Edexcel Chemistry Equilibrium 入门

## Edexcel化学 动态平衡 Le Chatelier Kc Kp 考点 Equilibrium is one of the most heavily examined topics in Edexcel A-Level Chemistry, typically accounting for 12-18% of the marks across Papers 1 and 2. Mastering dynamic equilibrium, Le Chatelier's Principle, and both Kc and Kp is non-negotiable for any student aiming for an A or A* grade. 化学平衡是Edexcel A-Level化学中考查最密集的章节之一，通常在Paper 1和Paper 2中占总分的12-18%。掌握动态平衡、勒夏特列原理以及Kc和Kp是任何冲A或A*学生的必修课。 This guide covers every concept, calculation, and exam technique you need — in both English and Chinese — so you can study efficiently and walk into the exam hall with confidence. 本文用中英双语覆盖了你需要的每一个概念、每一种计算和每一项应试技巧，让你高效备考，自信走进考场。 --- ### Part 1: Reversible Reactions and Dynamic Equilibrium 可逆反应与动态平衡 A reversible reaction is one that can proceed in both the forward and backward directions. It is represented by the ⇌ symbol (not →). 可逆反应是指既能正向进行也能逆向进行的反应，用⇌符号表示（而非→）。 A reversible reaction reaches **dynamic equilibrium** when the rate of the forward reaction exactly equals the rate of the backward reaction. At this point: 当可逆反应的正反应速率恰好等于逆反应速率时，体系达到了**动态平衡**。此时： - The concentrations of all reactants and products remain constant (but are NOT necessarily equal). 所有反应物和生成物的浓度保持恒定（但不一定相等）。 - The reaction has not stopped — both forward and backward reactions continue at the same rate. 反应并未停止——正逆反应以相同速率持续进行。 - Equilibrium can only be established in a **closed system** where no matter enters or escapes. 平衡只能在**封闭体系**中建立，物质无法进出。 Forward Reaction 正反应
Rate: initially fast, then slows
初始快，随后减慢
Reactants get used up over time
反应物随时间消耗 Backward Reaction 逆反应
Rate: initially zero, then speeds up
初始为零，随后加快
Products accumulate, enabling backward reaction
生成物累积后逆反应开始 At Equilibrium 平衡时
Forward rate = Backward rate
正逆速率相等
All concentrations remain constant
所有浓度保持恒定 **Why is it called "dynamic"? 为什么叫"动态"？** Because both reactions are still happening. It is not static — molecules are constantly reacting in both directions. The macroscopic properties (colour, pressure, concentration) appear unchanging, but at the molecular level, the system is highly active. 因为两个方向的反应仍在进行。它不是静态的——分子在两个方向上持续反应。宏观性质（颜色、压强、浓度）看起来不变，但在分子层面，体系是高度活跃的。 **Common Edexcel misconception alert 常见误解提醒**: Students often write "the reaction stops at equilibrium." This is wrong and will lose marks. The correct phrasing is: "the forward and backward reactions continue at equal rates." 学生常写"平衡时反应停止"。这是错误的，会丢分。正确表述是："正逆反应以相同速率持续进行。" --- ### Part 2: Le Chatelier's Principle 勒夏特列原理 "If a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to oppose that change." This is the single most important qualitative principle in the entire topic. "如果处于动态平衡的体系受到条件改变的影响，平衡位置会向对抗该改变的方向移动。"这是整个章节最重要的定性原则。 🔥 Increase Temperature 升温
Equilibrium shifts in the endothermic direction
平衡向吸热方向移动

Reason: The system absorbs the added heat by favouring the endothermic reaction. 体系通过倾向吸热反应来吸收外加的热量。 ❄️ Decrease Temperature 降温
Equilibrium shifts in the exothermic direction
平衡向放热方向移动

Reason: The system releases heat by favouring the exothermic reaction. 体系通过倾向放热反应来释放热量。 ⬆️ Increase Pressure 加压
Equilibrium shifts toward side with fewer gas moles
平衡向气体分子数少的一侧移动

Reason: Reducing the number of gas molecules reduces the pressure. 减少气体分子数可以降低压强。 ⬇️ Decrease Pressure 减压
Equilibrium shifts toward side with more gas moles
平衡向气体分子数多的一侧移动

Reason: Increasing the number of gas molecules increases the pressure. 增加气体分子数可以升高压强。 ➕ Increase Concentration of Reactant 增加反应物浓度
Equilibrium shifts to the right (product side)
平衡向右（生成物侧）移动

Reason: The system uses up the added reactant by making more product. 体系通过生成更多产物来消耗增加的反应物。 ⚡ Add a Catalyst 加催化剂
No shift in equilibrium position
平衡位置不变

Reason: Catalyst speeds up forward and backward reactions equally. Only the time to reach equilibrium decreases. 催化剂同等加速正逆反应。仅缩短达平衡时间。 **The Edexcel 3-point answer template 三步答题模板**: When asked "Explain what happens to the equilibrium when X is changed," always include: 当被问"解释当X改变时平衡会如何"时，务必包含以下三点： 1. **Direction**: "The equilibrium shifts to the left/right." 方向：平衡向左/右移动。 2. **Reason (ΔH or moles)**: "Because the forward reaction is exothermic/endothermic, OR because there are fewer/more moles of gas on the product side." 原因（ΔH或分子数）：因为正反应放热/吸热，或因为生成物侧气体分子数更少/更多。 3. **Link to Le Chatelier**: "This opposes the increase/decrease in temperature/pressure." 联系勒夏特列：这对抗了温度/压强的升高/降低。 This template consistently earns full marks on 3-4 mark explain questions. 这个模板在3-4分的解释题上稳定拿满。 --- ### Part 3: The Equilibrium Constant Kc 平衡常数Kc For a general reaction: aA + bB ⇌ cC + dD **Kc = [C]^c × [D]^d / [A]^a × [B]^b** Where [X] is the equilibrium concentration of X in mol dm^-3. 其中[X]是物质X的平衡浓度，单位为mol dm^-3。 **Crucial properties of Kc Kc的关键性质**: 1. Kc is constant at a given temperature. Changing temperature changes Kc. Kc在给定温度下为常数。改变温度会改变Kc。 2. Changing concentration or pressure does NOT change Kc — only the position of equilibrium shifts to restore the same Kc value. 改变浓度或压强不会改变Kc——只有平衡位置移动以恢复相同的Kc值。 3. Adding a catalyst does NOT change Kc. 加催化剂不改变Kc。 4. Kc >> 1: products are favoured at equilibrium. 生成物在平衡时占优势。 5. Kc << 1: reactants are favoured at equilibrium. 反应物在平衡时占优势。 6. **Solids and pure liquids do not appear** in the Kc expression — their concentration is effectively constant. **固体和纯液体不出现**在Kc表达式中——它们的浓度实际上恒定。 Worked Example 1 例题1: Basic Kc Calculation

For the reaction: H2(g) + I2(g) ⇌ 2HI(g) at 700 K
反应: H2(g) + I2(g) ⇌ 2HI(g)，温度700 K

At equilibrium in a 1.0 dm^3 vessel:
[H2] = 0.20 mol dm^-3, [I2] = 0.20 mol dm^-3, [HI] = 1.60 mol dm^-3

Step 1: Write the Kc expression. 写出Kc表达式。
Kc = [HI]^2 / ([H2] × [I2])

Step 2: Substitute values. 代入数值。
Kc = (1.60)^2 / (0.20 × 0.20) = 2.56 / 0.04 = 64

Step 3: Determine units. 确定单位。
Units = (mol dm^-3)^2 / (mol dm^-3)(mol dm^-3) = no units 无单位

Interpretation 解读: Kc = 64 >> 1, so products (HI) are strongly favoured. Kc = 64 >> 1，生成物(HI)占明显优势。 Worked Example 2 例题2: Kc with ICE table

For the reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g)
反应: PCl5(g) ⇌ PCl3(g) + Cl2(g)

Initial: 2.0 mol PCl5 in a 2.0 dm^3 vessel (→ 1.0 mol dm^-3). No PCl3 or Cl2 initially.
At equilibrium, 40% of PCl5 has dissociated.

ICE Table:
Species Initial Change Equilibrium PCl5 1.00 -0.40 0.60 PCl3 0 +0.40 0.40 Cl2 0 +0.40 0.40
Kc = [PCl3][Cl2] / [PCl5] = (0.40)(0.40) / 0.60 = 0.16 / 0.60 = 0.267 mol dm^-3 --- ### Part 4: Partial Pressure and Kp 分压与Kp When all species in an equilibrium are gases, we use Kp — the equilibrium constant expressed in terms of partial pressures rather than concentrations. 当平衡中的所有物质都是气体时，我们用Kp——以分压而非浓度表示的平衡常数。 **Key definitions 关键定义**: - **Mole fraction 摩尔分数** (χ): The proportion of a gas in a mixture. χ_A = moles of A / total moles of all gases. 气体A在混合物中的比例。χ_A = A的摩尔数 / 总气体摩尔数。 - **Partial pressure 分压** (p): The pressure a gas would exert if it alone occupied the container. p_A = χ_A × P_total. 气体A单独占据容器时产生的压强。p_A = χ_A × P_total。 For aA(g) + bB(g) ⇌ cC(g) + dD(g): **Kp = (pC)^c × (pD)^d / (pA)^a × (pB)^b** Worked Example 3 例题3: Kp Calculation

N2(g) + 3H2(g) ⇌ 2NH3(g) at 200 atm total pressure
反应: N2(g) + 3H2(g) ⇌ 2NH3(g)，总压200 atm

At equilibrium: n(N2) = 1.0 mol, n(H2) = 3.0 mol, n(NH3) = 2.0 mol
Total moles 总摩尔数 = 1.0 + 3.0 + 2.0 = 6.0 mol

Mole fractions 摩尔分数:
χ(N2) = 1.0/6.0 = 0.167
χ(H2) = 3.0/6.0 = 0.500
χ(NH3) = 2.0/6.0 = 0.333

Partial pressures 分压:
p(N2) = 0.167 × 200 = 33.3 atm
p(H2) = 0.500 × 200 = 100 atm
p(NH3) = 0.333 × 200 = 66.7 atm

Kp = (pNH3)^2 / (pN2 × pH2^3)
= (66.7)^2 / (33.3 × 100^3)
= 4449 / (33.3 × 1,000,000)
= 4449 / 33,300,000
= 1.34 × 10^-4 atm^-2 **Key exam tips for Kp questions Kp题关键应试技巧**: - Always check that mole fractions sum to 1.0 — if they don't, you've made an error. 务必检查摩尔分数之和是否为1.0——如果不是，说明你算错了。 - Kp units must be derived from the expression; never guess them. Kp单位必须从表达式推导；不要猜测。 - If the total pressure changes during the reaction (e.g., due to temperature change), you must use the final total pressure at equilibrium. 如果反应过程中总压改变（如温度变化），必须使用平衡时的最终总压。 - Heterogeneous equilibria: solids and pure liquids have constant "concentration" and are omitted from both Kc and Kp expressions. 非均相平衡：固体和纯液体"浓度"恒定，从Kc和Kp表达式中省略。 --- ### Part 5: Effect of Temperature on Kc and Kp 温度对Kc和Kp的影响 This is a conceptually important section that Edexcel likes to test. 这是概念上重要的一节，Edexcel喜欢考查。 Exothermic forward reaction 正反应放热 (ΔH < 0)

Increase T: Equilibrium shifts LEFT → Kc decreases
升温：平衡左移→Kc减小

Decrease T: Equilibrium shifts RIGHT → Kc increases
降温：平衡右移→Kc增大 Endothermic forward reaction 正反应吸热 (ΔH > 0)

Increase T: Equilibrium shifts RIGHT → Kc increases
升温：平衡右移→Kc增大

Decrease T: Equilibrium shifts LEFT → Kc decreases
降温：平衡左移→Kc减小 **The van't Hoff relationship 范特霍夫关系** (not required for calculation but useful for understanding): ln(K2/K1) = -(ΔH/R) × (1/T2 - 1/T1). This explains why a plot of ln K against 1/T gives a straight line with slope = -ΔH/R. ln(K2/K1) = -(ΔH/R) × (1/T2 - 1/T1)。这解释了为什么ln K对1/T作图得到一条斜率为-ΔH/R的直线。 **Exam application 考试应用**: If Edexcel gives you Kc values at two different temperatures, you can deduce whether the forward reaction is exothermic or endothermic without any calculation — just look at whether Kc increases or decreases with temperature. 如果Edexcel给你两个温度下的Kc值，你无需计算即可推断正反应是放热还是吸热——只需看Kc随温度是增大还是减小。 --- ### Part 6: Industrial Applications 工业应用 The Haber Process 哈伯法
N2(g) + 3H2(g) ⇌ 2NH3(g)
ΔH = -92 kJ mol^-1

Industrial conditions 工业条件:
Temperature: 400-450 °C
Pressure: 200 atm
Catalyst: Finely divided iron 铁粉

The forward reaction is exothermic. Low temperature favours high yield but makes the rate unacceptably slow. 200 atm favours the product side (4 moles → 2 moles of gas), but higher pressure would be too expensive. The iron catalyst allows a lower operating temperature while maintaining an acceptable rate. 正反应放热。低温有利于高产率但速率过慢无法接受。200 atm有利于生成物侧（4分子气体→2分子气体），但更高压强成本过高。铁催化剂允许在较低温度下维持可接受的速率。 The Contact Process 接触法
2SO2(g) + O2(g) ⇌ 2SO3(g)
ΔH = -197 kJ mol^-1

Industrial conditions 工业条件:
Temperature: 450 °C
Pressure: 1-2 atm
Catalyst: V2O5 (vanadium(V) oxide) 五氧化二钒

Why only 1-2 atm? The equilibrium already lies heavily to the right at atmospheric pressure — Kc is very large. Higher pressure would add cost without meaningful yield improvement. Furthermore, SO3 is then absorbed in concentrated H2SO4 to make oleum, which is diluted to make sulfuric acid. 为什么只需1-2 atm？常压下平衡已经强偏向右侧——Kc非常大。高压增加成本但产率提升有限。此外，SO3随后被浓硫酸吸收制成发烟硫酸，再稀释得到硫酸。 **The compromise question formula 折中条件题答题公式** (5-6 marks): Edexcel loves this question. Use this structure: Edexcel喜欢这道题。用这个结构： 1. State that low temperature increases yield (exothermic) but reduces rate. 说明低温增加产率（放热）但降低速率。 2. State that high pressure increases yield (fewer gas moles in products) but increases cost and safety risk. 说明高压增加产率（生成物气体分子少）但增加成本和安全风险。 3. Explain that the chosen temperature is a compromise between yield and rate. 解释所选温度是产率和速率之间的折中。 4. Explain that the chosen pressure balances yield against economic cost. 解释所选压强平衡了产率和经济成本。 5. Mention the catalyst speeds up both forward and backward reactions equally, allowing a lower temperature to be used. 提到催化剂同等加速正逆反应，允许使用更低温度。 --- ### Part 7: Common Edexcel Question Types 常见Edexcel题型 Type 1: Explain the equilibrium shift (3-4 marks) 解释平衡移动 (3-4分)

Q: "When the temperature is increased, explain what happens to the equilibrium yield of ammonia in the Haber Process."
问："升高温度时，解释哈伯法中氨的平衡产率会如何变化。"

Model answer 标准答案:
"The equilibrium shifts to the left (endothermic direction). The forward reaction is exothermic (ΔH = -92 kJ mol^-1), so the system opposes the temperature increase by favouring the endothermic backward reaction, which absorbs heat. Therefore, the equilibrium yield of ammonia decreases."
"平衡向左（吸热方向）移动。正反应放热(ΔH = -92 kJ mol^-1)，体系通过倾向吸热的逆反应来对抗温度升高，吸收热量。因此，氨的平衡产率下降。"

Note: The ΔH sign is explicitly stated — this is what earns the analysis mark. 注意：明确写出ΔH符号——这是拿分析分的关键。 Type 2: Kc calculation with ICE table (5-6 marks) 用ICE表格的Kc计算 (5-6分)

Q: "2.0 mol of ethyl ethanoate and 2.0 mol of water are mixed. At equilibrium, 0.60 mol of ethanoic acid is present. Calculate Kc."
CH3COOC2H5 + H2O ⇌ CH3COOH + C2H5OH (all in 1.0 dm^3)

Model approach 标准解法:
- Set up ICE table with 4 rows (Initial, Change, Equilibrium) 建立4行ICE表格
- Let x = amount of ethanoic acid formed = 0.60 mol 令x = 生成的乙酸量 = 0.60 mol
- Then ethanol = 0.60, ester remaining = 2.0 - 0.60 = 1.40, water = 2.0 - 0.60 = 1.40
- Kc = [CH3COOH][C2H5OH] / [CH3COOC2H5][H2O] = (0.60)^2 / (1.40)^2 = 0.184
- Units: all concentrations are mol dm^-3; numerator and denominator both (mol dm^-3)^2 → no units Type 3: Effect of catalyst (2 marks) 催化剂的影响 (2分)

Q: "Explain the effect of adding a catalyst on the equilibrium yield of the Contact Process."
问："解释添加催化剂对接触法平衡产率的影响。"

Model answer 标准答案:
"A catalyst has no effect on the position of equilibrium or on the equilibrium yield. It increases the rate of both the forward and backward reactions equally, so the time taken to reach equilibrium decreases, but the equilibrium composition remains unchanged."
"催化剂对平衡位置或平衡产率没有影响。它同等增加正反应和逆反应的速率，因此达到平衡的时间缩短，但平衡组成不变。"

Memorise this exact phrasing — "increases rate of BOTH forward and backward reactions EQUALLY." 背下这个精确表述——"同等增加正反应和逆反应的速率。" --- ### Part 8: Graph Analysis 图像分析 Edexcel frequently asks students to interpret concentration-time and rate-time graphs. Edexcel经常要求学生解读浓度-时间和速率-时间图像。 Concentration-time graph 浓度-时间图

Reactant curve: starts high, decreases, then flattens to a horizontal line at equilibrium.
反应物曲线：起始高，下降，然后到平衡时趋于水平。

Product curve: starts at zero, increases, then flattens at equilibrium.
生成物曲线：从零开始，上升，然后到平衡时趋于水平。

A horizontal line means equilibrium has been reached. 水平线意味着达到平衡。 Rate-time graph 速率-时间图

Forward rate curve: starts high, decreases, and meets the backward rate curve at equilibrium.
正反应速率曲线：起始高，下降，在平衡处与逆反应速率曲线相遇。

Backward rate curve: starts at zero, increases, and meets the forward rate curve at equilibrium.
逆反应速率曲线：从零开始，上升，在平衡处与正反应速率曲线相遇。

At equilibrium, the two curves merge into one horizontal line. 平衡时，两条曲线合并为一条水平线。 **When conditions change 条件改变时**: - A sudden increase in reactant concentration → forward rate jumps up, then both rates readjust to a new equilibrium. 反应物浓度突然增加→正反应速率跳升，然后两者重新调至新平衡。 - A temperature increase → both rates jump up, but the endothermic direction jumps more → curves readjust. 温度升高→两个速率都跳升，但吸热方向升得更多→曲线重新调整。 - Adding a catalyst → both rates jump up equally → faster to equilibrium, but same final position. 加催化剂→两个速率同等跳升→更快达平衡，但最终位置不变。 --- ### Part 9: Summary and Exam Checklist 总结与考前清单 Before the exam, make sure you can 考前确保你能做到:

✅ Define dynamic equilibrium and state its three characteristics. 定义动态平衡并说明其三大特征。
✅ State Le Chatelier's Principle and apply it to any change in temperature, pressure, or concentration. 陈述勒夏特列原理并应用于任何温度、压强或浓度的变化。
✅ Write Kc and Kp expressions for any reaction, excluding solids and pure liquids. 为任何反应写出Kc和Kp表达式，排除固体和纯液体。
✅ Construct ICE tables and solve for unknown concentrations. 构建ICE表格并求解未知浓度。
✅ Calculate mole fractions and partial pressures from given data. 从给定数据计算摩尔分数和分压。
✅ Derive the units of Kc and Kp from their expressions. 从表达式推导Kc和Kp的单位。
✅ Explain the Haber and Contact process conditions using Le Chatelier's Principle and economic reasoning. 用勒夏特列原理和经济理由解释哈伯法和接触法的条件。
✅ Interpret concentration-time and rate-time graphs, including responses to condition changes. 解读浓度-时间和速率-时间图像，包括对条件变化的响应。
✅ State precisely why a catalyst does NOT affect equilibrium position or constant. 精确说明为什么催化剂不影响平衡位置或常数。 --- Equilibrium is one of the most logical and rewarding topics in A-Level Chemistry. Once you understand that the system always acts to oppose any change, every prediction follows naturally. The calculations are systematic — master the ICE table and you will never lose a mark on Kc or Kp questions. Practise past papers, memorise the answer templates, and you will walk into the exam with confidence. 化学平衡是A-Level化学中最富有逻辑性、回报率最高的章节之一。一旦你理解了体系总是对抗任何改变，所有的预测都水到渠成。计算也是系统性的——掌握了ICE表格，你在Kc和Kp题上就不会丢分。多练真题，熟记答题模板，你就能自信地走进考场。 Good luck with your Edexcel Chemistry exam! 祝你的Edexcel化学考试顺利！ --- 🎓 Want personalised A-Level Chemistry help? 想要个性化A-Level化学辅导？
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