📚 CCEA Year 12 Chemistry Formulae & Theorems Quick Reference Handbook | CCEA 12年级化学公式定理速查手册
This quick reference handbook brings together the essential formulae, definitions, and quantitative relationships required for CCEA AS Level Chemistry. Every formula is presented with its standard notation and typical units, followed by a concise explanation of its application. Use this guide to support your problem-solving, revision, and practical write-ups throughout Year 12.
本速查手册汇集了 CCEA AS 化学所需的所有核心公式、定义与定量关系。每个公式均配有标准符号与常用单位，并附有简要的应用说明。在 12 年级的学习、复习与实验报告撰写中，这份指南将帮助你快速定位关键方程并理解其使用场景。
1. The Mole and Molar Mass | 摩尔与摩尔质量
r of moles n is calculated by dividing the mass m of a substance by its molar mass M: n = m / M. Molar mass has units of g mol⁻¹ and is numerically equal to the relative atomic or formula mass.
摩尔是物质的量的国际单位。1 摩尔恰好包含 6.02214076 × 10²³ 个基本单元（阿伏伽德罗常数 Nₐ）。物质的量 n 可通过质量 m 除以摩尔质量 M 计算：n = m / M。摩尔质量的单位是 g mol⁻¹，数值上等于相对原子质量或相对式量。
- Key formula: n = m / M (mass in g, molar mass in g mol⁻¹)
- Number of particles: N = n × Nₐ
- Molar volume of a gas at room temperature and pressure (RTP, 20 °C, 1 atm): approximately 24 dm³ mol⁻¹
- 核心公式：n = m / M（质量以 g 为单位，摩尔质量以 g mol⁻¹ 为单位）
- 粒子数目：N = n × Nₐ
- 常温常压下气体的摩尔体积（RTP, 20 °C, 1 atm）：约 24 dm³ mol⁻¹
2. Empirical and Molecular Formulae | 经验式与分子式
The empirical formula gives the simplest whole‑number ratio of atoms of each element in a compound. It is determined by converting percentage composition or measured masses to moles, then dividing by the smallest number of moles. The molecular formula is a whole‑number multiple of the empirical formula, found by comparing the molar mass of the compound with the empirical formula mass.
经验式表示化合物中各元素原子的最简整数比。通过将质量分数或测得的质量转化为物质的量，再除以最小物质的量即可求得经验式。分子式是经验式的整数倍，可通过比较化合物的摩尔质量与经验式质量确定。
- Moles of each element = mass (g) / Aᵣ (or percentage / Aᵣ)
- Ratio = divide by smallest mole value
- Molecular formula multiplier: n = Mᵣ (compound) / empirical formula mass
- 各元素物质的量 = 质量 (g) / Aᵣ（或质量分数 / Aᵣ）
- 比率 = 各值除以最小物质的量
- 分子式倍数因子：n = 化合物的 Mᵣ / 经验式质量
3. Gas Laws and the Ideal Gas Equation | 气体定律与理想气体方程
e relationship between pressure p, volume V, amount n and absolute temperature T for an ideal gas. R is the molar gas constant, 8.314 J K⁻¹ mol⁻¹ when p is in Pa and V in m³.
理想气体状态方程综合了玻意耳定律、查理定律和阿伏伽德罗定律：pV = nRT。它描述了理想气体的压强 p、体积 V、物质的量 n 与绝对温度 T 之间的关系。R 为摩尔气体常数，当 p 单位为 Pa、V 为 m³ 时 R = 8.314 J K⁻¹ mol⁻¹。
pV = nRT
| Symbol | Quantity | Typical units |
| p | Pressure | Pa (or kPa, atm – must convert) |
| V | Volume | m³ (1 m³ = 1000 dm³) |
| n | Amount of gas | mol |
| R | Molar gas constant | 8.314 J K⁻¹ mol⁻¹ |
| T | Absolute temperature | K (T/K = θ/°C + 273) |
4. Concentration and Volumetric Analysis | 浓度与滴定分析
e and V is the volume of solution in dm³. A solution of 1.00 mol dm⁻³ contains 1 mole of solute per cubic decimetre. In volumetric analysis, the unknown concentration is determined from a titration using the formula c₁V₁ / n₁ = c₂V₂ / n₂, where n₁ and n₂ are the stoichiometric coefficients in the balanced equation.
浓度 c 表示单位体积溶液中溶质的物质的量。基本公式为 c = n / V，其中 n 是溶质的物质的量，V 是溶液的体积（单位 dm³）。1.00 mol dm⁻³ 的溶液表示每立方分米含 1 摩尔溶质。在滴定分析中，未知浓度可通过 c₁V₁ / n₁ = c₂V₂ / n₂ 计算，式中 n₁ 和 n₂ 是配平方程中的化学计量数。
c = n / V and c₁V₁ / n₁ = c₂V₂ / n₂
- Volume must be in dm³: 1 dm³ = 1000 cm³
- For a solid dissolved to make a standard solution: n = m / M, then c = n / V
- In back titrations, calculate the amount of reagent that actually reacted with the analyte
- 体积必须使用 dm³：1 dm³ = 1000 cm³
- 配制标准溶液时：n = m / M，然后 c = n / V
- 返滴定中，先计算实际与分析物反应的试剂量
5. Atom Economy and Percentage Yield | 原子经济性与产率
Atom economy measures the proportion of reactant atoms that end up in the desired product. It is a way of assessing the 'greenness' of a reaction. Percentage yield compares the actual mass of product obtained to the maximum theoretical mass. Both are essential in evaluating the efficiency of a chemical process.
原子经济性衡量有多少反应物原子最终进入了目标产物，是评价反应“绿色程度”的指标。产率则是实际获得的产品质量与最大理论产量之比。两者都是评估化工过程效率的关键参数。
tom economy = (Molar mass of desired product / Sum of molar masses of all reactants) × 100%
Percentage yield = (Actual mass of product / Theoretical mass of product) × 100%
- High atom economy reduces waste and cost.
- Yield may be reduced by incomplete reaction, side reactions or losses during purification.
- 原子经济性高可减少废物和成本。
- 产率可能因反应不完全、副反应或纯化损失而降低。
6. Enthalpy Changes and Calorimetry | 焓变与量热法
h the surroundings is calculated using q = mcΔT, where m is the mass of the solution (or water), c is the specific heat capacity (usually 4.18 J g⁻¹ K⁻¹ for water), and ΔT is the temperature change. The molar enthalpy change is then ΔH = −q / n (exothermic if negative).
焓变 ΔH 是在恒压下化学反应吸收或释放的热量。量热法中，与环境交换的热量用 q = mcΔT 计算，其中 m 是溶液（或水）的质量，c 是比热容（水通常取 4.18 J g⁻¹ K⁻¹），ΔT 是温度变化。摩尔焓变则为 ΔH = −q / n（放热时为负值）。
q = mcΔT and ΔH = −q / n
- Units: q in J, ΔH in kJ mol⁻¹ (so convert J to kJ by dividing by 1000).
- Assume the density of dilute aqueous solutions is 1.00 g cm⁻³.
- Common ΔH values: standard enthalpy of combustion (ΔH°꜀), standard enthalpy of formation (ΔH°f), standard enthalpy of neutralisation (ΔH°ₙₑᵤₜ).
- 单位：q 为 J，ΔH 为 kJ mol⁻¹（需将 J 转换为 kJ，即除以 1000）。
- 假设稀水溶液的密度为 1.00 g cm⁻³。
- 常见 ΔH 值：标准燃烧焓 (ΔH°꜀)、标准生成焓 (ΔH°f)、标准中和焓 (ΔH°ₙₑᵤₜ)。
7. Hess’s Law and Bond Enthalpies | 赫斯定律与键能
alculated indirectly using enthalpy of formation or combustion data. Mean bond enthalpy is an average energy required to break one mole of a specific covalent bond in the gaseous state. Approximate ΔH can be estimated using the energy balance: ΔH ≈ Σ (bond enthalpies of bonds broken) − Σ (bond enthalpies of bonds formed).
赫斯定律指出，反应的总焓变只取决于始态和终态，与途径无关。利用该定律，可通过生成焓或燃烧焓数据间接计算 ΔH。平均键能是断裂 1 摩尔气态共价键所需的平均能量。近似 ΔH 可通过能量平衡估算：ΔH ≈ Σ（断裂键的键能总和）− Σ（形成键的键能总和）。
ΔH = ΣΔH°f (products) − ΣΔH°f (reactants)
ΔH ≈ ΣBE (bonds broken) − ΣBE (bonds formed)
- In an enthalpy cycle, all arrows point in the direction of the change being considered.
- Bond enthalpy calculations are approximate because mean bond enthalpies are averaged over many molecules.
- 在焓变循环中，箭头方向应与所考虑的变化方向一致。
- 由于平均键能是许多分子的平均值，键能计算只能给出近似结果。
8. Kinetics: Rate of Reaction | 动力学：反应速率
eactant or product per unit time. For a reactant A, the average rate = −Δ[A] / Δt; for a product B, rate = Δ[B] / Δt. The initial rate is often obtained from a concentration–time graph at t = 0.
化学反应速率衡量反应物转化为产物的快慢，可以用单位时间内反应物或产物浓度的变化表示。对于反应物 A，平均反应速率 = −Δ[A] / Δt；对于产物 B，速率 = Δ[B] / Δt。初始速率通常从浓度-时间曲线在 t = 0 处的斜率得到。
Rate = change in concentration / change in time
- Factors affecting rate: concentration, pressure (gases), temperature, surface area, catalyst.
- A catalyst provides an alternative reaction pathway with lower activation energy (Eₐ).
- Maxwell–Boltzmann distribution explains why temperature and catalysts increase rate.
- 影响速率的因素：浓度、压强（气体）、温度、表面积、催化剂。
- 催化剂提供活化能 (Eₐ) 更低的替代反应途径。
- 麦克斯韦-玻尔兹曼分布解释了温度与催化剂为何能提高反应速率。
9. Chemical Equilibrium and Kc | 化学平衡与 Kc
For a reversible reaction at equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium constant Kc is defined in terms of the concentrations of products and reactants raised to powers equal to their stoichiometric coefficients. Kc is constant at a given temperature; changing temperature alters Kc, while concentration or pressure changes shift the position of equilibrium but do not change Kc.
可逆反应达到平衡时，正逆反应速率相等。平衡常数 Kc 定义为产物浓度与反应物浓度以其化学计量数为指数的乘积之比。在给定温度下 Kc 是常数；改变温度会改变 Kc 的值，而浓度或压强的变化只会使平衡位置移动，不会改变 Kc。
For aA + bB ⇌ cC + dD, Kc = [C]c [D]d / [A]a [B]b
- The units of Kc depend on the stoichiometry of the reaction.
- Solid and pure liquid concentrations are not included in the expression.
- Le Chatelier’s principle predicts the direction of shift when a condition is changed.
- Kc 的单位取决于反应的化学计量数。
- 纯固体和纯液体的浓度不写入 Kc 表达式。
- 勒夏特列原理可预测条件改变时平衡移动的方向。
10. Acids, Bases, and pH | 酸、碱与 pH
⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. A neutral solution has [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³, giving pH 7. For strong monoprotic acids, [H⁺] equals the concentration of the acid.
pH 是氢离子浓度的对数标度：pH = −log₁₀[H⁺]。25 °C 时，水的离子积 Kₕ = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。中性溶液中 [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³，pH = 7。对于强一元酸，[H⁺] 即等于酸的浓度。
pH = −log₁₀[H⁺] and [H⁺] = 10⁻ᵖᴴ
Kₕ = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ (at 25 °C)
- pOH = −log₁₀[OH⁻]; pH + pOH = 14 at 25 °C.
- protic acids (e.g. H₂SO₄), [H⁺] = 2 × acid concentration (provided complete dissociation).
- pOH = −log₁₀[OH⁻]；25 °C 时 pH + pOH = 14。
- 对于强二元酸（如 H₂SO₄），[H⁺] = 2 × 酸的浓度（假设完全解离）。
11. Weak Acids and the Acid Dissociation Constant Ka | 弱酸与酸解离常数 Ka
A weak acid HA partially dissociates in water: HA ⇌ ion c, if the degree of dissociation is very small ([H⁺] ≈ [A⁻] and [HA] ≈ c), then [H⁺] ≈ √Ka × c. pKa = −log₁₀Ka; weaker acids have larger pKa values.
弱酸 HA 在水中部分解离：HA ⇌ H⁺ + A⁻。酸解离常数 Ka 是该解离反应的平衡常数：Ka = [H⁺][A⁻] / [HA]。对于浓度为 c 的弱酸，若解离度很小（[H⁺] ≈ [A⁻] 且 [HA] ≈ c），则 [H⁺] ≈ √Ka × c。pKa = −log₁₀Ka；弱酸的 pKa 值越大，酸性越弱。
Ka = [H⁺][A⁻] / [HA] and [H⁺] ≈ √Ka × c
- The approximation is valid only when c / Ka > 100 or [H⁺] < 5% of c.
- Ka and pKa are temperature‑dependent.
- 近似公式仅在 c / Ka > 100 或 [H⁺] 小于 c 的 5% 时成立。
- Ka 和 pKa 均随温度变化。
12. Buffer Solutions | 缓冲溶液
d and its conjugate base. The pH of an acidic buffer can be calculated using the Henderson– [HA]). The buffer is most effective when the ratio [A⁻]/[HA] is close to 1, i.e. pH ≈ pKa.
缓冲溶液能在加入少量酸或碱时抵抗 pH 的改变，通常由弱酸及其共轭碱组成。酸性缓冲液的 pH 可通过 Henderson–Hasselbalch 方程计算：pH = pKa + log₁₀([A⁻] / [HA])。当 [A⁻]/[HA] 接近 1，即 pH ≈ pKa 时，缓冲能力最强。
pH = pKa + log₁₀([A⁻] / [HA])
- Buffers can also be made from excess weak acid and a strong base (partial neutralisation).
- Biological examples include the carbonate buffer (H₂CO₃ / HCO₃⁻) controlling blood pH.
- 缓冲溶液也可由过量弱酸与强碱部分中和制得。
- 生物体系中的例子包括碳酸缓冲对 (H₂CO₃ / HCO₃⁻) 调控血液 pH。
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