📚 IB OCR Chemistry: Last-Minute Revision Notes | IB OCR 化学：考前冲刺笔记
As you prepare for your IB or OCR Chemistry exams, last-minute revision should focus on key concepts, common calculations, and exam-specific strategies. This guide consolidates the essential topics that frequently appear, highlighting both similarities and subtle differences between these two rigorous curricula.
在备考 IB 或 OCR 化学考试的最后冲刺阶段，应集中精力攻克关键概念、常见计算以及针对具体考试的策略。本指南整合了高频必考主题，并突出这两个严格课程体系之间的共性与细微差异。
1. Mole Calculations and Stoichiometry | 摩尔计算与化学计量
The mole (n) is the central unit; n = m / M where m is mass in g and M is molar mass in g mol⁻¹. Always begin with a fully balanced chemical equation to convert between moles of different substances.
摩尔(n)是核心单位；n = m/M，其中m是质量(g)，M是摩尔质量(g mol⁻¹)。务必从完全配平的化学方程式入手，才能在不同物质的摩尔数之间进行换算。
For solutions, use n = c × V, but ensure volume is in dm³. If given cm³, divide by 1000: V(dm³) = V(cm³)/1000. The same applies when calculating concentration from moles and volume.
对于溶液，使用 n = c × V，但要确认体积单位是 dm³。若体积是 cm³，则除以 1000：V(dm³) = V(cm³)/1000。在根据摩尔数和体积计算浓度时同样适用。
Ideal gas law: pV = nRT. Memorize the gas constant R = 8.31 J K⁻¹ mol⁻¹. Temperature must always be in Kelvin (K = °C + 273). Be ready to rearrange for molar mass or density.
理想气体状态方程：pV = nRT。牢记气体常数 R = 8.31 J K⁻¹ mol⁻¹。温度必须始终使用开尔文温度(K = °C + 273)。要能娴熟地变形公式以求解摩尔质量或密度。
Be aware of the different standard conditions: IB uses STP (0 °C, 100 kPa) with molar volume 22.7 dm³ mol⁻¹; OCR typically uses RTP (25 °C, 101 kPa) giving 24.0 dm³ mol⁻¹. Always check which condition is specified in the question.
注意不同体系的标况：IB 使用 STP（0 °C, 100 kPa），摩尔体积为 22.7 dm³ mol⁻¹；OCR 通常使用 RTP（25 °C, 101 kPa），摩尔体积为 24.0 dm³ mol⁻¹。务必核对题目指定的条件。
2. Atomic Structure and Electron Configuration | 原子结构与电子排布
Recall the subatomic particles: protons (relative mass 1, charge +1), neutrons (1, 0), electrons (1/1836, –1). Isotopes have the same number of protons but different numbers of neutrons, so chemical properties are identical while physical properties may differ.
熟记亚原子粒子：质子（相对质量 1，电荷 +1）、中子（1，0）、电子（1/1836，–1）。同位素具有相同的质子数但中子数不同，因此化学性质相同而物理性质可能不同。
Write electron configurations using s, p, d notation, for example 1s² 2s² 2p⁶ for Ne. For ions, add or remove electrons from the highest energy level (4s before 3d when filling, but remove 4s electrons first for transition metals).
用 s, p, d 符号书写电子排布，例如 Ne 为 1s² 2s² 2p⁶。对于离子，从最高能级得失电子（填充时 4s 先于 3d，但对过渡金属，电离时优先失去 4s 电子）。
IB expects orbital diagrams with arrows; OCR may ask you to deduce electron configurations from ionisation energies. Both syllabuses require you to explain trends in first ionisation energy across a period (general increase due to increasing nuclear charge and similar shielding) and down a group (decrease due to increased distance and shielding).
IB 要求画带箭头的轨道示意图；OCR 可能要求由电离能推断电子排布。两个课程体系都要求解释第一电离能在同周期的递变规律（核电荷增加、屏蔽相似，总体增大）以及同族递变（因距离和屏蔽增大而减小）。
Use the continuous nature of successive ionisation energies to predict the group of an element. A large jump indicates removal from a new, inner electron shell.
利用逐级电离能的连续变化推测元素所在主族。出现大幅跳跃说明开始从新的内层电子壳层移走电子。
3. Bonding, Structure and Molecular Shapes | 化学键、结构与分子形状
Ionic bonding arises from electrostatic attraction between oppositely charged ions, typically between a metal and a non-metal. Covalent bonding involves the sharing of electron pairs. Metallic bonding is the attraction between lattice of positive ions and delocalised electrons.
离子键由带相反电荷的离子之间的静电引力形成，通常存在于金属与非金属之间。共价键涉及电子对的共享。金属键则是正离子晶格与离域电子之间的引力。
Use VSEPR theory to predict molecular shapes: count the number of electron domains (bonding pairs + lone pairs) around the central atom. 2 domains → linear (180°), 3 → trigonal planar (120°), 4 with no lone pairs → tetrahedral (109.5°), 4 with one lone pair → trigonal pyramidal (107°), 4 with two lone pairs → bent (104.5°). Be comfortable drawing these with wedges and dashes.
用 VSEPR 理论预测分子形状：计算中心原子周围的电子域数（成键对 + 孤对电子）。2个域 → 线形(180°)，3个 → 平面三角形(120°)，4个、无孤对 → 正四面体形(109.5°)，4个、一对孤对 → 三角锥形(107°)，4个、两对孤对 → V形(104.5°)。要能熟练用楔形线画出示意图。
Electronegativity difference determines bond polarity. A large difference (>1.7) often indicates ionic character, while a difference below ~0.5 is non-polar covalent. Polarity of molecules depends on both bond polarity and symmetry: CO₂ is linear and symmetric, so overall non-polar; H₂O is bent, making it polar.
电负性差值决定键的极性。差值较大（>1.7）通常意味着离子性，差值小于约0.5则为非极性共价键。分子的极性取决于键的极性与对称性：CO₂ 线形对称，整体为非极性；H₂O 为 V形，呈极性。
IB may ask about formal charge and resonance structures; OCR highlights bond enthalpy and intermolecular forces (London, permanent dipole-dipole, hydrogen bonding). Hydrogen bonds occur when H is covalently bonded to N, O or F and is attracted to a lone pair on N, O or F of another molecule.
IB 中可能涉及形式电荷与共振式；OCR 侧重键能和分子间作用力（伦敦色散力、永久偶极-偶极作用和氢键）。当 H 与 N、O 或 F 以共价键结合，并被另一个分子的 N、O 或 F 上的孤对电子吸引时，形成氢键。
4. Thermochemistry and Hess’s Law | 热化学与盖斯定律
Exothermic reactions release energy (ΔH negative); endothermic reactions absorb energy (ΔH positive). Enthalpy change can be measured using q = mcΔT, where m is the mass of solution, c is specific heat capacity (usually 4.18 J g⁻¹ °C⁻¹ for aqueous solutions), and ΔT is temperature change.
放热反应释放能量(ΔH 为负值)；吸热反应吸收能量(ΔH 为正值)。可用 q = mcΔT 测量焓变，其中 m 为溶液质量，c 为比热容（水溶液通常取 4.18 J g⁻¹ °C⁻¹），ΔT 为温度变化。
Then ΔH = –q/n, where n is the moles of the limiting reactant. The negative sign for exothermic processes ensures ΔH < 0. Include the sign in your answer.
然后 ΔH = –q/n，n 为限量反应物的摩尔数。放热过程加上负号以保证 ΔH < 0。答案必须保留正负号。
Hess’s Law: the enthalpy change for a reaction is independent of the route taken. Use standard enthalpies of combustion or formation: ΔHr = Σ ΔHf(products) – Σ ΔHf(reactants). Alternatively, construct an energy cycle.
盖斯定律：一个反应的焓变与途径无关。利用标准燃烧焓或标准生成焓：ΔHr = Σ ΔHf(产物) – Σ ΔHf(反应物)。也可以构建能量循环图。
Mean bond enthalpies can provide an estimate of ΔH: ΔH ≈ Σ(bond enthalpies broken) – Σ(bond enthalpies formed). Remember that bond enthalpies are averages and only valid for gaseous species.
平均键焓可用来估算 ΔH：ΔH ≈ Σ(断裂键焓) – Σ(形成键焓)。注意键焓均为平均值，且仅适用于气态物种。
5. Kinetics and Rate Equations | 动力学与速率方程
Rate is defined as change in concentration per unit time. Factors affecting rate: concentration, pressure (for gases), surface area, temperature and catalysts. At the molecular level, rate depends on collision frequency and the fraction of collisions with energy ≥ activation energy (Ea).
速率定义为单位时间内浓度的变化。影响速率的因素：浓度、压强（气体）、表面积、温度及催化剂。在分子层面，速率取决于碰撞频率以及能量 ≥ 活化能(Ea)的碰撞比例。
The Maxwell-Boltzmann distribution shows that at higher temperatures, the curve flattens and shifts to the right, greatly increasing the proportion of molecules exceeding Ea. A catalyst provides an alternative pathway with lower Ea, so more molecules have sufficient energy without lowering temperature.
麦克斯韦-玻尔兹曼分布表明，高温下曲线变平且右移，超过 Ea 的分子比例大幅增加。催化剂提供一条低 Ea 的替代路径，使更多分子具备足够能量，而温度无需降低。
IB HL and OCR both cover rate equations: for a reaction aA + bB → products, rate = k[A]m[B]n. The orders m and n are determined by experiment, not by stoichiometry. The overall order is m + n. Use initial rates or half-life methods to find orders.
IB HL 和 OCR 均涉及速率方程：对于 aA + bB → 产物，速率 = k[A]m[B]n。级数 m、n 由实验确定，而非化学计量数。总级数为 m + n。常用初速率法或半衰期法确定级数。
For a first-order reaction, half-life t½ = ln 2 / k, which is constant. A plot of ln[A]t vs. time gives a straight line with slope –k. Be capable of sketching concentration–time graphs for zero, first and second order.
对于一级反应，半衰期 t½ = ln2 / k，且为常数。作 ln[A]t–时间图应得一条斜率为 –k 的直线。要能画出零级、一级和二级反应的浓度–时间曲线示意图。
6. Chemical Equilibria | 化学平衡
Dynamic equilibrium occurs when the rates of forward and reverse reactions are equal in a closed system. Le Chatelier’s principle states that a system at equilibrium will shift to partially counteract any imposed change in concentration, pressure or temperature.
在封闭体系中，当正逆反应速率相等时达到动态平衡。勒夏特列原理指出，平衡体系会通过部分抵消浓度、压力或温度的变化来移动平衡位置。
Increasing temperature favours the endothermic direction; increasing pressure favours the side with fewer gas moles. Catalysts do not alter the position of equilibrium – they only speed up the attainment of equilibrium.
升高温度有利于吸热方向；增大压强有利于气体分子总数较少的一侧。催化剂不改变平衡位置——它只加快达到平衡的速率。
The equilibrium constant Kc = [products] / [reactants], raised to the powers of the coefficients in the balanced equation. Kc is temperature-dependent only. A large Kc indicates equilibrium lies to the right (products favoured).
平衡常数 Kc = [产物] / [反应物]，各浓度以配平方程式中系数为指数。Kc 仅随温度改变。较大的 Kc 值说明平衡偏向右侧（产物占优势）。
OCR may require calculation of Kc including units; IB often uses Kc and sometimes Kp for gaseous equilibria. Kp is expressed in terms of partial pressures. The relationship Kp = Kc(RT)Δn may be useful, where Δn is the change in gas moles.
OCR 可能要求计算带单位的 Kc；IB 常用 Kc，有时对气态平衡使用 Kp。Kp 用分压表达。Kp = Kc(RT)Δn 这一关系可能派上用场，其中 Δn 为气体摩尔数的变化。
7. Acids, Bases and pH | 酸、碱与 pH
Bronsted-Lowry acids are proton donors; bases are proton acceptors. Strong acids (HCl, HNO₃, H₂SO₄) fully dissociate in water. Weak acids (CH₃COOH) partially dissociate, establishing an equilibrium with Ka = [H⁺][A⁻] / [HA].
布朗斯特-劳里酸是质子给体，碱是质子受体。强酸（HCl、HNO₃、H₂SO₄）在水中完全电离。弱酸（CH₃COOH）部分电离，并建立平衡，Ka = [H⁺][A⁻] / [HA]。
pH = –log₁₀[H⁺]; [H⁺] = 10⁻pH. For strong monoprotic acids, [H⁺] equals the acid concentration. For weak acids, use the approximation [H⁺] = √(Ka × [HA]) and always check that the approximation is valid (degree of dissociation < 5%).
pH = –log₁₀[H⁺]；[H⁺] = 10⁻pH。强一元酸中 [H⁺] 等于酸浓度。弱酸可用近似公式 [H⁺] = √(Ka × [HA])，并需验证近似是否成立（电离度 < 5%）。
The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), is invaluable for buffer calculations. Buffers resist changes in pH upon addition of small amounts of acid or base and are made from a weak acid and its conjugate base.
亨德森-哈塞尔巴尔赫方程 pH = pKa + log([A⁻]/[HA]) 对缓冲溶液计算至关重要。缓冲溶液能抵御外加少量酸碱引起的pH变化，由弱酸与其共轭碱组成。
When performing acid-base titrations, the equivalence point is where moles of H⁺ = moles of OH⁻. The pH at equivalence depends on the nature of the acid and base (strong-strong gives pH 7; weak-strong gives pH > 7). Know how to sketch pH curves and select suitable indicators.
进行酸碱滴定时，等当点处 H⁺ 的摩尔数等于 OH⁻ 的摩尔数。等当点的 pH 取决于酸碱本性（强强 pH=7；弱强 pH>7）。要会绘制 pH 曲线并选择合适的指示剂。
8. Redox and Electrochemistry | 氧化还原与电化学
Oxidation is loss of electrons; reduction is gain (OIL RIG). Oxidation numbers help identify redox processes: elements in their standard state = 0, O nearly always –2
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