temperature — it does not change with concentration, pressure, or catalyst. $K_{eq} = 1$ is a special case, not the rule. Adding a reactant changes $Q$ but $K_{eq}$ stays constant.$K_{eq}$ 在给定温度下是固定的——它不随浓度、压强或催化剂变化。$K_{eq} = 1$ 是特例，而非规律。加入反应物会改变 $Q$，但 $K_{eq}$ 保持不变。
For the dissolution equilibrium $\text{CaF}_2\text{(s)} \rightleftharpoons \text{Ca}^{2+}\text{(aq)} + 2\,\text{F}^-\text{(aq)}$, if $K_{sp} = 3.9 \times 10^{-11}$ and molar solubility $= s$, which expression is correct? 🇨🇦 SCH4U E3.4 / BC Chemistry 12对于溶解平衡 $\text{CaF}_2\text{(s)} \rightleftharpoons \text{Ca}^{2+}\text{(aq)} + 2\,\text{F}^-\text{(aq)}$，若 $K_{sp} = 3.9 \times 10^{-11}$，摩尔溶解度 $= s$，哪个表达式正确？🇨🇦 SCH4U E3.4 / BC Chemistry 12
Q9
$K_{sp} = s^3$
$K_{sp} = s \cdot 2s = 2s^2$
$K_{sp} = s \cdot (2s)^2 = 4s^3$
$K_{sp} = 2s^3$
If $s$ mol/L of CaF$_2$ dissolves: $[\text{Ca}^{2+}] = s$ and $[\text{F}^-] = 2s$ (stoichiometric ratio 1:2). $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3$. Note: $[\text{F}^-] = 2s$ is squared, giving $4s^2$.若 $s$ mol/L 的 CaF$_2$ 溶解：$[\text{Ca}^{2+}] = s$，$[\text{F}^-] = 2s$（化学计量比 1:2）。$K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3$。注意：$[\text{F}^-] = 2s$ 被平方，得 $4s^2$。
For CaF$_2$: 1 Ca$^{2+}$ and 2 F$^-$ per formula unit. If solubility $= s$: $[\text{Ca}^{2+}] = s$, $[\text{F}^-] = 2s$. $K_{sp} = s(2s)^2 = 4s^3$, not $s^3$ or $2s^2$ or $2s^3$. The coefficient on F$^-$ (which is 2) becomes part of the ion concentration AND the exponent in $K_{sp}$.对于 CaF$_2$：每个化学式单位有 1 个 Ca$^{2+}$ 和 2 个 F$^-$。若溶解度 $= s$：$[\text{Ca}^{2+}] = s$，$[\text{F}^-] = 2s$。$K_{sp} = s(2s)^2 = 4s^3$，而非 $s^3$、$2s^2$ 或 $2s^3$。F$^-$ 的系数（为 2）成为离子浓度的一部分，同时也是 $K_{sp}$ 中的指数。
In the Haber process, ammonia is condensed and removed continuously. This is an application of which principle, and what is the effect? 🇺🇸 NGSS HS-PS1-6 / 🇨🇦 BC Chemistry 12 / AB Chem 30在哈伯法中，氨被持续冷凝并移走。这是哪条原理的应用，效果如何？🇺🇸 NGSS HS-PS1-6 / 🇨🇦 BC Chemistry 12 / AB Chem 30
Q10
Le Chatelier: removal of product shifts reverse, decreasing total yield勒沙特列：移走产物使平衡向逆方向移动，降低总产率
Removing product increases $K_{eq}$, driving more conversion移走产物增大 $K_{eq}$，推动更多转化
The catalyst becomes more effective when NH$_3$ is removed移走 NH$_3$ 后催化剂变得更有效
Le Chatelier: removal of product shifts forward, increasing yield; $K_{eq}$ unchanged勒沙特列：移走产物使平衡向正方向移动，提高产率；$K_{eq}$ 不变
Removing NH$_3$ lowers $[\text{NH}_3]$, making $Q < K_{eq}$. By Le Chatelier (or $Q < K_{eq}$ criterion), the system shifts forward to restore equilibrium, producing more NH$_3$. Since temperature is unchanged, $K_{eq}$ is unchanged. Continuous removal keeps $Q < K_{eq}$ indefinitely, achieving far higher total yield than a single-pass equilibrium.移走 NH$_3$ 降低了 $[\text{NH}_3]$，使 $Q < K_{eq}$。根据勒沙特列原理（或 $Q < K_{eq}$ 标准），体系向正方向移动以恢复平衡，生成更多 NH$_3$。由于温度不变，$K_{eq}$ 不变。持续移走使 $Q < K_{eq}$ 无限期保持，实现远高于单次平衡的总体产率。
Removing a product shifts the equilibrium forward (toward more products), not reverse. $K_{eq}$ does not change (temperature is constant). The high total yield is achieved by continuously keeping the system out of equilibrium in the forward direction.移走产物使平衡向正方向移动（产生更多产物），而非逆方向。$K_{eq}$ 不变（温度恒定）。高总体产率是通过持续保持体系在正方向上偏离平衡而实现的。
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Next Steps后续单元

What This Feeds Into本单元的去向

Chemical equilibrium is the conceptual bridge between stoichiometry and the quantitative treatment of acids, bases, and solubility. Every subsequent topic — acid-base equilibria ($K_a$, $K_b$, buffers), redox half-reactions, solubility and precipitation, and electrochemical cells — uses the same $K_{eq}$ expression framework and the same ICE-table technique you practised here. The cross-references below point at the college-credit feeder and the preceding unit in this course.化学平衡是化学计量学与酸碱、溶解度定量处理之间的概念桥梁。每一个后续主题——酸碱平衡（$K_a$、$K_b$、缓冲溶液）、氧化还原半反应、溶解度与沉淀、电化学电池——都使用你在这里练习过的相同的 $K_{eq}$ 表达式框架和相同的 ICE 表格技术。以下链接指向大学学分衔接课程和本课程的前一个单元。

Within High School Chemistry.在 HS Chemistry 内部。

The preceding unit, Reaction Rates (Kinetics), explained why rates change with temperature and concentration — the microscopic foundation for Le Chatelier's temperature and concentration effects in this guide. The following unit, Acids, Bases, and pH (Unit 9 at the quantitative depth), applies the same $K_{eq}$ framework to weak acid/base ionisation ($K_a$, $K_b$, $K_w$) and buffer calculations — ICE-table skills from §6 are used directly. Solubility and Precipitation (part of Unit 8) uses $K_{sp}$ from §5. Electrochemistry (Unit 13) uses equilibrium concepts to derive the Nernst equation.前一个单元《反应速率（动力学）》解释了速率为何随温度和浓度变化——这为本指南中勒沙特列的温度和浓度效应提供了微观基础。后续单元《酸碱与 pH》（Unit 9，定量深度）将相同的 $K_{eq}$ 框架应用于弱酸/弱碱的电离（$K_a$、$K_b$、$K_w$）和缓冲液计算——直接使用来自 §6 的 ICE 表格技能。溶解度与沉淀（Unit 8 的一部分）使用来自 §5 的 $K_{sp}$。电化学（Unit 13）使用平衡概念推导能斯特方程。

College-credit feeder and cross-subject links.大学学分衔接与跨学科链接。

IB Chemistry HL · Reactivity 2: How Much, How Fast, How Far? (the college-credit feeder for equilibrium, $K_{eq}$, $K_{sp}$, Le Chatelier, and ICE tables at IB depth)IB Chemistry HL · Reactivity 2：多少、多快、多远？（化学平衡、$K_{eq}$、$K_{sp}$、勒沙特列原理与 ICE 表格在 IB 深度下的大学学分衔接）

If you are aiming for IB Chemistry HL or AP Chemistry, the equilibrium constant expressions, Le Chatelier analysis, ICE-table technique, and $K_{sp}$ calculations here are assumed from the first week of the college-credit course. IB Chemistry HL Reactivity 2 extends this with quantitative Gibbs energy ($\Delta G = -RT\ln K$), the Brønsted-Lowry acid-base equilibrium, and advanced buffer calculations. AP Chemistry Unit 7 (Equilibrium) and Unit 8 (Acids and Bases) build directly on the $K_{eq}$ and ICE-table foundation of this guide. The $Q$ vs $K_{eq}$ comparison is used to predict reaction direction in every subsequent quantitative unit.如果你目标是 IB Chemistry HL 或 AP Chemistry，这里的平衡常数表达式、勒沙特列分析、ICE 表格技术和 $K_{sp}$ 计算从大学学分课程的第一周就被默认掌握。IB Chemistry HL Reactivity 2 通过定量吉布斯能（$\Delta G = -RT\ln K$）、布朗斯特-劳里酸碱平衡和高级缓冲液计算来延伸这部分内容。AP Chemistry Unit 7（平衡）和 Unit 8（酸和碱）直接建立在本指南的 $K_{eq}$ 和 ICE 表格基础上。$Q$ 与 $K_{eq}$ 的比较在每一个后续定量单元中都用于预测反应方向。

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High School Chemistry · Chemical Equilibrium · v1高中化学 · 化学平衡 · v1