Tag: Chemistry

📚 IB Chemistry: Redox Processes | IB化学：氧化还原反应过程

Redox (reduction–oxidation) reactions involve the transfer of electrons between chemical species. These processes are fundamental to chemistry, biology and industry, from cellular respiration to battery technology. In IB Chemistry, students learn to identify, balance and apply redox reactions through the systematic use of oxidation numbers and half equations.

氧化还原反应指化学物种之间发生电子转移的反应。此类过程是化学、生物及工业的基础，涉及细胞呼吸到电池技术等方方面面。在 IB 化学中，学生通过系统地使用氧化数和半反应方程来识别、配平及应用氧化还原反应。

1. Oxidation and Reduction Definitions | 氧化与还原的定义

Oxidation is the loss of electrons, while reduction is the gain of electrons. The mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain) helps remember the electron-transfer definitions. Oxidation can also be described as an increase in oxidation number, and reduction as a decrease in oxidation number.

氧化是失电子，还原是得电子。记忆口诀 OIL RIG（Oxidation Is Loss, Reduction Is Gain）有助于牢记电子转移的定义。氧化也可描述为氧化数升高，还原描述为氧化数降低。

For example, when iron(III) ions react with iodide ions, iron(III) is reduced to iron(II) and iodide is oxidised to iodine.

例如，铁(III)离子与碘离子反应时，铁(III)被还原为铁(II)，碘离子被氧化为碘单质。

2. Oxidation Numbers | 氧化数的确定

Oxidation numbers are assigned by a set of rules: all free elements have oxidation number 0; a monatomic ion has an oxidation number equal to its charge; fluorine is always –1 in compounds; oxygen is usually –2 (except in peroxides where it is –1, and in OF₂ where it is +2); hydrogen is +1 when bonded to non-metals and –1 when bonded to metals; the sum of oxidation numbers equals the overall charge of the molecule or ion.

氧化数的确定遵循一系列规则：单质中元素的氧化数为 0；单原子离子的氧化数等于其电荷；氟在化合物中恒为 –1；氧通常为 –2（但在过氧化物中为 –1，在 OF₂ 中为 +2）；氢与非金属结合时为 +1，与金属结合时为 –1；分子或离子的氧化数总和等于其总电荷。

For example, in the dichromate ion Cr₂O₇²⁻, the overall charge is –2. Since each oxygen is –2, the seven oxygen atoms contribute –14. Therefore, the two chromium atoms must sum to +12, so each chromium has oxidation number +6.

例如，在重铬酸根离子 Cr₂O₇²⁻ 中，总电荷为 –2。每个氧为 –2，七个氧原子共贡献 –14。因此，两个铬原子必须合计为 +12，故每个铬的氧化数为 +6。

3. Recognising Redox Reactions | 识别氧化还原反应

A reaction is redox if the oxidation numbers of elements change between reactants and products. For example, in the reaction between copper(II) oxide and hydrogen, CuO + H₂ → Cu + H₂O, copper is reduced (+2 to 0) and hydrogen is oxidised (0 to +1). Identifying these changes allows us to determine which species is oxidised and which is reduced.

如果反应前后元素的氧化数发生变化，则该反应是氧化还原反应。例如，氧化铜与氢气的反应 CuO + H₂ → Cu + H₂O 中，铜被还原（从 +2 到 0），氢被氧化（从 0 到 +1）。通过识别氧化数变化，可判断哪种物质被氧化、哪种被还原。

Some reactions that appear to involve atom transfer are also redox. For instance, in Fe₂O₃ + 3CO → 2Fe + 3CO₂, iron is reduced and carbon is oxidised, even though no simple electron transfer is visible at first glance.

一些看似涉及原子转移的反应同样是氧化还原反应。例如，Fe₂O₃ + 3CO → 2Fe + 3CO₂ 中，铁被还原、碳被氧化，尽管乍看之下没有明显的电子转移。

4. Oxidising and Reducing Agents | 氧化剂与还原剂

An oxidising agent is a species that causes oxidation by being reduced itself; it accepts electrons. A reducing agent causes reduction by being oxidised itself; it donates electrons. Examples of strong oxidising agents include O₂, halogens, Fe³⁺, MnO₄⁻ and Cr₂O₇²⁻. Common reducing agents include H₂, Na, I⁻ and Fe²⁺.

氧化剂是能使其他物质氧化的物种，其自身被还原（得到电子）。还原剂是能使其他物质还原的物种，其自身被氧化（失去电子）。强氧化剂的例子包括 O₂、卤素、Fe³⁺、MnO₄⁻ 和 Cr₂O₇²⁻。常见还原剂包括 H₂、Na、I⁻ 和 Fe²⁺。

In the reaction between acidified permanganate and iron(II) sulfate, MnO₄⁻ is reduced to Mn²⁺, so permanganate is the oxidising agent; Fe²⁺ is oxidised to Fe³⁺, so iron(II) is the reducing agent.

在酸化高锰酸钾与硫酸亚铁的反应中，MnO₄⁻ 被还原为 Mn²⁺，因此高锰酸根是氧化剂；Fe²⁺ 被氧化为 Fe³⁺，因此亚铁离子是还原剂。

5. Writing Half Equations | 书写半反应方程

Half equations show oxidation or reduction separately. To balance half equations in acidic medium: balance atoms other than H and O, then balance O by adding H₂O, balance H by adding H⁺, and finally balance charge by adding electrons. In basic medium, use OH⁻ and H₂O, or add H⁺ first and then neutralise with OH⁻.

半反应方程单独表示氧化或还原。在酸性介质中配平半反应：先平衡除 H 和 O 外的原子，然后用 H₂O 平衡 O，用 H⁺ 平衡 H，最后用电子 e⁻ 平衡电荷。在碱性介质中则使用 OH⁻ 和 H₂O，也可先加 H⁺ 再中和。

Example, reduction of permanganate to manganese(II) in acid:

例如，酸性条件下高锰酸根被还原为锰(II)离子：

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Example, oxidation of iodide to iodine:

例如，碘离子被氧化为碘单质：

2I⁻ → I₂ + 2e⁻

6. Combining Half Equations | 合并半反应方程

To build a full redox equation, combine the oxidation and reduction half equations after multiplying by appropriate factors so that the number of electrons lost equals the number gained. For example, the oxidation of Fe²⁺ to Fe³⁺ is Fe²⁺ → Fe³⁺ + e⁻. Multiplying this by 5 and adding it to the acidic permanganate half equation gives:

要构造完整的氧化还原方程式，需将氧化半反应和还原半反应乘以适当系数后相加，使失去电子数等于得到电子数。例如，Fe²⁺ 氧化为 Fe³⁺ 的半反应为 Fe²⁺ → Fe³⁺ + e⁻。将其乘以 5，再与酸性高锰酸根半反应相加，得：

5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Check that atoms and charges are balanced: left side has 6 positive charges from 5Fe²⁺ and 8H⁺, plus 1 negative from MnO₄⁻, giving a net charge of +13; right side has 5Fe³⁺ (+15) and Mn²⁺ (+2), giving +17? Wait, careful: the net charge calculation should be +13 on both sides. Actually, the left net charge is 5(+2) + 8(+1) + 1(-1) = +17. Wait, 10+8-1=17. The right side: 5(+3) + 1(+2) = +17. Yes balanced. Need to correct this in text. Let’s write correctly.

检查原子和电荷是否守恒：左侧净电荷为 5×( +2) + 8×( +1) + 1×( –1) = +17；右侧为 5×( +3) + 1×( +2) = +17，电荷守恒。

In balanced equations, water molecules and H⁺ ions often appear as products or reactants depending on the acidity of the medium. In basic conditions, OH⁻ may be added to neutralise excess H⁺.

在配平方程中，水分子和 H⁺ 离子通常作为产物或反应物出现，具体取决于介质的酸碱性。在碱性条件下，可加入 OH⁻ 来中和多余的 H⁺。

7. Redox Titration and Stoichiometry | 氧化还原滴定与化学计量

Redox titrations use a known concentration of an oxidising or reducing agent to determine the amount of an analyte. Potassium permanganate is a common titrant because its purple colour disappears in the presence of reducing agents, so it acts as its own indicator. Starch is often used as an indicator for iodine titrations.

氧化还原滴定使用已知浓度的氧化剂或还原剂来测定分析物的含量。高锰酸钾是常用滴定剂，因为其紫色在还原剂存在时褪色，可作为自身指示剂。碘量法滴定中常使用淀粉作指示剂。

Stoichiometric calculations use the relationship n = c × V, then apply the mole ratio from the balanced redox equation. For example, 20.0 cm³ of 0.100 mol dm⁻³ iron(II) solution reacts with acidified permanganate according to the 5:1 ratio. The amount of Fe²⁺ is 0.00200 mol, so the amount of MnO₄⁻ is 0.000400 mol. If the titre volume is 10.0 cm³, the concentration of MnO₄⁻ is:

化学计量计算使用关系式 n = c × V，再应用配平氧化还原方程中物质的量的比例。例如，20.0 cm³ 的 0.100 mol dm⁻³ 亚铁溶液与酸化高锰酸钾按 5:1 比例反应。Fe²⁺ 的物质的量为 0.00200 mol，因此 MnO₄⁻ 的物质的量为 0.000400 mol。若滴定体积为 10.0 cm³，则 MnO₄⁻ 的浓度为：

c(MnO₄⁻) = 0.000400 mol / 0.0100 dm³ = 0.0400 mol dm⁻³

8. Electrochemical Cells | 电化学电池

Electrochemical cells convert chemical energy into electrical energy. In a galvanic cell, oxidation occurs at the anode (the negative electrode) and reduction occurs at the cathode (the positive electrode). The salt bridge completes the circuit by allowing ions to flow and maintains electrical neutrality in the half-cells.

电化学电池将化学能转化为电能。在原电池中，氧化发生在阳极（负极），还原发生在阴极（正极）。盐桥通过使离子流动来闭合电路，并维持各半电池中的电荷平衡。

The shorthand cell diagram follows the convention: anode half-cell on the left, cathode half-cell on the right, with a double vertical line for the salt bridge. For a zinc–copper cell, the diagram is:

电池简式的约定：阳极半电池写在左侧，阴极半电池写在右侧，双竖线代表盐桥。对于锌铜电池，简式为：

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

The electrode reaction at the zinc electrode is Zn(s) → Zn²⁺(aq) + 2e⁻; at the copper electrode, Cu²⁺(aq) + 2e⁻ → Cu(s).

锌电极上的电极反应为 Zn(s) → Zn²⁺(aq) + 2e⁻；铜电极上为 Cu²⁺(aq) + 2e⁻ → Cu(s)。

9. Standard Electrode Potentials | 标准电极电势

Standard electrode potential (E°) measures a half-cell’s tendency to be reduced under standard conditions (1 mol dm⁻³ solutions, 298 K, 1 atm gases). The more positive the E°, the stronger the oxidising agent. Standard hydrogen electrode (SHE) is assigned E° = 0.00 V and is used as the reference half-cell.

标准电极电势（E°）衡量半电池在标准条件（溶液 1 mol dm⁻³，298 K，气体 1 atm）下被还原的倾向。E° 越正，氧化能力越强。标准氢电极（SHE）的 E° 被规定为 0.00 V，作为参比半电池。

The overall cell potential is calculated as E°cell = E°cathode – E°anode. For the zinc–copper cell:

电池总电动势的计算式为 E°cell = E°cathode – E°anode。对于锌铜电池：

E°cell = (+0.34 V) – (–0.76 V) = +1.10 V

A positive E°cell indicates a spontaneous redox reaction under standard conditions.

若 E°cell 为正，则表明氧化还原反应在标准条件下可以自发进行。

10. Predicting Feasibility of Reactions | 预测反应能否进行

Using standard electrode potentials, we can predict whether a redox reaction will occur. Compare the two half reactions: the half reaction with the more positive E° will proceed as reduction, while the one with the less positive E° will proceed as oxidation. The resulting E°cell must be positive for the reaction to be feasible.

利用标准电极电势可以预测氧化还原反应能否发生。比较两个半反应：E° 较正的半反应发生还原，E° 较负的半反应发生氧化。最终计算得到的 E°cell 必须为正，反应才可行。

For example, acidified permanganate (E° = +1.51 V) can oxidise Fe²⁺ (Fe³⁺/Fe²⁺ E° = +0.77 V) because the overall cell potential is 1.51 – 0.77 = +0.74 V. Therefore, the reaction is spontaneous.

例如，酸化高锰酸根（E° = +1.51 V）能氧化 Fe²⁺（Fe³⁺/Fe²⁺ E° = +0.77 V），因为电池总电动势为 1.51 – 0.77 = +0.74 V，反应自发进行。

Keep in mind that a positive E° predicts thermodynamic feasibility, not the rate of reaction. Many thermodynamically feasible reactions are slow without a catalyst.

需要注意，E° 为正仅预测热力学可行性，并不代表反应速率。许多热力学可行的反应在无催化剂时速率很慢。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Hess’s Law Principles & Calculation Skills | IB化学：盖斯定律原理与计算技巧

Hess’s Law is one of the core ideas in IB Chemistry energetics. It states that the enthalpy change of a reaction depends only on the initial and final states, not on the route taken. This article explains why this rule works and shows you step-by-step methods to solve Hess’s Law problems.

盖斯定律是IB化学能量学中的核心概念之一。它指出：一个反应的焓变只取决于反应的始态和终态，而与反应途径无关。本文将从原理出发，手把手演示如何用盖斯定律进行准确计算。

1. What Is Hess’s Law? | 什么是盖斯定律？

Hess’s Law says: the enthalpy change of a reaction is the same whether the reaction takes place in one step or in several steps, provided the initial and final conditions are identical.

盖斯定律指出：在初始和最终条件相同的情况下，无论反应一步完成还是分几步完成，反应的总焓变都相同。

This law is also called the law of constant heat summation. It was proposed by the Russian chemist Germain Hess in 1840. For example, the enthalpy change of converting graphite to diamond can be calculated indirectly using combustion data, because direct conversion is difficult to perform in a laboratory.

这一定律又称“热量总和守恒定律”，由俄国化学家盖斯于1840年提出。例如，石墨转化为金刚石的焓变难以直接在实验室中测定，但可以通过燃烧焓数据间接计算。

2. Why Does the Route Not Matter? | 为什么反应途径不影响焓变？

Enthalpy H is a state function. A state function depends only on the current state of the system, not on how that state was reached. Therefore, the change ΔH = H(products) − H(reactants) only uses the start and end states.

焓H是一个状态函数。状态函数只由系统当前的状态决定，而不管系统是如何到达该状态的。因此，ΔH = H(生成物) − H(反应物) 只取决于始态和终态。

A useful analogy is hiking: the altitude difference between two peaks depends only on their heights, not on the trail you choose. Similarly, the enthalpy difference depends only on the initial and final chemical states.

一个有用的类比是登山：两座山峰之间的高度差只取决于它们的高度，而与所走的路线无关。同样，焓差也只取决于起始和最终的化学状态。

If two different routes gave different enthalpy changes, we could go one way and return the other to create or destroy energy. This would violate the

Published by TutorHao | IB Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Equilibrium Constant and Equilibrium Law Calculations in IB Chemistry HL | IB化学HL：平衡常数与平衡定律计算

The equilibrium constant is one of the most powerful concepts in chemistry. It tells us the extent to which a reaction proceeds, predicts the direction of a reaction, and enables quantitative analysis of equilibrium systems. In IB Chemistry HL, mastering equilibrium constant calculations is essential for success in both Paper 1 and Paper 2.

平衡常数是化学中最有力的概念之一。它告诉我们反应进行的程度，预测反应的方向，并能够对平衡体系进行定量分析。在IB化学HL中，掌握平衡常数的计算是Paper 1和Paper 2取得好成绩的关键。

1. The Concept of Chemical Equilibrium | 化学平衡的概念

A chemical equilibrium is established when the rates of the forward and reverse reactions become equal, and the concentrations of reactants and products remain constant over time. The system is said to be in a state of dynamic equilibrium because both forward and reverse reactions continue to occur at the same rate.

当正反应和逆反应的速率相等，反应物和生成物的浓度随时间保持不变时，就建立了化学平衡。该系统处于动态平衡状态，因为正反应和逆反应仍在以相同的速率持续进行。

• Equilibrium is only possible in closed systems.

  平衡只在封闭体系中才能实现。

• Equilibrium can be approached from either direction — starting with only reactants or only products.

  平衡可以从任意方向趋近——无论是只从反应物开始还是只从生成物开始。

• Equilibrium is dynamic, not static — molecules constantly interconvert.

  平衡是动态的而非静态的——分子在不断相互转化。

2. The Law of Mass Action | 质量作用定律

The law of mass action, formulated by Guldberg and Waage in 1864, states that at a given temperature, the rate of a chemical reaction is proportional to the product of the active masses (concentrations) of the reactants, each raised to the power of its stoichiometric coefficient. For a general reaction at equilibrium:

质量作用定律由古德贝格和瓦格于1864年提出，该定律指出在给定温度下，化学反应速率与各反应物活度（浓度）以其化学计量系数为幂次的乘积成正比。对于达到平衡的通用反应：

aA + bB ⇌ cC + dD

The equilibrium expression is written as:

平衡表达式写为：

K_c = [C]^c[D]^d / [A]^a[B]^b

• K_c is the equilibrium constant expressed in terms of molar concentrations (mol dm⁻³).

  K_c 是以物质的量浓度（mol dm⁻³）表示的平衡常数。

• Square brackets [ ] denote molar concentration.

  方括号 [ ] 表示摩尔浓度。

• Only aqueous species and gases appear in the equilibrium expression; pure solids and pure liquids are excluded.

  只有水溶液中的物种和气体出现在平衡表达式中；纯固体和纯液体不包括在内。

3. The Equilibrium Constant K_c | 平衡常数 K_c

The equilibrium constant K_c is a dimensionless quantity that characterises the position of equilibrium at a specific temperature. A large K_c (K_c >> 1) indicates that the equilibrium lies far to the right, favouring products. A small K_c (K_c << 1) indicates that the equilibrium lies to the left, favouring reactants.

平衡常数K_c是一个无量纲的量，它表征在特定温度下平衡的位置。大的K_c（K_c >> 1）表示平衡强烈偏向右边，有利于生成物；小的K_c（K_c << 1）表示平衡偏向左边，有利于反应物。

The equilibrium constant depends only on temperature. It is unaffected by changes in concentration, pressure, or the presence of a catalyst. This is a crucial point frequently tested in IB exams.

平衡常数仅取决于温度。它不受浓度、压力变化或催化剂存在的影响。这是IB考试中经常考查的一个关键点。

An exception: if the reaction is reversed, the new K_c is the reciprocal of the original. If the reaction is multiplied by a factor n, the new K_c is raised to the power n. If two reactions are added, their K values are multiplied.

一个例外：如果反应方向反转，新的K_c是原来K_c的倒数。如果反应方程乘以因子n，新的K_c将取n次幂。如果两个反应相加，它们的K值相乘。

4. K_p: The Equilibrium Constant for Gases | K_p：气态平衡常数

For reactions involving gases, the equilibrium constant can be expressed in terms of partial pressures. For the general reaction:

对于涉及气体的反应，平衡常数可以用分压来表示。对于通用反应：

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K_p = (P_C)^c(P_D)^d / (P_A)^a(P_B)^b

where P_A is the partial pressure of gas A. Partial pressure is calculated by multiplying the mole fraction of a gas by the total pressure of the system:

其中P_A是气体A的分压。分压通过将气体的摩尔分数乘以体系的总压来计算：

• Partial pressure of A = mole fraction of A × total pressure

  A的分压 = A的摩尔分数 × 总压

• P_A = (n_A/n_total) × P_total

  P_A = (n_A/n_total) × P_total

The relationship between K_p and K_c is given by:

K_p与K_c之间的关系为：

K_p = K_c(RT)^Δn

where Δn = (c + d) − (a + b), R is the ideal gas constant (8.31 J mol⁻¹ K⁻¹), and T is the temperature in kelvin.

其中 Δn = (c + d) − (a + b)，R是理想气体常数（8.31 J mol⁻¹ K⁻¹），T是以开尔文为单位的温度。

5. The Reaction Quotient Q | 反应商 Q

The reaction quotient Q has the same mathematical form as the equilibrium constant, but it is calculated using concentrations or pressures that are not necessarily at equilibrium. It is used to predict the direction in which a reaction will proceed to reach equilibrium.

反应商Q与平衡常数具有相同的数学形式，但它使用的是不一定处于平衡状态的浓度或压力。它用于预测反应向哪个方向进行以达到平衡。

• If Q < K: the reaction will proceed in the forward direction to form more products.

  如果Q < K：反应将向正方向进行，生成更多产物。

• If Q = K: the system is at equilibrium.

  如果Q = K：体系处于平衡状态。

• If Q > K: the reaction will proceed in the reverse direction to form more reactants.

  如果Q > K：反应将向逆方向进行，生成更多反应物。

6. Stoichiometric Setup for Equilibrium Calculations | 平衡计算的化学计量设列

Equilibrium calculation problems typically require an ICE table (Initial, Change, Equilibrium). This systematic approach is crucial for solving IB equilibrium problems accurately.

平衡计算问题通常需要使用ICE表格（初始、变化、平衡）。这种系统性的方法对于准确解决IB平衡问题至关重要。

The coefficients m, n, p, q are the stoichiometric coefficients from the balanced equation. The variable x represents the extent of reaction in terms of concentration change. Always use the mole ratio from the balanced equation to determine the change row.

系数m、n、p、q来自配平方程式的化学计量系数。变量x表示反应进行程度（以浓度变化表示）。始终使用配平方程式中的摩尔比来确定变化行。

7. Worked Example: Calculating K_c from Initial and Equilibrium Concentrations | 例题：从初始浓度和平衡浓度计算K_c

Problem: 0.50 mol of H₂ and 0.50 mol of I₂ are placed in a 1.00 dm³ flask at 700 K. At equilibrium, 0.78 mol of HI is present. Calculate K_c.

题目：将0.50 mol H₂和0.50 mol I₂放入1.00 dm³的烧瓶中，温度为700 K。平衡时，有0.78 mol HI存在。计算K_c。

Solution: The balanced equation is:

解答：配平的方程式为：

H₂(g) + I₂(g) ⇌ 2HI(g)

Since the volume is 1.00 dm³, initial concentrations are: [H₂]₀ = 0.50 mol dm⁻³, [I₂]₀ = 0.50 mol dm⁻³, [HI]₀ = 0. The equilibrium concentration of HI is 0.78 mol dm⁻³.

由于体积为1.00 dm³，初始浓度为：[H₂]₀ = 0.50 mol dm⁻³，[I₂]₀ = 0.50 mol dm⁻³，[HI]₀ = 0。HI的平衡浓度为0.78 mol dm⁻³。

From the stoichiometry, 2 moles of HI are produced for every 1 mole of H₂ consumed. Therefore:

根据化学计量关系，每消耗1摩尔H₂会产生2摩尔HI。因此：

[H₂] consumed = [I₂] consumed = 0.78/2 = 0.39 mol dm⁻³

[H₂] = 0.50 − 0.39 = 0.11 mol dm⁻³
[I₂] = 0.50 − 0.39 = 0.11 mol dm⁻³
[HI] = 0.78 mol dm⁻³

Now substitute into the equilibrium expression:

现在代入平衡表达式：

K_c = [HI]² / ([H₂][I₂]) = (0.78)² / (0.11 × 0.11) = 0.6084 / 0.0121 = 50.3

Thus K_c = 50.3 (no units). Note that the magnitude of K_c > 1 confirms the equilibrium favours products.

因此K_c = 50.3（无单位）。注意K_c > 1的量级确认平衡有利于生成物。

8. Worked Example: Calculating Extent of Reaction from K_c | 例题：从K_c计算反应进行程度

Problem: For the reaction N₂O₄(g) ⇌ 2NO₂(g), K_c = 0.210 at 300 K. If 1.00 mol of N₂O₄ is placed in a 2.00 dm³ container, calculate the concentration of NO₂ at equilibrium.

题目：对于反应N₂O₄(g) ⇌ 2NO₂(g)，在300 K时K_c = 0.210。如果将1.00 mol N₂O₄放入2.00 dm³容器中，计算平衡时NO₂的浓度。

Solution: Initial concentration of N₂O₄ = 1.00/2.00 = 0.500 mol dm⁻³. Let x be the concentration of N₂O₄ that decomposes. Then:

解答：N₂O₄的初始浓度 = 1.00/2.00 = 0.500 mol dm⁻³。设x为分解的N₂O₄的浓度。则：

[N₂O₄] = 0.500 − x
[NO₂] = 2x

Substituting into the equilibrium expression:

代入平衡表达式：

K_c = (2x)² / (0.500 − x) = 0.210

4x² = 0.210(0.500 − x)
4x² = 0.105 − 0.210x
4x² + 0.210x − 0.105 = 0

Using the quadratic formula:

使用二次方程求根公式：

x = [−0.210 + √(0.210² + 4 × 4 × 0.105)] / (2 × 4)

x = [−0.210 + √(0.0441 + 1.68)] / 8 = [−0.210 + 1.313] / 8 = 0.138

The positive root is taken. Therefore [NO₂] = 2x = 2 × 0.138 = 0.276 mol dm⁻³.

取正根。因此[NO₂] = 2x = 2 × 0.138 = 0.276 mol dm⁻³。

Note: when K is small and the initial concentration is large, you may approximate 0.500 − x ≈ 0.500, but the IB syllabus requires you to know when such approximation is valid and when it is not. Always check whether the approximation error is acceptable.

注意：当K值很小且初始浓度较大时，可以近似认为0.500 − x ≈ 0.500，但IB大纲要求你知道何时这种近似是有效的，何时不是。始终检查近似误差是否可接受。

9. Equilibrium Calculations with Halogens: A Classic IB Problem | 卤素平衡计算：经典IB题型

Consider the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with K_c = 0.0411 at a certain temperature. If 0.100 mol of PCl₅ is placed in a 1.00 dm³ flask, find the equilibrium concentrations of all species.

考虑反应PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)，在某温度下K_c = 0.0411。如果将0.100 mol PCl₅放入1.00 dm³烧瓶中，求所有物种的平衡浓度。

Solution: ICE Table (all concentrations in mol dm⁻³):

解答：ICE表格（所有浓度单位为mol dm⁻³）：

K_c = [PCl₃][Cl₂] / [PCl₅] = x² / (0.100 − x) = 0.0411

Using the quadratic formula:

使用二次方程求根公式：

x² + 0.0411x − 0.00411 = 0

x = [−0.0411 + √(0.00169 + 0.01644)] / 2 = [−0.0411 + 0.1346] / 2 = 0.04675

Therefore: [PCl₃] = [Cl₂] = 0.0468 mol dm⁻³, [PCl₅] = 0.100 − 0.0468 = 0.0532 mol dm⁻³.

因此：[PCl₃] = [Cl₂] = 0.0468 mol dm⁻³，[PCl₅] = 0.100 − 0.0468 = 0.0532 mol dm⁻³。

You can verify by substituting back: 0.0468²/0.0532 = 0.0412 ≈ 0.0411. ✔

你可以代回验证：0.0468²/0.0532 = 0.0412 ≈ 0.0411。✔

10. Factors Affecting the Equilibrium Constant | 影响平衡常数的因素

The only factor that changes the value of K_c or K_p is temperature. The effect follows the principle that equilibrium shifts in the endothermic direction when temperature increases, and in the exothermic direction when temperature decreases.

唯一能改变K_c或K_p数值的因素是温度。这种效应遵循如下原理：温度升高时平衡向吸热方向移动，温度降低时平衡向放热方向移动。

• For an endothermic forward reaction (ΔH > 0): increasing T increases K.

  对于正向吸热的反应（ΔH > 0）：升高温度会增大K。

• For an exothermic forward reaction (ΔH < 0): increasing T decreases K.

  对于正向放热的反应（ΔH < 0）：升高温度会减小K。

The van’t Hoff equation relates K to temperature quantitatively:

范特霍夫方程将K与温度定量关联：

ln(K₂/K₁) = −ΔH°/R × (1/T₂ − 1/T₁)

Although the van’t Hoff equation is not explicitly required in the IB HL syllabus for calculation, understanding the qualitative relationship between K and temperature is essential for multiple-choice and short-answer questions.

虽然IB HL大纲不要求用范特霍夫方程进行计算，但理解K与温度之间的定性关系对于选择题和简答题至关重要。

11. Common Pitfalls and Examination Tips | 常见错误与考试技巧

Students often lose marks in equilibrium calculations due to avoidable mistakes. Here are the most common pitfalls and how to avoid them:

学生在平衡计算中常常因可避免的错误而丢分。以下是最常见的陷阱以及如何避免它们：

• Omitting pure solids and liquids: Species in solid or pure liquid state are excluded from the equilibrium expression. Only gases and aqueous species appear in K in terms of concentration.

  忽略纯固体和纯液体：固态和纯液态物种不包括在平衡表达式中。在浓度平衡常数中，只有气体和水溶液物种出现。

• Incorrect stoichiometric coefficients: The powers in the equilibrium expression must match the coefficients in the balanced equation exactly.

  错误的化学计量系数：平衡表达式中的幂次必须与配平方程式中的系数完全一致。

• Forgetting units: In IB, K values are generally treated as dimensionless; however, you should be aware that the equilibrium expression yields units when the sum of coefficients differs between products and reactants.

  忘记单位：在IB中，K值通常被视为无量纲；然而，你应该意识到当产物和反应物的系数总和不同时，平衡表达式会产生单位。

• Choosing the wrong root of the quadratic: Always select the positive root that gives physically meaningful (positive) concentrations.

  选择二次方程错误的根：始终选择能给出物理上有意义（正值）浓度的正根。

• Confusing Q with K: Q is calculated at any point in time; K only at equilibrium. Using Q when K is required (or vice versa) is a fatal error.

  混淆Q与K：Q在任何时间点都可计算；K只在平衡时存在。在需要K时使用Q（或反之）是致命错误。

12. Summary and Examination Strategy | 总结与应试策略

Equilibrium constant calculations follow a predictable pattern. To maximise your marks, follow this systematic approach:

平衡常数的计算遵循可预测的规律。为了最大化你的分数，请遵循以下系统化步骤：

1. Write the balanced chemical equation.

   写出配平的化学方程式。

2. Identify the phase of each species — exclude solids and pure liquids from the K expression.

   确定每种物质的相态——从K表达式中排除固体和纯液体。

3. Write down the equilibrium expression — check the stoichiometric coefficients.

   写出平衡表达式——检查化学计量系数。

4. Set up an ICE table with all concentrations in the same units (usually mol dm⁻³).

   建立ICE表格，确保所有浓度使用相同单位（通常为mol dm⁻³）。

5. Substitute equilibrium concentrations into the K expression.

   将平衡浓度代入K表达式。

6. Solve for x — use the quadratic formula if necessary; choose the physically meaningful root.

   求解x——必要时使用二次方程求根公式；选择物理上有意义的根。

7. Verify your answer: substitute back into the equilibrium expression to confirm K matches.

   验证答案：代回平衡表达式以确认K值吻合。

8. State the answer with correct significant figures and appropriate units (if any).

   以正确的有效数字和适当的单位（如有）表述答案。

In Paper 2, equilibrium calculation questions typically carry 4–7 marks. The marks are allocated for each step: balanced equation, K expression, ICE setup, algebraic solution, and final answer with verification. Show all your working — partial credit is awarded generously.

在Paper 2中，平衡计算题通常占4–7分。分数分配到每个步骤：方程式、K表达式、ICE建立、代数求解以及带验证的最终答案。展示你所有的解题过程——部分分是会慷慨给出的。

Mastering equilibrium calculations requires practice. Work through as many past-paper questions as possible, paying attention to the mark schemes to understand how marks are allocated.

掌握平衡计算需要大量练习。尽量多做历年真题，关注评分标准以理解分数如何分配。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Core Concepts of Acid-Base Theories | IB化学：酸碱理论核心知识梳理

Acid-base chemistry is one of the most central topics in IB Chemistry, bridging stoichiometry, equilibrium, and organic reaction mechanisms. This article provides a systematic review of the core acid-base theories, from the classic Arrhenius model to the more general Brønsted-Lowry and Lewis frameworks, and highlights the key definitions, conjugate pairs, amphiprotic species, and acid strength concepts you need for your exams.

酸碱化学是IB化学中最核心的主题之一，它将化学计量、化学平衡和有机反应机理紧密联系起来。本文系统梳理了酸碱理论的核心内容，从经典的阿伦尼乌斯模型到更广泛的布朗斯特-劳里和路易斯框架，并重点讲解考试中必需的的定义、共轭对、两性物种和酸强度概念。

1. Arrhenius Theory | 阿伦尼乌斯理论

The Arrhenius theory defines acids as substances that produce H⁺ (or H₃O⁺) ions in aqueous solution, and bases as substances that produce OH⁻ ions in aqueous solution. For example, HCl dissociates to give H⁺ and Cl⁻, while NaOH dissociates to give Na⁺ and OH⁻.

阿伦尼乌斯理论将酸定义为在水中产生H⁺（或H₃O⁺）的物质，将碱定义为在水中产生OH⁻的物质。例如，HCl解离产生H⁺和Cl⁻，而NaOH解离产生Na⁺和OH⁻。

HA → H⁺ + A⁻     BOH → B⁺ + OH⁻

This theory is limited because it only applies to aqueous solutions and cannot explain the basicity of substances like NH₃, which does not contain OH⁻ ions but still acts as a base in water.

该理论仅适用于水溶液，无法解释像NH₃这类不含OH⁻但仍在水中显碱性的物质，因此具有很大的局限性。

2. Brønsted-Lowry Theory | 布朗斯特-劳里理论

The Brønsted-Lowry theory expands the definition: an acid is a proton (H⁺) donor, and a base is a proton acceptor. This theory works in any solvent and explains the behavior of NH₃ in water: NH₃ accepts a proton from H₂O to form NH₄⁺ and OH⁻.

布朗斯特-劳里理论扩展了定义：酸是质子的给予体，碱是质子的接受体。该理论适用于任何溶剂，并解释了NH₃在水中的行为：NH₃从H₂O接受一个质子生成NH₄⁺和OH⁻。

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

In this reaction, H₂O acts as an acid because it donates a proton, while NH₃ acts as a base because it accepts a proton. The reverse reaction is also acid-base: NH₄⁺ donates a proton and OH⁻ accepts it.

在该反应中，H₂O作为酸给出质子，NH₃作为碱接受质子。逆反应同样是酸碱反应：NH₄⁺给出质子，OH⁻接受质子。

3. Conjugate Acid-Base Pairs | 共轭酸碱对

A conjugate acid-base pair consists of two species that differ by one proton (H⁺). The acid has one more H⁺ than its conjugate base; the base has one fewer H⁺ than its conjugate acid.

共轭酸碱对由相差一个质子（H⁺）的两个物种构成。酸比其共轭碱多一个H⁺；碱比其共轭酸少一个H⁺。

HA ⇌ H⁺ + A⁻    (HA / A⁻ is a conjugate pair)

Common examples include: HCl/Cl⁻, NH₄⁺/NH₃, H₂O/OH⁻, and H₃O⁺/H₂O. In every acid-base equilibrium, there are two conjugate pairs.

常见例子有：HCl/Cl⁻、NH₄⁺/NH₃、H₂O/OH⁻和H₃O⁺/H₂O。在每个酸碱平衡中都存在两个共轭对。

For the equilibrium CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, the pairs are CH₃COOH/CH₃COO⁻ and H₃O⁺/H₂O.

对于平衡CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺，其共轭对分别是CH₃COOH/CH₃COO⁻和H₃O⁺/H₂O。

4. Amphiprotic Species | 两性物种

An amphiprotic species can act either as a proton donor (acid) or as a proton acceptor (base), depending on the reaction conditions. Water is the most important amphiprotic substance: it can accept a proton to form H₃O⁺, or donate a proton to form OH⁻.

两性物种既能作为质子给予体（酸），也能作为质子接受体（碱），具体取决于反应条件。水是最重要的两性物质：它既可以接受质子形成H₃O⁺，也可以给出质子形成OH⁻。

Other common amphiprotic species include HCO₃⁻, HSO₄⁻, H₂PO₄⁻, and HPO₄²⁻. For example, HCO₃⁻ reacts with HCl as a base: HCO₃⁻ + H⁺ → H₂CO₃, but reacts with NaOH as an acid: HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O.

其他常见的两性物种包括HCO₃⁻、HSO₄⁻、H₂PO₄⁻和HPO₄²⁻。例如，HCO₃⁻与HCl反应时作为碱：HCO₃⁻ + H⁺ → H₂CO₃；但与NaOH反应时作为酸：HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O。

5. Lewis Acid-Base Theory | 路易斯酸碱理论

The Lewis theory is the most general acid-base model. A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. This theory covers reactions that do not involve protons, such as the formation of complex ions.

路易斯理论是最广义的酸碱模型。路易斯酸是电子对的接受体，路易斯碱是电子对的给予体。该理论涵盖了不涉及质子的反应，例如配合离子的形成。

BF₃ + NH₃ → BF₃-NH₃

In this reaction, BF₃ accepts a lone pair from NH₃, so BF₃ is a Lewis acid and NH₃ is a Lewis base. Similarly, H⁺ is a Lewis acid (it accepts an electron pair), and OH⁻, CN⁻, and H₂O are Lewis bases.

在该反应中，BF₃接受来自NH₃的孤对电子，因此BF₃是路易斯酸，NH₃是路易斯碱。类似地，H⁺是路易斯酸（它接受电子对），而OH⁻、CN⁻和H₂O都是路易斯碱。

6. Acid and Base Strength | 酸和碱的强度

Strong acids and bases dissociate completely in aqueous solution. Common strong acids include HCl, HBr, HI, HNO₃, H₂SO₄, and HClO₄. Common strong bases include group 1 and group 2 hydroxides such as NaOH, KOH, and Ca(OH)₂.

强酸和强碱在水中完全解离。常见强酸包括HCl、HBr、HI、HNO₃、H₂SO₄和HClO₄。常见强碱包括第1族和第2族金属氢氧化物，如NaOH、KOH和Ca(OH)₂。

Weak acids and bases only partially dissociate in water. Examples include CH₃COOH, H₂CO₃, and NH₃. The strength of a weak acid is quantified by the acid dissociation constant Kₐ, while the strength of a weak base is quantified by K_b.

弱酸和弱碱在水中仅部分解离，例如CH₃COOH、H₂CO₃和NH₃。弱酸的强度用电离常数Kₐ定量表示，弱碱的强度用K_b定量表示。

CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺    Kₐ = [CH₃COO⁻][H₃O⁺]/[CH₃COOH]

A larger Kₐ indicates a stronger acid. Conversely, a larger K_b indicates a stronger base. For a conjugate pair, Kₐ × K_b = K_w = 1.0 × 10⁻¹⁴ at 25 °C.

Kₐ越大表示酸越强；同理，K_b越大表示碱越强。对于共轭酸碱对，Kₐ × K_b = K_w = 1.0 × 10⁻¹⁴（25 °C时）。

7. The pH Scale and pOH | pH标度和pOH

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration: pH = -log₁₀[H⁺]. Similarly, pOH = -log₁₀[OH⁻]. At 25 °C, pH + pOH = 14.

溶液的pH定义为氢离子浓度的负对数：pH = -log₁₀[H⁺]。类似地，pOH = -log₁₀[OH⁻]。在25 °C时，pH + pOH = 14。

Remember that pH is a logarithmic scale: a decrease in pH by 1 unit corresponds to a ten-fold increase in [H⁺]. This means pH 3 is ten times more acidic than pH 4, and pH 2 is one hundred times more acidic than pH 4.

记住pH是对数标度：pH每降低1个单位，[H⁺]增大10倍。这意味着pH 3的酸性是pH 4的10倍，pH 2的酸性是pH 4的100倍。

8. Strong vs. Weak Acid Calculations | 强酸与弱酸的计算

For a strong acid such as HCl with concentration c, [H⁺] = c, so pH = -log₁₀ c. For a strong base such as NaOH, [OH⁻] = c, so pOH = -log₁₀ c and pH = 14 – pOH.

对于浓度为c的强酸（如HCl），[H⁺] = c，因此pH = -log₁₀ c。对于强碱（如NaOH），[OH⁻] = c，因此pOH = -log₁₀ c，pH = 14 – pOH。

For a weak acid HA, the equilibrium must be considered. If the initial concentration is c and the degree of dissociation is x, then:

对于弱酸HA，需要考虑平衡。若初始浓度为c，解离度为x，则：

Kₐ = x²/(c – x) ≈ x²/c   (if x is much smaller than c)

Then [H⁺] = x, and pH = -log₁₀ x. Always check the approximation: if x/c is less than 5%, the approximation is usually considered valid in IB.

此时[H⁺] = x，pH = -log₁₀ x。始终要检验近似条件：在IB课程中，通常认为x/c小于5%时近似成立。

9. The Role of Water in Acid-Base Equilibria | 水在酸碱平衡中的作用

Water undergoes autoionization, producing H₃O⁺ and OH⁻. The ion product constant of water is K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C.

水发生自偶电离，产生H₃O⁺和OH⁻。水的离子积常数K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴（25 °C时）。

2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

Because the autoionization of water is endothermic, K_w increases with temperature. At temperatures above 25 °C, pH 7 is no longer neutral; neutrality occurs when pH = pOH, which is less than 7.

由于水的自偶电离是吸热过程，K_w随温度升高而增大。在高于25 °C时，pH 7不再代表中性；中性条件是pH = pOH，此时pH小于7。

10. pH Curves and Indicators | pH曲线与指示剂

A pH titration curve shows how the pH of a solution changes as a titrant is added. There are four main types: strong acid-strong base, strong acid-weak base, weak acid-strong base, and weak acid-weak base.

pH滴定曲线表示随着滴定剂的加入，溶液pH的变化。主要有四种类型：强酸-强碱、强酸-弱碱、弱酸-强碱和弱酸-弱碱。

• Strong acid + strong base: equivalence point pH = 7; steep vertical section around the equivalence point.

  强酸+强碱：等当点pH = 7；等当点附近有陡峭的垂直段。

• Strong acid + weak base: equivalence point pH < 7 (because the conjugate acid of the weak base hydrolyzes).

  强酸+弱碱：等当点pH < 7（因为弱碱的共轭酸发生水解）。

• Weak acid + strong base: equivalence point pH > 7 (because the conjugate base of the weak acid hydrolyzes).

  弱酸+强碱：等当点pH > 7（因为弱酸的共轭碱发生水解）。

An acid-base indicator is a weak acid (HIn) whose colour differs from its conjugate base (In⁻). The end point of a titration is chosen so that the indicator’s colour change range overlaps with the steep part of the pH curve.

酸碱指示剂是一种弱酸（HIn），其颜色与其共轭碱（In⁻）不同。选择指示剂时，应使其变色范围与pH曲线的陡峭部分重叠。

11. Buffer Solutions | 缓冲溶液

A buffer solution resists changes in pH when small amounts of acid or base are added. It typically contains a weak acid and its conjugate base (e.g., CH₃COOH / CH₃COO⁻) or a weak base and its conjugate acid (e.g., NH₃ / NH₄⁺).

缓冲溶液在加入少量酸或碱时能抵抗pH的变化。缓冲溶液通常由弱酸及其共轭碱（如CH₃COOH / CH₃COO⁻）或弱碱及其共轭酸（如NH₃ / NH₄⁺）组成。

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer:

亨德森-哈塞尔巴尔赫方程用于计算缓冲溶液的pH：

pH = pKₐ + log₁₀([A⁻]/[HA])

When [A⁻] = [HA], the buffer has maximum capacity and pH = pKₐ. Adding a small amount of H⁺ shifts the equilibrium to convert A⁻ to HA; adding OH⁻ converts HA to A⁻. As long as the reservoir of HA and A⁻ is not exhausted, pH changes are minimal.

当[A⁻] = [HA]时，缓冲容量最大，pH = pKₐ。加入少量H⁺会使平衡移动，将A⁻转化为HA；加入OH⁻则将HA转化为A⁻。只要HA和A⁻的储备未被耗尽，pH变化就很小。

12. Exam Tips and Common Pitfalls | 考试技巧与常见误区

In IB exams, students often confuse the terms “strong” and “concentrated.” Strength refers to the degree of dissociation, while concentration refers to the amount of solute per unit volume. A dilute strong acid can have a higher pH than a concentrated weak acid.

在IB考试中，学生常常混淆“强”和“浓”的概念。强度指解离程度，而浓度指单位体积内溶质的量。稀的强酸其pH可能大于浓的弱酸。

• Always identify conjugate pairs by looking for species that differ by exactly one H⁺.

  始终通过寻找相差恰好一个H⁺的物种来识别共轭对。

• Remember that amphiprotic species appear in the middle of a polyprotic acid sequence, such as HCO₃⁻.

  记住两性物种出现在多元酸序列的中间位置，例如HCO₃⁻。

• Do not use Kₐ for strong acids; strong acids have Kₐ values too large to measure and are assumed to dissociate fully.

  不要对强酸使用Kₐ；强酸的Kₐ值太大无法测量，通常假定完全解离。

• When calculating pH of a weak base, first find [OH⁻], then pOH, then pH using pH = 14 – pOH (at 25 °C).

  计算弱碱的pH时，先求[OH⁻]，再求pOH，最后用pH = 14 – pOH（25 °C）得到pH。

Understanding the relationships between Kₐ, K_b, K_w, pH, and pOH is essential for solving IB acid-base problems systematically. Always write the equilibrium first, then apply the appropriate approximation.

理解Kₐ、K_b、K_w、pH和pOH之间的关系，是系统解决IB酸碱问题的关键。务必先写出平衡方程式，再应用适当的近似方法。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Properties of Acids and Bases | IB化学：酸碱的性质

Acids and bases are fundamental to chemistry, influencing everything from industrial processes to biological systems. This article explores their key properties, definitions, and calculations as required by the IB Chemistry syllabus (CIE board).

酸碱是化学的基础，影响着从工业过程到生物系统的方方面面。本文根据IB化学教学大纲（CIE考试局）探讨它们的主要性质、定义和计算方法。

1. Defining Acids and Bases | 酸碱的定义

There are three main definitions of acids and bases that you need to know for IB Chemistry.

在IB化学中，你需要掌握酸碱的三种主要定义。

The Arrhenius definition states that an acid produces H⁺ ions in water, while a base produces OH⁻ ions.

阿伦尼乌斯定义指出，酸在水中产生H⁺离子，碱在水中产生OH⁻离子。

The Brønsted-Lowry definition is broader: an acid is a proton donor, and a base is a proton acceptor.

布朗斯特-劳里定义更为广泛：酸是质子给予体，碱是质子接受体。

The Lewis definition is the most general: an acid is an electron-pair acceptor, and a base is an electron-pair donor.

路易斯定义最为通用：酸是电子对接受体，碱是电子对给予体。

2. The pH Scale | pH标度

The pH scale expresses the concentration of hydrogen ions in a solution. It is defined as:

pH标度表示溶液中氢离子的浓度。其定义如下：

pH = -log₁₀[H⁺]

where [H⁺] is measured in mol dm⁻³. A neutral solution at 25°C has [H⁺] = 1.0 × 10⁻⁷ mol dm⁻³, so pH = 7.

其中[H⁺]的单位是mol dm⁻³。在25°C时，中性溶液的[H⁺] = 1.0 × 10⁻⁷ mol dm⁻³，因此pH = 7。

Similarly, pOH is defined as pOH = -log₁₀[OH⁻]. At 25°C, pH + pOH = 14 for aqueous solutions.

类似地，pOH定义为pOH = -log₁₀[OH⁻]。在25°C时，水溶液的pH + pOH = 14。

3. Strong and Weak Acids and Bases | 强酸与弱酸、强碱与弱碱

Acids and bases can be classified as strong or weak depending on their degree of dissociation in water.

根据在水中的电离程度，酸和碱可分为强与弱。

Strong acids, such as HCl, HNO₃ and H₂SO₄, dissociate completely in water. For example, HCl → H⁺ + Cl⁻.

强酸如HCl、HNO₃和H₂SO₄在水中完全电离。例如，HCl → H⁺ + Cl⁻。

Weak acids, such as CH₃COOH and H₂CO₃, dissociate only partially. The equilibrium is written with a ⇌ sign: CH₃COOH ⇌ H⁺ + CH₃COO⁻.

弱酸如CH₃COOH和H₂CO₃仅部分电离，其平衡用⇌表示：CH₃COOH ⇌ H⁺ + CH₃COO⁻。

Strong bases, such as NaOH and KOH, dissociate fully, releasing OH⁻ ions. Weak bases, such as NH₃, only partly accept protons from water: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

强碱如NaOH和KOH完全电离，释放OH⁻离子。弱碱如NH₃仅部分从水中接受质子：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。

4. Conjugate Acid-Base Pairs | 共轭酸碱对

In the Brønsted-Lowry theory, an acid and its conjugate base differ by one proton (H⁺).

在布朗斯特-劳里理论中，酸与其共轭碱相差一个质子（H⁺）。

For example, in the reaction HCl + H₂O ⇌ H₃O⁺ + Cl⁻, the acid HCl donates a proton to water, forming H₃O⁺. The conjugate pairs are HCl/Cl⁻ and H₃O⁺/H₂O.

例如，在反应HCl + H₂O ⇌ H₃O⁺ + Cl⁻中，酸HCl将质子给予水，生成H₃O⁺。共轭对为HCl/Cl⁻和H₃O⁺/H₂O。

The stronger the acid, the weaker its conjugate base, and vice versa.

酸越强，其共轭碱越弱，反之亦然。

5. Acid-Base Reactions and Ionic Equations | 酸碱反应与离子方程式

Acid-base reactions are proton-transfer reactions. The simplest neutralisation reaction is between hydrogen ions and hydroxide ions:

酸碱反应是质子转移反应。最简单的中和反应是氢离子与氢氧根离子的反应：

H⁺(aq) + OH⁻(aq) → H₂O(l)

When writing ionic equations, species that do not change, known as spectator ions, are omitted. For example, for HCl + NaOH, the net ionic equation is the one above.

写离子方程式时，不参加反应、未发生变化的离子（称为旁观离子）被省略。例如，对于HCl + NaOH，净离子方程式就是上面的方程式。

6. The Ionic Product of Water, Kw | 水的离子积Kw

Water undergoes self-ionisation: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). The equilibrium constant for this process is called the ionic product of water, Kw.

水会发生自耦电离：H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)。该过程的平衡常数称为水的离子积Kw。

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 25°C

Kw increases with temperature because self-ionisation is endothermic. Therefore pH of pure water is 7 only at 25°C; at higher temperatures it is lower, but the solution remains neutral.

Kw随温度升高而增大，因为自耦电离是吸热的。因此，纯水的pH仅在25°C时为7；在更高温度下pH更低，但溶液仍为中性。

7. pH of Strong Acids and Bases | 强酸强碱的pH计算

For a strong acid with concentration c, since it dissociates completely, [H⁺] = c and pH = -log₁₀ c.

对于浓度为c的强酸，由于完全电离，[H⁺] = c，因此pH = -log₁₀ c。

For a strong base with concentration c, [OH⁻] = c, pOH = -log₁₀ c, and pH = 14 – pOH at 25°C.

对于浓度为c的强碱，[OH⁻] = c，pOH = -log₁₀ c，在25°C时pH = 14 – pOH。

Example: Calculate pH of 0.05 mol dm⁻³ H₂SO₄. Since H₂SO₄ is diprotic, [H⁺] = 2 × 0.05 = 0.10 mol dm⁻³, so pH = -log₁₀(0.10) = 1.00.

例如：计算0.05 mol dm⁻³ H₂SO₄的pH。由于H₂SO₄是二元酸，[H⁺] = 2 × 0.05 = 0.10 mol dm⁻³，因此pH = -log₁₀(0.10) = 1.00。

8. pH of Weak Acids | 弱酸的pH计算

For a weak acid HA, the equilibrium is HA ⇌ H⁺ + A⁻. If the initial concentration is c₀ and the acid dissociation constant is Kₐ, we can approximate [

Published by TutorHao | IB Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: The pH Scale and Acidity | IB化学：pH标度与酸碱度

The pH scale is one of the most practical tools in chemistry. It allows us to describe the acidity or alkalinity of a solution using a single number, rather than expressing concentrations that may vary over many orders of magnitude. In IB Chemistry, understanding pH is essential for acid-base equilibria, titrations, buffer systems, and many everyday applications.

pH标度是化学中最实用的工具之一。它用单一数值来描述溶液的酸碱性，而不用去处理可相差许多数量级的浓度数值。在IB化学中，理解pH对酸碱平衡、滴定、缓冲体系以及许多日常应用都至关重要。

1. Acid-Base Strength vs. Concentration | 酸碱强度与浓度的区别

Acid strength describes how completely an acid dissociates in water. A strong acid, such as HCl, dissociates almost fully: HCl → H⁺ + Cl⁻. A weak acid, such as ethanoic acid (CH₃COOH), only partially dissociates: CH₃COOH ⇌ CH₃COO⁻ + H⁺. Strength is an intrinsic property of the acid.

酸碱强度描述的是酸在水中解离的完全程度。强酸如HCl几乎完全解离：HCl → H⁺ + Cl⁻。弱酸如乙酸CH₃COOH仅部分解离：CH₃COOH ⇌ CH₃COO⁻ + H⁺。强度是酸本身的内在性质。

Concentration, on the other hand, refers to the amount of acid dissolved in a given volume of solution. A concentrated weak acid can have a lower pH than a dilute strong acid, or vice versa, depending on conditions. IB Chemistry requires you to separate these ideas clearly when predicting pH values.

而浓度是指在一定体积溶液中所溶解酸量的多少。浓的弱酸的pH可能比稀的强酸更低，也可能更高，具体取决于条件。IB化学要求你清晰区分这两个概念，以便正确预测pH值。

2. The Definition of pH | pH的定义

The pH scale was introduced by Sørensen in 1909. It is defined as the negative logarithm to base 10 of the hydrogen ion concentration:

pH标度由Sørensen于1909年提出。其定义为氢离子浓度的负十进制对数：

pH = −log₁₀[H⁺]

Here [H⁺] is measured in mol/dm³. A ten-fold increase in hydrogen ion concentration decreases pH by one unit. For a solution with [H⁺] = 1.0 × 10⁻⁴ mol/dm³, pH = 4.00.

其中[H⁺]的单位是mol/dm³。氢离子浓度每增大10倍，pH就减少1个单位。若[H⁺] = 1.0 × 10⁻⁴ mol/dm³，则pH = 4.00。

Because logarithms compress a wide range, pH allows chemists to compare very dilute hydrogen ion concentrations conveniently. On calculators, the pH is normally rounded to two decimal places.

由于对数将宽范围数值压缩，pH使化学家可以方便地比较极稀的氢离子浓度。计算时pH通常保留两位小数。

3. The pH Scale and Its Range | pH标度及其范围

At 25 °C, pure water has [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/dm³, giving pH = 7.00. Solutions with pH below 7 are acidic; pH above 7 are basic. pH = 7 is neutral at this temperature.

在25 °C时，纯水中[H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/dm³，即pH = 7.00。pH低于7的溶液为酸性；pH高于7的溶液为碱性；在此温度下pH = 7为中性。

Although the common scale is 0 to 14, pH values can fall outside this range. A concentrated strong acid can have pH below 0, and a concentrated strong base can have pH above 14. The scale is simply a convention, not a physical limit.

虽然常见pH范围是0到14，但pH值也可能超出此范围。浓强酸的pH可能低于0，浓强碱的pH可能高于14。这个标度只是惯例，并不是物理限制。

Temperature affects pH because the ionisation of water is endothermic. At higher temperatures, Kw increases and the neutral pH becomes lower than 7. IB exam questions usually state 25 °C unless otherwise specified.

温度会影响pH，因为水的电离是吸热过程。温度升高时，Kw增大，中性的pH会低于7。IB考试题目通常会注明25 °C，除非另有说明。

4. Calculating pH, pOH, and Kw | 计算pH、pOH与Kw

Water autoionises according to the equilibrium: H₂O ⇌ H⁺ + OH⁻. The ion product constant is:

水存在自偶电离平衡：H₂O ⇌ H⁺ + OH⁻。其离子积常数为：

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C

Taking negative logarithms of both sides gives a useful relation:

对等式两边同时取负对数，可得一个常用关系式：

pOH = −log₁₀[OH⁻] and pH + pOH = 14.00

For a solution with [OH⁻] = 1.0 × 10⁻³ mol/dm³, pOH = 3.00, so pH = 11.00. You can also find [H⁺] by [H⁺] = Kw / [OH⁻], then calculate pH directly.

对于[OH⁻] = 1.0 × 10⁻³ mol/dm³的溶液，pOH = 3.00，因此pH = 11.00。你也可以先用[H⁺] = Kw / [OH⁻]求出[H⁺]，再直接计算pH。

5. Strong vs. Weak Acids and Bases | 强酸弱酸、强碱弱碱

Strong acids and bases dissociate fully in aqueous solution. Common strong acids include HCl, HBr, HI, HNO₃, H₂SO₄ and HClO₄. Common strong bases include NaOH, KOH and Ba(OH)₂. Their equilibrium positions lie far to the right.

强酸和强碱在水中完全解离。常见强酸包括HCl、HBr、HI、HNO₃、H₂SO₄和HClO₄。常见强碱包括NaOH、KOH和Ba(OH)₂。它们的平衡位置远远偏向右边。

Weak acids and bases only partially dissociate. Ethanoic acid, carbonic acid, ammonia and amines fall into this category. Their strength is described by equilibrium constants:

弱酸和弱碱仅部分解离。乙酸、碳酸、氨和胺类都属于此类。它们的强度用平衡常数来描述：

Ka = [H⁺][A⁻] / [HA] and Kb = [BH⁺][OH⁻] / [B]

Larger Ka values mean stronger acids, and larger Kb values mean stronger bases. It is common to use pKa = −log₁₀Ka; a smaller pKa corresponds to a stronger acid.

Ka值越大，酸越强；Kb值越大，碱越强。通常也使用pKa = −log₁₀Ka；pKa越小，酸越强。

6. pH of Strong Acids and Strong Bases | 强酸和强碱的pH

For a strong monoprotic acid that fully dissociates, the hydrogen ion concentration equals the initial acid concentration. Thus pH = −log₁₀[acid]. For example, 0.020 mol/dm³ HCl has [H⁺] = 0.020 mol/dm³, so pH = 1.70.

对于完全解离的强一元酸，氢离子浓度等于初始酸浓度。因此pH = −log₁₀[酸]。例如，0.020 mol/dm³ HCl的[H⁺] = 0.020 mol/dm³，所以pH = 1.70。

Strong bases require an extra step. First find [OH⁻], then pOH, then pH = 14.00 − pOH. For 0.010 mol/dm³ NaOH, [OH⁻] = 0.010 mol/dm³, pOH = 2.00, and pH = 12.00.

强碱则需要多一步：先求[OH⁻]，再求pOH，最后用pH = 14.00 − pOH。对于0.010 mol/dm³ NaOH，[OH⁻] = 0.010 mol/dm³，pOH = 2.00，因此pH = 12.00。

For dibasic bases like Ba(OH)₂, be careful: each mole of Ba(OH)₂ produces two moles of OH⁻. A 0.0050 mol/dm³ Ba(OH)₂ solution has [OH⁻] = 0.010 mol/dm³ and pH = 12.00.

对于Ba(OH)₂这类二元碱要小心：每摩尔Ba(OH)₂产生两摩尔OH⁻。0.0050 mol/dm³ Ba(OH)₂溶液中的[OH⁻] = 0.010 mol/dm³，pH = 12.00。

7. pH of Weak Acids: Using Ka | 弱酸的pH：使用Ka

Weak acids do not dissociate fully, so pH cannot be found directly from concentration. Instead, we use an ICE table and Ka. For a generic weak acid HA:

弱酸不会完全解离，因此不能直接从浓度求pH。我们需要使用ICE表（初始-变化-平衡）和Ka。对一般弱酸HA：

HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻] / [HA]

If the initial concentration is c and the extent of dissociation is x, then at equilibrium [H⁺] = x, [A⁻] = x, and [HA] ≈ c. Because x is very small for a weak acid, the approximation Ka ≈ x² / c is often valid.

若初始浓度为c，解离度为x，则平衡时[H⁺] = x，[A⁻] = x，[HA] ≈ c。由于弱酸的x很小，近似式Ka ≈ x² / c通常成立。

Solving for x gives [H⁺] = √(Ka × c). For 0.100 mol/dm³ ethanoic acid with Ka = 1.8 × 10⁻⁵ mol/dm³:

解出x得[H⁺] = √(Ka × c)。对于0.100 mol/dm³乙酸，Ka = 1.8 × 10⁻⁵ mol/dm³：

[H⁺] = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³ mol/dm³, pH = 2.87

Notice this pH is higher than that of a strong acid of the same concentration. The approximation fails when the acid is very dilute or Ka is large; then the quadratic equation must be used.

注意：相同浓度下，这个pH高于强酸的pH。当酸极稀或Ka较大时，该近似不再成立，此时必须使用二次方程精确求解。

8. pH of Salt Solutions: Hydrolysis | 盐溶液的pH：水解

Salts are produced in neutralisation reactions, but their solutions are not always neutral. This is due to hydrolysis: the ions react with water to generate H⁺ or OH⁻.

盐产生于中和反应，但其溶液并不总是中性。这是因为水解：离子与水反应生成H⁺或OH⁻。

A salt from a weak acid and a strong base, such as sodium ethanoate, gives a basic solution. The ethanoate ion hydrolyses:

由弱酸和强碱生成的盐，如乙酸钠，其溶液呈碱性。乙酸根离子发生水解：

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

The equilibrium constant for this hydrolysis is Kb = Kw / Ka. Since Ka of ethanoic acid is small, Kb is not negligible, and OH⁻ is produced.

该水解反应的平衡常数Kb = Kw / Ka。由于乙酸的Ka较小，Kb不可忽略，因此会产生OH⁻。

Conversely, a salt from a strong acid and a weak base, such as ammonium chloride, gives an acidic solution. Salts from strong acids and strong bases, such as NaCl, do not hydrolyse and remain neutral at 25 °C.

相反，由强酸和弱碱生成的盐，如氯化铵，其溶液呈酸性。由强酸和强碱生成的盐，如NaCl，不发生水解，在25 °C时保持中性。

9. Buffer Solutions and pH Control | 缓冲溶液与pH控制

A buffer solution resists changes in pH when small amounts of acid or base are added. It contains a weak acid and its conjugate base, or a weak base and its conjugate acid. Blood is a natural buffer system based on H₂CO₃ / HCO₃⁻.

缓冲溶液在加入少量酸或碱时能抵抗pH变化。它含有弱酸及其共轭碱，或弱碱及其共轭酸。血液就是基于H₂CO₃ / HCO₃⁻的天然缓冲体系。

For an acidic buffer made from HA and A⁻, the Henderson-Hasselbalch equation is used:

对于由HA和A⁻组成的酸性缓冲溶液，可使用Henderson-Hasselbalch方程：

pH = pKa + log₁₀([A⁻] / [HA])

When [A⁻] = [HA], pH = pKa. Adding a small amount of H⁺ shifts the equilibrium to convert A⁻ into HA; adding OH⁻ converts HA into A⁻. The ratio [A⁻]/[HA] changes only slightly, so pH remains nearly constant.

当[A⁻] = [HA]时，pH = pKa。加入少量H⁺会使平衡移动，将A⁻转化为HA；加入OH⁻则将HA转化为A⁻。比值[A⁻]/[HA]仅发生微小变化，因此pH几乎不变。

To prepare a buffer with a desired pH, choose a weak acid whose pKa is close to that pH and adjust the ratio of salt to acid. Buffer capacity depends on the total concentration of the acid-conjugate base pair.

要配制特定pH的缓冲溶液，应选择pKa接近所需pH的弱酸，并调节盐与酸的比例。缓冲容量取决于酸-共轭碱对的总浓度。

10. pH Indicators and Titrations | pH指示剂与滴定

Indicators are weak acids whose conjugate base has a different colour. For a typical indicator HIn, the equilibrium is HIn ⇌ H⁺ + In⁻. When pH changes, the ratio of [In⁻]/[HIn] changes, producing a colour change over a narrow pH range.

指示剂是共轭碱具有不同颜色的弱酸。常见指示剂HIn存在平衡：HIn ⇌ H⁺ + In⁻。当pH变化时，[In⁻]/[HIn]的比值会改变，从而在较窄的pH范围内产生颜色变化。

In an acid-base titration, pH changes dramatically near the equivalence point. The indicator must change colour within the steep section of the titration curve. Choosing the wrong indicator leads to inaccurate results.

在酸碱滴定中，接近等当点时pH发生急剧变化。指示剂的变色范围必须落在滴定曲线陡峭段内。选择错误的指示剂会导致不准确的结果。

11. Applications and Biological Importance | 应用与生物学意义

The pH scale has practical importance in medicine, agriculture and industry. Enzymes in the human body work best within narrow pH ranges; for example, pepsin in the stomach functions optimally around pH 2, while trypsin in the intestine prefers pH around 8.

pH标度在医学、农业和工业中具有实际意义。人体内的酶只在较窄的pH范围内发挥最佳活性：例如，胃中的胃蛋白酶在pH约为2时活性最高，而肠道中的胰蛋白酶则偏好pH约为8的环境。

Soils with extreme pH are unsuitable for most crops: acidic soils are often treated with limestone (CaCO₃) to raise pH. Aquaculture and fish farming also monitor pH closely because fish are sensitive to even small shifts in water acidity.

pH极端不适的土壤不适合大多数农作物：通常用石灰石CaCO₃处理酸性土壤以提高pH。水产养殖也密切监测pH，因为鱼类对水中酸性的微小变化都非常敏感。

From laboratory calculations to real-world systems, the pH scale connects the microscopic concentration of H⁺ to macroscopic outcomes. Mastering pH, pOH, Kw and Ka gives you the tools to analyse any acid-base equilibrium you meet in IB Chemistry.

从实验室计算到现实系统，pH标度将微观的H⁺浓度与宏观结果联系起来。掌握pH、pOH、Kw和Ka，将使你具备分析IB化学中任何酸碱平衡的能力。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Lewis Acid–Base Theory: Key Concepts for IB Chemistry HL | 路易斯酸碱理论要点：IB化学HL精讲

The Lewis theory of acids and bases, proposed by Gilbert N. Lewis in 1923, offers a broader and more general framework than the Arrhenius and Brønsted–Lowry models. Instead of focusing on protons or hydroxide ions, this theory centres on the transfer of electron pairs, making it essential for understanding a vast range of chemical reactions, including those in non-aqueous solvents and coordination chemistry.

路易斯酸碱理论由吉尔伯特·N·路易斯于1923年提出，比阿伦尼乌斯理论和布朗斯特-洛瑞理论更为广泛和普适。该理论不再聚焦于质子或氢氧根离子，而是以电子对的转移为核心，使其成为理解非水溶剂中反应和配位化学等广泛领域的关键工具。

1. The Lewis Definition | 路易斯定义

A Lewis acid is defined as any species that can accept an electron pair. A Lewis base is any species that can donate an electron pair. The product formed when a Lewis acid and a Lewis base react is called a Lewis adduct or Lewis acid–base complex.

路易斯酸定义为任何能够接受电子对的物种；路易斯碱定义为任何能够给出电子对的物种。路易斯酸与路易斯碱反应生成的产物称为路易斯加合物或路易斯酸碱配合物。

In the reaction between BF₃ and NH₃, we can identify the electron-pair acceptor and donor clearly:

在BF₃与NH₃的反应中，我们可以清晰地识别电子对的受体与给体：

BF₃ + NH₃ → F₃B←NH₃

Here, the boron atom in BF₃ has only six valence electrons and therefore lacks a full octet. It accepts the lone pair from nitrogen. BF₃ is the Lewis acid; NH₃ is the Lewis base.

在这里，BF₃中的硼原子只有六个价电子，未达到八隅体结构，因此它接受来自氮的孤对电子。BF₃是路易斯酸，NH₃是路易斯碱。

Key point: the arrow in the adduct F₃B←NH₃ points from the donor (N) towards the acceptor (B), from the Lewis base to the Lewis acid.

要点：加合物F₃B←NH₃中的箭头从给体（N）指向受体（B），即从路易斯碱指向路易斯酸。

2. Comparing Lewis with Brønsted–Lowry | 路易斯理论与布朗斯特-洛瑞理论的比较

All Brønsted–Lowry acids are Lewis acids, because a proton (H⁺) is an electron-pair acceptor. However, many Lewis acids are not Brønsted–Lowry acids — the latter model is a subset of the former. Similarly, all Brønsted–Lowry bases are Lewis bases, because any species that accepts a proton must donate an electron pair to form the H–B bond.

所有布朗斯特-洛瑞酸都是路易斯酸，因为质子（H⁺）是电子对受体。然而，许多路易斯酸并不是布朗斯特-洛瑞酸——后者只是前者的一个子集。同样地，所有布朗斯特-洛瑞碱都是路易斯碱，因为任何能够接受质子的物种必定要给出电子对来形成H–B键。

• Arrhenius model: limited to aqueous solutions; acid produces H⁺, base produces OH⁻.

  阿伦尼乌斯模型：仅限于水溶液；酸产生H⁺，碱产生OH⁻。

• Brønsted–Lowry model: applies to proton transfer in any solvent.

  布朗斯特-洛瑞模型：适用于任何溶剂中的质子转移。

• Lewis model: applies to any electron-pair transfer, including reactions with no proton involvement at all.

  路易斯模型：适用于任何电子对转移，包括完全不涉及质子的反应。

Consider the reaction between CaO and SO₃ to form calcium sulfate:

考虑CaO与SO₃反应生成硫酸钙：

CaO + SO₃ → CaSO₄

Here, O²⁻ donates an electron pair to S⁶⁺. No proton transfer occurs. This reaction would be difficult to classify under Brønsted–Lowry theory but is straightforward under Lewis theory.

此反应中O²⁻向S⁶⁺给出电子对，没有质子转移发生。用布朗斯特-洛瑞理论难以分类，但用路易斯理论则一目了然。

3. Identifying Lewis Acids and Bases | 识别路易斯酸与碱

To identify Lewis acids and bases efficiently, students should look for common structural indicators.

为了高效识别路易斯酸与碱，学生应寻找常见的结构特征。

Lewis acids include: species with incomplete octets (BF₃, AlCl₃, FeCl₃), positively charged ions or electron-deficient cations (H⁺, Ag⁺, Cu²⁺, Fe³⁺), and molecules with polar multiple bonds where the central atom is electron-deficient (CO₂, SO₂, SO₃).

路易斯酸包括：具有不完整八隅体的物种（BF₃、AlCl₃、FeCl₃）、带正电荷的离子或缺电子阳离子（H⁺、Ag⁺、Cu²⁺、Fe³⁺），以及中心原子缺电子的极性多重键分子（CO₂、SO₂、SO₃）。

Lewis bases include: all anions (Cl⁻, OH⁻, CN⁻, F⁻), molecules with lone pairs (NH₃, H₂O, CO, R–OH), and species with pi-electron systems that can donate electron density (C₆H₆, alkenes).

路易斯碱包括：所有阴离子（Cl⁻、OH⁻、CN⁻、F⁻）、含孤对电子的分子（NH₃、H₂O、CO、R–OH），以及能够提供电子密度的π电子体系物种（C₆H₆、烯烃）。

For CO₂, the carbon atom is bonded to two highly electronegative oxygen atoms. The electron density around carbon is heavily withdrawn, exposing a positive region that can accept electron pairs from Lewis bases such as OH⁻.

对于CO₂，碳原子与两个高电负性的氧原子成键，碳周围的电子密度被严重拉走，暴露出一个正电区域，可以接受来自OH⁻等路易斯碱的电子对。

4. The Lewis Adduct and Dative Covalent Bonds | 路易斯加合物与配位共价键

The bond formed between a Lewis acid and a Lewis base is called a coordinate covalent bond or dative covalent bond. In such a bond, both shared electrons originate from the same atom — the Lewis base. Once formed, there is no distinction between a dative bond and an ordinary covalent bond in terms of bond strength or length.

路易斯酸与路易斯碱之间形成的键称为配位共价键。在这类键中，两个共享电子均来自同一原子——即路易斯碱。一旦形成，配位键与普通共价键在键能和键长上并无区别。

A typical example is the reaction between ammonia and boron trifluoride:

典型的例子是氨与三氟化硼的反应：

NH₃ + BF₃ → H₃N→BF₃

In H₃N→BF₃, the N–B bond is formed using the lone pair on nitrogen. The arrow in the structural formula points towards the acceptor atom.

在H₃N→BF₃中，N–B键由氮上的孤对电子形成。结构式中的箭头指向受体原子。

Another familiar example is the formation of the hydroxonium ion (H₃O⁺). Water donates one lone pair to a proton, forming a dative bond. Similarly, when ammonia accepts a proton, the ammonium ion (NH₄⁺) is formed with one dative bond.

另一个熟悉的例子是水合氢离子（H₃O⁺）的形成。水向质子给出一个孤对电子，形成配位键。类似地，氨接受质子时，铵离子（NH₄⁺）形成并含有一个配位键。

In coordination complexes such as [Cu(H₂O)₆]²⁺, each water ligand donates an electron pair to the Cu²⁺ ion. The Cu²⁺ ion acts as a Lewis acid and each H₂O acts as a Lewis base. This framework is central to transition metal chemistry.

在配位配合物如[Cu(H₂O)₆]²⁺中，每个水配体向Cu²⁺离子给出一个电子对。Cu²⁺离子作为路易斯酸，每个H₂O作为路易斯碱。这一框架是过渡金属化学的核心。

5. HSAB Principle: Hard and Soft Acids and Bases | 硬软酸碱（HSAB）原理

The HSAB principle, introduced by Ralph Pearson, classifies Lewis acids and bases as ‘hard’ or ‘soft’ based on their size, charge density and polarisability. Hard species are small, highly charged and not very polarisable. Soft species are large, low-charge and highly polarisable.

HSAB原理由拉尔夫·皮尔逊提出，根据体积大小、电荷密度和极化性将路易斯酸与碱分为”硬”和”软”两类。硬物种体积小、电荷高、不易极化；软物种体积大、电荷低、高度可极化。

The key rule is: hard acids prefer to bind with hard bases, and soft acids prefer to bind with soft bases. This preference is governed primarily by electrostatic interactions for hard species and by covalent interactions for soft species.

核心规则是：硬酸偏好与硬碱结合，软酸偏好与软碱结合。这种偏好对于硬物种主要由静电作用主导，对于软物种主要由共价作用主导。

This principle explains many observations in chemistry. For example, Ag⁺ (a soft acid) forms a stable complex with I⁻ (a soft base) rather than with F⁻. In medicine, the toxicity of heavy-metal ions such as Hg²⁺ is related to their affinity for soft sulphur-containing sites in proteins.

该原理解释了化学中的许多现象。例如，Ag⁺（软酸）与I⁻（软碱）形成稳定配合物而非与F⁻结合。在医学中，Hg²⁺等重金属离子的毒性与其对蛋白质中含硫软位点的亲和力有关。

6. Strengths of Lewis Acids and Bases | 路易斯酸碱的强度

The strength of a Lewis acid is determined by its ability to attract an electron pair. For example, the Lewis acidity of boron trihalides follows the order: BF₃ < BCl₃ < BBr₃. This may seem counterintuitive on the basis of electronegativity. The more electronegative fluorine should withdraw more electron density from boron, making BF₃ the most electron-deficient and thus the most acidic. However, in practice BF₃ is the weakest Lewis acid among the boron trihalides. This is because the halogen lone pairs can donate electron density into the empty p-orbital of boron, forming p–π back-bonding. Fluorine's 2p orbitals match well with boron's 2p orbital, providing the most effective stabilisation.

路易斯酸的强度由其吸引电子对的能力决定。例如，三卤化硼的路易斯酸性顺序为：BF₃ < BCl₃ < BBr₃。这看起来似乎与电负性直觉相悖。电负性更强的氟应该从硼上拉走更多电子密度，使BF₃最缺电子、酸性最强。然而实践中BF₃是三卤化硼中最弱的路易斯酸。原因是卤素的孤对电子可以向硼的空p轨道提供电子密度，形成p–π反馈键。氟的2p轨道与硼的2p轨道匹配良好，提供了最有效的稳定化作用。

For Lewis bases, gas-phase basicity indicates that NH₃ is a stronger Lewis base than PH₃ because nitrogen is more electronegative and holds its lone pair more tightly. However, in aqueous solution the trend can reverse due to solvation effects.

对于路易斯碱，气相碱性表明NH₃比PH₃是更强的路易斯碱，因为氮的电负性更高，能更紧密地持有孤对电子。然而在水溶液中，由于溶剂化效应，该趋势可能发生反转。

7. Applications of Lewis Acid–Base Theory | 路易斯酸碱理论的应用

Lewis acid–base theory is indispensable for understanding organic reaction mechanisms. In electrophilic addition to alkenes, the electrophile (E⁺) acts as a Lewis acid and the alkene pi-bond acts as a Lewis base. In nucleophilic substitution, the nucleophile is the Lewis base and the substrate carbon acts as the Lewis acid.

路易斯酸碱理论是理解有机反应机理不可或缺的工具。在烯烃的亲电加成中，亲电体（E⁺）作为路易斯酸，烯烃的π键作为路易斯碱。在亲核取代反应中，亲核体是路易斯碱，底物的碳是路易斯酸。

A common illustrative example is the reaction of aluminium chloride with an acyl chloride in Friedel–Crafts acylation:

一个常见的示例是Friedel-Crafts酰基化中氯化铝与酰氯的反应：

CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻

Here, AlCl₃ accepts a chloride ion (a Lewis base), generating the acylium ion CH₃CO⁺, which is a strong electrophile in aromatic substitution.

此反应中AlCl₃接受氯离子（路易斯碱），生成酰基正离子CH₃CO⁺，后者是芳香取代反应中的强亲电体。

Lewis acid–base theory also explains the catalytic role of transition metals in industrial processes. In heterogeneous catalysis, reactant molecules donate electron density to the metal surface, weakening their internal bonds and lowering the activation energy. In enzyme catalysis, metal ions in active sites act as Lewis acids to stabilise negatively charged transition states.

路易斯酸碱理论还解释了过渡金属在工业催化的作用。在多相催化中，反应物分子向金属表面提供电子密度，削弱其内部键合并降低活化能。在酶催化中，活性位点中的金属离子作为路易斯酸来稳定带负电荷的过渡态。

8. Common Pitfalls and Exam Tips | 常见误区与考试提示

Students often confuse the Lewis definition with the oxidation-state concept. Remember: Lewis acidity depends on electron-pair acceptance, not on oxidation state. For example, CO₂ has carbon in the +4 oxidation state, but it is a Lewis acid because the carbon can accept an electron pair from a base.

学生常将路易斯定义与氧化态概念混淆。记住：路易斯酸性取决于接受电子对的能力，而非氧化态。例如CO₂中碳为+4氧化态，但它是路易斯酸，因为碳可以接受来自碱的电子对。

Another common error is claiming that all metal cations are strong Lewis acids without considering charge density. A small, highly charged cation such as Al³⁺ is a stronger Lewis acid than a large, singly charged cation such as Na⁺.

另一个常见错误是认为所有金属阳离子都是强路易斯酸，而未考虑电荷密度。小体积高电荷的阳离子如Al³⁺比大体积单价阳离子Na⁺是更强的路易斯酸。

In exam questions, students may be asked to identify the Lewis acid and base in a given reaction. The safest strategy is to follow the electrons: locate the species that provides the lone pair and the species that accepts it.

在考题中，学生常被要求识别给定反应中的路易斯酸和碱。最稳妥的策略是追踪电子：找到提供孤对电子的物种和接受孤对电子的物种。

When drawing reaction mechanisms, always use a curved arrow. The arrow must originate from the lone pair of the Lewis base and point towards the atom of the Lewis acid that forms the new bond.

绘制反应机理时，务必使用弯箭头。箭头必须从路易斯碱的孤对电子出发，指向路易斯酸中形成新键的原子。

Finally, do not claim that H⁺ exists as a bare proton in aqueous solution. It is always hydrated to form H₃O⁺ or more complex species such as H₅O₂⁺.

最后，不要声称H⁺在水溶液中以裸质子形式存在。它总是水合形成H₃O⁺或更复杂的物种如H₅O₂⁺。

9. Worked Example | 例题精解

Consider the reaction: Ag⁺ + 2NH₃ → [Ag(NH₃)₂]⁺

考虑反应：Ag⁺ + 2NH₃ → [Ag(NH₃)₂]⁺

Identify the Lewis acid, the Lewis base and the dative bond.

请识别路易斯酸、路易斯碱和配位键。

Solution: Ag⁺ is the Lewis acid because it accepts electron pairs from two ammonia molecules. Each NH₃ is a Lewis base because nitrogen donates its lone pair to Ag⁺. Each Ag–N bond is a dative covalent bond.

解答：Ag⁺是路易斯酸，因为它接受来自两个氨分子的电子对。每个NH₃是路易斯碱，因为氮向Ag⁺给出孤对电子。每个Ag–N键都是配位共价键。

In this complex, silver has a coordination number of 2 with a linear geometry. The dative bonds are equivalent in strength and length to any ordinary covalent bond of the same type.

在该配合物中，银的配位数为2，呈直线型几何构型。两个配位键在强度和键长上与同类型的普通共价键完全等同。

10. Summary | 总结

The Lewis acid–base theory extends far beyond proton transfer, providing a unified framework for understanding acids and bases in both aqueous and non-aqueous systems, organic and inorganic chemistry, and catalysis. Mastering the ability to identify electron-pair donors and acceptors is essential for success in IB Chemistry HL.

路易斯酸碱理论远远超越了质子转移的范畴，为理解水溶液和非水体系、有机化学和无机化学以及催化反应中的酸碱行为提供了统一框架。掌握识别电子对给体和受体的能力是IB化学HL取得成功的关键。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Comparison of Major Acid-Base Theories | IB化学：酸碱理论主要学说比较

Acid–base theories are fundamental to chemistry, yet no single definition explains every reaction. In IB Chemistry, you are expected to understand and apply three major models — Arrhenius, Brønsted–Lowry, and Lewis — and to recognise why each was proposed and where it succeeds or fails.

酸碱理论是化学的基础，但没有任何一个定义能解释所有反应。在IB化学中，你需要理解并运用三大主要模型——阿伦尼乌斯、布仑斯特-劳里和路易斯——并认识到它们各自为何被提出、在何处有效、在何处失效。

1. The Need for Multiple Theories | 1. 多种理论存在的必要性

Early acid–base ideas were based on observable properties such as sour taste, colour changes in dyes, and reaction with carbonates. These observations led to empirical definitions, but they could not explain reactions in non-aqueous solvents or in gases.

早期酸碱概念建立在可观察的性质上，如酸味、染料颜色变化以及与碳酸盐的反应。这些观察产生了经验性定义，但不能解释非水溶剂中或气相中的反应。

As chemists studied more reactions, they realised that acidity and basicity are not absolute properties of substances alone; they depend on the reaction context. This is why multiple theories exist.

随着化学家研究更多反应，他们意识到酸性和碱性并非物质自身的绝对属性，而是取决于反应环境。这正是多种理论并存的原因。

2. Arrhenius Theory | 2. 阿伦尼乌斯理论

In 1887, Svante Arrhenius defined an acid as a substance that produces hydrogen ions (H⁺) in aqueous solution, and a base as a substance that produces hydroxide ions (OH⁻) in aqueous solution.

1887年，斯万特·阿伦尼乌斯将酸定义为在水溶液中产生氢离子（H⁺）的物质，将碱定义为在水溶液中产生氢氧根离子（OH⁻）的物质。

For example, HCl dissolves in water to give H⁺ and Cl⁻, while NaOH gives Na⁺ and OH⁻. The neutralisation reaction is written simply as H⁺ + OH⁻ → H₂O.

例如，HCl溶于水产生H⁺和Cl⁻，而NaOH产生Na⁺和OH⁻。中和反应可简单写为 H⁺ + OH⁻ → H₂O。

HCl(aq) → H⁺(aq) + Cl⁻(aq)

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

This theory is simple and useful for common aqueous solutions, especially in introductary stoichiometry and titration calculations.

该理论简单实用，适用于常见水溶液，尤其是基础化学计量学和滴定计算。

3. Limitations of Arrhenius Theory | 3. 阿伦尼乌斯理论的局限

The Arrhenius theory fails outside water. For example, ammonia acting as a base in water is explained, but much of the reaction happens because NH₃ accepts a proton from water rather than releasing OH⁻ directly.

阿伦尼乌斯理论在水之外失效。例如，氨在水中表现为碱，但该反应实际上是因为NH₃从水中接受质子，而不是直接解离出OH⁻。

Also, many bases that do not contain hydroxide ion, such as ammonia, cannot be classified by the Arrhenius definition.

此外，许多不含氢氧根离子的碱，如氨，无法用阿伦尼乌斯定义归类。

• Acidic or basic behaviour is limited to aqueous solution.

  酸碱行为仅限于水溶液。

• Only one type of base (OH⁻ producer) is allowed.

  只允许一种碱（产生OH⁻的物质）。

• Cannot explain reactions in liquid ammonia, sulphuric acid, or the gas phase.

  无法解释液氨、液态硫酸或气相中的反应。

4. Brønsted–Lowry Theory | 4. 布仑斯特-劳里理论

In 1923, Johannes Brønsted and Thomas Lowry independently defined an acid as a proton (H⁺) donor and a base as a proton acceptor. This definition works in any solvent, not just water.

1923年，约翰内斯·布仑斯特与托马斯·劳里独立将酸定义为质子（H⁺）给予体，将碱定义为质子接受体。该定义适用于任何溶剂，而不仅是水。

In the reaction between ammonia and water, NH₃ accepts a proton from H₂O, so NH₃ is the base and H₂O is the acid.

在氨与水的反应中，NH₃从H₂O获得质子，因此NH₃是碱，H₂O是酸。

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

Here, water is amphiprotic because it can act as both an acid and a base.

这里水是两性物质，因为它既可作为酸也可作为碱。

5. Conjugate Acid–Base Pairs | 5. 共轭酸碱对

A Brønsted–Lowry acid–base reaction always involves two conjugate pairs. When an acid donates a proton, it becomes its conjugate base; when a base accepts a proton, it becomes its conjugate acid.

布仑斯特-劳里酸碱反应始终涉及两对共轭对。当酸失去质子时，变成其共轭碱；当碱获得质子时，变成其共轭酸。

For a general acid HA, the equilibrium is:

对于一般酸HA，平衡可表示为：

HA(aq) + H₂O(l) ⇌ A⁻(aq) + H₃O⁺(aq)

The pair HA/A⁻ is a conjugate pair, and H₃O⁺/H₂O is another conjugate pair.

HA/A⁻是一个共轭对，H₃O⁺/H₂O是另一个共轭对。

Strong acids have weak conjugate bases, and weak acids have relatively stronger conjugate bases.

强酸的共轭碱很弱，而弱酸的共轭碱相对较强。

6. Amphiprotic Species | 6. 两性物质

An amphiprotic species can donate or accept a proton, meaning it has both a conjugate acid and a conjugate base. Hydrogen carbonate ion (HCO₃⁻) is a classic example.

两性物质既能给出质子也能接受质子，这意味着它既有共轭酸也有共轭碱。碳酸氢根离子（HCO₃⁻）是典型例子。

As an acid: HCO₃⁻(aq) + H₂O(l) ⇌ CO₃²⁻(aq) + H₃O⁺(aq)

作为酸： HCO₃⁻(aq) + H₂O(l) ⇌ CO₃²⁻(aq) + H₃O⁺(aq)

As a base: HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq)

作为碱： HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq)

Water, HSO₄⁻, H₂PO₄⁻, and amino acids can also be amphiprotic.

水、HSO₄⁻、H₂PO₄⁻以及氨基酸也是两性物质。

7. Lewis Theory | 7. 路易斯理论

Gilbert Lewis proposed an even broader theory: a Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor.

吉尔伯特·路易斯提出了更广泛的理论：路易斯酸是电子对接受体，路易斯碱是电子对给予体。

This definition includes reactions with no proton transfer. For example, boron trifluoride reacts with ammonia:

该定义包含了没有质子转移的反应。例如，三氟化硼与氨的反应：

BF₃(g) + NH₃(g) → F₃B–NH₃(s)

BF₃ has an incomplete octet and accepts a lone pair from NH₃, so BF₃ is a Lewis acid and NH₃ is a Lewis base.

BF₃具有不完全的八隅体，接受NH₃的孤对电子，因此BF₃是路易斯酸，NH₃是路易斯碱。

All Brønsted–Lowry acids are also Lewis acids because donating a proton involves accepting an electron pair from the base.

所有布仑斯特-劳里酸同时也是路易斯酸，因为给出质子意味着接受来自碱的电子对。

8. Lewis Theory and Complex Ions | 8. 路易斯理论与配合离子

The Lewis definition is especially useful for transition-metal complex ions. A central metal cation acts as a Lewis acid, while ligands act as Lewis bases because they donate lone pairs.

路易斯定义尤其适用于过渡金属配合离子。中心金属阳离子作为路易斯酸，配体因提供孤对电子而作为路易斯碱。

Cu²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq)

Similarly, Fe³⁺ accepts six lone pairs from water or cyanide ions to form hydrated or cyano complexes.

类似地，Fe³⁺接受来自水或氰离子的六对孤对电子，形成水合或氰配合物。

• Lewis acids often include metal ions, BF₃, AlCl₃, and SO₃.

  路易斯酸常包括金属离子、BF₃、AlCl₃和SO₃。

• Lewis bases include NH₃, H₂O, OH⁻, and halide ions.

  路易斯碱包括NH₃、H₂O、OH⁻和卤离子。

• Coordinate covalent bonds formed this way are central to complex ion chemistry.

  由此形成的配位共价键是配合离子化学的核心。

9. Comparison of the Three Theories | 9. 三种理论比较

The table below summarises the key differences.

下表总结了关键区别。

Example classification of the same species under each theory:

同一物种在不同理论下的分类示例：

• HCl: Arrhenius acid, Brønsted–Lowry acid, Lewis acid.

  HCl：阿伦尼乌斯酸、布仑斯特-劳里酸、路易斯酸。

• NaOH: Arrhenius base; OH⁻ is a Brønsted–Lowry base and a Lewis base.

  NaOH：阿伦尼乌斯碱；OH⁻是布仑斯特-劳里碱和路易斯碱。

• BF₃: Not an Arrhenius or Brønsted–Lowry acid (in the classical sense), but a Lewis acid.

  BF₃：不是阿伦尼乌斯酸，也不是经典意义上的布仑斯特-劳里酸，但是路易斯酸。

10. Choosing the Right Theory in IB Exams | 10. IB考试中如何选择正确理论

In IB Chemistry multiple-choice and short-answer questions often ask you to identify conjugate pairs or classify a reaction using a specific theory.

在IB化学选择题和简答题中，常要求你识别共轭对，或使用特定理论对反应进行分类。

When the question mentions ‘proton donor/acceptor’, you must apply the Brønsted–Lowry definition. When it mentions ‘electron-pair donor/acceptor’, use Lewis theory.

当题目提到“质子给予体/接受体”时，必须使用布仑斯特-劳里定义；当提到“电子对给予体/接受体”时，使用路易斯理论。

For acid–base equilibria and pH calculations, Brønsted–Lowry theory is the default. For complex-ion formation and non-proton transfer reactions, Lewis theory is necessary.

对于酸碱平衡和pH计算，布仑斯特-劳里理论是默认选择。对于配合离子形成和无质子转移的反应，路易斯理论是必需的。

A common exam trap is calling NH₃ a base because it ‘produces OH⁻’ — better to say it accepts a proton from water, producing OH⁻ as a result.

一个常见考试陷阱是将NH₃称为“产生OH⁻的碱”——更准确的说法是它从水分子获得质子，从而生成OH⁻。

11. Relationship Between the Theories | 11. 三种理论之间的联系

Arrhenius theory is a subset of Brønsted–Lowry theory, which in turn is a subset of Lewis theory.

阿伦尼乌斯理论是布仑斯特-劳里理论的子集，而布仑斯特-劳里理论又是路易斯理论的子集。

Every Arrhenius acid is a Brønsted–Lowry acid, and every Brønsted–Lowry acid is a Lewis acid. The reverse is not true.

所有阿伦尼乌斯酸都是布仑斯特-劳里酸，所有布仑斯特-劳里酸都是路易斯酸；反之则不然。

For example, H⁺ is both a proton and an empty orbital capable of accepting an electron pair, so it fits all three definitions. BF₃ has no protons, so it only fits the Lewis definition.

例如，H⁺既是质子，又具有能接受电子对的空轨道，因此符合全部三种定义。BF₃没有质子，因此只符合路易斯定义。

Arrhenius ⊂ Brønsted–Lowry ⊂ Lewis

This hierarchy helps explain why Lewis theory is the most powerful, yet also the least intuitive for simple aqueous acid–base reactions.

这一层次关系有助于理解为何路易斯理论最强大，但对简单的溶液酸碱反应却最不直观。

12. Key Takeaways for Your Revision | 12. 复习要点总结

You should be able to define and compare the three theories with examples and limitations.

你应该能够定义三种理论，并举例说明它们的异同与局限。

• Arrhenius: aqueous H⁺ / OH⁻; limited to water.

  阿伦尼乌斯：水溶液中的H⁺/OH⁻；局限在水溶液。

• Brønsted–Lowry: proton transfer; works in any solvent.

  布仑斯特-劳里：质子转移；适用于任何溶剂。

• Lewis: electron-pair transfer; broadest definition, covers complex ions and gas-phase reactions.

  路易斯：电子对转移；定义最广，涵盖配合离子和气相反应。

• Always identify conjugate acid–base pairs in equilibria.

  在平衡问题中始终识别共轭酸碱对。

• Use the theory that matches the question’s language: ‘proton’ → Brønsted–Lowry; ‘electron pair’ → Lewis.

  选择与题目语言匹配的理论：出现“质子”用布仑斯特-劳里；出现“电子对”用路易斯。

Mastering these distinctions will help you solve IB questions confidently and avoid common misconceptions.

掌握这些区别将帮助你自信地解决IB题目，并避免常见误解。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Core Concepts of Chemical Equilibrium | IB化学：化学平衡核心考点梳理

Chemical equilibrium is one of the most fundamental topics in IB Chemistry. It explains how reversible reactions reach a state where the rates of the forward and reverse reactions are equal, and it forms the basis for understanding acid-base, redox, and solubility equilibria. This article summarizes the core concepts and problem-solving strategies you need for your exams.

化学平衡是IB化学中最基础且最重要的主题之一。它解释了可逆反应如何达到正逆反应速率相等的状态，也是理解酸碱平衡、氧化还原平衡和溶解平衡的基础。本文为你梳理考试中必备的核心概念与解题策略。

1. Dynamic Equilibrium | 动态平衡

A reversible reaction reaches dynamic equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant, but both reactions are still occurring continuously. Macroscopic properties such as color, pressure, and concentration no longer change.

当正反应速率等于逆反应速率时，可逆反应达到动态平衡。平衡时，反应物和产物的浓度保持不变，但正逆反应仍在持续进行。宏观性质如颜色、压力和浓度不再变化。

• Equilibrium is dynamic at the molecular level, not static.
• It can only occur in a closed system.
• Equilibrium can be approached from either direction.

• 平衡在分子层面是动态的，而不是静态的。
• 平衡只能在封闭体系中建立。
• 平衡可以从正反应或逆反应方向达到。

2. Equilibrium Constant Kc | 平衡常数 Kc

For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentrations is written as:

对于一般反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数写作：

Kc = [C]ᶜ [D]ᵈ / ([A]ᵃ [B]ᵇ)

Square brackets denote molar concentrations at equilibrium. Pure solids and pure liquids are omitted from the expression. The value of Kc depends only on temperature, not on initial concentrations or the presence of a catalyst.

方括号表示平衡时的摩尔浓度。纯固体和纯液体不出现在表达式中。Kc 只取决于温度，与初始浓度或催化剂的存在无关。

• Kc > 10³: equilibrium lies far to the right (product-favoured).
• Kc < 10⁻³: equilibrium lies far to the left (reactant-favoured).
• 10⁻³ < Kc < 10³: significant amounts of both reactants and products.

• Kc > 10³：平衡强烈向右移动，有利于产物。
• Kc < 10⁻³：平衡强烈向左移动，有利于反应物。
• 10⁻³ < Kc < 10³：反应物和产物都有显著存在。

3. Kc and Kp for Gases | 气体反应的 Kc 与 Kp

For reactions involving gases, the equilibrium constant can also be expressed using partial pressures. For aA + bB ⇌ cC + dD, the pressure equilibrium constant is:

对于涉及气体的反应，平衡常数也可以用分压表示。对于 aA + bB ⇌ cC + dD，压力平衡常数为：

Kp = (P_C)ᶜ (P_D)ᵈ / ((P_A)ᵃ (P_B)ᵇ)

Partial pressures are usually measured in atm or kPa. For the same reaction at the same temperature, Kp and Kc are related by the equation:

分压通常以 atm 或 kPa 为单位。在同一温度下，同一反应的 Kp 与 Kc 的关系为：

Kp = Kc (RT)^Δn_gas

where Δn_gas = (c + d) − (a + b), which is the change in the number of moles of gas. R is the ideal gas constant, and T is the absolute temperature.

其中 Δn_gas = (c + d) − (a + b)，即气体物质的量变化值。R 是理想气体常数，T 是绝对温度。

4. Reaction Quotient Q | 反应商 Q

The reaction quotient Q has the same mathematical form as Kc, but uses concentrations at any moment, not necessarily at equilibrium. Comparing Q with Kc tells us which direction the reaction will shift to reach equilibrium.

反应商 Q 与 Kc 的数学表达式相同，但使用的是任意时刻的浓度，不一定处于平衡。比较 Q 与 Kc 的大小可判断反应将向哪个方向移动以达到平衡。

5. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. This principle allows us to predict the qualitative effect of changes in concentration, pressure, and temperature.

勒夏特列原理指出：当动态平衡受到条件改变的干扰时，平衡位置会向抵消该改变的方向移动。该原理帮助我们定性预测浓度、压力和温度变化的影响。

• Adding a reactant shifts equilibrium to the right.
• Removing a product also shifts equilibrium to the right.
• Changes in pressure only affect reactions with unequal numbers of gas moles.
• Temperature changes alter the value of K itself.

• 增加反应物，平衡向右移动。
• 减少产物，平衡向右移动。
• 压力变化只影响反应前后气体物质的量不等的反应。
• 温度变化会改变 K 值本身。

6. Effect of Concentration Changes | 浓度变化的影响

When the concentration of a species is changed, Q no longer equals Kc. The system responds by shifting in the direction that consumes the added species or replenishes the removed species. For example, in the reaction N₂ + 3H₂ ⇌ 2NH₃, adding N₂ increases Q, causing a shift to the right to produce more NH₃.

当某物种的浓度改变时，Q 不再等于 Kc。体系会朝着消耗新增物种或补充被移除物种的方向移动。例如，在 N₂ + 3H₂ ⇌ 2NH₃ 中，增加 N₂ 会使 Q 暂时变小（或增加分母），促使平衡向右移动以生成更多 NH₃。

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

If NH₃ is removed continuously from the system, the equilibrium shifts right, increasing the yield of ammonia. This is the principle behind industrial ammonia synthesis.

如果不断从体系中移走 NH₃，平衡将向右移动，提高氨的产率。这是工业合成氨的原理之一。

7. Effect of Pressure and Volume | 压力与体积的影响

For reactions involving gases, changing pressure or volume affects equilibrium only if Δn_gas ≠ 0. Increasing pressure (decreasing volume) shifts the equilibrium toward the side with fewer moles of gas. Decreasing pressure (increasing volume) shifts it toward the side with more moles of gas.

对于涉及气体的反应，只有当 Δn_gas ≠ 0 时，改变压力或体积才会影响平衡。增大压力（减小体积）会使平衡向气体物质的量较少的一侧移动；减小压力（增大体积）则向气体物质的量较多的一侧移动。

• N₂ + 3H₂ ⇌ 2NH₃: increasing pressure favours NH₃ (4 mol → 2 mol).
• PCl₅ ⇌ PCl₃ + Cl₂: increasing pressure favours PCl₅ (1 mol → 2 mol).
• Adding an inert gas at constant volume does not shift equilibrium.

• N₂ + 3H₂ ⇌ 2NH₃：增大压力有利于 NH₃（4 mol → 2 mol）。
• PCl₅ ⇌ PCl₃ + Cl₂：增大压力有利于 PCl₅（1 mol → 2 mol）。
• 恒容条件下加入惰性气体不会使平衡移动。

Note: Adding an inert gas at constant pressure increases the total volume, thereby decreasing the partial pressures of reactants and products. If Δn_gas ≠ 0, the equilibrium will shift in the direction that produces more gas moles.

注意：恒压条件下加入惰性气体会增大总体积，从而降低反应物和产物的分压。若 Δn_gas ≠ 0，平衡将向产生更多气体物质的量的方向移动。

8. Effect of Temperature and van’t Hoff Equation | 温度的影响与范特霍夫方程

Temperature is the only factor that changes the value of K. For an exothermic forward reaction, increasing temperature shifts equilibrium to the left and decreases Kc. For an endothermic forward reaction, increasing temperature shifts equilibrium to the right and increases Kc.

温度是唯一能改变 K 值的因素。对于正向放热的反应，升高温度会使平衡向左移动并减小 Kc；对于正向吸热的反应，升高温度会使平衡向右移动并增大 Kc。

The temperature dependence of K is given by the van’t Hoff equation:

K 对温度的依赖关系由范特霍夫方程给出：

ln(K₂/K₁) = -ΔH°/R × (1/T₂ – 1/T₁)

where ΔH° is the standard enthalpy change of the reaction, R = 8.314 J mol⁻¹ K⁻¹, and T₁ and T₂ are absolute temperatures. This equation is used to calculate K at a different temperature if K at one temperature and ΔH° are known.

其中 ΔH° 是反应的标准焓变，R = 8.314 J mol⁻¹ K⁻¹，T₁ 和 T₂ 是绝对温度。已知某一温度下的 K 和 ΔH° 时，可直接用该方程计算另一温度下的 K。

9. Catalysts and Equilibrium | 催化剂与平衡

A catalyst speeds up both the forward and reverse reactions equally, so it does not change the position of equilibrium. It only helps the system reach equilibrium faster. Catalysts do not affect Kc or Kp values.

催化剂同时等程度地加快正逆反应速率，因此不改变平衡位置。它只是帮助体系更快地达到平衡。催化剂不影响 Kc 或 Kp 值。

• Catalysts lower the activation energy for both directions.
• Catalysts do not alter the yield at equilibrium.
• Catalysts are important for industrial processes to reduce time and energy costs.

• 催化剂降低正逆两个方向的活化能。
• 催化剂不改变平衡时的产率。
• 催化剂在工业中用于缩短时间、降低能耗。

10. Equilibrium Calculations using ICE Table | 用 ICE 表进行平衡计算

The ICE table (Initial, Change, Equilibrium) is a systematic method for equilibrium problems. For the reaction A + B ⇌ C, assume initial concentrations [A]₀, [B]₀, [C]₀, and let x be the change in concentration of C. Then the change in A is −x, and for B is −x (stoichiometric coefficients must be accounted for).

ICE 表（初始、变化、平衡）是解决平衡问题的系统方法。对于反应 A + B ⇌ C，设初始浓度为 [A]₀、[B]₀、[C]₀，令 x 表示 C 的浓度变化，则 A 的变化为 −x，B 的变化为 −x（需考虑化学计量系数）。

Substitute the equilibrium row into the Kc expression and solve for x. For quadratic equations, use the quadratic formula. Always check which root is chemically meaningful (concentrations cannot be negative).

将平衡行代入 Kc 表达式并解出 x。若出现二次方程，使用求根公式。务必检查哪个根在化学上有意义（浓度不能为负）。

Kc = x / ((a – x)(b – x))

11. Heterogeneous Equilibria and Ksp | 多相平衡与溶度积

For heterogeneous equilibria involving solids or pure liquids, the activity of pure solids and liquids is defined as 1, so they are omitted from equilibrium expressions. For example, in CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium constant is simply Kp = P(CO₂).

对于涉及固体或纯液体的多相平衡，纯固体和纯液体的活度定义为 1，因此不出现在平衡表达式中。例如，CaCO₃(s) ⇌ CaO(s) + CO₂(g) 的平衡常数简化为 Kp = P(CO₂)。

The solubility product Ksp is the equilibrium constant for dissolving a sparingly soluble salt. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq):

溶度积 Ksp 是难溶盐溶解过程的平衡常数。对于 AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)：

Ksp = [Ag⁺][Cl⁻]

Ksp can be used to predict precipitation: if Q > Ksp, precipitation occurs; if Q < Ksp, the solution is unsaturated and no precipitate forms.

Ksp 可用于判断沉淀的生成：若 Q > Ksp，则产生沉淀；若 Q < Ksp，则溶液未饱和，无沉淀生成。

12. Key Exam Tips and Common Mistakes | 考试要点与常见错误

In IB exams, students often lose marks because of missing units, forgetting to omit solids/liquids, or using initial concentrations instead of equilibrium concentrations. Always write the K expression before plugging in numbers, and state whether Kc has units (the IB usually accepts units only if derived from the expression).

在IB考试中，学生常因漏写单位、忘记省略固体/液体，或将初始浓度当作平衡浓度而失分。代入数值前务必先写出 K 的表达式，并说明 Kc 是否有单位（IB通常只有在由表达式导出时才承认单位）。

• Always use the correct stoichiometric coefficients as exponents.
• Check the direction: if Q < K, forward shift; if Q > K, reverse shift.
• Only temperature can change K.
• Do not include solids, pure liquids, or inert gases in K expressions.

• 始终使用正确的化学计量系数作为指数。
• 判断方向：Q < K 正向移动；Q > K 逆向移动。
• 只有温度能改变 K。
• 固体、纯液体、惰性气体不写入 K 表达式。

Published by TutorHao | IB Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Core Concepts of Chemical Kinetics | IB化学：化学动力学核心知识梳理

Chemical kinetics is the branch of chemistry that deals with the rates of chemical reactions and the mechanisms by which they occur. In IB Chemistry, understanding kinetics allows you to predict how fast reactions proceed, interpret rate data, and explain how different conditions affect reaction speed.

化学动力学是化学中研究化学反应速率以及反应发生机理的分支。在IB化学中，理解动力学能够帮助你预测反应进行的快慢、解读速率数据，并解释不同条件如何影响反应速率。

1. Rate of Reaction | 反应速率

The rate of a reaction is defined as the change in concentration of a reactant or product per unit time. It is usually expressed in mol dm⁻³ s⁻¹. For a reactant A, the rate is negative because its concentration decreases; for a product, it is positive.

反应速率定义为反应物或生成物浓度随时间的变化量，通常以 mol dm⁻³ s⁻¹ 为单位。对于反应物 A，其浓度减小，速率取负值；对于生成物，速率取正值。

For a general reaction aA + bB → cC + dD, the rate can be written as:

对于一般反应 aA + bB → cC + dD，速率可以写为：

Rate = −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt = (1/d)Δ[D]/Δt

The average rate is calculated over a time interval, while the instantaneous rate is the slope of the tangent at a specific time on a concentration–time graph.

平均速率是在一段时间间隔内计算的，而瞬时速率是浓度-时间图上某一点切线的斜率。

2. Rate Expression and Order of Reaction | 速率表达式与反应级数

The rate expression (or rate law) relates the rate of a reaction to the concentrations of reactants, each raised to a power. For the reaction aA + bB → products, the experimental rate expression is:

速率表达式（或速率定律）将反应速率与各反应物浓度的幂次关联起来。对于反应 aA + bB → 产物，实验测得的速率表达式为：

Rate = k[A]^m[B]ⁿ

Here, k is the rate constant, m is the order with respect to A, and n is the order with respect to B. The overall order is m + n. The orders are not necessarily equal to the stoichiometric coefficients; they must be determined experimentally.

其中 k 是速率常数，m 是相对于 A 的反应级数，n 是相对于 B 的反应级数。总反应级数为 m + n。反应级数不一定等于化学计量系数，必须通过实验确定。

• Zero order: Rate = k (constant), independent of reactant concentration.
• 零级反应：Rate = k（常数），与反应物浓度无关。
• First order: Rate = k[A], rate proportional to [A].
• 一级反应：Rate = k[A]，速率与 [A] 成正比。
• Second order: Rate = k[A]², or Rate = k[A][B].
• 二级反应：Rate = k[A]²，或 Rate = k[A][B]。

3. Determining Reaction Orders | 测定反应级数

Reaction orders are determined by the initial rates method or by using integrated rate laws and concentration–time graphs. In the initial rates method, you change the concentration of one reactant while keeping others constant and compare the initial rates.

反应级数可通过初始速率法或利用积分速率定律和浓度-时间图来确定。在初始速率法中，保持其他反应物浓度不变，只改变一种反应物的浓度，并比较初始速率。

If doubling [A] doubles the rate while [B] is held constant, the reaction is first order in A. If doubling [A] quadruples the rate, it is second order in A. If the rate remains unchanged, it is zero order in A.

如果保持 [B] 不变，将 [A] 加倍使速率加倍，则反应对 A 是一级；如果将 [A] 加倍使速率变为四倍，则对 A 是二级；如果速率不变，则对 A 是零级。

For a first-order reaction, a graph of ln[A] versus t gives a straight line with slope −k. For a second-order reaction, a graph of 1/[A] versus t is linear with slope k.

对于一级反应，ln[A] 对 t 作图得到斜率为 −k 的直线；对于二级反应，1/[A] 对 t 作图呈直线，斜率为 k。

4. The Rate Constant k and Units | 速率常数k及其单位

The rate constant k is a proportionality constant that depends on temperature and the presence of a catalyst. It is constant for a given reaction at a fixed temperature. The units of k depend on the overall order of the reaction.

速率常数 k 是一个比例常数，取决于温度和催化剂的存在。在固定温度下，对给定反应它是一个常数。k 的单位取决于反应的总级数。

For a reaction with overall order n, the units of k are (mol dm⁻³)¹⁻ⁿ s⁻¹. For a zero-order reaction, k has units mol dm⁻³ s⁻¹. For first order, k has units s⁻¹. For second order, k has units dm³ mol⁻¹ s⁻¹.

对于总级数为 n 的反应，k 的单位为 (mol dm⁻³)¹⁻ⁿ s⁻¹。零级反应的 k 单位是 mol dm⁻³ s⁻¹；一级反应是 s⁻¹；二级反应是 dm³ mol⁻¹ s⁻¹。

5. Collision Theory and Activation Energy | 碰撞理论与活化能

Collision theory states that for a reaction to occur, particles must collide with sufficient energy and correct orientation. Only a small fraction of collisions are effective enough to lead to product formation.

碰撞理论认为，反应发生的条件是粒子必须具有足够的能量并以正确的取向碰撞。只有一小部分碰撞有效到足以生成产物。

The minimum energy required for an effective collision is called the activation energy, Eₐ. This energy is needed to break or weaken existing bonds and form the transition state (activated complex).

有效碰撞所需的最小能量称为活化能 Eₐ。该能量用于断裂或削弱已有化学键，并形成过渡态（活化配合物）。

Rate ∝ collision frequency × fraction of effective collisions

速率 ∝ 碰撞频率 × 有效碰撞比例

6. Maxwell–Boltzmann Distribution | 麦克斯韦-玻尔兹曼分布

The Maxwell–Boltzmann distribution shows the spread of energy among particles in a sample at a given temperature. The area under the curve represents the total number of particles, and the shaded region beyond Eₐ represents particles with energy equal to or greater than the activation energy.

麦克斯韦-玻尔兹曼分布显示在给定温度下样品中粒子能量的分布。曲线下的面积代表粒子总数，超过 Eₐ 的阴影区域代表能量等于或大于活化能的粒子。

Increasing the temperature shifts the distribution curve to the right and flattens it, increasing the fraction of particles with energy above Eₐ. This explains why reaction rates increase sharply with temperature.

升高温度会使分布曲线向右移动并变得平坦，从而增加能量高于 Eₐ 的粒子比例。这就解释了为什么反应速率随温度升高而急剧增加。

Adding a catalyst lowers Eₐ, allowing a larger fraction of particles to have sufficient energy, thereby increasing the rate without changing the temperature.

加入催化剂会降低 Eₐ，使更大比例的粒子具有足够能量，从而在温度不变的条件下提高速率。

7. Factors Affecting Reaction Rate | 影响反应速率的因素

Key factors include concentration, temperature, surface area, pressure (for gases), and catalysts. Each affects the rate by altering collision frequency or the fraction of effective collisions.

关键因素

Published by TutorHao | IB Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Rate Equations and Reaction Mechanism Analysis | IB化学（HL）：速率方程与反应机理分析

The rate equation is a mathematical expression that links the rate of a chemical reaction to the concentrations of the species involved. In IB Chemistry HL, you are expected to determine rate equations from experimental data, interpret concentration–time graphs, and connect the rate-determining step of a proposed mechanism to the observed rate equation.

速率方程是将化学反应速率与参与反应的物种浓度关联起来的数学表达式。在 IB 化学 HL 中，你需要能够根据实验数据确定速率方程、解读浓度–时间图，并将所提出机理的决速步与观测到的速率方程联系起来。

1. The Rate Equation | 速率方程的基本形式

For a general reaction aA + bB → products, the rate equation is often written as:

对于一般反应 aA + bB → 产物，速率方程通常写作：

rate = k[A]ᵐ[B]ⁿ

Here, k is the rate constant, [A] and [B] are concentrations in mol dm⁻³, and m and n are the orders of reaction with respect to A and B. The overall order is m + n.

其中 k 是速率常数，[A] 和 [B] 的单位为 mol dm⁻³，m 和 n 分别是反应对 A 和 B 的分级数。总级数为 m + n。

It is essential to understand that m and n are empirical quantities. They cannot be deduced from the stoichiometric coefficients in the balanced equation; they must be found from experimental data.

必须明确的是，m 和 n 是经验量。它们不能由配平方程式中的化学计量系数推导出来，而必须通过实验数据确定。

2. Orders of Reaction | 反应级数的意义

The order with respect to a reactant tells you how the initial rate changes when the concentration of that reactant changes, with all other concentrations held constant.

对某一反应物的级数表示在其他浓度保持不变时，改变该反应物浓度会如何影响初始速率。

• Zero order: changing the concentration has no effect on the rate.
• 零级：改变浓度对速率没有影响。
• First order: doubling the concentration doubles the rate; the rate is directly proportional to concentration.
• 一级：浓度加倍，速率加倍；速率与浓度成正比。
• Second order: doubling the concentration quadruples the rate; the rate is proportional to the square of concentration.
• 二级：浓度加倍，速率变为原来的 4 倍；速率与浓度的平方成正比。

3. Determining Orders from Initial Rates | 通过初始速率法确定级数

The initial rates method involves running several experiments with different initial concentrations and measuring the initial rate of each. Comparing experiments where only one concentration changes at a time allows you to identify each order.

初始速率法是在若干次实验中采用不同的初始浓度，并测量每次反应的初始速率。通过比较只改变一种浓度的实验，就可以确定各个级数。

Worked example: For the reaction A + B → products, the following initial rates were measured.

示例：对于反应 A + B → 产物，测得如下初始速率。

Comparing experiments 1 and 2, doubling [A] while keeping [B] constant quadruples the rate. Therefore the order with respect to A is 2. Comparing experiments 1 and 3, doubling [B] leaves the rate unchanged, so the order with respect to B is 0. The rate equation is rate = k[A]².

比较实验 1 和实验 2，在 保持 [B] 不变 的条件下将 [A] 加倍，速率变为原来的 4 倍，因此对 A 的级数为 2。比较实验 1 和实验 3，将 [B] 加倍而速率不变，因此对 B 的级数为 0。速率方程为 rate = k[A]²。

4. Concentration–Time Graphs and Integrated Rate Laws | 浓度–时间图与积分速率方程

Concentration–time graphs can also reveal the order of a reaction. For different orders, different functions of concentration give straight lines when plotted against time.

浓度–时间图同样可以揭示反应级数。对于不同级数，将浓度的不同函数对时间作图可得直线。

• Zero order: [A] versus t gives a straight line with slope −k.
• 零级：[A] 对 t 作图得直线，斜率为 −k。
• First order: ln[A] versus t gives a straight line with slope −k.
• 一级：ln[A] 对 t 作图得直线，斜率为 −k。
• Second order: 1/[A] versus t gives a straight line with slope k.
• 二级：1/[A] 对 t 作图得直线，斜率为 k。

The integrated rate law for a first-order reaction is:

一级反应的积分速率方程为：

ln[A] = −kt + ln[A]₀

where [A]₀ is the initial concentration. This equation has the form y = mx + c, so a plot of ln[A] against t gives a straight line.

其中 [A]₀ 为初始浓度。该方程具有 y = mx + c 的形式，因此 ln[A] 对 t 作图得到直线。

5. Half-Life | 半衰期

The half-life, t½, is the time taken for the concentration of a reactant to fall to half its original value.

半衰期 t½ 是反应物浓度降到初始值一半所需的时间。

• For a first-order reaction, t½ is independent of concentration and is given by t½ = ln2 / k.
• 对于一级反应，t½ 与浓度无关，且 t½ = ln2 / k。
• For a zero-order reaction, t½ = [A]₀ / (2k), so the half-life decreases as the initial concentration decreases.
• 对于零级反应，t½ = [A]₀ / (2k)，因此半衰期随初始浓度降低而减小。
• For a second-order reaction, t½ = 1 / (k[A]₀), so the half-life increases as the reaction proceeds.
• 对于二级反应，t½ = 1 / (k[A]₀)，因此半衰期随反应进行而增大。

In first-order reactions, equal successive half-lives over equal time intervals are a useful diagnostic sign.

在一级反应中，等时间间隔内出现连续等长的半衰期是一个很有用的判据。

6. Units of the Rate Constant | 速率常数的单位

The units of k depend on the overall order of the reaction and can be derived from the rate equation:

k 的单位取决于反应总级数，可以通过速率方程推导：

• Zero order: rate has units mol dm⁻³ s⁻¹, so k has units mol dm⁻³ s⁻¹.
• 零级：速率的单位为 mol dm⁻³ s⁻¹，因此 k 的单位为 mol dm⁻³ s⁻¹。
• First order: k has units s⁻¹.
• 一级：k 的单位为 s⁻¹。
• Second order: k has units dm³ mol⁻¹ s⁻¹.
• 二级：k 的单位为 dm³ mol⁻¹ s⁻¹。

General rule: units of k = (mol dm⁻³)^(1−n) s⁻¹ for an nth-order overall reaction.

通用规则：对于总级数为 n 的反应，k 的单位 = (mol dm⁻³)^(1−n) s⁻¹。

7. Temperature Dependence and the Arrhenius Equation | 温度依赖性与阿伦尼乌斯方程

The rate constant k changes with temperature. The Arrhenius equation describes this relationship:

速率常数 k 随温度变化。阿伦尼乌斯方程描述了这一关系：

k = A exp(−Eₐ / RT)

Here, A is the Arrhenius pre-exponential factor, Eₐ is the activation energy in J mol⁻¹, R is the gas constant (8.31 J K⁻¹ mol⁻¹), and T is the absolute temperature in kelvin.

其中 A 为阿伦尼乌斯指前因子，Eₐ 为活化能（单位为 J mol⁻¹），R 为气体常数（8.31 J K⁻¹ mol⁻¹），T 为热力学温度（单位为 K）。

Taking natural logarithms gives the linear form:

取自然对数得到线性形式：

ln k = −Eₐ / RT + ln A

Thus, a plot of ln k against 1/T should be a straight line with slope −Eₐ/R. This allows the activation energy to be determined experimentally.

因此，ln k 对 1/T 作图应为直线，斜率为 −Eₐ/R。这样就可以通过实验测定活化能。

8. Reaction Mechanisms and Elementary Steps | 反应机理与基元步骤

A reaction mechanism is a sequence of elementary steps that describes the pathway from reactants to products. Each elementary step has a molecularity, which is the number of species that must collide in that step.

反应机理是描述从反应物到产物路径的一系列基元步骤。每个基元步骤都有其分子性，即该步骤中必须发生碰撞的物种数目。

• Unimolecular step: a single molecule rearranges or decomposes; rate = k[A].
• 单分子步骤：单个分子发生重排或分解；rate = k[A]。
• Bimolecular step: two species collide; rate = k[A][B] (or k[A]² if both are the same species).
• 双分子步骤：两个物种碰撞；rate = k[A][B]（若两个物种相同则为 k[A]²）。
• Termolecular step: three species collide; these are rare because three-body collisions are uncommon.
• 三分子步骤：三个物种碰撞；由于三体碰撞罕见，这种步骤很少见。

Mechanisms must be consistent with the experimentally determined rate equation. They can never be proven absolutely, but they can be disproved if they predict a different rate law.

机理必须与实验测定的速率方程一致。机理永远无法被绝对证明，但如果它预测出的速率定律不同，则可以被否定。

9. The Rate-Determining Step | 决速步

The rate-determining step (RDS) is the slowest step in a reaction mechanism. The overall rate of reaction is determined by this slowest step.

决速步（RDS）是反应机理中最慢的一步。整个反应的速率由这一步决定。

For a two-step mechanism:

对于两步机理：

Step 1 (slow): NO₂ + NO₂ → NO₃ + NO

Step 2 (fast): NO₃ + CO → NO₂ + CO₂

The rate law is predicted from the slow step: rate = k[NO₂]². This matches the experimental second-order behaviour in NO₂. The CO concentration does not appear because CO participates only after the rate-determining step.

速率定律由慢步骤预测：rate = k[NO₂]²。这与实验上对 NO₂ 的二级行为吻合。CO 的浓度没有出现，因为 CO 在决速步之后才参与。

10. Reaction Intermediates and Catalysts in Mechanisms | 反应中间体与催化剂在机理中的作用

A reaction intermediate is a species that is produced in one elementary step and consumed in a later step. It does not appear in the overall balanced equation. Intermediates often appear in rate equations if they are formed before or during the slow step.

反应中间体是在某个基元步骤中生成、并在后续步骤中被消耗的物种。它不会出现在总反应方程式中。如果中间体在慢步骤之前或慢步骤中生成，它们通常会出现在速率方程中。

Example of a mechanism with an intermediate:

涉及中间体的机理示例：

Step 1 (fast): A → I

Step 2 (slow): I + B → products

If step 1 reaches equilibrium quickly, [I] can be expressed in terms of [A] using the equilibrium constant K = [I]/[A]. Then rate = k₂[I][B] = k₂K[A][B], so the observed rate law is first order in A and first order in B.

如果第 1 步快速达到平衡，可以用平衡常数 K = [I]/[A] 将 [I] 表示为 [A] 的函数。则 rate = k₂[I][B] = k₂K[A][B]，所以观测到的速率定律对 A 为一级、对 B 为一级。

A catalyst provides an alternative pathway with a lower activation energy. It is consumed and then regenerated in a later step, so it does not appear in the overall equation. Catalysts often appear in the rate law if they are involved in the rate-determining step.

催化剂提供了活化能更低的替代反应路径。它在某一步被消耗，在后续步骤中重新生成，因此不出现在总反应方程式中。若催化剂参与决速步，它通常会出现在速率定律中。

11. Worked Example: From Data to Mechanism | 例题：从数据到机理

Consider the reaction H₂(g) + 2ICl(g) → I₂(g) + 2HCl(g). The experimental rate equation is rate = k[H₂][ICl]. A proposed mechanism is:

考虑反应 H₂(g) + 2ICl(g) → I₂(g) + 2HCl(g)。实验速率方程为 rate = k[H₂][ICl]。提出以下机理：

Step 1 (slow): H₂ + ICl → HI + HCl

Step 2 (fast): HI + ICl → I₂ + HCl

The predicted rate law from step 1 is rate = k[H₂][ICl], which matches the experimental rate equation. The intermediate HI is produced in the slow step and consumed in the fast step. This mechanism is consistent with the experimental observations.

由第 1 步预测的速率定律为 rate = k[H₂][ICl]，与实验速率方程一致。中间体 HI 在慢步骤中生成，在快步骤中消耗。该机理与实验观测一致。

If a mechanism predicts a rate law different from the experimental one, the mechanism must be rejected or modified.

如果某个机理预测的速率定律与实验不一致，则该机理必须被否定或修正。

12. Summary of Key Points | 核心要点总结

• Rate orders are determined experimentally, never from stoichiometric coefficients.
• 级数由实验确定，绝不能从化学计量系数得出。
• The rate-determining step controls the overall rate; its rate law must match the experimental rate law.
• 决速步控制总反应速率；其速率定律必须与实验速率定律一致。
• Concentration–time plots and the method of initial rates are essential tools for finding orders and k.
• 浓度–时间图和初始速率法是确定级数和 k 的重要工具。
• The Arrhenius equation connects the rate constant to temperature and activation energy.
• 阿伦尼乌斯方程将速率常数与温度和活化能联系起来。
• Intermediates and catalysts can be identified in mechanisms and often appear in the rate equation when they participate in or precede the slow step.
• 中间体和催化剂可以在机理中被识别；当它们参与决速步或在决速步之前生成时，通常会出现在速率方程中。

Mastering rate equations and reaction mechanisms is central to HL Kinetics. Practice deriving rate laws from data and checking each proposed mechanism step by step.

掌握速率方程和反应机理是 HL 化学动力学部分的核心。请多练习从数据推导速率定律，并逐步检查每个提出的机理。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Collision Theory and Reaction Rates Explained | IB化学：碰撞理论与反应速率关系解析

Collision theory is a fundamental model used to explain how chemical reactions occur and why their rates vary under different conditions. It states that for a reaction to take place, reactant particles must collide with sufficient energy and the correct orientation. This article explores the key concepts of collision theory and its direct relationship with reaction rates, as required for IB Chemistry.

碰撞理论是解释化学反应如何发生以及反应速率在不同条件下为何变化的基本模型。该理论认为，反应要发生，反应物粒子必须具有足够的能量并以正确的取向发生碰撞。本文将深入探讨碰撞理论的核心概念及其与反应速率的直接关系，以满足IB化学课程的要求。

1. What Is Collision Theory? | 什么是碰撞理论？

Collision theory proposes that chemical reactions only occur when reactant particles collide with one another. However, not every collision leads to a reaction. Only collisions that possess enough kinetic energy and have a suitable geometry are effective. The rate of a reaction is therefore proportional to the frequency of effective collisions per unit time.

碰撞理论认为，化学反应只有在反应物粒子相互碰撞时才会发生。然而，并非每一次碰撞都能引发反应。只有具备足够动能并具有合适几何取向的碰撞才是有效碰撞。因此，反应速率与单位时间内有效碰撞的频率成正比。

Reaction rate ∝ effective collision frequency
反应速率 ∝ 有效碰撞频率

2. Conditions for a Successful Collision | 有效碰撞的条件

For a collision to result in a chemical reaction, two essential conditions must be satisfied. First, the colliding particles must have total kinetic energy equal to or greater than the activation energy (Eₐ). Second, the particles must collide with the correct orientation so that the appropriate atoms come into contact with each other.

要使一次碰撞能够引发化学反应，必须满足两个基本条件。第一，碰撞粒子的总动能必须大于或等于活化能（Eₐ）。第二，粒子必须以正确的取向碰撞，使相关原子能够相互接触。

These two conditions explain why many collisions in a gas or solution are unsuccessful. At room temperature, billions of collisions occur every second, yet only a tiny fraction possess enough energy and the right alignment to break and form bonds.

这两个条件解释了为什么气体或溶液中的许多碰撞都是不成功的。在室温下，每秒发生数十亿次碰撞，但只有极少数碰撞拥有足够的能量和正确的排列方式来断裂和形成化学键。

3. Activation Energy and the Energy Barrier | 活化能与能垒

Activation energy is the minimum energy that reactant particles must possess before they can successfully undergo a chemical reaction. It represents the energy barrier that must be overcome to break existing bonds and initiate the formation of new bonds. A reaction with a high activation energy tends to be slow because very few particles have enough energy to overcome the barrier.

活化能是反应物粒子成功进行化学反应所必须拥有的最低能量。它代表必须克服的能量障碍，以断裂已有的化学键并开始形成新的化学键。活化能较高的反应通常较慢，因为只有很少的粒子拥有足够的能量来跨越该能垒。

The relationship between energy and reaction progress is often shown in an energy profile diagram. The highest point on the curve corresponds to the transition state, where bonds are partially broken and partially formed. The difference between the reactants’ energy and the transition state energy equals the activation energy of the forward reaction.

能量与反应进程之间的关系通常用能量剖面图表示。曲线上的最高点对应过渡态，此时化学键部分断裂、部分形成。反应物能量与过渡态能量之差等于正反应的活化能。

4. Molecular Orientation and the Steric Factor | 分子取向与空间因子

Even if particles have sufficient kinetic energy, they must also approach each other in the correct geometric arrangement. For example, in the reaction between NO₂ and CO, the nitrogen atom of NO₂ must collide with the carbon atom of CO. If the oxygen atom of NO₂ hits the carbon atom instead, no reaction occurs, even at high energy.

即使粒子具有足够的动能，它们还必须以正确的几何方式相互接近。例如，在NO₂与CO的反应中，NO₂的氮原子必须与CO的碳原子碰撞。如果NO₂的氧原子撞到碳原子，即使能量很高，也不会发生反应。

The fraction of collisions that occur with the correct orientation is called the steric factor, often denoted by P. For simple reactions involving small molecules, P may be near 1, but for complex molecules, P can be very small, reducing the effective collision rate dramatically.

以正确取向发生的碰撞比例称为空间因子，通常用P表示。对于涉及小分子的简单反应，P可能接近1；但对于复杂分子，P可能非常小，从而显著降低有效碰撞速率。

5. The Maxwell–Boltzmann Distribution | 麦克斯韦-玻尔兹曼分布

In a sample of gas or liquid, individual particles have different kinetic energies. The Maxwell–Boltzmann distribution shows the spread of these energies at a given temperature. The curve begins at the origin, rises to a peak, and then decreases exponentially at higher energies. The area under the curve represents the total number of particles.

在气体或液体样品中，单个粒子具有不同的动能。麦克斯韦-玻尔兹曼分布显示了在给定温度下这些能量的分布范围。曲线从原点开始，上升到峰值，然后在较高能量处指数下降。曲线下的面积代表粒子总数。

Only particles with energy equal to or greater than Eₐ can react. On the distribution curve, this is represented by the shaded area to the right of the Eₐ mark. The size of this area determines the fraction of particles that can undergo an effective collision at any instant.

只有能量大于或等于Eₐ的粒子才能发生反应。在分布曲线上，这用Eₐ标记右侧的阴影区域表示。该区域的大小决定了在任何瞬间能够发生有效碰撞的粒子比例。

6. Temperature and Reaction Rate | 温度与反应速率

Raising the temperature increases the average kinetic energy of particles, causing the Maxwell–Boltzmann curve to shift to the right and become shorter and flatter. The most important effect is that the fraction of particles exceeding Eₐ increases significantly, even with a modest temperature rise. This greatly increases the rate of effective collisions.

升高温度会增加粒子的平均动能，使麦克斯韦-玻尔兹曼曲线向右移动，并变得更低更扁平。最重要的是，即使温度升高幅度不大，超过Eₐ的粒子比例也会显著增加。这大大增加了有效碰撞的速率。

As a rough rule, many reactions double or triple their rates for every 10 °C increase in temperature. Collision theory explains this by emphasizing that the rate depends not on the total collision frequency alone, but on the proportion of collisions that are energetic enough to overcome the activation barrier.

粗略而言，许多反应每升高10 °C，速率会加倍或增加两倍。碰撞理论通过强调速率不仅仅取决于总碰撞频率，还取决于具有足够能量跨越活化能垒的碰撞比例，来解释这一现象。

7. Concentration and Pressure Effects | 浓度与压力影响

Increasing the concentration of reactants in solution raises the number of particles per unit volume. This leads to a higher collision frequency and therefore a higher effective collision rate. Similarly, for gaseous reactions, increasing pressure compresses the gas, bringing particles closer together and increasing the collision frequency.

提高溶液中反应物的浓度会增加单位体积内的粒子数。这导致更高的碰撞频率，从而提高有效碰撞速率。类似地，对于气体反应，增加压力会压缩气体，使粒子更靠近，从而增加碰撞频率。

It is important to note that concentration changes do not alter the activation energy or the energy distribution; they only change how often particles encounter each other. Consequently, the rate increases linearly in many simple reactions, consistent with the proportionality between rate and collision frequency.

需要注意的是，浓度变化不会改变活化能或能量分布；它只改变粒子相互遇见的频率。因此，在许多简单反应中，速率呈线性增加，这与速率和碰撞频率成正比的关系一致。

8. Surface Area and Heterogeneous Reactions | 表面积与非均相反应

For reactions involving solids, the reaction occurs at the surface. If the solid is broken into smaller pieces, its total surface area increases, exposing more particles to the other reactant. This increases the frequency of collisions between the solid and the surrounding reactant molecules, thus speeding up the reaction.

对于涉及固体的反应，反应发生在固体的表面。如果将固体磨成更小的颗粒，其总表面积增大，使更多粒子暴露于其他反应物。这增加了固体与周围反应物分子之间的碰撞频率，从而加快反应速率。

Powdered calcium carbonate reacts with hydrochloric acid much faster than a single large lump of the same mass. The collision theory explains this by considering the number of reactive sites available for collisions per unit time. A larger surface area provides more sites for effective collisions.

粉末状碳酸钙与盐酸的反应比相同质量的一大块碳酸钙快得多。碰撞理论通过考虑单位时间内可供碰撞的反应位点数来解释这一点。更大的表面积提供了更多有效碰撞的位点。

9. Catalysts and Alternative Pathways | 催化剂与替代路径

A catalyst increases the rate of a reaction by providing an alternative pathway with a lower activation energy. This means that at a given temperature, a much larger fraction of particles have enough energy to overcome the lower barrier. The catalyst does not affect the collision frequency or the Maxwell–Boltzmann distribution; it changes the minimum energy required.

催化剂通过提供一条活化能更低的替代路径来提高反应速率。这意味着在给定温度下，有更大比例的粒子拥有足够的能量来跨越更低的能垒。催化剂不影响碰撞频率或麦克斯韦-玻尔兹曼分布；它改变的是所需的最低能量。

For example, in the decomposition of hydrogen peroxide, manganese(IV) oxide catalyzes the reaction by lowering Eₐ from about 75 kJ mol⁻¹ to roughly 50 kJ mol⁻¹. This dramatically increases the number of effective collisions and hence the rate, while the catalyst itself is regenerated at the end.

例如，在过氧化氢分解中，二氧化锰通过将Eₐ从约75 kJ mol⁻¹降低到约50 kJ mol⁻¹来催化反应。这显著增加了有效碰撞次数，从而提高了速率，而催化剂本身在反应结束时被再生。

10. The Role of the Rate Constant and the Arrhenius Equation | 速率常数与阿伦尼乌斯方程的作用

Collision theory provides a qualitative picture, but the Arrhenius equation gives a quantitative link between temperature, activation energy, and the rate constant k. The equation is usually written as:

碰撞理论提供了定性的图像，而阿伦尼乌斯方程则给出了温度、活化能与速率常数k之间的定量联系。该方程通常写作：

k = A e⁻ᵉᵃ/ᴿᵀ

Here, A is the frequency factor related to the collision frequency and steric factor, Eₐ is the activation energy, R is the universal gas constant (8.31 J K⁻¹ mol⁻¹), and T is the absolute temperature. Taking natural logarithms gives a linear form used to determine Eₐ from experimental data.

其中，A是与碰撞频率和空间因子相关的频率因子，Eₐ是活化能，R是通用气体常数（8.31 J K⁻¹ mol⁻¹），T是绝对温度。取自然对数可以得到线性形式，用于从实验数据中确定Eₐ。

The exponential term e⁻ᵉᵃ/ᴿᵀ represents the fraction of molecules with energy equal to or greater than Eₐ. Increasing T or decreasing Eₐ makes this term larger, directly explaining why both temperature rises and catalysts accelerate reactions.

指数项e⁻ᵉᵃ/ᴿᵀ代表能量大于或等于Eₐ的分子比例。增大T或减小Eₐ会使这一项变大，这直接解释了为什么升高温度和加入催化剂都能加速反应。

11. Limitations of Collision Theory | 碰撞理论的局限性

Collision theory is a powerful model, but it has limitations. It treats molecules as simple spheres and ignores the detailed internal motions and electronic rearrangements that occur during a collision. Some reactions, such as those involving large biological molecules, require very specific orientations that the simple steric factor cannot fully capture.

碰撞理论是一个强大的模型，但也有其局限性。它将分子视为简单的球体，忽略了碰撞过程中发生的详细内部运动和电子重排。某些反应，尤其是涉及大生物分子的反应，需要非常特定的取向，简单的空间因子无法完全描述。

Additionally, the theory assumes that all collisions with sufficient energy and orientation lead to products. In reality, some collisions may form an unstable complex that returns to reactants. Transition state theory, which focuses on the activated complex and the energy barrier, provides a more detailed picture of the reaction path.

此外，该理论假设所有具有足够能量和正确取向的碰撞都会生成产物。实际上，一些碰撞可能形成不稳定的复合物并返回反应物。过渡态理论关注活化复合物和能垒，为反应路径提供了更详细的描述。

12. Practical Applications in IB Chemistry | IB化学中的实际应用

Understanding collision theory helps IB students predict and compare reaction rates. When asked to explain why a reaction speeds up under certain conditions, the standard answer involves three factors: more frequent collisions, a higher proportion of successful collisions, or a lower activation energy. Always separate these ideas clearly in exam responses.

理解碰撞理论有助于IB学生预测和比较反应速率。当被要求解释为什么反应在特定条件下加速时，标准答案涉及三个因素：更频繁的碰撞、更高比例的有效碰撞，或更低的活化能。在考试作答中务必清晰区分这些概念。

For example, increasing temperature affects both the frequency and the energy of collisions, but the dominant effect is the exponential increase in the fraction of particles above Eₐ. Adding a catalyst, on the other hand, does not increase collision frequency but instead lowers the energy threshold. A precise explanation using collision theory demonstrates a thorough understanding of chemical kinetics.

例如，升高温度同时影响碰撞频率和碰撞能量，但主导效应是能量高于Eₐ的粒子比例的指数增加。另一方面，加入催化剂并不增加碰撞频率，而是降低能量阈值。用碰撞理论进行精确解释，能够展示对化学动力学的深入理解。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Entropy and Spontaneity in IB Chemistry HL | IB化学（HL）：熵与自发反应判断要点

In IB Chemistry Higher Level, predicting whether a reaction is spontaneous requires more than just enthalpy change. The key lies in entropy and the total entropy change of the universe. This article covers the essential points you need to master: what entropy is, how to calculate entropy changes, and how to use Gibbs free energy to judge spontaneity.

在 IB 化学高级水平（HL）中，判断一个反应是否自发仅看焓变远远不够，关键在于熵以及宇宙总熵变。本文涵盖你需要掌握的核心要点：什么是熵、如何计算熵变，以及如何利用吉布斯自由能判断自发性。

1. What Is Entropy? | 什么是熵？

Entropy, symbol S, is a measure of the dispersal of energy or the number of possible microstates available to a system. In simple terms, it describes the degree of disorder or randomness of particles.

熵，符号为 S，是能量分散程度或系统可能微观状态数的量度。简单来说，它描述粒子无序性（混乱度）的大小。

A system with more possible arrangements has higher entropy. Gases have much higher entropy than liquids and solids. For example, solid NaCl has low entropy, while NaCl dissolved in water has higher entropy because the ions are spread throughout the solution.

具有更多可能排列方式的系统具有更高的熵。气体的熵远高于液体和固体。例如，固体 NaCl 的熵较低，而 NaCl 溶于水后熵较高，因为离子分散在整个溶液中。

The third law of thermodynamics states that the entropy of a perfectly ordered pure crystal at absolute zero (0 K) is zero. This allows chemists to define absolute standard molar entropies, S°, measured in J K⁻¹ mol⁻¹.

热力学第三定律指出，在绝对零度（0 K）下，完美有序的纯晶体熵为零。这使化学家能够定义绝对标准摩尔熵 S°，单位为 J K⁻¹ mol⁻¹。

2. Predicting the Sign of Entropy Change | 判断熵变的正负

Although exact calculation is useful, IB exam questions often ask you to predict whether ΔS is positive or negative. Check three major factors.

虽然精确计算很有用，但 IB 考试题常要求你判断 ΔS 是正还是负。可关注三大因素。

• Change in number of gas molecules: More gas molecules means higher entropy. Example: 2H₂O₂(aq) → 2H₂O(l) + O₂(g), gas is produced, so ΔS is positive.
• 气体分子数变化：气体分子数增加，熵增大。例如：2H₂O₂(aq) → 2H₂O(l) + O₂(g)，生成了气体，所以 ΔS 为正。
• Change of state: Melting and vaporization increase entropy; condensation and freezing decrease entropy.
• 状态变化：熔化、气化使熵增大；凝结、凝固使熵减小。
• Dissolving a solid or gas: Dissolving a solid in water usually increases entropy; dissolving a gas decreases entropy because gas molecules become confined in solution.
• 溶解过程：固体溶于水通常使熵增大；气体溶于水使熵减小，因为气体分子被限制在溶液中。

In the reaction N₂(g) + 3H₂(g) → 2NH₃(g), four moles of gas become two moles of gas. The number of gas molecules decreases, so the sign of ΔS for the system is negative.

在反应 N₂(g) + 3H₂(g) → 2NH₃(g) 中，四摩尔气体变为两摩尔气体，气体分子数减少，所以系统熵变 ΔS 为负。

3. Calculating Standard Entropy Change | 计算标准熵变

For a chemical reaction, the standard entropy change of the system is calculated from standard molar entropies:

对于化学反应，系统的标准熵变由标准摩尔熵计算：

ΔS° = ΣS°(products) − ΣS°(reactants)

Remember to multiply each S° value by its stoichiometric coefficient. The result is in J K⁻¹ mol⁻¹ (per mole of reaction as written).

记得将每个 S° 值乘以其化学计量系数。结果的单位为 J K⁻¹ mol⁻¹（按所写反应式的每摩尔反应计）。

Example: Calculate ΔS° for CaCO₃(s) → CaO(s) + CO₂(g), given S° values in J K⁻¹ mol⁻¹: CaCO₃ = 92.9, CaO = 39.7, CO₂ = 213.7.

示例：计算 CaCO₃(s) → CaO(s) + CO₂(g) 的 ΔS°，已知各物质的 S°（J K⁻¹ mol⁻¹）：CaCO₃ = 92.9，CaO = 39.7，CO₂ = 213.7。

ΔS° = (39.7 + 213.7) − (92.9) = 160.5 J K⁻¹ mol⁻¹

The positive value matches the prediction that one mole of gas is produced from a solid, increasing disorder.

正值符合预测：固体产生了一摩尔气体，无序度增大。

4. Total Entropy Change and the Second Law | 总熵变与热力学第二定律

The second law of thermodynamics states that in any spontaneous process, the entropy of the universe increases. The universe includes the system and the surroundings.

热力学第二定律指出，任何自发过程中，宇宙的熵增大。宇宙包括系统和环境。

ΔS_total = ΔS_system + ΔS_surroundings

For a spontaneous change, ΔS_total > 0. This rule is absolute; there are no exceptions.

自发变化要求 ΔS_total > 0。这一规则是绝对的，没有例外。

At constant pressure, the entropy change of the surroundings is related to the enthalpy change of the system:

在恒压下，环境的熵变与系统的焓变有关：

ΔS_surroundings = −ΔH_system / T

An exothermic reaction (ΔH < 0) releases heat into the surroundings, increasing the entropy of the surroundings. An endothermic reaction (ΔH > 0) absorbs heat from the surroundings, decreasing the entropy of the surroundings.

放热反应（ΔH < 0）向环境释放热量，使环境熵增大；吸热反应（ΔH > 0）从环境吸收热量，使环境熵减小。

5. Gibbs Free Energy and Spontaneity | 吉布斯自由能与自发性

Combining the equations gives the Gibbs free energy change:

将以上方程合并可得吉布斯自由能变：

ΔG = ΔH − TΔS

This is effectively a measure of the total entropy change of the universe, expressed in energy units. At constant temperature and pressure:

这实际上是宇宙总熵变的能量表达形式。在恒温恒压下：

注意：ΔG < 0 并不等于反应快。热力学只判断可能性，动力学决定速度。

Note: ΔG < 0 does not mean the reaction is fast. Thermodynamics judges possibility; kinetics determines rate.

ΔG = ΔG° + RT ln Q

At equilibrium, ΔG = 0 and Q = K, giving:

在平衡时，ΔG = 0 且 Q = K，由此可得：

ΔG° = −RT ln K

6. How ΔH and ΔS Signs Control Spontaneity | ΔH 与 ΔS 的正负如何控制自发性

The sign of ΔG depends on the signs of ΔH and ΔS, as well as the temperature T. This gives four general cases.

ΔG 的符号取决于 ΔH 与 ΔS 的正负，以及温度 T。这给出了四种常见情形。

例：水结冰 H₂O(l) → H₂O(s) 放热（ΔH < 0）且无序度减小（ΔS < 0）。低温时 |TΔS| 较小，ΔG < 0，所以低温下自发；高温时 |TΔS| 超过 ΔH，ΔG > 0，所以高温下不自发。

Example: For freezing of water H₂O(l) → H₂O(s), the process is exothermic (ΔH < 0) and has decreasing disorder (ΔS < 0). At low temperature, |TΔS| is small so ΔG < 0 and it is spontaneous; at high temperature, |TΔS| dominates and ΔG > 0, so it becomes non-spontaneous.

7. Calculating the Temperature of Transition | 计算转变温度

When ΔH and ΔS have the same sign, spontaneity changes at the temperature where ΔG = 0:

当 ΔH 与 ΔS 同号时，在 ΔG = 0 的温度处自发性发生转变：

T = ΔH / ΔS

For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178 kJ mol⁻¹ and ΔS° = +160.5 J K⁻¹ mol⁻¹. Convert ΔH to J: 178000 J mol⁻¹.

对于 CaCO₃(s) → CaO(s) + CO₂(g)，ΔH° = +178 kJ mol⁻¹，ΔS° = +160.5 J K⁻¹ mol⁻¹。将 ΔH 转换为 J：178000 J mol⁻¹。

T = 178000 / 160.5 ≈ 1109 K

Above about 1109 K, the decomposition of calcium carbonate becomes spontaneous because the positive TΔS term overcomes the positive ΔH. This is why limestone decomposes only when heated strongly.

大约在 1109 K 以上，碳酸钙分解变为自发，因为正的 TΔS 项超过了正的 ΔH。这就是为什么石灰石只有在强热条件下才分解。

Be careful with units: ΔH is usually given in kJ mol⁻¹, while ΔS is in J K⁻¹ mol⁻¹. Convert one of them before calculating T or ΔG.

务必注意单位：ΔH 通常以 kJ mol⁻¹ 给出，而 ΔS 以 J K⁻¹ mol⁻¹ 给出。计算 T 或 ΔG 之前须先统一单位。

8. Worked Example: Is the Reaction Spontaneous? | 综合例题：反应是否自发？

Consider the reaction 2NO(g) + O₂(g) → 2NO₂(g) at 298 K. Data: ΔH° = −114 kJ mol⁻¹; ΔS° = −146 J K⁻¹ mol⁻¹.

考虑反应 2NO(g) + O₂(g) → 2NO₂(g)，温度为 298 K。数据：ΔH° = −114 kJ mol⁻¹；ΔS° = −146 J K⁻¹ mol⁻¹。

Step 1: Convert ΔS to kJ: −0.146 kJ K⁻¹ mol⁻¹.

第一步：将 ΔS 换算为 kJ：−0.146 kJ K⁻¹ mol⁻¹。

ΔG = −114 − (298)(−0.146) = −114 + 43.5 = −70.5 kJ mol⁻¹

Because ΔG < 0, the reaction is spontaneous at 298 K. In this case, ΔH is negative but ΔS is negative, so the reaction becomes less spontaneous at higher temperatures. It would be non-spontaneous above T = 114 / 0.146 ≈ 781 K.

因为 ΔG < 0，该反应在 298 K 下自发。此例中 ΔH 为负而 ΔS 为负，因此温度升高时自发性降低。当温度高于 T = 114 / 0.146 ≈ 781 K 时变为不自发。

9. Common Exam Mistakes and Tips | 常见错误与考试技巧

• Unit errors: Mixing kJ and J is the most common mistake. Always convert ΔS to kJ or ΔH to J before using ΔG = ΔH − TΔS.
• 单位错误：混用 kJ 和 J 是最常见错误。使用 ΔG = ΔH − TΔS 前，务必将 ΔS 换成 kJ 或将 ΔH 换成 J。
• Confusing ΔG and ΔG°: ΔG° refers to standard conditions (1 bar, 1 mol dm⁻³, usually 298 K). ΔG at non-standard conditions requires the reaction quotient Q.
• 混淆 ΔG 与 ΔG°：ΔG° 指标准状况（1 bar，1 mol dm⁻³，通常 298 K）。非标准状况下的 ΔG 需要用到反应商 Q。
• Forgetting stoichiometric coefficients: When calculating ΔS° or ΔH°, multiply each substance’s value by its coefficient in the balanced equation.
• 忽略化学计量系数：计算 ΔS° 或 ΔH° 时，必须将各物质的数值乘以其在平衡方程式中的系数。
• Thinking spontaneous means fast: A diamond is thermodynamically unstable relative to graphite at room temperature, but the conversion is extremely slow. Spontaneity says nothing about rate.
• 认为自发等于快速：在室温下，金刚石相对石墨在热力学上不稳定，但转化极慢。自发性与反应速率无关。

10. Entropy in Electrochemistry and Equilibrium | 电化学与平衡中的熵

In electrochemistry, the relationship ΔG° = −nFE° connects Gibbs free energy to cell potential. Therefore, a positive E° indicates a spontaneous redox reaction.

在电化学中，关系式 ΔG° = −nFE° 将吉布斯自由能与电池电动势联系起来。因此，正的 E° 表明氧化还原反应自发。

Also, since ΔG° = −RT ln K, a large negative ΔG° corresponds to an equilibrium constant much greater than 1, meaning the reaction strongly favours products.

同时，由于 ΔG° = −RT ln K，越负的 ΔG° 对应越大的平衡常数 K，说明反应强烈偏向生成物。

When answering exam questions, identify whether you are asked about the system, the surroundings, or the total entropy. Many marks are lost by describing only ΔS_system when the question asks for the second law of thermodynamics and ΔS_total.

答题时，先确认题目问的是系统熵、环境熵还是总熵。很多同学只讨论 ΔS_system 而忽略题目要求的是热力学第二定律与 ΔS_total，因而失分。

11. Quick Revision Checklist | 快速复习清单

• 熵的定义与单位 J K⁻¹ mol⁻¹。Definition and unit of entropy.
• 从分子数、状态、溶解等定性判断 ΔS 正负。Qualitative prediction of ΔS sign.
• 利用 ΔS° = ΣS°(产物) − ΣS°(反应物) 计算。Calculation using ΔS°.
• 热力学第二定律：ΔS_total > 0 为自发。Second law: ΔS_total > 0 for spontaneous change.
• ΔS_surroundings = −ΔH/T。Surroundings entropy change.
• ΔG = ΔH − TΔS 及正负判据。Gibbs free energy criterion.
• 四类 ΔH/ΔS 组合与温度的关系。Four combinations of ΔH and ΔS.
• T = ΔH/ΔS 求转变温度。Transition temperature.
• ΔG° = −RT ln K 与电化学关系。Link to equilibrium and electrochemistry.

By mastering entropy and Gibbs free energy, you can reliably predict spontaneity in any IB HL thermodynamics question. Remember to check units, use the correct equations, and always connect the calculation back to the fundamental rule: the entropy of the universe must increase.

掌握熵与吉布斯自由能后，你就能可靠地判断 IB HL 热力学题中任何反应的自发性。切记检查单位、选用正确公式，并将计算始终回归到基本规律：宇宙的总熵必须增大。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry HL: Mastering Energy Cycle Diagrams | IB化学HL：能量循环图解题思路

Energy cycles are among the most powerful tools in IB Chemistry HL for calculating enthalpy changes that cannot be measured directly. By visually arranging formation, ionisation, dissociation, and lattice enthalpy changes, students can apply Hess’s Law with confidence and precision.

能量循环图是 IB 化学 HL 中最强大的工具之一，用于计算无法直接测量的焓变。通过将生成焓、电离能、解离能和晶格焓等变化可视化排列，学生可以自信且精确地运用盖斯定律。

1. What Is an Energy Cycle? | 什么是能量循环图？

An energy cycle is a closed loop of enthalpy changes that connects reactants to products through alternative pathways. Because enthalpy is a state function, the total enthalpy change for the overall reaction is identical regardless of the route taken, as long as the initial and final states are the same.

能量循环图是一个由焓变组成的闭合回路，通过不同路径将反应物与产物连接起来。由于焓是状态函数，只要始态和终态相同，无论采取哪条路径，总反应的焓变都是相同的。

The power of an energy cycle lies in its ability to convert a difficult or impossible measurement into a set of known or measurable steps. In HL, you will most often encounter Born-Haber cycles for ionic compounds and simple Hess cycles for formation or combustion reactions.

能量循环图的强大之处在于，它能把一个困难甚至无法测量的量转化为一系列已知或可测量的步骤。在 HL 课程中，最常见的是离子化合物的玻恩-哈伯循环，以及生成反应或燃烧反应的简单盖斯循环。

2. The Core Principle: Hess’s Law | 核心原理：盖斯定律

Hess’s Law states that the enthalpy change of a reaction depends only on the initial and final states, not on the pathway taken. This principle is the foundation of every energy cycle you will ever draw.

盖斯定律指出，反应焓变只取决于始态和终态，而与反应路径无关。这一原理是你绘制所有能量循环图的基础。

For a general cycle, if the clockwise enthalpy changes sum to ΔH₁ and the anticlockwise sum to ΔH₂, then ΔH₁ + ΔH₂ = 0 when the cycle is closed. This allows you to set up an equation with one unknown value.

对于一个一般循环，如果顺时针方向焓变总和为 ΔH₁，逆时针方向总和为 ΔH₂，则闭合循环满足 ΔH₁ + ΔH₂ = 0。这使你可以建立含有一个未知量的方程。

ΔH₁ + ΔH₂ = 0

In a diagram, you should always label arrows with enthalpy values and their sign. Remember: if an arrow points upward, the process is endothermic (positive ΔH); if it points downward, the process is exothermic (negative ΔH).

在图中，你应当始终用焓值和正负号标注箭头。请记住：箭头向上表示吸热过程（ΔH 为正）；箭头向下表示放热过程（ΔH 为负）。

3. Recognising Ionic Compounds: Born-Haber Cycles | 识别离子化合物：玻恩-哈伯循环

Born-Haber cycles are specialised energy cycles that describe the formation of an ionic compound, such as NaCl or MgO, from its elements in their standard states. They are required knowledge for HL students under the structure and reactivity topic.

玻恩-哈伯循环是专门描述离子化合物（如 NaCl 或 MgO）由标准态元素生成的能量循环。这是 HL 学生在“结构与反应性”专题中必须掌握的内容。

The key steps in a Born-Haber cycle include atomisation of elements, ionisation of the metal, electron affinity of the non-metal, and formation of the ionic lattice. Each step is associated with a specific enthalpy change.

玻恩-哈伯循环中的关键步骤包括：元素的原子化、金属的电离、非金属的电子亲和能，以及离子晶格的形成。每一步都对应一个特定的焓变。

4. Representing Electron Affinity and Ionisation Energy | 电子亲和能与电离能的表示

Ionisation energy is always endothermic, because energy is required to remove an electron from an atom or ion. In a Born-Haber cycle, ionisation steps are drawn as upward arrows.

电离能总是吸热的，因为从原子或离子中移走电子需要能量。在玻恩-哈伯循环中，电离步骤用向上的箭头表示。

Electron affinity is often exothermic for the first electron in many non-metals, because energy is released when an electron is added. In the IB data booklet, however, values are often given with a sign; you must check whether the sign is already included in the provided value.

对于许多非金属元素，得到第一个电子时通常放热，因此第一电子亲和能为负值。但在 IB 数据手册中，给出的数值有时已包含符号，因此你必须确认所提供的数值是否已经带符号。

M(g) → M⁺(g) + e⁻ ΔH = +IE₁

X(g) + e⁻ → X⁻(g) ΔH = EA₁ (usually negative)

A common trick in exam questions is to give EA as a positive number while showing the process as exothermic. In that case, you must insert a negative sign when placing it in the cycle. Conversely, if the question gives EA as negative, use it directly.

考试中常见的陷阱是：给出的 EA 为正数，但过程实际放热。此时在循环中必须加上负号；反之，若题目给出的 EA 为负数，则直接使用。

5. Lattice Enthalpy: Definitions and Signs | 晶格焓：定义与符号

Lattice enthalpy is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions. This is the definition used in the IB data booklet for lattice formation enthalpy.

晶格焓是指从气态离子形成一摩尔离子化合物时的焓变。IB 数据手册中采用的是晶格生成焓的定义。

Formation of a lattice from gaseous ions is exothermic, because electrostatic attractions release energy. Therefore ΔH_lat for formation is negative. In contrast, lattice dissociation enthalpy is the energy needed to break one mole of solid into gaseous ions, and is positive.

从气态离子形成晶格是放热的，因为静电吸引释放能量，因此晶格生成焓为负。与此相反，晶格解离焓是将一摩尔固体拆成气态离子所需的能量，其值为正。

M⁺(g) + X⁻(g) → MX(s) ΔH_lat = negative

Always check which definition the examiner is using. In IB HL multiple-choice and paper 2 questions, the phrase “lattice enthalpy” usually refers to the formation of the lattice, so a negative sign is expected unless stated otherwise.

始终注意考官使用哪个定义。在 IB HL 选择题和 Paper 2 中，“晶格焓”通常指晶格生成焓，因此除非另有说明，一般预期为负值。

6. Constructing a Born-Haber Cycle Step by Step | 逐步构建玻恩-哈伯循环

To construct a clear Born-Haber cycle, begin with the elements in their standard states at the bottom-left of the diagram. Then follow the steps upward to the gaseous atoms, ions, and finally down to the ionic lattice.

要绘制清晰的玻恩-哈伯循环，首先在图的左下角画出标准态元素，然后沿步骤向上到达气态原子、气态离子，最后向下到达离子晶格。

Follow this reliable sequence for a simple metal halide MX:

对于简单的金属卤化物 MX，可遵循以下可靠顺序：

• Convert metal M(s) to M(g): add ΔH_atom(M)
• Convert ½X₂(g) to X(g): add ½ ΔH_atom(X₂), often called bond dissociation per mole of atoms
• Ionise M(g) to M⁺(g): add IE₁(M)
• Add electron to X(g): add EA₁(X)
• Combine gaseous ions to form MX(s): add ΔH_lat(negative)

Each step should be represented as one arrow. The vertical order of the arrows is important: the sum of upward steps must match the sum of downward steps when closed by the formation enthalpy of MX(s) from elements.

每个步骤应表示为一个箭头。箭头的纵向顺序很重要：当通过 MX(s) 的元素生成焓闭合循环时，所有向上步骤之和必须等于所有向下步骤之和。

ΔH_f(MX) = ΔH_atom(M) + ½ΔH_atom(X₂) + IE₁(M) + EA₁(X) + ΔH_lat

Once the diagram is complete, you can rearrange this equation to solve for any missing value.

一旦循环图绘制完成，你就可以重排这个方程来求算任何缺失的量。

7. Solving for Unknown Values | 求算未知量

The most common exam question asks you to calculate lattice enthalpy from a Born-Haber cycle. Follow the algebraic method: write the full equation, substitute known values, and isolate the unknown.

最常见的考题要求你从玻恩-哈伯循环计算晶格焓。采用代数方法：写出完整方程，代入已知数值，然后分离未知量。

Suppose you have the following data for a hypothetical salt M⁺X⁻:

假设你拥有以下假想盐 M⁺X⁻ 的数据：

Substitute into the equation:

代入方程：

-450 = 150 + 120 + 500 + (-300) + ΔH_lat

-450 = 470 + ΔH_lat

ΔH_lat = -920 kJ mol⁻¹

Always include the sign in your final answer. A lattice enthalpy of -920 kJ mol⁻¹ indicates the exothermic formation of the lattice from gaseous ions.

最终答案中务必包含正负号。晶格焓为 -920 kJ mol⁻¹ 表示离子化合物从气态离子生成时放热。

8. Enthalpy of Solution and Hydration | 溶液焓与水合焓

Energy cycles also appear when studying the enthalpy of solution, ΔH_sol. This can be broken into lattice dissociation and hydration of gaseous ions. The cycle connects the solid ionic compound, its gaseous ions, and the dissolved hydrated ions.

在研究溶液焓 ΔH_sol 时也会用到能量循环。溶液焓可以分解为晶格解离和气态离子的水合两个部分。循环连接固体离子化合物、气态离子和水合离子。

The relationship is:

关系式为：

ΔH_sol = -ΔH_lat + ΣΔH_hyd

Here, -ΔH_lat is the lattice dissociation enthalpy (positive), and ΣΔH_hyd is the sum of hydration enthalpies of the cation and anion. Hydration enthalpies are exothermic, so they are negative values.

其中，-ΔH_lat 是晶格解离焓（正值），ΣΔH_hyd 是阳离子与阴离子水合焓之和。水合焓放热，因此为负值。

When solving problems, note that ΔH_sol can be positive or negative depending on the balance between lattice energy and hydration energy. This explains why some salts feel cold when dissolving.

解题时注意，ΔH_sol 可以为正也可以为负，取决于晶格能和水合能的平衡。这也解释了为什么有些盐溶解时会感觉冰凉。

9. Using Energy Cycles to Compare Lattice Enthalpies | 用能量循环比较晶格焓

Energy cycles help you rationalise trends in lattice enthalpy. For example, lattice enthalpy magnitude increases with higher ionic charges and smaller ionic radii, because electrostatic attraction is stronger.

能量循环帮助你合理理解晶格焓的变化趋势。例如，晶格焓的绝对值随离子电荷升高和离子半径减小而增大，因为静电吸引更强。

Consider MgO versus NaCl. MgO has Mg²⁺ and O²⁻ ions with smaller radii, so its lattice enthalpy is much more exothermic than NaCl. A Born-Haber cycle for MgO requires additional ionisation and electron affinity steps.

以 MgO 与 NaCl 为例。MgO 含有半径更小的 Mg²⁺ 和 O²⁻ 离子，因此其晶格焓的放热程度远高于 NaCl。MgO 的玻恩-哈伯循环需要额外的电离能和电子亲和能步骤。

For oxides, the second electron affinity of oxygen is endothermic because adding an electron to a negative ion requires energy to overcome electrostatic repulsion. This is a common feature in cycles for oxides and is often tested in HL questions.

对于氧化物，氧的第二电子亲和能是吸热的，因为向负离子添加电子需要克服静电排斥。这是氧化物循环中的常见特征，也是 HL 考试常考的内容。

10. Common Mistakes and Pitfalls | 常见错误与陷阱

Many students lose marks in energy cycle questions due to avoidable sign errors. One classic mistake is treating lattice dissociation enthalpy as if it were lattice formation enthalpy, and then using the wrong sign.

许多学生在能量循环题中因可避免的符号错误而失分。一个经典错误是把晶格解离焓当成晶格生成焓，因而使用错误的符号。

Another common pitfall is forgetting to halve bond dissociation energies when the element is diatomic. For example, the atomisation enthalpy of chlorine in the Born-Haber cycle uses ½Cl₂(g) → Cl(g), so you must use half the bond dissociation enthalpy.

另一个常见陷阱是忘记在双原子元素中取键解离能的一半。例如，在玻恩-哈伯循环中，氯的原子化是 ½Cl₂(g) → Cl(g)，因此必须使用键解离能的一半。

Students also misread electron affinity values. If the data booklet or question gives a positive EA₁ even though the process is exothermic, you must change the sign in the equation. Do not blindly copy the number.

学生也常误读电子亲和能数值。如果数据手册或题目给出正 EA₁，但过程实际放热，那么在方程中必须改变符号。不要盲目照抄数字。

Finally, always draw the cycle before starting algebra. A correct diagram helps you see the sign and direction of every step, reducing the risk of arithmetic mistakes.

最后，在做代数运算之前一定要先画循环图。正确的图能帮你判断每一步的符号和方向，降低计算出错的风险。

11. Exam Tips for HL Paper 2 and 3 | HL Paper 2 和 3 的考试技巧

In Paper 2, Born-Haber cycle questions usually require you to define terms, write equations for each step, and calculate lattice enthalpy. Read the definition carefully: the examiner may ask for “lattice dissociation enthalpy” or “lattice formation enthalpy.”

在 Paper 2 中，玻恩-哈伯循环题通常要求你定义术语、写出每一步的方程，并计算晶格焓。请仔细阅读题目：考官可能要求的是“晶格解离焓”或“晶格生成焓”。

Show your working clearly. Even if the final numerical answer is wrong, you can still earn method marks for correctly substituting values and setting up the equation. In IB exams, method marks are often more generous than answer marks.

清晰展示你的计算过程。即使最终数值错误，你仍可能因正确代入数值和建立方程而获得步骤分。在 IB 考试中，步骤分通常比答案分更宽松。

For Paper 3, you may be asked to compare experimental and theoretical lattice enthalpies using a Born-Haber cycle. A large difference suggests significant covalent character in the ionic compound, as predicted by Fajans’ rules.

在 Paper 3 中，你可能需要比较实验晶格焓与理论晶格焓。若两者差异较大，则表明离子化合物具有显著的共价性，这可以用法扬斯规则解释。

Use your time wisely: if a cycle question seems long, write the master equation first. Then allocate values to each term. This approach keeps your thinking organised and saves repeated reading of the question.

合理利用时间：如果循环题看起来较长，先写出主方程，再把数值分配到每一项。这种方法能让思路保持清晰，避免反复读题。

12. Worked Example | 例题精解

Let’s work through a full problem similar to an IB HL short-answer question. Consider the ionic compound RbF. The formation enthalpy is -558 kJ mol⁻¹. The atomisation enthalpy of Rb(s) is +81 kJ mol⁻¹. The atomisation enthalpy of F₂(g) is +79 kJ mol⁻¹ for one mole of F atoms. The first ionisation energy of Rb is +403 kJ mol⁻¹. The electron affinity of F is -328 kJ mol⁻¹. Calculate the lattice enthalpy of RbF.

让我们完整解答一道类似 IB HL 简答题的题目。考虑离子化合物 RbF。生成焓为 -558 kJ mol⁻¹。Rb(s) 的原子化焓为 +81 kJ mol⁻¹。F₂(g) 的原子化焓（按一摩尔 F 原子计）为 +79 kJ mol⁻¹。Rb 的第一电离能为 +403 kJ mol⁻¹。F 的电子亲和能为 -328 kJ mol⁻¹。计算 RbF 的晶格焓。

Write the cycle equation for an alkali metal fluoride:

写出碱金属氟化物的循环方程：

ΔH_f(RbF) = ΔH_atom(Rb) + ΔH_atom(F) + IE₁(Rb) + EA₁(F) + ΔH_lat

Substitute the known values:

代入已知数值：

-558 = 81 + 79 + 403 + (-328) + ΔH_lat

Simplify the right-hand side:

化简右侧：

81 + 79 + 403 – 328 = 235

Thus:

因此：

-558 = 235 + ΔH_lat

ΔH_lat = -793 kJ mol⁻¹

The lattice enthalpy of RbF is -793 kJ mol⁻¹. This large negative value reflects the strong electrostatic attraction between Rb⁺ and F⁻ ions in the crystal lattice.

RbF 的晶格焓为 -793 kJ mol⁻¹。这一较大的负值反映了 Rb⁺ 与 F⁻ 离子在晶格中强烈的静电吸引。

From this example, you can see the importance of consistent signs and careful arithmetic. Always check each number and its sign before moving to the final calculation.

从这个例子可以看出，保持符号一致和计算仔细非常重要。在进行最终计算前，务必检查每个数值及其符号。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Bond Enthalpy Concepts and Calculating Enthalpy Changes of Reactions | IB化学：键焓概念与反应焓变计算

Bond enthalpy is one of the most practical tools in thermochemistry. It allows chemists to estimate the enthalpy change of a reaction without performing experiments, simply by considering the energy required to break bonds and the energy released when new bonds form.

键焓是热化学中最实用的工具之一。它使化学家无需进行实验，只需考虑断裂化学键所需的能量和形成新化学键时释放的能量，即可估算反应焓变。

1. What Is Bond Enthalpy? | 什么是键焓？

Bond enthalpy, also called bond dissociation enthalpy, is the energy required to break one mole of a specific covalent bond in the gaseous state, producing gaseous atoms or radicals. It is always a positive value because breaking bonds requires energy input.

键焓，又称键解离焓，是指在气态下断裂一摩尔特定共价键所需的能量，生成气态原子或自由基。由于断键需要输入能量，键焓始终为正值。

For example, the bond enthalpy of the H–H bond is +436 kJ mol⁻¹. This means that to break one mole of H–H bonds in H₂ gas, 436 kJ of energy must be absorbed.

例如，H–H 键的键焓为 +436 kJ mol⁻¹。这意味着要断裂 H₂ 气体中一摩尔 H–H 键，必须吸收 436 kJ 的能量。

Key conditions for bond enthalpy values:

键焓值的关键条件：

• All species must be in the gaseous state.
• All species must be in the gaseous state.
• The bond is broken homolytically, meaning each atom retains one electron from the shared pair.

• 所有物质必须处于气态。
• 键的断裂是均裂，即每个原子保留共用电子对中的一个电子。

2. Average Bond Enthalpies vs. Exact Bond Enthalpies | 平均键焓与精确键焓

The C–H bond enthalpy in methane (CH₄) is not exactly the same as the C–H bond enthalpy in ethane (C₂H₆). The exact value depends on the molecular environment surrounding the bond. Because of this, chemists use average bond enthalpies, which are mean values calculated from a range of compounds.

甲烷（CH₄）中的 C–H 键焓与乙烷（C₂H₆）中的 C–H 键焓并不完全相同。精确值取决于键周围的分子环境。因此，化学家使用平均键焓，即从多种化合物中计算得出的平均值。

For example, the average C–H bond enthalpy is approximately +412 kJ mol⁻¹, while the actual dissociation enthalpy for the first C–H bond in methane is +438 kJ mol⁻¹. Average bond enthalpies are useful for estimation but not for exact calculations.

例如，C–H 键的平均键焓约为 +412 kJ mol⁻¹，而甲烷中第一个 C–H 键的实际解离焓为 +438 kJ mol⁻¹。平均键焓适用于估算，但不适用于精确计算。

This is why bond enthalpy calculations are described as estimates. In IB Chemistry, you are expected to know that results obtained from average bond enthalpies are approximate.

这就是为什么键焓计算被描述为估算。在 IB 化学中，你需要知道根据平均键焓得到的结果是近似值。

3. The Relationship Between Bond Enthalpy and Bond Strength | 键焓与键强度的关系

Bond enthalpy is a direct measure of bond strength. A higher bond enthalpy means a stronger bond. Triple bonds are stronger than double bonds, which are stronger than single bonds.

键焓是键强度的直接量度。键焓越高，键越强。三键强于双键，双键强于单键。

For example:

例如：

C–C: +347 kJ mol⁻¹ < C=C: +612 kJ mol⁻¹ < C≡C: +838 kJ mol⁻¹

This trend exists because multiple bonds involve more shared electron pairs, resulting in greater electrostatic attraction between the nuclei and the shared electrons.

这一趋势存在的原因是多键涉及更多共享电子对，导致原子核与共享电子之间的静电引力更强。

Bond length also follows an inverse relationship with bond enthalpy. Shorter bonds are generally stronger because the nuclei are closer to the shared electrons, increasing attraction.

键长与键焓呈反比关系。较短的键通常更强，因为原子核更接近共享电子，引力增大。

4. Using Bond Enthalpies to Calculate Reaction Enthalpy | 用键焓计算反应焓变

For a gaseous reaction, the enthalpy change can be estimated using the following principle: energy must be absorbed to break bonds in the reactants, and energy is released when new bonds form in the products.

对于气相反应，焓变可通过以下原理估算：断裂反应物中的化学键需要吸收能量，而在产物中形成新键时会释放能量。

ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed)

ΔH = Σ(断裂键的键焓) − Σ(形成键的键焓)

In words: the enthalpy change of reaction equals the total energy required to break all reactant bonds minus the total energy released when all product bonds form.

用语言表述：反应焓变等于断裂所有反应物化学键所需的总能量减去形成所有产物化学键时释放的总能量。

If the energy needed to break bonds is greater than the energy released by forming bonds, ΔH is positive and the reaction is endothermic. If the opposite is true, ΔH is negative and the reaction is exothermic.

如果断键所需能量大于成键释放能量，则 ΔH 为正值，反应为吸热反应。如果相反，则 ΔH 为负值，反应为放热反应。

5. Step-by-Step Example: Combustion of Hydrogen | 分步示例：氢气的燃烧

Consider the reaction: H₂(g) + ½O₂(g) → H₂O(g). Let us estimate ΔH using bond enthalpies.

考虑反应：H₂(g) + ½O₂(g) → H₂O(g)。让我们用键焓估算 ΔH。

Step 1: Identify all bonds in reactants and products.

步骤 1：找出反应物和产物中的所有化学键。

Reactants: one H–H bond (436 kJ mol⁻¹) and half an O=O bond (½ × 498 kJ mol⁻¹).

反应物：一个 H–H 键（436 kJ mol⁻¹）和半个 O=O 键（½ × 498 kJ mol⁻¹）。

Products: two O–H bonds (2 × 463 kJ mol⁻¹).

产物：两个 O–H 键（2 × 463 kJ mol⁻¹）。

ΔH = [436 + (½ × 498)] − [2 × 463]

ΔH = [436 + 249] − [926] = 685 − 926 = −241 kJ mol⁻¹

The calculated value is −241 kJ mol⁻¹, which is close to the actual enthalpy of formation of gaseous water (−242 kJ mol⁻¹). This demonstrates the usefulness of the bond enthalpy method.

计算值为 −241 kJ mol⁻¹，非常接近气态水的实际生成焓（−242 kJ mol⁻¹）。这证明了键焓方法的实用性。

6. Step-by-Step Example: Complete Combustion of Methane | 分步示例：甲烷的完全燃烧

Consider the reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).

考虑反应：CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)。

Step 1: Count bonds in reactants.

步骤 1：数出反应物中的化学键。

• 4 C–H bonds in CH₄: 4 × 412 = 1648 kJ mol⁻¹
• 2 O=O bonds in O₂: 2 × 498 = 996 kJ mol⁻¹

• CH₄ 中的 4 个 C–H 键：4 × 412 = 1648 kJ mol⁻¹
• O₂ 中的 2 个 O=O 键：2 × 498 = 996 kJ mol⁻¹

Total energy to break bonds = 1648 + 996 = 2644 kJ mol⁻¹.

断键总能量 = 1648 + 996 = 2644 kJ mol⁻¹。

Step 2: Count bonds in products.

步骤 2：数出产物中的化学键。

• 2 C=O bonds in CO₂: 2 × 799 = 1598 kJ mol⁻¹
• 4 O–H bonds in 2H₂O: 4 × 463 = 1852 kJ mol⁻¹

• CO₂ 中的 2 个 C=O 键：2 × 799 = 1598 kJ mol⁻¹
• 2H₂O 中的 4 个 O–H 键：4 × 463 = 1852 kJ mol⁻¹

Total energy released = 1598 + 1852 = 3450 kJ mol⁻¹.

成键释放总能量 = 1598 + 1852 = 3450 kJ mol⁻¹。

ΔH = 2644 − 3450 = −806 kJ mol⁻¹

The experimental value for this reaction is approximately −890 kJ mol⁻¹. The difference arises because average bond enthalpies are used, and the C=O bond in CO₂ is particularly strong and sensitive to its environment.

该反应的实验值约为 −890 kJ mol⁻¹。差异源于使用了平均键焓，且 CO₂ 中的 C=O 键特别强，对环境敏感。

7. Bond Enthalpy vs. Hess’s Law | 键焓与盖斯定律的比较

Hess’s Law and bond enthalpy calculations are two different methods for finding reaction enthalpy, and IB Chemistry requires you to understand the distinction between them.

盖斯定律和键焓计算是求反应焓变的两种不同方法，IB 化学要求你理解它们之间的区别。

Hess’s Law uses known enthalpy changes of formation or combustion and works with exact values. It is reliable and accurate, provided the data are reliable.

盖斯定律使用已知的生成焓或燃烧焓数据，处理的是精确值。只要数据可靠，就是可靠且准确的。

Bond enthalpy calculations use average values from different compounds, so they are estimates. They are less accurate but conceptually powerful because they reveal the physical meaning of bond breaking and forming.

键焓计算使用来自不同化合物的平均值，因此属于估算。它们不太准确，但在概念上更强，因为它们揭示了断键和成键的物理意义。

8. Common Mistakes in Bond Enthalpy Calculations | 键焓计算中的常见错误

Students frequently make several errors when performing bond enthalpy calculations. Being aware of these will help you avoid them in the IB exam.

学生在进行键焓计算时经常犯一些错误。了解这些错误将帮助你在 IB 考试中避免它们。

Mistake 1: Forgetting to multiply bond enthalpy by the number of bonds present. For example, one mole of CH₄ contains four C–H bonds, not one.

错误 1：忘记将键焓乘以化学键的数目。例如，一摩尔 CH₄ 含有四个 C–H 键，而不是一个。

Mistake 2: Using the wrong sign. Bond breaking is endothermic (positive), and bond forming is exothermic (negative). The formula ΔH = bonds broken − bonds formed already accounts for the signs. Do not add them.

错误 2：使用错误的符号。断键是吸热的（正值），成键是放热的（负值）。公式 ΔH = 断键 − 成键已经考虑了符号。不要把两者相加。

Mistake 3: Ignoring the state of matter. Bond enthalpy calculations are only valid when all reactants and products are in the gaseous state. If a substance is liquid or solid, additional phase-change enthalpy must be considered, and the simple bond enthalpy method is no longer directly applicable.

错误 3：忽略物质的状态。只有在所有反应物和产物均为气态时，键焓计算才有效。如果物质是液态或固态，则需要考虑额外的相变焓，简单的键焓方法将不再直接适用。

Mistake 4: Confusing average bond enthalpy with standard enthalpy of atomisation. Atomisation refers to the formation of atoms from elements, while bond enthalpy refers to breaking a specific bond within a molecule.

错误 4：混淆平均键焓与标准原子化焓。原子化焓指从元素形成原子，而键焓指断裂分子中特定化学键。

9. Worked Example with Multiple Bonds | 含多重键的计算示例

Consider the hydrogenation of ethene: C₂H₄(g) + H₂(g) → C₂H₆(g).

考虑乙烯的氢化反应：C₂H₄(g) + H₂(g) → C₂H₆(g)。

Step 1: Draw the structural formulas to identify all bonds.

步骤 1：画出结构式，找出所有化学键。

C₂H₄ has one C=C bond and four C–H bonds. C₂H₆ has one C–C bond and six C–H bonds. H₂ has one H–H bond.

C₂H₄ 有一个 C=C 键和四个 C–H 键。C₂H₆ 有一个 C–C 键和六个 C–H 键。H₂ 有一个 H–H 键。

Step 2: Calculate energy to break bonds.

步骤 2：计算断键能量。

1 × C=C (612) + 4 × C–H (4 × 412) + 1 × H–H (436) = 612 + 1648 + 436 = 2696 kJ mol⁻¹

Step 3: Calculate energy released on forming bonds.

步骤 3：计算成键释放的能量。

1 × C–C (347) + 6 × C–H (6 × 412) = 347 + 2472 = 2819 kJ mol⁻¹

ΔH = 2696 − 2819 = −123 kJ mol⁻¹

The negative sign indicates the hydrogenation of ethene is exothermic. The experimental value is approximately −137 kJ mol⁻¹, confirming the estimate is reasonable.

负号表明乙烯加氢是放热反应。实验值约为 −137 kJ mol⁻¹，证实该估算是合理的。

10. Exam Tips for IB Chemistry | IB 化学考试技巧

In the IB exam, bond enthalpy questions are usually straightforward once you master the method. Here are some tips to maximise your marks.

在 IB 考试中，一旦你掌握了方法，键焓问题通常相对简单。以下是一些帮你拿满分的技巧。

• Always draw or visualise the structural formula before counting bonds. This reduces errors.
• Write down the full equation including the multiplication factors before substituting numbers.
• State that the answer is an estimate because average bond enthalpies are used.
• Check whether all species are in the gaseous state. Mention this in your answer if relevant.
• Include units (kJ mol⁻¹) in every step and in the final answer.

• 在数键之前，务必画出或想象结构式。这能减少错误。
• 在代入数字之前，写下包含所有乘数的完整算式。
• 注明答案为估算值，因为使用的是平均键焓。
• 检查所有物质是否均为气态。如有必要，在答案中说明这一点。
• 每一步和最终答案都要带上单位（kJ mol⁻¹）。

Also, be prepared to explain why bond enthalpy calculations give approximate values. The key reason is that bond enthalpies are averaged from many different molecules, so the exact bond strength in a particular molecule may differ.

此外，准备好解释为什么键焓计算给出近似值。关键原因是键焓是从许多不同分子中取平均得到的，因此特定分子中化学键的确切强度可能不同。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Orbital Hybridisation and Molecular Shapes in IB Chemistry HL | IB化学HL：轨道杂化与分子形状

In IB Chemistry Higher Level, understanding the three-dimensional arrangement of atoms within molecules is essential. Two complementary models—Valence Shell Electron Pair Repulsion (VSEPR) theory and the concept of orbital hybridisation—work together to explain and predict molecular geometry with remarkable accuracy. This article explores the fundamental relationship between hybridised atomic orbitals and molecular shapes, equipping you with the tools needed to excel in this core topic.

在IB化学高级水平课程中，理解分子内原子的三维排列至关重要。价层电子对互斥（VSEPR）理论与轨道杂化概念这两个互补模型相辅相成，能够以惊人的准确性解释并预测分子几何构型。本文深入探讨杂化原子轨道与分子形状之间的基本关系，为你掌握这一核心考点提供有力工具。

1. Why Hybridisation? The Limitations of Simple Atomic Orbitals | 为何需要杂化？简单原子轨道的局限性

The ground-state electron configuration of carbon is 1s²2s²2p², which suggests that carbon should form only two covalent bonds (using its two unpaired 2p electrons) and that the bond angles should be 90°. This prediction fails spectacularly: methane (CH₄) is tetrahedral with four equivalent C–H bonds arranged at 109.5°. Clearly, a new model is required.

碳的基态电子构型是1s²2s²2p²，这似乎意味着碳只能形成两条共价键（利用其两个未配对的2p电子），且键角应为90°。然而这一预测与事实严重不符：甲烷（CH₄）呈正四面体形，具有四条完全等价的C–H键，键角为109.5°。显然，我们需要一种全新的模型来解释这一现象。

Orbital hybridisation, proposed by Linus Pauling in 1931, resolves this issue by mathematically mixing atomic orbitals of similar energy to produce a set of equivalent hybrid orbitals. These hybrid orbitals have directional characteristics that match the observed molecular geometry.

由莱纳斯·鲍林于1931年提出的轨道杂化理论解决了这一难题：将能量相近的原子轨道进行数学上的混合，产生一组等价杂化轨道。这些杂化轨道具有与实验观测到的分子几何构型相匹配的方向性特征。

2. The sp³ Hybridisation: Tetrahedral Geometry | sp³杂化：正四面体几何

When one 2s orbital and three 2p orbitals hybridise, four equivalent sp³ orbitals are produced. These orbitals are oriented towards the vertices of a regular tetrahedron, with bond angles of 109.5°. This is the most common type of hybridisation for carbon atoms in organic molecules.

当一个2s轨道与三个2p轨道杂化时，产生四个等价的sp³杂化轨道。这些轨道指向正四面体的四个顶点，键角为109.5°。这是有机分子中碳原子最常见的杂化类型。

Consider methane: the central carbon atom undergoes sp³ hybridisation. Each of the four sp³ orbitals overlaps with the 1s orbital of a hydrogen atom, forming four equivalent C–H sigma (σ) bonds. Other examples include ethane (C₂H₆), ammonia (NH₃, bond angle 107°), and water (H₂O, bond angle 104.5°).

以甲烷为例：中心碳原子发生sp³杂化，四个sp³杂化轨道分别与氢原子的1s轨道重叠，形成四条等价的C–H σ键。其他实例包括乙烷（C₂H₆）、氨（NH₃，键角107°）和水（H₂O，键角104.5°）。

3. The sp² Hybridisation: Trigonal Planar Geometry | sp²杂化：平面三角形几何

When one 2s orbital combines with only two 2p orbitals, three equivalent sp² orbitals are generated. These lie in a single plane, pointing towards the corners of an equilateral triangle, with bond angles of 120°. One unhybridised 2p orbital remains perpendicular to the plane.

当一个2s轨道仅与两个2p轨道组合时，产生三个等价的sp²杂化轨道。这些轨道位于同一平面内，指向等边三角形的三个顶点，键角为120°。剩餘一个未参与杂化的2p轨道垂直于该平面。

Ethene (C₂H₄) is the classic example. Each carbon atom uses its three sp² orbitals to form σ bonds with two hydrogen atoms and the other carbon atom. The remaining unhybridised 2p orbitals on adjacent carbon atoms overlap sideways to form a pi (π) bond. The C=C double bond thus consists of one σ bond and one π bond.

乙烯（C₂H₄）是经典实例。每个碳原子用其三个sp²轨道与两个氢原子及另一个碳原子形成σ键。相邻碳原子上剩余的未杂化2p轨道侧向重叠形成π键。C=C双键因此由一条σ键和一条π键组成。

4. The sp Hybridisation: Linear Geometry | sp杂化：直线形几何

When one 2s orbital mixes with only one 2p orbital, two equivalent sp hybrid orbitals are produced. These are arranged linearly at 180° to each other. Two unhybridised 2p orbitals remain, both perpendicular to the molecular axis.

当一个2s轨道仅与一个2p轨道混合时，产生两个等价的sp杂化轨道。它们呈180°直线排列。剩余两个未杂化的2p轨道均垂直于分子轴。

Ethyne (C₂H₂) exemplifies this geometry. Each carbon atom forms two σ bonds (one to hydrogen and one to the adjacent carbon) using its sp orbitals. The two pairs of p orbitals on the two carbon atoms form two π bonds, resulting in a triple bond: one σ and two π components. The H–C≡C–H molecule is perfectly linear.

乙炔（C₂H₂）完美体现了这种几何构型。每个碳原子利用其sp轨道形成两条σ键（一条连接氢原子，一条连接相邻碳原子）。两个碳原子上两对p轨道形成两条π键，从而构成三键：一条σ键和两条π键。H–C≡C–H分子呈完美直线形。

5. Summary of the First Three Hybridisation Types | 前三种杂化类型总结

The table below summarises the relationship between hybridisation, electron-domain geometry, and molecular shape according to VSEPR theory:

下表总结了杂化类型、电子域几何构型与VSEPR理论中分子形状之间的关系：

For each hybridisation type, all hybrid orbitals are equivalent in energy and bond length. The number of hybrid orbitals always equals the number of atomic orbitals that were mixed.

对于每种杂化类型，所有杂化轨道在能量和键长上完全等价。杂化轨道的数目始终等于参与混合的原子轨道数目。

6. Identifying Hybridisation from Electron-Domain Geometry | 从电子域几何判断杂化

A reliable method for determining the hybridisation of a central atom is to first count its electron domains using VSEPR theory. An electron domain is any region of electron density: a single bond, a double bond, a triple bond, or a lone pair each counts as one domain.

判断中心原子杂化类型的一个可靠方法是：首先利用VSEPR理论计算其电子域数。电子域是任何电子密度区域：单键、双键、三键或孤对电子均各算作一个电子域。

• 2 electron domains → sp hybridisation → linear geometry

  2个电子域 → sp杂化 → 直线形

• 3 electron domains → sp² hybridisation → trigonal planar geometry

  3个电子域 → sp²杂化 → 平面三角形

• 4 electron domains → sp³ hybridisation → tetrahedral geometry

  4个电子域 → sp³杂化 → 正四面体

This method works because each hybrid orbital is associated with exactly one electron domain. A double bond, for instance, occupies one hybrid orbital as a sigma bond, even though it also contains a π component formed from an unhybridised p orbital.

这个方法是有效的，因为每个杂化轨道恰好对应一个电子域。例如，双键虽包含π成分（由未杂化的p轨道形成），但作为σ键仅占据一个杂化轨道。

7. The Role of π Bonds in Hybridisation | π键在杂化中的作用

It is crucial to recognise that π bonds are formed only by unhybridised p orbitals. This fundamental principle means that the presence of a double or triple bond directly indicates that the central atom has at least one unhybridised p orbital.

必须认识到，π键仅由未杂化的p轨道形成。这一基本原理意味着：双键或三键的存在直接表明中心原子至少有一个未杂化的p轨道。

For carbon, this yields a simple rule: single bonds alone correspond to sp³ hybridisation, one double bond corresponds to sp², and one triple bond (or two double bonds, such as in CO₂) corresponds to sp. The number of π bonds is exactly equal to the number of unhybridised p orbitals.

对于碳而言，这给出了一个简单规则：仅含单键对应sp³杂化；含一个双键对应sp²杂化；含一个三键（或两个双键，如CO₂）对应sp杂化。π键的数量恰好等于未杂化p轨道的数量。

C (single bonds only) → sp³
C (one double bond) → sp²
C (one triple bond or two double bonds) → sp

8. Lone Pairs and Distorted Geometries | 孤对电子与几何构型的畸变

Lone pairs of electrons occupy hybrid orbitals and therefore participate in the hybridisation scheme. However, they do not define the molecular shape—only the electron-domain geometry. The molecular shape is determined by the positions of the atoms alone.

孤对电子占据杂化轨道，因此参与杂化方案。然而它们不决定分子形状——只决定电子域几何构型。分子形状仅由原子的位置决定。

Ammonia (NH₃) is a textbook case. The nitrogen atom has four electron domains (three N–H bonds and one lone pair) and is therefore sp³ hybridised. However, because the lone pair exerts a slightly greater repulsion than a bonding pair, the bond angles are compressed from 109.5° to 107°.

氨（NH₃）是教科书级的实例。氮原子具有四个电子域（三条N–H键和一对孤对电子），因此发生sp³杂化。但由于孤对电子产生的斥力略大于成键电子对，键角从109.5°被压缩至107°。

Water goes further: with two lone pairs, the O–H bond angle is reduced to 104.5°. The order of repulsion strength is: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

水更为显著：由于具有两对孤对电子，O–H键角被减小至104.5°。斥力大小的顺序为：孤对–孤对 > 孤对–成键 > 成键–成键。

9. sp³d and sp³d² Hybridisation: Expanding Beyond the Octet | sp³d与sp³d²杂化：超越八隅体的扩展

Elements in Period 3 and beyond can utilise empty d orbitals to accommodate more than four electron domains. Phosphorus pentachloride (PCl₅) undergoes sp³d hybridisation, adopting a trigonal bipyramidal geometry with bond angles of 90° and 120°.

第三周期及以后的元素可以利用空的d轨道容纳四个以上的电子域。五氯化磷（PCl₅）发生sp³d杂化，采用三角双锥几何构型，键角分别为90°和120°。

Sulfur hexafluoride (SF₆) demonstrates sp³d² hybridisation, resulting in an octahedral geometry with all F–S–F bond angles at 90°. The 3s, 3p, and 3d orbitals of sulfur combine to form six equivalent hybrid orbitals.

六氟化硫（SF₆）展示了sp³d²杂化，形成正八面体几何构型，所有F–S–F键角均为90°。硫的3s、3p和3d轨道组合形成六个等价的杂化轨道。

10. Predicting Molecular Shape: A Step-by-Step Approach | 预测分子形状：分步方法

A systematic approach ensures accurate determination of both hybridisation and molecular shape for any given species:

系统化的方法能够确保准确确定任何给定物种的杂化类型和分子形状：

• Step 1: Draw the Lewis (electron dot) structure of the molecule, showing all valence electrons.

  第一步：画出分子的路易斯（电子点）结构，标示所有价电子。

• Step 2: Count the total number of electron domains around the central atom (bonds + lone pairs).

  第二步：计算中心原子周围的电子域总数（键数 + 孤对电子数）。

• Step 3: Assign the hybridisation based on the electron-domain count.

  第三步：根据电子域数量确定杂化类型。

• Step 4: Apply VSEPR theory to determine the electron-domain geometry, then refine this to the molecular shape by considering only the positions of atoms (ignore lone pairs when naming the shape).

  第四步：应用VSEPR理论确定电子域几何，然后仅根据原子位置（命名形状时忽略孤对电子）确定分子形状。

Lewis structure → Electron domains → Hybridisation → VSEPR → Molecular shape

This method works for all molecules in the IB HL syllabus, including those with expanded octets, and forms the foundation for understanding the connection between structure and reactivity.

这一方法适用于IB高级水平教学大纲中的所有分子，包括具有扩展八隅体的分子，并且为理解结构与反应性之间的联系奠定基础。

11. Sigma and Pi Bonding: The Full Picture | σ键与π键：完整的图景

A single bond always consists of exactly one σ bond, formed by the head-on overlap of two hybrid orbitals (or one hybrid orbital and one s orbital). A double bond contains one σ bond and one π bond. A triple bond contains one σ bond and two mutually perpendicular π bonds.

单键始终由一条σ键组成，由两个杂化轨道（或一个杂化轨道与一个s轨道）的端对端重叠形成。双键包含一条σ键和一条π键。三键包含一条σ键和两条相互垂直的π键。

Every σ bond framework corresponds to the hybridised orbitals, while the unhybridised p orbitals are entirely responsible for π bonding. This separation explains why rotation around a C=C double bond is restricted—rotation would require breaking the π bond—whereas rotation around a C–C single bond is generally free.

每个σ键骨架对应杂化轨道，而未杂化p轨道完全负责π键的形成。这种区分解释了为什么C=C双键周围的旋转受到限制——旋转需要破坏π键——而C–C单键周围的旋转通常是自由的。

12. Examination Tips and Common Pitfalls | 考试技巧与常见误区

IB Chemistry HL examinations frequently test hybridisation and molecular shape through data-based questions and structural analysis. To maximise your marks, pay attention to the following:

IB化学高级水平考试经常通过基于数据和结构分析的问题来考查杂化和分子形状。为了获得高分，请注意以下几点：

• Always count electron domains, not just bonding pairs. A double bond counts as one domain, not two.

  始终计算电子域，而非仅计算成键电子对。双键算作一个电子域，而非两个。

• Do not confuse electron-domain geometry with molecular shape. Water is “bent,” not “tetrahedral,” even though it is sp³ hybridised.

  不要混淆电子域几何与分子形状。水是”角形”，不是”正四面体形”，尽管它是sp³杂化。

• Use the correct notation: superscripts for hybrid orbitals (sp³, sp², sp) are essential.

  使用正确的记号：杂化轨道的上标（sp³、sp²、sp）至关重要。

• Memorise bond angles for the key geometries: 180°, 120°, 109.5°, and the distorted values for lone-pair systems.

  熟记关键几何构型的键角：180°、120°、109.5°，以及含孤对电子体系的畸变键角。

• When discussing hybridisation, always mention the involvement of d orbitals for Period 3 elements with five or six electron domains.

  在讨论杂化时，对于具有五个或六个电子域的第三周期元素，务必提及d轨道的参与。

Mastering orbital hybridisation not only secures examination marks but also provides deep insight into why molecules adopt the shapes they do—a concept that underpins stereochemistry, intermolecular forces, and reaction mechanisms throughout the IB HL curriculum.

掌握轨道杂化不仅能在考试中获得分数，还能深入理解分子为何呈现其特定的形状——这一概念支撑着整个IB高级水平课程中的立体化学、分子间作用力和反应机理。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry: Core Concepts of Energetics and Thermochemistry | IB化学：能量学与热化学核心知识

Energetics and thermochemistry form the backbone of understanding chemical reactions from an energy perspective. In IB Chemistry, this topic connects quantitative calculations with the fundamental question: how much energy is absorbed or released when chemical bonds are broken and formed? This article covers every core concept you need for exams, from enthalpy changes to Hess’s law, with clarity and precision.

能量学与热化学是从能量角度理解化学反应的核心。在IB化学中，这一主题将定量计算与一个基本问题紧密联系：当化学键断裂和形成时，体系吸收或释放多少能量？本文以清晰而精准的方式涵盖考试所需的全部核心概念，从焓变到赫斯定律。

1. System, Surroundings and Energy Transfer | 体系、环境与能量传递

In thermochemistry, the system is the part of the universe we are studying (e.g., the reactants and products in a flask), while the surroundings include everything else, such as the solvent, the flask, and the air. Energy can be transferred between the system and surroundings as heat (q) or work (w), but in a constant-pressure calorimeter, the heat change is equal to the enthalpy change.

在热化学中，体系（system）是我们研究的宇宙部分（例如烧瓶中的反应物和产物），而环境（surroundings）包括其他一切，如溶剂、烧瓶和空气。能量以热（q）或功（w）的形式在体系与环境之间传递。在恒压量热计中，热量变化等于焓变。

• Exothermic process: energy is released from system to surroundings (ΔH negative).

  放热过程：能量从体系释放到环境（ΔH为负）。

• Endothermic process: energy is absorbed from surroundings into system (ΔH positive).

  吸热过程：能量从环境吸收进入体系（ΔH为正）。

2. Enthalpy and Enthalpy Change | 焓与焓变

Enthalpy (H) is the total heat content of a system at constant pressure. We cannot measure absolute enthalpy, so chemists measure enthalpy change (ΔH): the difference between the enthalpy of products and reactants. For chemical reactions, ΔH is expressed in kJ per mole of reaction (kJ mol⁻¹).

焓（H）是体系在恒压下的总热含量。我们无法测量焓的绝对值，因此化学家测量焓变（ΔH），即产物与反应物焓的差值。对于化学反应，ΔH以每摩尔反应千焦（kJ mol⁻¹）表示。

ΔH = H(products) − H(reactants)

If ΔH < 0, the reaction is exothermic; if ΔH > 0, it is endothermic. The standard condition is 298 K and 100 kPa, indicated by the superscript θ: ΔH°.

若ΔH < 0，反应为放热；若ΔH > 0，则为吸热。标准状态为298 K和100 kPa，用上标θ表示：ΔH°。

3. Energy Profiles and Activation Energy | 能量曲线图与活化能

Energy profile diagrams show the energy pathway of a reaction. The reactants start at one energy level; the products end at another. The difference in height represents ΔH. The highest point along the curve is the transition state, and the energy required to reach it from the reactants is the activation energy (Eₐ).

能量曲线图显示反应的能量路径。反应物起始于某一能量水平，产物结束于另一水平。高度差代表ΔH。曲线最高点是过渡态，从反应物达到该点所需的能量就是活化能（Eₐ）。

• Exothermic profile: products lower than reactants; Eₐ appears on the “hump” going forward.

  放热曲线：产物低于反应物；正向的“峰”上显示Eₐ。

• Endothermic profile: products higher than reactants; reverse activation energy is larger.

  吸热曲线：产物高于反应物；逆向活化能更大。

Catalysts provide an alternative pathway with a lower activation energy, without changing ΔH.

催化剂提供了一条活化能更低的替代途径，但不改变ΔH。

4. Standard Enthalpy Changes of Reaction, Formation and Combustion | 标准反应焓变、生成焓变与燃烧焓变

Exam questions often refer to three specific types of standard enthalpy changes:

考试题目常涉及三种特定的标准焓变类型：

The standard enthalpy change of formation of an element in its most stable form is zero.

最稳定形态元素的标准生成焓变为零。

5. Bond Enthalpy and Average Bond Enthalpy | 键焓与平均键焓

Chemical reactions involve breaking bonds (endothermic) and forming bonds (exothermic). The bond enthalpy is the energy required to break one mole of a specific covalent bond in gaseous species. Because bond strengths vary slightly between compounds, tables use average bond enthalpies obtained from many compounds.

化学反应涉及断裂化学键（吸热）和形成化学键（放热）。键焓是在气态物种中断裂1摩尔特定共价键所需的能量。由于键强度在不同化合物中略有差异，表中通常使用从多种化合物获得的平均键焓。

ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed)

This method is useful for estimating ΔH when no calorimetry data is available, but it is only approximate because bonds in real molecules are influenced by their chemical environment.

该方法在没有量热数据时可用来估算ΔH，但仅为近似值，因为真实分子中的键会受化学环境影响。

6. Hess’s Law | 赫斯定律

Hess’s law states that the enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same. Since enthalpy is a state function, we can add enthalpy changes from multiple steps to find the overall ΔH.

赫斯定律指出，只要初始状态和最终状态相同，反应的焓变与路径无关。由于焓是状态函数，我们可以把多个步骤的焓变相加得到总ΔH。

Two common applications in IB exams:

IB考试中两种常见应用：

• Using ΔH°f: ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants)

  利用标准生成焓：ΔH°rxn = ΣΔH°f(产物) − ΣΔH°f(反应物)

• Using ΔH°c: ΔH°rxn = ΣΔH°c(reactants) − ΣΔH°c(products)

  利用标准燃烧焓：ΔH°rxn = ΣΔH°c(反应物) − ΣΔH°c(产物)

Always construct a Hess cycle or energy cycle diagram before solving, and pay attention to balancing coefficients.

解题前务必先画出赫斯循环或能量循环图，并注意配平系数。

7. Calorimetry and Experimental Determination of ΔH | 量热法与ΔH的实验测定

In a constant-pressure calorimeter, the heat absorbed or released by a reaction is calculated from the temperature change of the surrounding water. Key equations:

在恒压量热计中，反应吸收或释放的热量根据周围水的温度变化来计算。关键公式：

q = mcΔT

where q is heat (J), m is mass of water (g), c is specific heat capacity (4.18 J g⁻¹ K⁻¹ for water), and ΔT is the temperature change in K.

其中q为热量(J)，m为水的质量(g)，c为比热容（水的比热容为4.18 J g⁻¹ K⁻¹），ΔT为以K为单位的温度变化。

To convert to molar enthalpy change:

换算成摩尔焓变：

ΔH = −q / n

The negative sign accounts for the direction of heat flow relative to the system. In practice, experimental values are often less exothermic than theoretical values due to heat loss and incomplete reactions.

负号表示热量相对于体系流动的方向。实际实验中，由于热量损失和反应不完全，实验值通常比理论值放热更少。

8. Limitations and Errors in Calorimetry | 量热法的局限性与误差

Students should be able to evaluate experimental design. Major sources of error include:

学生应能评估实验设计。主要误差来源包括：

Using a polystyrene cup, stirring, and recording maximum or minimum temperatures precisely can reduce these errors. For combustion, a bomb calorimeter is more accurate.

使用聚苯乙烯泡沫杯、搅拌以及精确记录最高或最低温度可以减少这些误差。对于燃烧反应，弹式量热计更为准确。

9. Representing ΔH in Chemical Equations and Energy Cycles | 在化学方程式和能量循环中表示ΔH

Thermochemical equations include the enthalpy change alongside the balanced equation. For example:

热化学方程在配平方程旁边同时给出焓变。例如：

2H₂(g) + O₂(g) → 2H₂O(l)　ΔH° = −572 kJ mol⁻¹

Note that the “mol⁻¹” refers to the moles of reaction as written, not to a single substance. If coefficients are halved, ΔH is also halved. Energy cycles (Hess cycles) combine level diagrams showing formation, combustion, or direct transformation pathways.

注意这里的“mol⁻¹”指的是所写方程对应的反应摩尔数，而不是某一物质的摩尔数。如果系数减半，ΔH也减半。能量循环（赫斯循环）将生成、燃烧或直接转化路径的能级图组合在一起。

10. Predicting Enthalpy Changes from Bond Data | 利用键能数据预测焓变

When solving bond enthalpy problems, always:

在解答键焓问题时，务必：

• Draw the Lewis structures of all reactants and products.

  画出所有反应物和产物的路易斯结构。

• Count all bonds broken and all bonds formed.

  数清所有断裂和形成的化学键数量。

• Use average bond enthalpies in kJ mol⁻¹; multiply by the number of moles of each bond.

  使用平均键焓（单位kJ mol⁻¹），并乘以每种键的物质的量。

• Apply the formula: ΔH = Σ(broken) − Σ(formed).

  套用公式：ΔH = Σ(断裂) − Σ(形成)。

Remember that this method assumes all species are in the gaseous state, so it may not match exact experimental values for reactions involving liquids or solids.

请记住，此方法假设所有物质均为气态，因此对于涉及液体或固体的反应，结果可能与精确实验值不完全一致。

11. Common Exam Pitfalls | 常见考试误区

Here are the most frequent mistakes students make in this topic:

以下是学生在这一主题中最常见的错误：

• Confusing ΔH°f and ΔH°c sign conventions in Hess cycles.

  在赫斯循环中混淆ΔH°f和ΔH°c的正负号惯例。

• Forgetting to multiply ΔH by stoichiometric coefficients.

  忘记将ΔH乘以化学计量系数。

• Using the wrong mass in q = mcΔT (must be the mass of water, not the reactant).

  在q = mcΔT中使用了错误的质量（必须是水的质量，而不是反应物的质量）。

• Omitting units or writing kJ instead of kJ mol⁻¹.

  遗漏单位或写成kJ而不是kJ mol⁻¹。

• Stating that bond breaking is exothermic (it is endothermic).

  错误地认为断裂化学键是放热的（实际上是吸热的）。

Always reflect on whether the sign of ΔH makes physical sense: combustion is always exothermic, while decomposing a stable compound is usually endothermic.

始终反思ΔH的符号是否具有物理意义：燃烧总是放热，而分解稳定化合物通常是吸热。

12. Linking Energetics to the IB Syllabus and Beyond | 将能量学与IB大纲及更广内容联系起来

Energetics appears in multiple IB topics: for example, bond enthalpies support the study of reaction kinetics (activation energy), and Hess’s law underpins thermodynamic calculations in organic chemistry and electrochemistry. Being fluent in this topic also prepares you for higher-level concepts such as entropy and Gibbs free energy, which are introduced in the IB HL syllabus.

能量学出现在IB多个主题中：例如，键焓支持反应动力学（活化能）的研究，赫斯定律为有机化学和电化学中的热力学计算奠定基础。熟练掌握这一主题还能为你学习IB高级水平课程中引入的熵和吉布斯自由能等更高阶概念做好准备。

Practical tip: practice drawing Hess cycles from different data sets, and always label the direction of arrows (up for formation, down for combustion) to avoid sign errors.

实用建议：练习从不同数据集绘制赫斯循环，并始终标注箭头方向（生成向上，燃烧向下），以避免符号错误。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Intermolecular Forces and Their Effects on Material Properties | IB化学：分子间作用力对物质性质的影响

Intermolecular forces (IMFs) are the attractive forces that exist between individual molecules, as opposed to the intramolecular bonds that hold atoms together within a molecule. These forces, though far weaker than covalent or ionic bonds, determine nearly every physical property of molecular substances — from whether a liquid boils at 20 °C or 200 °C, to whether two substances can mix, to the complex folding of proteins in your body.

分子间作用力（IMFs）是存在于单个分子之间的吸引力，与将分子内原子结合在一起的化学键（分子内作用力）相对。这些力虽然远弱于共价键或离子键，却几乎决定了分子物质的每一种物理性质——从液体是在 20 °C 还是 200 °C 沸腾，到两种物质能否互溶，再到你体内蛋白质的复杂折叠。

For IB Chemistry (Topic 2 and 4), mastering IMFs is essential not only for explaining trends in boiling points, melting points, and solubility, but also for answering structured questions that require comparison and justification. In this article, we will systematically examine the three main types of intermolecular forces, their relative strengths, and how they manifest in observable material properties.

对于IB化学（Topic 2 和 Topic 4）而言，掌握分子间作用力不仅对解释沸点、熔点和溶解度的趋势至关重要，也直接关系到需要比较与论证的结构化问答题。在本文中，我们将系统地考察三类主要的分子间作用力、它们的相对强度，以及它们如何在可观测的物质性质中体现出来。

1. The Three Types of Intermolecular Forces | 三类分子间作用力

Intermolecular forces can be classified into three categories: London dispersion forces (also called instantaneous dipole-induced dipole forces), permanent dipole-dipole forces, and hydrogen bonding. All molecular substances experience London dispersion forces, but only polar molecules experience dipole-dipole forces and hydrogen bonding.

分子间作用力可分为三类：伦敦色散力（又称瞬时偶极-诱导偶极力）、永久偶极-偶极力和氢键。所有分子物质都会产生伦敦色散力，但只有极性分子才会产生偶极-偶极力和氢键。

A common IB exam misconception is that hydrogen bonding is a type of covalent bond. It is not — hydrogen bonding is an intermolecular force, albeit the strongest one. It occurs when a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) is attracted to a lone pair of electrons on a neighbouring electronegative atom.

IB考试中一个常见的误区是认为氢键是一种共价键。事实并非如此——氢键是一种分子间作用力，只不过是最强的分子间作用力。它发生在与高电负性原子（N、O 或 F）以共价键相连的氢原子，被邻近电负性原子上的孤对电子所吸引的情形。

2. London Dispersion Forces: The Universal Attraction | 伦敦色散力：普遍的吸引力

London dispersion forces arise from instantaneous dipoles. At any given instant, the electron cloud of a molecule may be unevenly distributed, creating a temporary dipole. This temporary dipole induces a dipole in a neighbouring molecule, resulting in a net attraction.

伦敦色散力源于瞬时偶极。在任何给定瞬间，分子的电子云都可能分布不均，从而形成瞬时偶极。该瞬时偶极会在邻近分子中诱导出偶极，最终产生净吸引力。

The strength of London dispersion forces depends primarily on two factors: the number of electrons in the molecule and the surface area available for contact. Larger molecules with more electrons have larger, more polarisable electron clouds, leading to stronger dispersion forces.

伦敦色散力的强度主要取决于两个因素：分子中的电子总数和可供接触的表面积。较大的分子拥有更多电子，其电子云更大且更易极化，因此色散力更强。

For example, among the noble gases, the boiling point increases steadily from helium (4.2 K) to xenon (165 K) as the number of electrons increases from 2 to 54. Similarly, in the halogen series, fluorine boils at −188 °C while iodine boils at 184 °C — a difference of over 370 °C due entirely to increasing London dispersion forces.

例如，在稀有气体中，沸点从氦（4.2 K）到氙（165 K）稳步上升，其电子数从 2 增加到 54。同样地，在卤素系列中，氟的沸点为 −188 °C，而碘的沸点为 184 °C——相差超过 370 °C，这完全归因于伦敦色散力的增强。

Boiling point trend: F₂ (−188 °C) < Cl₂ (−34 °C) < Br₂ (59 °C) < I₂ (184 °C)

Another important consideration is molecular shape. For isomeric hydrocarbons, the more branched isomer has a lower boiling point because branching reduces the surface area over which dispersion forces can act. For instance, 2,2-dimethylpropane (neopentane, b.p. 9.5 °C) boils at a much lower temperature than pentane (b.p. 36.1 °C), even though both have the formula C₅H₁₂.

另一个重要因素是分子形状。对于同分异构的烃类，支链越多的异构体沸点越低，因为支链化减少了色散力可以作用的表面积。例如，2,2-二甲基丙烷（新戊烷，沸点 9.5 °C）的沸点远低于正戊烷（沸点 36.1 °C），尽管两者的分子式都是 C₅H₁₂。

3. Permanent Dipole-Dipole Forces: The Polar Attraction | 永久偶极-偶极力：极性吸引

When a molecule contains polar bonds that do not cancel out symmetrically, the molecule possesses a permanent dipole. The positive end of one polar molecule is electrostatically attracted to the negative end of another, giving rise to dipole-dipole forces.

当分子含有极性键且这些极性键不能对称抵消时，该分子就具有永久偶极。一个极性分子的正端会受到另一个极性分子负端的静电吸引，从而产生偶极-偶极力。

Molecules such as HCl (δ⁺ on H, δ⁻ on Cl) and SO₂ exhibit permanent dipole-dipole interactions. Evidence for these forces comes from comparing boiling points of similar-sized molecules: HCl (b.p. −85 °C) boils much higher than F₂ (b.p. −188 °C), despite both having roughly comparable numbers of electrons. The extra 100 °C is due to dipole-dipole attractions in HCl.

像 HCl（H 端为 δ⁺，Cl 端为 δ⁻）和 SO₂ 这样的分子表现出永久偶极-偶极相互作用。这些力存在的证据来自比较大小相近分子的沸点：HCl（沸点 −85 °C）的沸点远高于 F₂（沸点 −188 °C），尽管两者的电子数大致相当。多出的约 100 °C 正是 HCl 中偶极-偶极力作用的结果。

It is important to recognise that polar molecules also experience London dispersion forces; dipole-dipole forces are an additional contribution. The total intermolecular attraction in a polar substance is the sum of all these contributions.

需要认识到，极性分子同样会产生伦敦色散力；偶极-偶极力只是额外的贡献。极性物质中总的分子间吸引力是所有这些贡献之和。

4. Hydrogen Bonding: A Special Case | 氢键：一种特殊情况

Hydrogen bonding is a particularly strong type of dipole-dipole interaction. It occurs when hydrogen is covalently bonded to nitrogen, oxygen, or fluorine — atoms that are small and highly electronegative. The resulting bond is so polarised that the hydrogen atom, essentially a bare proton with very little electron shielding, can be strongly attracted to a lone pair on another electronegative atom.

氢键是一种特别强的偶极-偶极相互作用。它发生在氢与氮、氧或氟——这些体积小且电负性高的原子——共价相连时。由此产生的键极性极强，氢原子实际上是一个几乎没有电子屏蔽的裸露质子，因此能强烈吸引另一电负性原子上的孤对电子。

Three conditions must be satisfied for hydrogen bonding to occur: (1) a hydrogen atom attached to N, O, or F; (2) a lone pair of electrons on N, O, or F of a neighbouring molecule; (3) the two atoms involved must be close in space. The hydrogen bond can be represented as follows:

氢键的形成必须满足三个条件：（1）氢原子与 N、O 或 F 相连；（2）相邻分子的 N、O 或 F 原子上有孤对电子；（3）两个相关原子在空间上足够接近。氢键可表示为：

δ⁻ O–H ··· O δ⁻ (where ··· represents the hydrogen bond | 其中 ··· 代表氢键)

A classic IB example is water. Each water molecule can form up to four hydrogen bonds: two via its hydrogen atoms and two via the lone pairs on oxygen. This extensive network of hydrogen bonds explains why water has such an unusually high boiling point (100 °C) compared to H₂S (b.p. −60 °C), even though sulfur is heavier and has more electrons.

一个经典的IB例子是水。每个水分子最多可以形成四个氢键：两个通过其氢原子，两个通过氧上的孤对电子。这种广泛的氢键网络解释了为什么水的沸点异常之高（100 °C），而 H₂S 的沸点却只有 −60 °C——尽管硫更重且电子更多。

5. Boiling and Melting Points: The Direct Consequence | 沸点与熔点：最直接的后果

The boiling point of a liquid is the temperature at which its vapour pressure equals atmospheric pressure. To boil, molecules must overcome all intermolecular forces and escape into the gas phase. Therefore, stronger IMFs lead to higher boiling points.

液体的沸点是指其蒸气压等于大气压时的温度。要使液体沸腾，分子必须克服所有分子间作用力并逃逸到气相中。因此，分子间作用力越强，沸点越高。

Consider the hydrides of Group 15–17 elements. If only London dispersion forces mattered, boiling points would increase smoothly with molar mass down each group. However, NH₃, H₂O, and HF all show anomalously high boiling points compared to their heavier congeners:

考虑第15–17族元素的氢化物。如果只有伦敦色散力起作用，各族的沸点应随摩尔质量自上而下平滑上升。然而，NH₃、H₂O 和 HF 与它们更重的同族化合物相比，沸点都异常高：

These anomalies are direct evidence that hydrogen bonding dramatically elevates boiling points. The positive δ⁺ region of one molecule strongly interacts with the δ⁻ region of another, far exceeding what dispersion forces alone would predict.

这些反常现象直接证明了氢键能显著提高沸点。一个分子的 δ⁺ 区域与另一个分子的 δ⁻ 区域强烈相互作用，远远超过了仅凭色散力所预测的值。

Melting points follow similar logic: solids require energy to overcome the ordered lattice of intermolecular forces. However, the correlation is less straightforward because packing efficiency and molecular symmetry also matter. For example, despite weaker IMFs, highly symmetrical molecules may pack more efficiently in the solid state, raising their melting points disproportionately.

熔点遵循类似的逻辑：固体需要能量来克服分子间作用力所维持的有序晶格。然而，这种相关性不那么直接，因为堆积效率与分子对称性也很重要。例如，尽管分子间作用力较弱，高度对称的分子在固态中可能堆积得更紧密，从而使熔点不成比例地升高。

6. Vapour Pressure and Volatility | 蒸气压与挥发性

Vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid at a given temperature. Molecules at the surface of a liquid with sufficient kinetic energy can escape into the gas phase. Stronger intermolecular forces make escape more difficult, resulting in lower vapour pressure and lower volatility.

蒸气压是在给定温度下，与液体处于平衡状态的蒸气所施加的压力。液体表面具有足够动能的分子可以逃逸到气相中。分子间作用力越强，逃逸越困难，蒸气压越低，挥发性越低。

Substances with high volatility, such as ethanol and acetone, have relatively weak IMFs and high vapour pressures at room temperature. Conversely, water, with its extensive hydrogen bonding, has a much lower vapour pressure. This is why wet clothes dry more slowly in humid conditions — the vapour pressure of water is lower when the air already contains significant water vapour.

乙醇和丙酮等高挥发性物质具有相对较弱的分子间作用力和较高的室温蒸气压。相反，水由于广泛的氢键作用，蒸气压要低得多。这就是为什么在潮湿条件下湿衣服干得更慢——当空气中已含有大量水蒸气时，水的蒸气压更低。

An IB exam question may ask you to compare the vapour pressures of two liquids. The correct approach is to identify which has stronger IMFs and then state that the weaker-IMF substance has higher vapour pressure and evaporates more readily at the same temperature.

IB考试题可能会要求你比较两种液体的蒸气压。正确的思路是：先判断哪种物质的分子间作用力更强，然后指出分子间作用力较弱的物质在相同温度下蒸气压更高、蒸发更快。

7. Solubility: Like Dissolves Like | 溶解度：相似相溶

Solubility is governed by the principle “like dissolves like” — that is, substances with similar types and strengths of intermolecular forces tend to be mutually soluble. Polar solutes dissolve in polar solvents because the solute-solvent dipole interactions compensate for the energy required to separate solute particles and solvent molecules.

溶解度的核心原则是”相似相溶”——即分子间作用力类型和强度相似的物质倾向于互溶。极性溶质溶于极性溶剂，是因为溶质-溶剂之间的偶极相互作用足以补偿分离溶质粒子和溶剂分子所需的能量。

Water, being highly polar and capable of hydrogen bonding, dissolves ionic compounds (via ion-dipole interactions), polar covalent compounds, and other hydrogen-bonding molecules such as ethanol and glucose. In contrast, non-polar substances like oil and grease, which interact only through London dispersion forces, are immiscible with water.

水是高度极性的溶剂，且能形成氢键，因此可以溶解离子化合物（通过离子-偶极相互作用）、极性共价化合物以及乙醇、葡萄糖等能形成氢键的分子。相反，仅通过伦敦色散力相互作用的非极性物质（如油和油脂）与水不互溶。

This principle also explains why I₂, a non-polar halogen, dissolves readily in non-polar solvents like hexane (giving a purple solution) but only sparingly in water (giving a brown solution). Similarly, the solubility of alcohols in water decreases as the hydrocarbon chain lengthens: methanol, ethanol, and propanol are fully miscible, but octanol is only slightly soluble.

这一原则也解释了为什么非极性的卤素单质 I₂ 易溶于己烷等非极性溶剂（得到紫色溶液），而在水中仅微溶（得到棕色溶液）。同样，醇类在水中的溶解度随碳链增长而下降：甲醇、乙醇和丙醇完全互溶，但辛醇仅微溶。

8. Viscosity and Surface Tension | 黏度与表面张力

Viscosity is a measure of a liquid’s resistance to flow. Liquids with strong intermolecular forces have higher viscosity because the molecules resist sliding past one another. Glycerol, with its three hydroxyl groups capable of extensive hydrogen bonding, is syrupy and viscous. In contrast, hexane, which experiences only dispersion forces, flows freely.

黏度是液体对流动阻力的度量。分子间作用力强的液体黏度更高，因为分子抵抗彼此滑移。甘油含有三个可形成广泛氢键的羟基，因此呈糖浆状且黏稠。相反，仅有色散力的己烷流动性很好。

Viscosity also depends on temperature: heating a liquid imparts kinetic energy to its molecules, partially overcoming IMFs and reducing viscosity. This is why cooking oil thins when heated and honey flows more easily when warmed.

黏度还取决于温度：加热液体赋予分子更多动能，部分克服分子间作用力，从而降低黏度。这就是为什么食用油加热后变稀，蜂蜜温热后更容易流动。

Surface tension is the energy required to increase the surface area of a liquid by a unit amount. Molecules at the surface experience a net inward pull because they are attracted to molecules below and beside them, but not above (in the gas phase). Stronger IMFs produce higher surface tension. Water, with hydrogen bonding, has a remarkably high surface tension — which is why small insects can walk on water and why water forms nearly spherical droplets.

表面张力是将液体表面积增加单位面积所需的能量。液体表面的分子受到来自下方和侧方分子的净向内拉力，因为上方（气相中）没有分子可供吸引。分子间作用力越强，表面张力越大。水由于氢键作用，表面张力异常之高——这就是为什么小昆虫能在水面行走，为什么水会形成近乎球形的水滴。

Surface tension: H₂O > ethanol > hexane

9. Biological Significance of Hydrogen Bonding | 氢键的生物学意义

Hydrogen bonding plays an indispensable role in biological systems. The double helix structure of DNA is stabilised by hydrogen bonds between complementary base pairs: adenine pairs with thymine via two hydrogen bonds, while cytosine pairs with guanine via three hydrogen bonds. This specific base pairing allows the genetic code to be faithfully replicated.

氢键在生物系统中扮演着不可或缺的角色。DNA 双螺旋结构由互补碱基之间的氢键来稳定：腺嘌呤与胸腺嘧啶通过两个氢键配对，而胞嘧啶与鸟嘌呤通过三个氢键配对。这种特异性碱基配对使遗传密码能够被忠实复制。

Protein structure also relies on hydrogen bonding: the α-helix and β-pleated sheet secondary structures are held together by hydrogen bonds between amino acid residues. When these bonds are disrupted — for example, by high temperature or extreme pH — the protein denatures and loses its biological function.

蛋白质结构同样依赖氢键：α-螺旋和β-折叠片层二级结构由氨基酸残基之间的氢键维系。当这些氢键被破坏时——例如高温或极端pH条件下——蛋白质变性并失去其生物功能。

The anomalous expansion of water upon freezing is also a consequence of hydrogen bonding. In ice, water molecules arrange into an open hexagonal lattice held by hydrogen bonds, making ice less dense than liquid water. This is why ice floats — a property essential for aquatic life in cold climates, as the insulating ice layer on the surface protects the water below from freezing solid.

水在结冰时体积膨胀的反常现象同样是氢键的结果。在冰中，水分子排列成由氢键维系的开放式六方晶格，使冰的密度小于液态水。这就是为什么冰会漂浮——这一特性对寒冷气候中的水生生物至关重要，因为表面的冰层具有保温作用，可防止下方的水体完全冻结。

10. Comparing Intermolecular Forces in Exam Problems | 考试问题中的分子间作用力比较

IB examination questions frequently ask students to compare and explain the relative strengths of intermolecular forces. A systematic approach is to follow this three-step framework:

IB考试题目经常要求学生比较和解释分子间作用力的相对强弱。一个系统化的方法是遵循以下三步框架：

• Step 1 | 第一步：Draw or visualise the molecular structures to determine polarity and presence of N–H, O–H, or F–H bonds. | 画出或想象分子结构，判断极性以及是否存在 N–H、O–H 或 F–H 键。
• Step 2 | 第二步：Identify all types of IMFs present (dispersion always, plus dipole-dipole, plus possibly hydrogen bonding). | 确定存在的所有分子间作用力类型（色散力总有，再加上偶极-偶极力，以及可能的氢键）。
• Step 3 | 第三步：Rank the substances based on the type and number of IMFs, considering also molecular size for dispersion forces. | 根据分子间作用力的类型和数量对物质进行排序，同时考虑分子大小对色散力的影响。

For example, consider CH₄, NH₃, and H₂O, all with similar molar masses (16–18 g mol⁻¹). CH₄ is non-polar and experiences only London dispersion forces. NH₃ is polar, with hydrogen bonding (one lone pair on N, three N–H bonds). H₂O is polar, with hydrogen bonding, and can form twice as many hydrogen bonds per molecule as NH₃. Therefore, the boiling point order is CH₄ (−164 °C) < NH₃ (−33 °C) < H₂O (100 °C).

例如，考虑摩尔质量相近（16–18 g mol⁻¹）的 CH₄、NH₃ 和 H₂O。CH₄ 是非极性分子，仅产生伦敦色散力。NH₃ 是极性分子，能形成氢键（N 上有一个孤对电子，三个 N–H 键）。H₂O 是极性分子，能形成氢键，且每个分子能形成的氢键数量是 NH₃ 的两倍。因此沸点顺序为 CH₄（−164 °C）< NH₃（−33 °C）< H₂O（100 °C）。

A subtle point that students often miss: when comparing two substances of very different molar masses, London dispersion forces can outweigh weak permanent dipole-dipole forces. For example, butanone (a polar ketone, M = 72) boils at 80 °C, while hexane (non-polar, M = 86) boils at 69 °C — the extra dispersion force in hexane nearly compensates for the dipole interactions in butanone.

一个学生经常忽略的微妙之处：当比较摩尔质量相差悬殊的两种物质时，伦敦色散力可能超过较弱的永久偶极-偶极力。例如，丁酮（极性酮，M = 72）沸点为 80 °C，而己烷（非极性，M = 86）沸点为 69 °C——己烷中多出的色散力几乎抵消了丁酮中的偶极相互作用。

11. Conclusion | 总结

Intermolecular forces, though individually weak, collectively determine the physical properties of molecular substances. London dispersion forces are universal and increase with electron count and surface area; permanent dipole-dipole forces add an extra attraction between polar molecules; and hydrogen bonding, the strongest IMF, explains the anomalous properties of water, ammonia, and hydrogen fluoride, as well as the intricate structures of biological macromolecules.

分子间作用力虽然单独看来很微弱，但它们共同决定了分子物质的物理性质。伦敦色散力具有普遍性，随电子数和表面积的增加而增强；永久偶极-偶极力在极性分子之间提供额外吸引力；而氢键作为最强的分子间作用力，解释了水、氨和氟化氢的反常性质，以及生物大分子的精密结构。

When approaching IB questions on this topic, always begin by identifying the types of molecules involved, then ranking their intermolecular forces systematically, and finally connecting these forces to the property being asked about — boiling point, vapour pressure, solubility, viscosity, or surface tension. With this structured strategy, you will be well-prepared to score full marks on intermolecular forces questions.

在解答IB相关问题时，始终从判断分子类型入手，然后系统地排序其分子间作用力强弱，最后将这些力与所问的性质——沸点、蒸气压、溶解度、黏度或表面张力——联系起来。有了这种结构化的策略，你就能在分子间作用力相关的题目中获得满分。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 IB Chemistry HL: First-Row Transition Elements | IB化学HL：第一过渡系元素要点

The first-row transition elements (Sc to Zn) are a cornerstone of IB Chemistry HL. This article consolidates the essential concepts—electronic configurations, variable oxidation states, complex ion formation, colour, magnetism, catalysis, and standard electrode potentials—that you need for exam success.

第一过渡系元素（Sc至Zn）是IB化学HL的基石内容。本文系统整合了核心考点——电子构型、可变氧化态、配离子形成、颜色、磁性、催化作用及标准电极电位——助你在考试中脱颖而出。

1. Electronic Configurations | 电子构型

The first transition series runs from Scandium (Z=21) to Zinc (Z=30). A key feature is the filling of the 3d subshell after the 4s subshell. For example, Titanium has the configuration [Ar] 3d²4s², while Chromium is [Ar] 3d⁵4s¹ and Copper is [Ar] 3d¹⁰4s¹—both exceptions arise from the extra stability of half-filled and fully-filled d subshells.

第一过渡系从钪（Z=21）延伸至锌（Z=30）。关键特征在于3d亚层先于4s亚层填充的次序。例如，钛的构型为 [Ar] 3d²4s²，而铬为 [Ar] 3d⁵4s¹、铜为 [Ar] 3d¹⁰4s¹——这两种例外源于半充满和全充满d亚层的额外稳定性。

When forming ions, the 4s electrons are removed first. Hence Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵, not the reverse. This is a common exam trap.

形成离子时，4s电子优先失去。因此Fe²⁺为 [Ar] 3d⁶，Fe³⁺为 [Ar] 3d⁵，而非相反顺序。这是常见的考试陷阱。

2. Variable Oxidation States | 可变氧化态

Transition metals exhibit multiple oxidation states because the 3d and 4s electrons have comparable energies. The maximum oxidation state increases across the series up to Mn (+7), after which the 3d electrons become increasingly tightly bound and harder to remove.

过渡金属呈现多种氧化态，因为3d与4s电子的能量相近。最高氧化态沿周期递增至Mn（+7），此后3d电子被束缚得越来越紧，难以继续失去。

The most stable oxidation states follow a pattern: Sc³⁺, Ti⁴⁺, V⁵⁺, Cr³⁺ or Cr⁶⁺, Mn²⁺, Fe²⁺/Fe³⁺, Co²⁺, Ni²⁺, Cu²⁺, Zn²⁺. Note that Sc³⁺ and Zn²⁺ have no unpaired d-electrons (d⁰ and d¹⁰ respectively), so they are not considered typical transition metals and their compounds are colourless and diamagnetic.

最稳定氧化态呈现一定规律：Sc³⁺、Ti⁴⁺、V⁵⁺、Cr³⁺或Cr⁶⁺、Mn²⁺、Fe²⁺/Fe³⁺、Co²⁺、Ni²⁺、Cu²⁺、Zn²⁺。注意Sc³⁺和Zn²⁺没有未配对d电子（分别为d⁰和d¹⁰），因此它们不被视为典型过渡金属，其化合物无色且抗磁性。

Higher oxidation states tend to form covalent compounds or oxoanions (e.g. VO₃⁻, CrO₄²⁻, MnO₄⁻), whereas lower oxidation states form ionic compounds. This explains why V₂O₅ and KMnO₄ are molecular/ionic covalent species rather than simple ionic solids.

高氧化态倾向于形成共价化合物或含氧酸根（如VO₃⁻、CrO₄²⁻、MnO₄⁻），而低氧化态形成离子化合物。这解释了为何V₂O₅和KMnO₄是共价型物种而非简单离子固体。

The reducing power of M²⁺ species decreases across the series, while oxidising power of higher states increases. For example, Fe²⁺ is a mild reducing agent, but Cu²⁺ is a mild oxidising agent.

M²⁺物种的还原性沿周期递减，而高氧化态的氧化性递增。例如，Fe²⁺是温和还原剂，而Cu²⁺是温和氧化剂。

3. Complex Ion Formation | 配离子形成

A complex ion consists of a central transition metal ion surrounded by ligands—species that donate lone pairs via dative (coordinate) covalent bonds. Common ligands include H₂O, NH₃, Cl⁻, CN⁻ and OH⁻.

配离子由中心过渡金属离子及周围配体构成——配体通过配位共价键提供孤对电子。常见配体包括H₂O、NH₃、Cl⁻、CN⁻和OH⁻。

Coordination number is the number of dative bonds formed. [Cu(H₂O)₆]²⁺ has coordination number 6 (octahedral), [CoCl₄]²⁻ has coordination number 4 (tetrahedral), and [CuCl₄]²⁻ is square planar. Bidentate ligands such as ethanedioate (C₂O₄²⁻) and 1,2-diaminoethane (en) form chelate rings—five- or six-membered—which greatly enhance stability.

配位数是形成的配位键数目。[Cu(H₂O)₆]²⁺配位数为6（八面体），[CoCl₄]²⁻配位数为4（四面体），[CuCl₄]²⁻为平面正方形。双齿配体如乙二酸根（C₂O₄²⁻）和1,2-二氨基乙烷（en）形成五元或六元螯合环，大幅提高稳定性。

The chelate effect arises because each bidentate ligand replaces two monodentate ligands, increasing the number of particles and thus increasing entropy (ΔS > 0), making ΔG more negative.

螯合效应源于每个双齿配体取代两个单齿配体，使粒子数增加、熵增大（ΔS > 0），从而使ΔG更负。

ΔG = ΔH − TΔS

Ligand substitution reactions are common: [Cu(H₂O)₆]²⁺ (pale blue) + 4Cl⁻ ⇌ [CuCl₄]²⁻ (yellow-green) + 6H₂O. The equilibrium shifts according to concentration and ligand strength.

配体取代反应非常常见：[Cu(H₂O)₆]²⁺（浅蓝色）+ 4Cl⁻ ⇌ [CuCl₄]²⁻（黄绿色）+ 6H₂O。平衡随浓度和配体强度移动。

4. Colour and d-d Transitions | 颜色与d-d跃迁

Transition metal complexes are coloured because of electronic transitions between split d-orbitals. In an octahedral field, the five degenerate d-orbitals split into two sets: the higher-energy e_g set (dₓ²₋ᵧ², d_z²) and the lower-energy t₂g set (dₓᵧ, dₓ_z, d_yz). The energy gap is called the crystal field splitting energy, Δ₀.

过渡金属配合物之所以有颜色，是因为电子在分裂后的d轨道之间跃迁。在八面体场中，五个简并d轨道分裂为两组：高能e_g组（dₓ²₋ᵧ²、d_z²）和低能t₂g组（dₓᵧ、dₓ_z、d_yz）。能隙称为晶体场分裂能Δ₀。

When white light passes through a complex, photons whose energy matches Δ₀ are absorbed, promoting an electron from t₂g to e_g. The transmitted/reflected light is the complementary colour. For example, [Cu(H₂O)₆]²⁺ absorbs red-orange light and appears blue.

当白光通过配合物时，能量恰好等于Δ₀的光子被吸收，将电子从t₂g激发至e_g。透射/反射光呈现互补色。例如，[Cu(H₂O)₆]²⁺吸收红橙色光，因而呈现蓝色。

Key factors affecting Δ₀ and hence colour:

• Nature of ligand—strong field ligands (CN⁻, CO) cause larger splitting than weak field ligands (Cl⁻, H₂O). The spectrochemical series ranks ligands: I⁻ < Cl⁻ < H₂O < NH₃ < en < CN⁻ < CO.
• 配体性质——强场配体（CN⁻、CO）比弱场配体（Cl⁻、H₂O）产生更大的分裂。光谱化学序列：I⁻ < Cl⁻ < H₂O < NH₃ < en < CN⁻ < CO。
• Oxidation state of metal—higher oxidation states increase Δ₀ because the metal ion is smaller and attracts ligands more strongly.
• 金属氧化态——氧化态越高，金属离子越小、对配体吸引力越强，Δ₀越大。
• Coordination number/geometry—octahedral splitting differs from tetrahedral. Tetrahedral splitting Δ_t ≈ 4/9 Δ₀, so tetrahedral complexes are often more intensely coloured.
• 配位数/几何构型——八面体分裂与四面体不同。四面体分裂Δ_t ≈ 4/9 Δ₀，因此四面体配合物颜色往往更深。

Ions with d⁰ or d¹⁰ configurations cannot undergo d-d transitions and are therefore colourless (e.g. Sc³⁺, Zn²⁺).

d⁰或d¹⁰构型的离子不能发生d-d跃迁，因此无色（如Sc³⁺、Zn²⁺）。

5. Magnetism | 磁性

Complexes containing unpaired d-electrons are paramagnetic—attracted by an external magnetic field. Complexes with all electrons paired are diamagnetic—weakly repelled by a magnetic field.

含有未配对d电子的配合物具有顺磁性——被外部磁场吸引。所有电子均配对的配合物具有抗磁性——被磁场微弱排斥。

In octahedral complexes, the d⁴–d⁷ configurations can be either high-spin or low-spin depending on whether Δ₀ is smaller or larger than the pairing energy (P):

在八面体配合物中，d⁴–d⁷构型可为高自旋或低自旋，取决于Δ₀与配对能（P）的相对大小：

• Weak field ligands → small Δ₀ → Δ₀ < P → high-spin (Hund's rule favoured)
• 弱场配体 → Δ₀较小 → Δ₀ < P → 高自旋（洪特规则优先）
• Strong field ligands → large Δ₀ → Δ₀ > P → low-spin (pairing favoured)
• 强场配体 → Δ₀较大 → Δ₀ > P → 低自旋（配对优先）

For d⁶, [Fe(CN)₆]⁴⁻ is low-spin (diamagnetic) because CN⁻ is a strong field ligand, while [Fe(H₂O)₆]²⁺ is high-spin (paramagnetic) because H₂O is a weak field ligand. The number of unpaired electrons determines the magnetic moment, measurable experimentally.

对于d⁶构型，[Fe(CN)₆]⁴⁻因CN⁻是强场配体而为低自旋（抗磁性），而[Fe(H₂O)₆]²⁺因H₂O为弱场配体而为高自旋（顺磁性）。未配对电子数决定磁矩，可通过实验测量。

6. Catalysis | 催化作用

Transition metals and their compounds are excellent catalysts because of their ability to adopt multiple oxidation states and provide vacant d-orbitals for substrate binding.

过渡金属及其化合物是优良催化剂，因为其能呈现多种氧化态并为底物结合提供空d轨道。

Homogeneous catalysis occurs when the catalyst is in the same phase as the reactants. A classic IB case is the Fe²⁺/Fe³⁺ catalysed decomposition of H₂O₂:

均相催化指催化剂与反应物处于同一相。IB经典案例是Fe²⁺/Fe³⁺催化H₂O₂分解：

2Fe²⁺ + H₂O₂ + 2H⁺ → 2Fe³⁺ + 2H₂O

2Fe³⁺ + H₂O₂ → 2Fe²⁺ + O₂ + 2H⁺

The catalyst cycles between two oxidation states, providing a lower-energy alternative pathway. Similarly, I⁻ catalyses H₂O₂ decomposition via I₂/I⁻ interconversion.

催化剂在两种氧化态之间循环，提供了更低能量的替代路径。类似地，I⁻通过I₂/I⁻互变催化H₂O₂分解。

Heterogeneous catalysis involves a solid catalyst with gaseous or liquid reactants. The metal surface adsorbs reactant molecules, weakening their bonds via d-orbital interactions. Examples include:

多相催化涉及固体催化剂与气态/液态反应物。金属表面吸附反应物分子，通过d轨道相互作用削弱其化学键。例如：

• Fe (Haber process, N₂ + 3H₂ ⇌ 2NH₃)
• Fe（哈伯法，N₂ + 3H₂ ⇌ 2NH₃）
• Ni (hydrogenation of alkenes, e.g. margarine production)
• Ni（烯烃氢化，如人造黄油生产）
• V₂O₅ (Contact process, 2SO₂ + O₂ ⇌ 2SO₃—V⁵⁺/V⁴⁺ redox cycling)
• V₂O₅（接触法，2SO₂ + O₂ ⇌ 2SO₃——V⁵⁺/V⁴⁺氧化还原循环）
• Pt/Pd/Rh (catalytic converters: 2CO + 2NO → 2CO₂ + N₂)
• Pt/Pd/Rh（三元催化器：2CO + 2NO → 2CO₂ + N₂）

Catalyst poisons, such as sulfur compounds in the Haber process, block active sites and reduce catalytic efficiency—this is why the Contact process operates before the Haber process in industry.

催化剂毒物（如哈伯法中的硫化物）会堵塞活性位点、降低催化效率——因此工业上接触法需先于哈伯法进行。

7. Standard Electrode Potentials | 标准电极电位

Transition metals show a wide range of standard electrode potentials (E° values), reflecting their ability to participate in redox reactions. The E° value depends on the sum of ionisation energies, hydration energies and sublimation enthalpies.

过渡金属呈现范围广泛的标准电极电位（E°值），反映其参与氧化还原反应的能力。E°值取决于电离能、水合能和升华焓的总和。

For example, E°(Fe³⁺/Fe²⁺) = +0.77 V, E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V. These values predict that Zn will displace Cu²⁺ from solution:

例如，E°(Fe³⁺/Fe²⁺) = +0.77 V，E°(Cu²⁺/Cu) = +0.34 V，E°(Zn²⁺/Zn) = −0.76 V。这些数值预测Zn能从溶液中置换Cu²⁺：

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)    E°cell = +1.10 V

Because each transition metal has multiple oxidation states, each has multiple E° values. Disproportionation occurs when the E° for reduction to a lower state is more positive than the E° for oxidation to a higher state. A classic example: Cu⁺ disproportionates because E°(Cu²⁺/Cu⁺) = +0.15 V and E°(Cu⁺/Cu) = +0.52 V result in a positive overall cell potential for:

由于每种过渡金属有多种氧化态，因此也有多个E°值。歧化反应发生在还原为低价态的E°比氧化为高价态的E°更正时。经典案例：Cu⁺歧化，因为E°(Cu²⁺/Cu⁺) = +0.15 V且E°(Cu⁺/Cu) = +0.52 V，导致以下反应总电池电位为正：

2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)    E°cell = +0.37 V

Similarly, Fe³⁺ does not disproportionate because E°(Fe³⁺/Fe²⁺) = +0.77 V is lower than E°(Fe²⁺/Fe) = −0.44 V, so Fe²⁺ is stable in aqueous solution.

类似地，Fe³⁺不会歧化，因为E°(Fe³⁺/Fe²⁺) = +0.77 V低于E°(Fe²⁺/Fe) = −0.44 V，所以Fe²⁺在水溶液中是稳定的。

8. Trends Across the Series | 同周期趋势

Several periodic trends are exam-relevant:

以下同周期趋势是考试重点：

• Atomic radius decreases slightly across the series (Sc→Cu) due to increasing nuclear charge without significant shielding by d-electrons, then increases at Zn.
• 原子半径从Sc到Cu缓慢减小，因为核电荷增加而d电子屏蔽效应有限，到Zn时增大。
• First ionisation energy increases gradually, with slight irregularities at Cr and Cu due to stable d⁵ and d¹⁰ configurations.
• 第一电离能逐渐增大，在Cr和Cu处因d⁵和d¹⁰稳定构型出现微小不规则。
• Density and melting point are high due to strong metallic bonding involving d-electrons; Zn is lower because its d-shell is full and contributes less to bonding.
• 密度和熔点较高，因为d电子参与强金属键；Zn较低，因d壳层已满、对成键贡献小。
• E°(M²⁺/M) becomes more positive across the series (less negative), reflecting decreasing tendency to lose electrons.
• E°(M²⁺/M)沿周期变正（负值减小），反映失电子趋势减弱。

9. Common Exam Questions and Pitfalls | 常见考题与易错点

Q1: Explain why transition metals form coloured compounds.
Answer: d-orbital splitting → absorption of visible light → d-d transition → complementary colour observed. Mention Δ₀, ligand field and d-electron count.

问1：解释过渡金属化合物为何有颜色。
答案：d轨道分裂→吸收可见光→d-d跃迁→观察到互补色。需提及Δ₀、配体场和d电子数。

Q2: Why is Sc not a typical transition metal?
Answer: Sc³⁺ has a d⁰ configuration—no unpaired electrons, no variable oxidation states (only +3), colourless compounds, and no catalytic activity.

问2：为何Sc不是典型过渡金属？
答案：Sc³⁺为d⁰构型——无未配对电子、无可变氧化态（仅+3）、化合物无色、无催化活性。

Q3: Distinguish between high-spin and low-spin [Fe(CN)₆]³⁻.
Answer: CN⁻ is a strong field ligand, Δ₀ > P, so electrons pair in t₂g: low-spin, 1 unpaired electron, paramagnetic.

问3：区分[Fe(CN)₆]³⁻的高自旋与低自旋。
答案：CN⁻是强场配体，Δ₀ > P，电子在t₂g中配对：低自旋，1个未配对电子，顺磁性。

Q4: Write the electron configuration of Cr²⁺.
Answer: Cr = [Ar] 3d⁵4s¹; remove 4s first: Cr²⁺ = [Ar] 3d⁴.

问4：写出Cr²⁺的电子构型。
答案：Cr = [Ar] 3d⁵4s¹；先失去4s：Cr²⁺ = [Ar] 3d⁴。

Common mistakes include forgetting 4s-removed-first, confusing high/low spin with d-electron count, and failing to state both absorption and complementary colour in colour questions.

常见错误包括忘记4s先失去、混淆高/低自旋与d电子数，以及在颜色问题中只提吸收不提互补色。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）