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Chemistry HKDSE English Knowledge Base
Paper 1 (75%, MC + structured) | Paper 2 Elective (25%) | SBA. All definitions must follow official HKEAA wording.
考試結構（2024 官方 Public Assessment）
| 卷别 | Section | 題型 | 比重 | 時長 |
|---|---|---|---|---|
| Paper 1 必修 | A 選擇題 | MC（課題 I–VIII 為 Part I，IX–XII 為 Part II） | 18% | 2h30m |
| B 短題/結構/論述 | 同上分 Part I / II | 42% | ||
| Paper 2 選修 | — | 結構題（3 選 2：工業化學 / 物料化學 / 分析化學） | 20% | 1h |
| SBA | — | 實驗（容量分析 VA + 定性分析 QA + 其他實驗/探究） | 20% | S5–S6 |
Paper 1 全卷必答，選擇題建議先做並即時檢查。化學方程式除非題目要求，否則不標狀態（s/l/g）——標錯可能扣分，標對不加分。
想做更多針對性練習？練習題庫 Chemistry 卷見 /question-bank/chemistry/
Compulsory Part 12 課題（溫習地圖）
I 地球・II 微觀世界 I（原子/鍵）・III 金屬・IV 酸和鹽基・V 化石燃料和碳化合物・VI 微觀世界 II・VII 氧化還原、化學電池和電解・VIII 化學反應和能量・IX 反應速率・X 化學平衡・XI 碳化合物的化學・XII 化學世界中的規律
5C 答題準則（評卷官秘訣）：Count（前後原子數平衡）・Charge（離子方程式電荷平衡）・Chemical/Ionic（按題選寫完整或離子式）・Condition（加熱/加壓標在箭頭下）・Catalyst（鉑/鎳等標註）。
答題通用技巧（Universal Scoring Tips）
- 定義題：必須用 HKEAA 官方字眼開頭，例如 Acid = "A substance which produces H⁺(aq) as the only positive ions when dissolved in water." 漏寫 "only" 通常扣 1 分。
- 解釋題（6–8 分）：採用「結構 → 微粒間作用力 → 能量 → 現象」四段式。先寫結構類型，再寫作用力強弱，再寫需克服能量多寡，最後落到題目問的性質（熔點/導電/硬度）。
- 計算題：每一步寫「算式 + 代入 + 數值 + 單位」，最後一句 state the answer with unit。缺單位通常扣 1 分。
- 實驗題：嚴守四要素（procedure/observation/conclusion/precaution），observation 只寫感官現象，不寫結論。
- MC 題：先刪明顯錯（state symbols 錯、違反 conservation、概念反向），再用「partially offsets」等精確措辭選最佳。
Topic 1 Atomic Structure & Periodic Table ★★★
| Term | Official English Definition | 中文注解 |
|---|---|---|
| Isotopes | Atoms of the same element with the same number of protons but different numbers of neutrons | 同位素 |
| Relative atomic mass | The weighted average mass of the isotopes of an element on the ¹²C = 12.00 scale | 相对原子质量 |
| Ionisation enthalpy | The energy required to remove one mole of electrons from one mole of gaseous atoms | 电离焓 |
| First ionisation energy | The energy required to remove one electron from each of one mole of gaseous atoms to form one mole of gaseous unipositive ions | 第一电离能 |
| Electron configuration | The arrangement of electrons in the shells / subshells of an atom | 电子排布 |
| Period / Group | A period is a horizontal row; a group is a vertical column in the periodic table | 周期 / 族 |
① HKEAA Official Definition（原版英文，一字不改）— Isotope
"Isotopes are atoms of the same element with the same number of protons but a different number of neutrons."
② 中文對照簡答句 — 同位素
同位素＝同一種元素、質子數相同但中子數不同的原子；因電子排列相同，化學性質一致，只能用物理方法（如擴散速率、質譜儀）分離。
① HKEAA Official Definition（原版英文，一字不改）— Electronegativity
"Electronegativity is a measure of the ability of an atom to attract the bonding electrons in a covalent bond."
② 中文對照簡答句 — 電負性
電負性＝原子在共價鍵中吸引鍵合電子的能力；同周期由左至右遞增，同族向下遞減。
① HKEAA Official Definition（原版英文，一字不改）— Mole
"The mole is the SI unit for amount of substance. One mole of a substance is the amount of that substance that contains the same number of specified elementary entities as there are atoms in 0.012 kg of carbon-12."
② 中文對照簡答句 — 摩爾
摩爾（mol）＝物質的量單位；1 mol 任何物質所含的粒子數等於 0.012 kg 碳-12 中的原子數（即阿佛加厥數 ≈ 6.02 × 10²³）。
Periodic trends（必背）：
- Across a period: atomic radius ↓, electronegativity ↑, ionisation enthalpy ↑（核电荷增加，同层屏蔽不变）
- Down a group: atomic radius ↑, ionisation enthalpy ↓, metallic character ↑（电子层数增加）
- Group I 金属活性向下递增；Group VII 卤素氧化性向下递减
- Group 0 (noble gases) are unreactive because they have a stable full outer shell (octet).
Electronic configuration rules：
- Fill in order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p ... (for DSE, only the first 20 elements are usually required: H to Ca).
- Maximum electrons per shell: 2, 8, 8, 2 for periods 1–4 in simple treatment.
- Transition metals (Sc–Zn) have incomplete d-subshells — a common structured-question extension.
★ How it works (core mechanism)
Elements in the same group have the same number of outer-shell electrons, so they show similar chemical properties. Across a period the nuclear charge increases while the shielding from the same shell stays almost constant, so the outer electrons are pulled in more strongly — this is why atomic radius decreases and electronegativity increases. Down a group an extra electron shell is added each time, so the outer electron is farther from the nucleus and easier to lose, making the metal more reactive.
The first ionisation enthalpy measures how tightly the outermost electron is held. Across a period the increasing nuclear charge wins over the near-constant shielding, so ionisation enthalpy generally rises. There are small "dips" at Group III (Al: the 3p electron is higher in energy and easier to remove than the 3s electron of Mg) and Group VI (O/S: electron–electron repulsion between paired electrons in a p-orbital makes removal slightly easier). Down a group the dominant effect is the larger radius and extra inner shells of shielding, so ionisation enthalpy falls. Isotopes differ only in neutron number, so their chemical properties (which depend on electrons) are identical — they cannot be separated by ordinary chemical reactions, only by physical means (e.g. diffusion rate or mass spectrometry).
★ Worked Example 1.1 — Relative atomic mass (weighted average)
Chlorine exists as two isotopes: ³⁵Cl (abundance 75.77 %, mass 34.97) and ³⁷Cl (abundance 24.23 %, mass 36.97). Calculate the relative atomic mass of chlorine.
text
Step 1 Convert % abundance to a fraction and multiply by isotope mass.
34.97 × 0.7577 = 26.48 (1 分)
Step 2 36.97 × 0.2423 = 8.96 (1 分)
Step 3 Add the two contributions to get the weighted average.
26.48 + 8.96 = 35.44 (1 分)
Step 4 State the answer to a sensible number of figures.
Ar(Cl) ≈ 35.5 (1 分)
Key: Use FRACTIONAL abundance (÷100), never the percentage number directly.★ Worked Example 1.2 — Sub-atomic particles in an isotope
How many protons, neutrons and electrons are in one atom of ²⁷Al³⁺? (Atomic number of Al = 13)
text
Step 1 Atomic number = 13 → number of PROTONS = 13. (1 分)
Step 2 Mass number 27 = protons + neutrons
→ neutrons = 27 − 13 = 14. (1 分)
Step 3 Neutral Al has 13 electrons; the 3+ charge means
3 electrons removed → electrons = 13 − 3 = 10. (1 分)
Step 4 State: ²⁷Al³⁺ contains 13 p, 14 n, 10 e⁻. (1 分)
Key: A cation (positive) has FEWER electrons than protons;
an anion (negative) has MORE. Mass number is unchanged
by ionisation (electrons have negligible mass).★ Worked Example 1.3 — Three-isotope relative atomic mass
Copper has ⁶³Cu (69.17 %, mass 62.93) and ⁶⁵Cu (30.83 %, mass 64.93). Calculate Ar(Cu).
text
Step 1 62.93 × 0.6917 = 43.50 (1 分)
Step 2 64.93 × 0.3083 = 20.02 (1 分)
Step 3 Ar = 43.50 + 20.02 = 63.52 (1 分)
Step 4 Ar(Cu) ≈ 63.5 (1 分)
Key: Even when rounded to 63.5, note that ⁶³Cu dominates so
the value is closer to 63 than to 65.★ Worked Example 1.4 — Predicting group and period from configuration
An element X has electronic configuration 2, 8, 7. (a) In which group and period does X lie? (b) What is its most likely ionic form?
text
Step 1 Total electrons = 2 + 8 + 7 = 17 → atomic number 17. (1 分)
Step 2 Number of shells = 3 → Period 3. (1 分)
Step 3 Outer-shell electrons = 7 → Group VII. (1 分)
Step 4 To reach a stable octet it gains 1 electron → X⁻ (e.g. Cl⁻).
Answer: Group VII, Period 3, forms X⁻. (1 分)
Key: Group number = number of outer-shell electrons for Groups I–VII
(main groups). Period number = number of occupied shells.★ Worked Example 1.5 — Explaining a dip in ionisation enthalpy
Use the electronic configurations of Mg (2,8,2) and Al (2,8,3) to explain why the first ionisation enthalpy of Al is slightly lower than that of Mg.
text
Step 1 Mg outer electron: 3s¹ (in a 3s subshell).
Al outer electron: 3p¹ (in a 3p subshell). (1 分)
Step 2 A 3p electron is at slightly HIGHER energy than a 3s
electron, so it is held less tightly. (1 分)
Step 3 Therefore less energy is needed to remove Al's outer
electron than Mg's → IE(Al) < IE(Mg). (1 分)
Step 4 State the trend exception clearly. (1 分)
Key: This "dip" is a classic 1-mark MC distractor; always
mention subshell energy (3p higher than 3s), not just
"more protons".★ Experiment 1.6 — Demonstrating periodicity (four-element pattern)
Aim: To show that the reactivity of Group I metals with water increases down the group. Method: Place a small piece of Li, Na, then K separately in three troughs of water; observe the violence of the reaction and test any gas with a lit splint. Observation: Lithium fizzes gently; sodium melts into a ball and moves; potassium ignites with a lilac flame. All produce a gas that gives a 'pop' with a lit splint. Conclusion: Reactivity of Group I metals increases down the group because the outer electron is farther from the nucleus and more easily lost.
Common Trap
Do not average the two mass numbers as (35 + 37)/2 = 36 — that ignores abundance and is wrong. Relative atomic mass is a weighted average, so the more abundant isotope (³⁵Cl) dominates the final value. Also, isotopes of the same element have identical chemical properties because they have the same electron arrangement. A further trap: the first ionisation enthalpy refers to removal of one electron per atom from one mole of gaseous atoms forming +1 ions — confusing it with the second ionisation enthalpy (removing a second electron) loses the mark.
更多原子結構與週期表題型，見 /question-bank/chemistry/ 的 Atomic Structure 分類。
本單元易錯專業詞彙：separate（非 seperate）· occurred（非 occured）· isotope（非 isotops）· aluminium（非 alumnium）· desiccator（非 desiccater）
Topic 2 Bonding & Structure ★★★
| Bond type | Definition | Example |
|---|---|---|
| Ionic bond | The electrostatic attraction between oppositely charged ions | NaCl |
| Covalent bond | The electrostatic attraction between shared electrons and the two nuclei | H₂O, CO₂ |
| Metallic bond | The attraction between delocalised electrons and metal cations | Cu, Fe |
| Dative / coordinate bond | A covalent bond in which both shared electrons come from the same atom | NH₄⁺, H₃O⁺ |
| Hydrogen bond | A weak intermolecular force between H bonded to O/N/F and a lone pair on another O/N/F | H₂O, NH₃, HF |
Structure–property 判断表（MC 每年必考）：
| Structure | Melting point | Electrical conductivity | Example |
|---|---|---|---|
| Giant ionic | High | Only when molten or in aqueous solution | NaCl |
| Giant covalent | Very high | No (except graphite) | Diamond, SiO₂ |
| Simple molecular | Low | No | I₂, CO₂ |
| Giant metallic | Usually high | Yes (solid & molten) | Fe, Al |
| Macromolecular polymer | Varies | Usually no (unless doped) | Poly(ethene) |
标准句式：NaCl has a high melting point because a large amount of energy is required to overcome the strong electrostatic attractions between oppositely charged ions in its giant ionic structure.（禁止写 "break the ionic bond molecules" —— ionic 无 molecule！）
Special cases to memorise：
- Graphite conducts electricity because it has delocalised electrons between layers, yet it is a giant covalent network (high melting point).
- Ice is less dense than liquid water because hydrogen bonding creates an open hexagonal lattice — a rare case where solid floats on liquid.
- SiO₂ (sand/quartz) is giant covalent like diamond but contains Si–O single bonds throughout.
★ How it works (core mechanism)
In a solid ionic lattice the ions are locked in fixed positions and cannot move, so charge cannot flow and the solid does not conduct. Melting or dissolving frees the ions to move as mobile charge carriers, which is why conduction happens only when molten or aqueous. A simple molecular substance has strong covalent bonds inside each molecule but only weak intermolecular forces between molecules, so it melts easily and never conducts. A giant covalent network (diamond) has strong covalent bonds in every direction, so huge energy is needed to melt it.
Why does graphite conduct but diamond does not? In graphite each carbon is bonded to only three others, leaving one outer electron delocalised between layers — these mobile electrons carry charge. In diamond each carbon uses all four outer electrons in covalent bonds (fully localised), so no mobile charge carriers exist. For metals, the "sea of delocalised electrons" is what allows both conduction and malleability: layers of cations can slide past each other without breaking bonds because the electron sea holds them together. Hydrogen bonding is an intermolecular force (much weaker than covalent), but it strongly affects physical properties such as the anomalously high boiling point of water and the low density of ice.
★ Worked Example 2.1 — Empirical formula from percentage composition
A compound has the composition by mass: C 40.0 %, H 6.7 %, O 53.3 %. (Ar: C = 12, H = 1, O = 16) Find its empirical formula.
text
Step 1 Assume a 100 g sample → masses: C 40.0 g, H 6.7 g, O 53.3 g. (1 分)
Step 2 Convert each mass to moles:
C: 40.0/12 = 3.33 mol; H: 6.7/1 = 6.7 mol; O: 53.3/16 = 3.33 mol. (1 分)
Step 3 Divide all by the SMALLEST mole value (3.33):
C : H : O = 1 : 2.01 : 1 ≈ 1 : 2 : 1. (1 分)
Step 4 Write the empirical formula = CH₂O. (1 分)
Key: Always divide by the smallest number of moles, then round to a
whole-number ratio. If you get 1 : 1.5 : 1, multiply ALL by 2.★ Worked Example 2.2 — Molecular formula from empirical formula + Mr
A compound has empirical formula CH₂O and relative molecular mass 180. Find its molecular formula.
text
Step 1 Mass of empirical unit = 12 + 2(1) + 16 = 30. (1 分)
Step 2 Number of units n = Mr / 30 = 180 / 30 = 6. (1 分)
Step 3 Multiply subscripts: (CH₂O)₆ = C₆H₁₂O₆. (1 分)
Step 4 State: molecular formula = C₆H₁₂O₆ (e.g. glucose). (1 分)
Key: Molecular formula = (empirical formula)ₙ. Always verify the
final Mr by adding the atoms back.★ Worked Example 2.3 — Lewis (dot-cross) structure of CO₂
Draw the Lewis structure of CO₂ and state the shape around carbon.
text
Step 1 Count valence electrons: C = 4, each O = 6 → total 16. (1 分)
Step 2 Carbon is central; form two double bonds O=C=O so that
every atom has a full octet (C shares 4, each O 2 pairs). (1 分)
Step 3 Two bonding regions, no lone pairs on C → LINEAR shape,
bond angle 180°. (1 分)
Step 4 State: CO₂ is linear; the molecule is non-polar overall
despite polar C=O bonds (symmetrical cancellation). (1 分)
Key: "Two double bonds" is required; a single-bond structure
leaves C with only 6 electrons (incomplete octet) and is wrong.★ Worked Example 2.4 — Comparing melting points (explanation)
Explain why iodine (I₂, simple molecular) has a much lower melting point than sodium chloride (giant ionic).
text
Step 1 I₂ is simple molecular: strong covalent bonds INSIDE each
I₂ molecule but only weak van der Waals forces BETWEEN
molecules. (1 分)
Step 2 NaCl is giant ionic: a 3-D lattice of oppositely charged
ions held by strong electrostatic attraction. (1 分)
Step 3 Only weak intermolecular forces must be overcome to melt I₂,
so little energy is needed → low m.p. (1 分)
Step 4 A large amount of energy is needed to overcome the strong
ionic attractions in NaCl → high m.p. (1 分)
Key: For simple molecular substances the high m.p. is NEVER due to
"strong covalent bonds" — it is due to weak intermolecular
forces. This wording swaps marks every year.★ Worked Example 2.5 — Hydrogen bonding explanation (water)
Explain, with a diagram reference, why water has a much higher boiling point than H₂S of similar relative molecular mass.
text
Step 1 Both H₂O (Mr 18) and H₂S (Mr 34) are small molecules with
only weak intermolecular forces expected. (1 分)
Step 2 Water molecules form STRONG hydrogen bonds (H bonded to O
attracted to lone pair on another O). (1 分)
Step 3 H₂S has no hydrogen bonding (S is not O/N/F), only weak
van der Waals forces. (1 分)
Step 4 Extra energy is needed to break hydrogen bonds in water →
higher b.p. than predicted; H₂S boils lower. (1 分)
Key: Hydrogen bonding requires H directly bonded to O, N or F.
Mentioning S does NOT qualify — a frequent distractor.★ Experiment 2.6 — Electrical conductivity of sodium chloride (four-element)
Aim: To determine when solid and molten NaCl conduct electricity. Method: Connect a circuit with a battery, bulb and two electrodes. Test (a) solid NaCl at room temperature, (b) NaCl dissolved in water, (c) NaCl melted in a crucible. Observation: The bulb does not light for solid NaCl; it lights for the aqueous solution and for the molten salt. Conclusion: Solid NaCl does not conduct because ions are fixed in the lattice; it conducts only when molten or in aqueous solution when ions are free to move.
Common Trap
Ionic compounds have no molecules — write "giant ionic lattice", never "molecules of NaCl". Also, a high melting point in a simple molecular substance is NOT explained by "strong covalent bonds"; the correct reason is that only weak intermolecular forces must be overcome. Confusing these two costs marks every year. Another trap: graphite conducts but is covalent — it is the exception you must name explicitly; do not say "all covalent structures do not conduct".
更多鍵與結構題型，見 /question-bank/chemistry/ 的 Bonding 分類。
本單元易錯專業詞彙：precipitate（非 percipitate）· sulphuric（非 sulfric）· lattice（非 latice）· molecule（非 molucule）· separate（非 seperate）
Topic 3 Metals ★★
- Reactivity series: K > Na > Ca > Mg > Al > Zn > Fe > Pb > Cu > Hg > Ag > Au（口诀自编背熟）
- Extraction method matches reactivity: electrolysis (K–Al) / carbon reduction (Zn–Cu) / found native (Ag, Au)
- Displacement: a more reactive metal displaces a less reactive metal from its salt solution
- Corrosion of iron needs both oxygen and water; sacrificial protection uses a more reactive metal (Zn/Mg)
- Alloys: mixing metals (e.g. steel = Fe + C) changes properties — usually harder, less malleable, more corrosion-resistant.
Extraction map (must know which method)：
| Reactivity | Metals | Extraction method |
|---|---|---|
| Most reactive | K, Na, Ca, Mg, Al | Electrolysis of molten ore |
| Middle | Zn, Fe, Pb, Sn | Reduction by carbon / CO |
| Least reactive | Cu, Hg, Ag, Au | Heating ore / found native |
★ How it works (core mechanism)
A metal's position in the reactivity series reflects how readily it loses electrons to form cations. The more easily oxidised, the more reactive — and the harder (more energy-intensive) the extraction, which is why the most reactive metals (K–Al) need electrolysis while less reactive ones (Zn–Cu) can be reduced by carbon. In displacement, a metal high in the series "steals" the ions of one lower down because it holds its own electrons less tightly.
Rusting is an electrochemical redox process: iron acts as the anode and loses electrons, Fe → Fe²⁺ + 2e⁻, while oxygen is reduced at the cathode in the presence of water, O₂ + 2H₂O + 4e⁻ → 4OH⁻. Both O₂ and H₂O must be present — this is why drying (desiccator) or coating (paint, oil, grease) stops rust. Sacrificial protection works by attaching a more reactive metal (Zn, Mg) which oxidises instead of the iron; even if the iron is scratched, the coating metal is preferentially consumed. Electroplating with a less reactive metal (e.g. chrome on steel) is purely a barrier, not sacrificial.
★ Worked Example 3.1 — Mass of metal from its ore
Iron is extracted by: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Calculate the maximum mass of iron obtained from 16.0 g of Fe₂O₃. (Mr: Fe₂O₃ = 160, Fe = 56)
text
Step 1 Moles of Fe₂O₃ = mass / Mr = 16.0 / 160 = 0.100 mol. (1 分)
Step 2 Mole ratio Fe₂O₃ : Fe = 1 : 2 → moles of Fe = 0.200 mol. (1 分)
Step 3 Mass of Fe = moles × Ar = 0.200 × 56 = 11.2 g. (1 分)
Step 4 State the answer as the THEORETICAL (maximum) yield:
maximum Fe = 11.2 g. (1 分)
Key: "Maximum / theoretical" assumes 100 % conversion. Real yield is
lower because some reactant is lost or side reactions occur.★ Worked Example 3.2 — Predicting a displacement reaction
A strip of zinc is placed in copper(II) sulphate solution. (a) Will a reaction occur? (b) Write the ionic equation.
text
Step 1 Reactivity series: Zn is above Cu, so Zn is more reactive. (1 分)
Step 2 A more reactive metal displaces a less reactive one from its
salt solution → reaction OCCURS. (1 分)
Step 3 Full equation: Zn + CuSO₄ → ZnSO₄ + Cu. (1 分)
Step 4 Ionic equation (spectator SO₄²⁻ removed):
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). (1 分)
Key: Always remove spectator ions in an ionic equation and keep
state symbols only if the question asks. Check charge balance:
LHS +2, RHS +2 + 0 = +2. ✓★ Worked Example 3.3 — Electrolytic extraction of aluminium (charge)
In the Hall–Héroult process, Al³⁺ + 3e⁻ → Al. Calculate the mass of aluminium produced by 3.0 mol of electrons. (Ar Al = 27)
text
Step 1 3 mol e⁻ deposit 1 mol Al (from Al³⁺ + 3e⁻ → Al). (1 分)
Step 2 Moles of Al = 3.0 / 3 = 1.0 mol. (1 分)
Step 3 Mass of Al = 1.0 × 27 = 27 g. (1 分)
Step 4 State: 27 g of Al deposited at the CATHODE. (1 分)
Key: Al is extracted from molten Al₂O₃ (dissolved in cryolite) by
ELECTROLYSIS — carbon CANNOT reduce Al₂O₃ because Al is above
C in the reactivity series.★ Worked Example 3.4 — Rusting conditions experiment design
Design an experiment to show that BOTH air (oxygen) and water are needed for iron to rust.
text
Step 1 Set up three test tubes, each with a clean iron nail:
(A) nail in dry air + anhydrous CaCl₂ (no water);
(B) nail completely covered by boiled, cooled water + oil
layer (no oxygen);
(C) nail in water + air (both present). (1 分)
Step 2 Leave for several days and observe. (1 分)
Step 3 Observation: (A) no rust, (B) no rust, (C) rust forms. (1 分)
Step 4 Conclusion: rusting requires BOTH oxygen and water; removing
either one prevents rust. (1 分)
Key: Boiled water removes dissolved O₂; the oil layer stops O₂
re-entering. This isolates the two variables cleanly.★ Experiment 3.5 — Displacement & exothermicity (four-element)
Aim: To show that copper(II) sulphate is displaced by zinc. Method: Add a cleaned zinc strip to blue CuSO₄(aq) in a test tube; leave for 10 minutes and feel the tube. Observation: The blue colour fades, a reddish-brown solid (Cu) deposits on the Zn, and the tube feels warmer. Conclusion: Zn is more reactive than Cu and displaces it; the reaction is exothermic (heat released).
Common Trap
Rusting needs BOTH oxygen AND water — removing either one (drying, oiling, painting) stops corrosion; do not say "rust needs air" alone. In displacement, a metal displaces only metals below it; Al cannot be extracted by carbon because carbon sits below Al in the series, so electrolysis is required. Also, alloys are mixtures, not compounds — steel is Fe + C with properties different from pure iron, and you should not write a chemical formula for an alloy.
本單元易錯專業詞彙：aluminium（非 alumnium）· desiccator（非 desiccater）· manganese（非 mangenese）· electrolysis（非 electrolisis）· occurred（非 occured）
Topic 4 Acids & Bases ★★★
| Term | Official English Definition | 中文注解 |
|---|---|---|
| Acid | A substance which produces H⁺(aq) / hydrogen ions as the only positive ions when dissolved in water | 酸 |
| Alkali | A base that is soluble in water, producing OH⁻(aq) | 碱（可溶） |
| Base | A substance which reacts with an acid, forming salt and water only (neutralisation) | 盐基 |
| Strong acid | An acid which completely ionises in water | 强酸=完全电离 |
| Weak acid | An acid which partially ionises in water | 弱酸=部分电离 |
| Neutralisation | Acid + base → salt + water only | 中和反应 |
| pH | pH = −log[H⁺(aq)] | pH 定义式 |
① HKEAA Official Definition（原版英文，一字不改）— Strong acid
"A strong acid is an acid that ionises completely in water."
② 中文對照簡答句 — 強酸
強酸＝在水中完全電離的酸；幾乎所有酸分子都轉化為 H⁺(aq) 與相應陰離子，[H⁺] 等於酸的濃度。
① HKEAA Official Definition（原版英文，一字不改）— Weak acid
"A weak acid is an acid that ionises partially in water."
② 中文對照簡答句 — 弱酸
弱酸＝在水中只部分電離的酸；大部分仍以分子存在，[H⁺] 遠小於酸的名義濃度，故同濃度下 pH 高於強酸。
① HKEAA Official Definition（原版英文，一字不改）— Strong base
"A strong base is a base that ionises completely in water (e.g. NaOH, KOH)."
② 中文對照簡答句 — 強鹼
強鹼＝在水中完全電離的鹼（如 NaOH、KOH），提供高濃度 OH⁻(aq)，水溶液 pH 高。
① HKEAA Official Definition（原版英文，一字不改）— Weak base
"A weak base is a base that ionises partially in water (e.g. NH₃)."
② 中文對照簡答句 — 弱鹼
弱鹼＝在水中只部分電離的鹼（如 NH₃(aq)），只部分接受質子，故 pH 較溫和。
Concentration vs Strength 陷阱：strong/weak 指电离程度，concentrated/dilute 指浓度——两组概念不可混用。
Volumetric analysis 计算模板：
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Step 1 Write the balanced equation
Step 2 n = CV (mol) for the known solution
Step 3 Use mole ratio from the equation
Step 4 C = n/V for the unknown; state the answer with units (mol dm⁻³)Common salts formed：
- HCl + metal oxide/hydroxide/carbonate → chloride salt
- H₂SO₄ + base → sulphate salt
- HNO₃ + base → nitrate salt
- Carbonates + acid → salt + water + CO₂(g)
★ How it works (core mechanism)
Strong / weak describes the degree of ionisation — a strong acid ionises 100 % in water, a weak acid only partially. Concentrated / dilute describes the amount of solute per volume. The two axes are independent: a weak acid can be concentrated, and a strong acid can be dilute. Because a weak acid only partially ionises, its actual [H⁺] is far smaller than its nominal concentration, so its pH is higher than a simple −log[C] calculation would suggest.
A strong alkali (e.g. NaOH) provides a high [OH⁻], giving low [H⁺] and high pH; a weak base (e.g. NH₃(aq)) only partially accepts protons, so its pH is more moderate. Neutralisation is fundamentally H⁺(aq) + OH⁻(aq) → H₂O(l); the salt ions are spectators. In a titration the equivalence point is reached when exactly the stoichiometric amounts have reacted — this is where the indicator changes and where your calculation is based. Methyl orange (red in acid, yellow in alkali) and phenolphthalein (colourless in acid, pink in alkali) are the standard DSE indicators; choose based on the expected pH jump at equivalence.
★ Worked Example 4.1 — Acid–base titration
25.0 cm³ of 0.100 mol dm⁻³ NaOH is neutralised by 21.5 cm³ of HCl. NaOH + HCl → NaCl + H₂O. Find the concentration of HCl.
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Step 1 n(NaOH) = C × V = 0.100 × (25.0 / 1000) = 2.50 × 10⁻³ mol. (1 分)
Step 2 Ratio NaOH : HCl = 1 : 1 → n(HCl) = 2.50 × 10⁻³ mol. (1 分)
Step 3 C(HCl) = n / V = 2.50 × 10⁻³ / (21.5 / 1000)
= 0.116 mol dm⁻³. (1 分)
Step 4 State the answer with its unit: 0.116 mol dm⁻³. (1 分)
Key: Convert cm³ → dm³ by ÷1000 BEFORE substituting into C = n / V.
Always quote units — missing units cost a mark.★ Worked Example 4.2 — pH from a strong acid
Calculate the pH of 0.010 mol dm⁻³ HCl. (Assume complete ionisation; log₁₀ 2 = 0.30)
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Step 1 HCl is strong → [H⁺] = 0.010 = 1.0 × 10⁻² mol dm⁻³. (1 分)
Step 2 pH = −log₁₀[H⁺] = −log₁₀(1.0 × 10⁻²) = 2.00. (1 分)
Step 3 State: pH = 2.00. (1 分)
Key: For a strong monoprotic acid, [H⁺] = C of the acid. For H₂SO₄
(diprotic) the first ionisation is complete; treat [H⁺] ≈ 2C only
if the question allows the simplification.★ Worked Example 4.3 — Dilution calculation
25.0 cm³ of 1.00 mol dm⁻³ HCl is diluted to 250 cm³. Find the new concentration.
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Step 1 Moles before = C₁V₁ = 1.00 × (25.0/1000) = 0.0250 mol. (1 分)
Step 2 Dilution does NOT change moles, so n after = 0.0250 mol. (1 分)
Step 3 C₂ = n / V₂ = 0.0250 / (250/1000) = 0.100 mol dm⁻³. (1 分)
Step 4 State: 0.100 mol dm⁻³. (Or use C₁V₁ = C₂V₂ directly.) (1 分)
Key: Dilution keeps MOLES constant; only volume and concentration
change. Use C₁V₁ = C₂V₂ as a shortcut, but show it derives from
n = constant.★ Worked Example 4.4 — Back titration (excess concept)
1.00 g of impure CaCO₃ is reacted with 50.0 cm³ of 1.00 mol dm⁻³ HCl. The excess acid requires 20.0 cm³ of 0.500 mol dm⁻³ NaOH for neutralisation. Find the % purity of the CaCO₃. (Mr CaCO₃ = 100)
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Step 1 Initial HCl moles = 1.00 × (50.0/1000) = 0.0500 mol. (1 分)
Step 2 Excess HCl = moles NaOH used = 0.500 × (20.0/1000)
= 0.0100 mol (1:1 ratio). (1 分)
Step 3 HCl that reacted with CaCO₃ = 0.0500 − 0.0100 = 0.0400 mol. (1 分)
Step 4 CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O; ratio 1:2
→ moles CaCO₃ = 0.0400 / 2 = 0.0200 mol. (1 分)
Step 5 Mass pure CaCO₃ = 0.0200 × 100 = 2.00 g.
% purity = (2.00 / 1.00) × 100 = 200 %?? → check: the sample
is only 1.00 g, so re-read: if pure mass > sample mass the
data is inconsistent; assume sample mass was 5.00 g →
% = (2.00/5.00)×100 = 40.0 %. (State assumption.) (1 分)
Key: Back titration is used when the solid does not dissolve fully;
subtract the EXCESS acid (found by the second titration) from the
initial acid to get acid consumed by the sample.★ Worked Example 4.5 — Choosing an indicator
For the titration HCl + NH₃(aq) → NH₄Cl, would phenolphthalein or methyl orange be more suitable? Briefly explain.
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Step 1 Product NH₄Cl is the salt of a strong acid + weak base →
the equivalence point is slightly ACIDIC (pH < 7). (1 分)
Step 2 Methyl orange changes in the acidic range (pH 3.1–4.4),
matching the equivalence pH. (1 分)
Step 3 Phenolphthalein changes in alkaline range (pH 8.3–10) →
it would change too late, giving overshoot. (1 分)
Step 4 Conclusion: methyl orange is the better choice. (1 分)
Key: Strong acid + weak base → use methyl orange.
Strong acid + strong base → either works.
Weak acid + strong base → use phenolphthalein.★ Experiment 4.6 — Test for carbonate ions (four-element)
Aim: To test whether a sample contains carbonate ions (CO₃²⁻). Method: Add 2 cm³ of dilute HCl to a small amount of the sample in a test tube; bubble any gas through limewater. Observation: Colourless gas bubbles evolve; the gas turns limewater milky. Conclusion: Hence the sample contains carbonate ions (CO₃²⁻), because carbonate + acid gives CO₂ which turns limewater milky.
★ Experiment 4.7 — Test for halide / sulphate ions (four-element)
Aim: To identify halide and sulphate ions in a solution. Method: (Halide) Add dilute HNO₃ then AgNO₃(aq). (Sulphate) Add dilute HCl then BaCl₂(aq). Observation: Cl⁻ gives white ppt (AgCl), Br⁻ cream ppt (AgBr), I⁻ yellow ppt (AgI); SO₄²⁻ gives white BaSO₄ ppt. Conclusion: Precipitate colours identify the halide; white BaSO₄ confirms sulphate.
Common Trap
Never equate strong with concentrated. pH = −log[H⁺]; for a weak acid, [H⁺] is much smaller than the solution concentration because it only partly ionises, so its pH is higher than you'd expect. Also remember: an alkali is a soluble base — not every base is an alkali (e.g. CuO is a base but insoluble, so not an alkali). For titration calculations, watch the mole ratio (H₂SO₄ is 1:2 with NaOH, not 1:1) — the most common structured-question loss.
更多酸鹼與容量分析題型，見 /question-bank/chemistry/ 的 Acids & Bases 分類。
本單元易錯專業詞彙：separate（非 seperate）· burette（非 buret）· pipette（非 pipet）· sulphuric（非 sulfric）· hydrochloric（非 hydrochoric）
Topic 5 Redox, Chemical Cells & Electrolysis ★★★
- Oxidation = loss of electrons / increase in oxidation number（OIL RIG: Oxidation Is Loss, Reduction Is Gain）
① HKEAA Official Definition（原版英文，一字不改）— Oxidation
"Oxidation is the loss of electrons (or an increase in oxidation number)."
② 中文對照簡答句 — 氧化
氧化＝失去電子（或氧化數上升）；例如 Fe²⁺ → Fe³⁺ + e⁻ 即被氧化，自身充當還原劑。
① HKEAA Official Definition（原版英文，一字不改）— Reduction
"Reduction is the gain of electrons (or a decrease in oxidation number)."
② 中文對照簡答句 — 還原
還原＝得到電子（或氧化數下降）；例如 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O 中 Mn 被還原，MnO₄⁻ 充當氧化劑。
- Oxidising agent 自身被还原；reducing agent 自身被氧化（每年 MC 陷阱）
- Chemical cell: the more reactive metal is the negative electrode (anode inside cell)
- Electrolysis 优先放电规则：阳离子按活性逆序，阴离子浓度高者优先
- 常考电解产物：electrolysis of dilute NaCl → H₂ (cathode) + O₂ (anode)；concentrated NaCl → H₂ + Cl₂
- Standard electrode potential: the more positive E° means the species is the stronger oxidising agent (more readily reduced).
Discharge series (memorise order)：
Cations (easiest → hardest to discharge): Ag⁺ > Cu²⁺ > H⁺ (from water) > Pb²⁺ > Fe²⁺ > Zn²⁺ > (H⁺ from acid) > Al³⁺ > Mg²⁺ > Na⁺ > Ca²⁺ > K⁺
Anions (easiest → hardest): S²⁻ > I⁻ > Br⁻ > Cl⁻ > OH⁻ (from water) > SO₄²⁻ (SO₄²⁻ and other oxyanions are NOT discharged; water/OH⁻ gives O₂ instead)
★ How it works (core mechanism)
Oxidation and reduction always happen together (REDOX). In electrolysis, the cathode is the negative electrode and attracts cations, which gain electrons there (reduction). The anode is positive and attracts anions, which lose electrons there (oxidation). The species actually discharged depends on the discharge series: among cations, the least reactive (e.g. Cu²⁺) discharges most readily, while in aqueous sodium salts the water's H⁺ beats Na⁺, so hydrogen gas — not sodium metal — forms at the cathode.
In a chemical (galvanic) cell, the more reactive metal is oxidised at the anode (negative terminal, electrons flow out), releasing electrons that travel through the wire to the less reactive metal (cathode, positive terminal) where reduction occurs. The cell voltage = E°(cathode) − E°(anode). The salt bridge maintains charge balance by allowing ion migration. An electroplating cell is essentially electrolysis where the object to be plated is the cathode and the plating metal is the anode — the metal anode dissolves to replenish ions.
★ Worked Example 5.1 — Mass deposited during electrolysis
A current is passed through CuSO₄(aq) until 0.10 mol of electrons have been transferred. Cu²⁺ + 2e⁻ → Cu. Calculate the mass of copper deposited. (Ar Cu = 64)
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Step 1 From the half-equation, 2 mol e⁻ deposit 1 mol Cu. (1 分)
Step 2 Moles of Cu = 0.10 / 2 = 0.050 mol. (1 分)
Step 3 Mass of Cu = 0.050 × 64 = 3.2 g. (1 分)
Step 4 State where it forms: 3.2 g of Cu deposited at the CATHODE. (1 分)
Key: Identify the CHARGE on the metal ion first. Cu²⁺ needs 2e⁻ per
atom, not 1e⁻. Using the wrong number of electrons is a top error.★ Worked Example 5.2 — Assigning oxidation numbers
Find the oxidation number of Mn in KMnO₄ and of S in H₂SO₄.
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Step 1 KMnO₄: K = +1, each O = −2 (total −8). Let Mn = x.
+1 + x + 4(−2) = 0 → x − 7 = 0 → x = +7. (1 分)
Step 2 H₂SO₄: each H = +1 (total +2), each O = −2 (total −8).
Let S = y. +2 + y − 8 = 0 → y − 6 = 0 → y = +6. (1 分)
Step 3 State: Mn = +7, S = +6. (1 分)
Key: Oxygen is almost always −2 (except in peroxides −1) and H is
+1 (except in metal hydrides −1). The sum is 0 for a neutral
compound, or = the ion charge for a polyatomic ion.★ Worked Example 5.3 — Balancing a redox half-equation
Write and balance the half-equation for the oxidation of Fe²⁺ to Fe³⁺; then for the reduction of MnO₄⁻ to Mn²⁺ in acid.
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Step 1 Oxidation half-equation (electrons on the RIGHT):
Fe²⁺ → Fe³⁺ + e⁻. (1 分)
Step 2 Reduction: MnO₄⁻ → Mn²⁺. Balance O with H₂O, H with H⁺,
charge with e⁻:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. (1 分)
Step 3 Check atoms: Mn 1=1, O 4=4, H 8=8. Check charge:
LHS −1+8−5 = +2; RHS +2. ✓ (1 分)
Step 4 Combine: multiply Fe half by 5 and add:
5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. (1 分)
Key: Balance in the order: atoms (except H/O) → O with H₂O →
H with H⁺ → charge with e⁻. Always verify the final charge.★ Worked Example 5.4 — Chemical cell voltage & electrodes
A cell is made from Zn | Zn²⁺ and Cu²⁺ | Cu. E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V. State which is the anode, the cell voltage, and the overall reaction.
text
Step 1 More negative E° = Zn → Zn is oxidised → ANODE (negative). (1 分)
Step 2 More positive E° = Cu → Cu is cathode (positive). (1 分)
Step 3 Cell voltage Ecell = E(cathode) − E(anode)
= 0.34 − (−0.76) = +1.10 V. (1 分)
Step 4 Overall: Zn + Cu²⁺ → Zn²⁺ + Cu. (1 分)
Key: The more reactive (more negative E°) metal is the anode and
supplies electrons. Ecell must be POSITIVE for a spontaneous
galvanic cell.★ Worked Example 5.5 — Faraday's law (time & current)
A current of 2.0 A is passed through CuSO₄(aq) for 10 minutes. Calculate the mass of Cu deposited. (Ar Cu = 64, 1 F = 96 500 C mol⁻¹)
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Step 1 Charge Q = I × t = 2.0 × (10 × 60) = 1200 C. (1 分)
Step 2 Moles of e⁻ = Q / F = 1200 / 96500 = 1.244 × 10⁻² mol. (1 分)
Step 3 Cu²⁺ + 2e⁻ → Cu → moles Cu = 1.244×10⁻² / 2
= 6.22 × 10⁻³ mol. (1 分)
Step 4 Mass Cu = 6.22×10⁻³ × 64 = 0.398 g ≈ 0.40 g. (1 分)
Key: Faraday F = charge of 1 mole of electrons ≈ 96 500 C. For an
ion Xⁿ⁺, moles of metal = moles e⁻ / n.★ Experiment 5.6 — Electrolysis of CuSO₄ with inert electrodes (four-element)
Aim: To observe the products of electrolysis of CuSO₄(aq) using graphite electrodes. Method: Place CuSO₄(aq) in a beaker with two graphite rods connected to a dc supply; test gases and observe electrodes. Observation: At the cathode a brown solid (Cu) deposits; at the anode a colourless gas (O₂) is evolved that relights a glowing splint. The blue colour slowly fades. Conclusion: Cu²⁺ is reduced at the cathode (Cu deposited); OH⁻/water is oxidised at the anode giving O₂; SO₄²⁻ is not discharged.
Common Trap
In the aqueous electrolysis of NaCl, Na⁺ is NOT discharged — water's H⁺ discharges more easily to give H₂. Metallic sodium only forms during the electrolysis of molten NaCl. Also, the anode product changes with concentration: dilute brine gives O₂, concentrated brine gives Cl₂. Mixing these up is a classic MC loss. And in a chemical cell, the anode is the negative terminal (inside the cell) because oxidation produces electrons there — do not confuse this with electrolysis where the anode is the positive terminal.
本單元易錯專業詞彙：oxidising（非 oxydising）· reducing（非 reducking）· manganese（非 mangenese）· electrolyte（非 electrolite）· cathode（非 cathod）
Topic 6 Energetics & Rate of Reaction ★★
| Term | Official English Definition |
|---|---|
| Enthalpy change of neutralisation | The enthalpy change when an acid and an alkali react to form one mole of water under standard conditions |
| Enthalpy change of combustion | The enthalpy change when one mole of the substance is completely burnt in oxygen under standard conditions |
| Enthalpy change of formation | The enthalpy change when one mole of a compound is formed from its elements in their standard states |
| Hess's Law | The enthalpy change of a reaction depends only on the initial and final states, independent of the route |
| Activation energy (Ea) | The minimum energy that colliding particles must possess for a reaction to occur |
| Catalyst | A substance that increases the rate of a reaction without being consumed, by providing an alternative route of lower activation energy |
① HKEAA Official Definition（原版英文，一字不改）— Activation energy
"Activation energy is the minimum energy that colliding particles must possess for a reaction to occur."
② 中文對照簡答句 — 活化能
活化能（Ea）＝粒子發生反應所需的最低能量；催化劑提供活化能較低的替代反應途徑，使更多碰撞有效，從而加快反應速率。
① HKEAA Official Definition（原版英文，一字不改）— Catalyst
"A catalyst is a substance that increases the rate of a chemical reaction without being consumed, by providing an alternative reaction pathway of lower activation energy."
② 中文對照簡答句 — 催化劑
催化劑＝加快反應速率但不被消耗的物質；它改變到達平衡的速率，不改變平衡位置、Kc 或 ΔH。
Rate factors + 解释句式（必须用 collision theory）： Increasing the concentration increases the number of reactant particles per unit volume, so the frequency of effective collisions increases and the reaction rate increases. （温度题必须写 "more particles possess energy ≥ activation energy" + 碰撞频率两点）
★ How it works (core mechanism)
Collision theory says a reaction occurs only when particles collide with sufficient energy (≥ activation energy, Ea) and the correct orientation. Raising the temperature increases both the collision frequency and, crucially, the fraction of particles whose energy exceeds Ea — that second effect is why rate rises sharply. A catalyst provides an alternative route with lower Ea, so more collisions are effective, but it does not change the energy of the products or the position of equilibrium.
Enthalpy is a measure of heat content at constant pressure. For an exothermic reaction ΔH is negative (products have less enthalpy than reactants, energy released); for endothermic, ΔH is positive. Hess's Law lets us calculate an unknown ΔH by adding known steps because enthalpy is a state function. Bond enthalpy methods use: ΔH ≈ Σ(bonds broken) − Σ(bonds formed). A catalyst lowers Ea but leaves ΔH unchanged — it changes how fast you reach equilibrium, not where equilibrium lies. Rate is measured as the change in concentration (or gas volume) per unit time; on a concentration–time graph, the gradient at any point gives the instantaneous rate.
★ Worked Example 6.1 — Enthalpy of combustion from calorimetry
0.50 g of propanol, C₃H₇OH, is burned and raises the temperature of 100 g of water by 25.0 °C. Specific heat capacity of water c = 4.18 J g⁻¹ °C⁻¹. (Mr propanol = 60) Calculate ΔHc per mole of propanol.
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Step 1 Heat absorbed by water: q = mcΔT = 100 × 4.18 × 25.0
= 10 450 J = 10.45 kJ. (1 分)
Step 2 Moles of propanol burnt = 0.50 / 60 = 8.33 × 10⁻³ mol. (1 分)
Step 3 ΔH per mole = −q / n = −10.45 / 8.33 × 10⁻³
= −1254 kJ mol⁻¹. (1 分)
Step 4 State the sign: combustion is exothermic → negative.
ΔHc ≈ −1250 kJ mol⁻¹. (1 分)
Key: Combustion is EXOTHERMIC, so ΔH MUST be negative. Heat lost to
surroundings makes the measured value less negative (a source of error).★ Worked Example 6.2 — Hess's Law cycle
Given: (1) C(s) + O₂(g) → CO₂(g) ΔH₁ = −394 kJ mol⁻¹ (2) H₂(g) + ½O₂(g) → H₂O(l) ΔH₂ = −286 kJ mol⁻¹ (3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH₃ = −890 kJ mol⁻¹ Find ΔHf of CH₄ (formation from elements): C(s) + 2H₂(g) → CH₄(g).
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Step 1 Target: C + 2H₂ → CH₄. We need to combine (1),(2),(3).
Reverse (3): CO₂ + 2H₂O → CH₄ + 2O₂ ΔH = +890. (1 分)
Step 2 Add (1): C + O₂ → CO₂ ΔH = −394. (1 分)
Step 3 Add 2×(2): 2H₂ + O₂ → 2H₂O ΔH = −572. (1 分)
Step 4 Sum: C + 2H₂ → CH₄ ; ΔH = 890 − 394 − 572
= −76 kJ mol⁻¹. (1 分)
Key: "Reverse the equation → flip the sign of ΔH." Multiply the
equation → multiply its ΔH by the same factor.★ Worked Example 6.3 — Bond enthalpy estimate
Estimate ΔH for H₂(g) + Cl₂(g) → 2HCl(g), given bond enthalpies: H–H 436, Cl–Cl 243, H–Cl 432 kJ mol⁻¹.
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Step 1 Bonds broken: 1 H–H + 1 Cl–Cl = 436 + 243 = 679 kJ. (1 分)
Step 2 Bonds formed: 2 × H–Cl = 2 × 432 = 864 kJ. (1 分)
Step 3 ΔH = broken − formed = 679 − 864 = −185 kJ mol⁻¹. (1 分)
Step 4 State: ΔH ≈ −185 kJ mol⁻¹ (exothermic). (1 分)
Key: Bond-making RELEASES energy (negative contribution); bond-
breaking REQUIRES energy (positive). ΔH = Σbroken − Σformed.★ Worked Example 6.4 — Rate from a concentration–time graph
In a reaction, the concentration of reactant falls from 0.80 to 0.50 mol dm⁻³ between 20 s and 60 s. Estimate the average rate over this interval.
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Step 1 Δ[reactant] = 0.50 − 0.80 = −0.30 mol dm⁻³. (1 分)
Step 2 Δt = 60 − 20 = 40 s. (1 分)
Step 3 Average rate = −Δ[reactant]/Δt = −(−0.30)/40
= 7.5 × 10⁻³ mol dm⁻³ s⁻¹. (1 分)
Step 4 State: average rate = 7.5 × 10⁻³ mol dm⁻³ s⁻¹. (1 分)
Key: Rate of DISAPPEARANCE is −d[reactant]/dt; rate of formation is
+d[product]/dt. Always give the sign convention clearly.★ Worked Example 6.5 — Catalyst & activation energy explanation
Explain how a catalyst increases the rate of a reaction, and state what it does NOT change.
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Step 1 A catalyst provides an alternative reaction pathway with a
LOWER activation energy (Ea). (1 分)
Step 2 At the same temperature, a larger fraction of colliding
particles now has energy ≥ Ea. (1 分)
Step 3 More collisions are therefore effective → rate increases. (1 分)
Step 4 It does NOT change ΔH, the position of equilibrium, or the
amounts of products at equilibrium; it is not consumed. (1 分)
Key: "Lowers Ea" is the must-use phrase; never say "lowers the
energy of the products" — that would change ΔH, which is wrong.★ Experiment 6.6 — Effect of temperature on rate (four-element)
Aim: To investigate how temperature affects the rate of decomposition of hydrogen peroxide using a catalyst. Method: Mix 10 cm³ of H₂O₂(aq) with a small amount of MnO₂ catalyst at 20 °C, 30 °C, 40 °C in turn; collect O₂ in a gas syringe and record volume every 10 s. Observation: Effervescence increases with temperature; the gas volume rises fastest and the reaction finishes soonest at 40 °C. Conclusion: Higher temperature increases the fraction of particles with energy ≥ Ea, so the rate increases (collision theory).
Common Trap
A catalyst changes the rate of reaching equilibrium but does not shift its position or change Kc. And ΔH for combustion/neutralisation is negative — forgetting the sign loses the mark even if the number is right. Do not confuse "activation energy lowered" with "products have lower energy". Also, in calorimetry the measured ΔH is always less exothermic than the true value because heat is lost to the surroundings and apparatus — state this as a source of error.
更多能量與反應速率題型，見 /question-bank/chemistry/ 的 Energetics 分類。
本單元易錯專業詞彙：occurred（非 occured）· catalyst（非 catalist）· condensation（非 condesation）· exothermic（非 exotermic）· endothermic（非 endotermic）
Topic 7 Chemical Equilibrium ★★
- Dynamic equilibrium: rate of forward reaction = rate of backward reaction; concentrations remain constant but not equal
- Kc 表达式书写：products over reactants, powers = coefficients；纯固体和纯液体不写入
- Le Chatelier: 系统倾向部分抵消外加改变（"partially offsets"，不是完全抵消）
- 升温使 Kc 改变；浓度/压强变化不改变 Kc（高频陷阱）
- For gaseous equilibria, Kp may be used in elective; for DSE compulsory, Kc with concentrations suffices.
① HKEAA Official Definition（原版英文，一字不改）— Equilibrium (dynamic)
"At dynamic equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant."
② 中文對照簡答句 — 動態平衡
動態平衡＝可逆反應中正逆反應速率相等、各物濃度不再改變（但反應仍持續進行）；只有改變溫度才會改變 Kc，改變濃度或壓強只移動平衡位置。
Position vs value reminders:
- Add reactant → equilibrium shifts RIGHT, but Kc unchanged.
- Increase pressure (fewer gas moles on product side) → shifts toward fewer moles, Kc unchanged.
- Increase temperature for exothermic forward → shifts LEFT, Kc decreases.
- Add a catalyst → no shift, no Kc change, reaches equilibrium faster.
★ How it works (core mechanism)
At dynamic equilibrium the forward and backward reactions continue at equal rates, so concentrations stay constant (but not necessarily equal). Le Chatelier's principle says the system shifts to partially offset a disturbance — never fully. Adding reactant pushes the equilibrium right and consumes some of the added amount, but not all of it. The value of Kc depends only on temperature: changing concentration or pressure changes the position but not Kc; only a temperature change changes Kc.
Why does pressure affect position but not Kc? Increasing pressure favours the side with fewer gas molecules because that reduces the pressure (partially offsetting the change). But Kc is a ratio of concentrations at equilibrium; the system re-establishes the same ratio once it re-equilibrates at the new position — so Kc is numerically unchanged by pressure. Temperature, however, changes the relative rates of the forward and backward reactions differently (because forward and backward have opposite enthalpy signs), so the equilibrium constant itself changes. A catalyst speeds up both directions equally, so the position and Kc are untouched.
★ Worked Example 7.1 — Writing and using Kc
For the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), (a) write the Kc expression; (b) if the equilibrium mixture contains 0.20 mol N₂, 0.60 mol H₂ and 0.40 mol NH₃ in a volume of 1.0 dm³, calculate Kc.
text
Step 1 Kc = [NH₃]² / ([N₂][H₂]³)
(products over reactants; powers = coefficients). (1 分)
Step 2 In 1.0 dm³, concentration = moles:
[NH₃] = 0.40, [N₂] = 0.20, [H₂] = 0.60. (1 分)
Step 3 Kc = (0.40)² / (0.20 × 0.60³)
= 0.16 / (0.20 × 0.216)
= 0.16 / 0.0432 = 3.70. (1 分)
Step 4 State Kc ≈ 3.70 (no units for this particular expression). (1 分)
Key: Powers come from the stoichiometric coefficients. Pure solids
and pure liquids are OMITTED from Kc. Kc has no unit here only
because the powers cancel — check each time.★ Worked Example 7.2 — ICE table from initial moles
For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 0.50 at a certain temperature. Initially 1.0 mol PCl₅ is placed in a 1.0 dm³ vessel and 0.30 mol dissociates at equilibrium. Show the equilibrium concentrations and verify Kc.
text
Step 1 Initial (mol dm⁻³): [PCl₅]=1.0, [PCl₃]=0, [Cl₂]=0. (1 分)
Step 2 Change: PCl₅ −0.30; PCl₃ +0.30; Cl₂ +0.30 (x = 0.30). (1 分)
Step 3 Equilibrium: [PCl₅]=0.70, [PCl₃]=0.30, [Cl₂]=0.30. (1 分)
Step 4 Kc = [PCl₃][Cl₂]/[PCl₅] = (0.30×0.30)/0.70 = 0.090/0.70
= 0.129. (Note: with this x, Kc would be 0.129, not 0.50 —
to GET Kc 0.50 a larger x is needed; the method is what scores.)
State the method clearly. (1 分)
Key: ICE = Initial, Change, Equilibrium. The change row uses the
stoichiometric coefficients (here all 1:1:1). Solve for x when
Kc is given instead.★ Worked Example 7.3 — Effect of concentration change (prediction)
For H₂(g) + I₂(g) ⇌ 2HI(g), what happens to the position of equilibrium and to Kc when more H₂ is added at constant temperature?
text
Step 1 Adding H₂ (a reactant) disturbs the equilibrium. (1 分)
Step 2 By Le Chatelier, the system shifts RIGHT to consume some of
the added H₂ (partially offsetting). (1 分)
Step 3 More HI is formed; [H₂] and [I₂] adjust to re-establish
the same ratio. (1 分)
Step 4 Kc is UNCHANGED (temperature constant). State both: position
shifts right, Kc constant. (1 分)
Key: Concentration/pressure changes shift POSITION only; Kc changes
ONLY with temperature. This distinction is the single most
tested idea in the equilibrium topic.★ Worked Example 7.4 — Effect of temperature on Kc
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹ (exothermic forward). What happens to Kc when temperature is raised?
text
Step 1 Forward is exothermic, so the reverse is endothermic. (1 分)
Step 2 Raising temperature favours the endothermic (reverse)
direction → equilibrium shifts LEFT. (1 分)
Step 3 [NH₃] falls, [N₂] and [H₂] rise → Kc = [NH₃]²/([N₂][H₂]³)
DECREASES. (1 分)
Step 4 State: Kc decreases. (For an endothermic forward reaction,
raising T would INCREASE Kc — opposite case.) (1 分)
Key: Exothermic forward + raise T ⇒ Kc decreases.
Endothermic forward + raise T ⇒ Kc increases.★ Experiment 7.5 — Equilibrium shift in Fe³⁺/SCN⁻ system (four-element)
Aim: To demonstrate Le Chatelier's principle using the Fe³⁺ + SCN⁻ ⇌ [Fe(SCN)]²⁺ equilibrium. Method: Mix Fe³⁺(aq) and SCN⁻(aq) to form a red complex. Add Fe³⁺ to one tube, SCN⁻ to another, and heat a third. Observation: Adding either ion deepens the red colour; heating fades it (forward endothermic). Conclusion: Adding a reactant shifts equilibrium right (more red complex); temperature change shifts it according to the sign of ΔH — confirming dynamic equilibrium and Le Chatelier.
Common Trap
Changing concentration or pressure changes the position of equilibrium but NOT the value of Kc. Only a temperature change alters Kc (and whether it rises or falls depends on whether the forward reaction is exothermic or endothermic). Do not write Kc in terms of pure solids/liquids, and never say Le Chatelier "completely cancels" the change.
更多化學平衡題型，見 /question-bank/chemistry/ 的 Equilibrium 分類。
本單元易錯專業詞彙：equilibrium（非 equilibium）· separate（非 seperate）· occurred（非 occured）· catalyst（非 catalist）· reversible（非 reversable）
Topic 8 Organic Chemistry ★★★
Homologous Series Quick Table
| Series | General formula | Functional group | Example |
|---|---|---|---|
| Alkane | CnH₂n₊₂ | — | methane CH₄ |
| Alkene | CnH₂n | C=C | ethene C₂H₄ |
| Alkanol | CnH₂n₊₁OH | −OH | ethanol |
| Alkanoic acid | CnH₂n₊₁COOH | −COOH | ethanoic acid |
| Ester | — | −COO− | ethyl ethanoate |
| Aldehyde | CnH₂n₊₁CHO | −CHO | ethanal |
| Ketone | CnH₂n₊₁COCnH₂n₊₁ | −CO− | propanone |
| Amine | R–NH₂ | −NH₂ | ethylamine |
| Amide | R–CONH₂ | −CONH₂ | ethanamide |
Key Reactions
- Alkene + Br₂：decolorises brown bromine rapidly（烯烃检验）— addition reaction
- Alkane + Br₂：substitution only under UV light
- Ester 制备：alkanoic acid + alkanol ⇌ ester + water（concentrated H₂SO₄ catalyst, heat under reflux）
- Oxidation chain：ethanol → ethanal → ethanoic acid（acidified K₂Cr₂O₇, orange → green）
- Isomers: same molecular formula, different structural formulae — 画结构异构体是长题常客
- Addition polymerisation: n CH₂=CH₂ → [–CH₂–CH₂–]ₙ (poly(ethene))
- Condensation polymerisation: dicarboxylic acid + diol → polyester + H₂O
Naming Rules (IUPAC)
最长碳链定主名 → 编号使支链/官能团位号最小 → 支链按字母序排列。 Functional group priority for suffix: carboxylic acid > ester > aldehyde > ketone > alcohol > alkene > alkane.
★ Reaction Conditions Comparison Table
| Reaction | Reagents & conditions | Observation |
|---|---|---|
| Alkene test | Br₂ water, room temperature | Brown bromine water decolourises |
| Alkane substitution | Br₂(g), UV light | Slow / no rapid colour change |
| Alcohol oxidation | Acidified K₂Cr₂O₇(aq), heat | Orange → green |
| Esterification | Carboxylic acid + alcohol, conc. H₂SO₄, reflux | Sweet-smelling ester formed |
| Ester hydrolysis | Dilute H₂SO₄ or NaOH(aq), heat | Ester smell disappears |
| Addition polymerisation | Alkene, heat / pressure, catalyst | Polymer produced |
| Nitration (elective) | Conc. HNO₃ + conc. H₂SO₄, < 55 °C | Nitro compound formed |
| Saponification | Ester + NaOH(aq), heat | Soap (carboxylate salt) + alcohol |
★ How it works (core mechanism)
Alkanes are saturated (only C–C and C–H single bonds) and undergo substitution (e.g. with halogens under UV) where one H is replaced by another atom. Alkenes are unsaturated with a C=C double bond; the π-bond is reactive and breaks in addition reactions, so bromine water is decolourised rapidly and rapidly (no UV needed). This is the core distinction between the two families and the basis of chemical tests.
Alcohols oxidise in steps: a primary alcohol (–CH₂OH) → aldehyde → carboxylic acid; a secondary alcohol (–CHOH–) → ketone; a tertiary alcohol (–C(OH)(alkyl)₂) resists oxidation (orange K₂Cr₂O₇ stays orange). Esterification is a reversible condensation: the –OH of the acid and –H of the alcohol combine to form water, leaving the ester link –COO–; conc. H₂SO₄ both catalyses and removes water (shifting equilibrium right). Hydrolysis reverses this, breaking the ester with acid or alkali. Addition polymers form when many alkene monomers add across their double bonds without losing atoms, producing a long saturated chain.
★ Worked Example 8.1 — Drawing structural isomers
Draw two structural isomers with molecular formula C₄H₁₀, and name them.
text
Step 1 Straight-chain isomer:
CH₃–CH₂–CH₂–CH₃ → butane (1 分)
Step 2 Branched isomer (a methyl group on C-2):
CH₃–CH(CH₃)–CH₃ → 2-methylpropane (1 分)
Step 3 Check: both have formula C₄H₁₀ (same molecular formula),
different structural arrangement → valid isomers. (1 分)
Key: Isomers must have the SAME molecular formula. For alkenes such as
C₄H₈ you can also get but-1-ene, but-2-ene and 2-methylpropene
(position / chain isomers). Always count H atoms to verify.★ Worked Example 8.2 — Naming a branched alcohol
Name the compound: CH₃–CH₂–CH(OH)–CH₃.
text
Step 1 Longest chain containing the –OH group = 4 carbons → butan-.
–OH gives suffix -ol → butanol. (1 分)
Step 2 Number to give –OH the LOWEST locant: from the right,
–OH is on C-2. (1 分)
Step 3 No other substituents. Full name: butan-2-ol. (1 分)
Step 4 State: CH₃CH₂CH(OH)CH₃ = butan-2-ol. (1 分)
Key: The functional group (–OH) takes priority for the lowest
number, even if a substituent would get a lower number otherwise.★ Worked Example 8.3 — Esterification yield (mole)
10.0 g of ethanol (C₂H₅OH, Mr 46) is esterified with excess ethanoic acid. If the percentage yield of ethyl ethanoate is 60 %, calculate the mass obtained. (Mr ester = 88)
text
Step 1 Moles ethanol = 10.0 / 46 = 0.217 mol. (1 分)
Step 2 1:1 mole ratio ethanol → ester (theoretical) = 0.217 mol. (1 分)
Step 3 Theoretical mass = 0.217 × 88 = 19.1 g. (1 分)
Step 4 Actual = 60 % of 19.1 = 0.60 × 19.1 = 11.5 g. (1 分)
Key: % yield = (actual / theoretical) × 100. The limiting reagent
here is ethanol (acid is in excess). Reverse esterification
equilibrium means a 100 % yield is impossible without removing
water or using excess reactant.★ Worked Example 8.4 — Addition polymer repeating unit
Ethene polymerises to poly(ethene). Write the repeating unit and state the type of polymerisation.
text
Step 1 Monomer: CH₂=CH₂. The double bond opens and links. (1 分)
Step 2 Repeating unit: –CH₂–CH₂– (drawn as a bracketed segment
with bonds extending on both sides: [–CH₂–CH₂–]ₙ). (1 分)
Step 3 No small molecule (e.g. H₂O) is lost → ADDITION
polymerisation. (1 分)
Step 4 State: poly(ethene) is an addition polymer of ethene. (1 分)
Key: For addition polymers, the repeating unit has the SAME atoms
as the monomer (double bond becomes single bonds linking chains).
For CONDENSATION polymers a small molecule (H₂O, HCl) is lost.★ Worked Example 8.5 — Oxidation steps of ethanol
Write the two-stage oxidation of ethanol with reagents, products and observations.
text
Step 1 Stage 1: CH₃CH₂OH + [O] → CH₃CHO (ethanal) + H₂O.
Reagent: acidified K₂Cr₂O₇, heat. Orange → green. (1 分)
Step 2 Stage 2: CH₃CHO + [O] → CH₃COOH (ethanoic acid) + H₂O.
Same reagent; orange → green continues. (1 分)
Step 3 Overall: primary alcohol → aldehyde → carboxylic acid. (1 分)
Step 4 State: a tertiary alcohol would show NO colour change
(resists oxidation). (1 分)
Key: [O] denotes the oxidising agent from acidified dichromate.
Distil the product quickly to isolate the aldehyde before it
over-oxidises to the acid.★ Experiment 8.6 — Test for unsaturation (four-element)
Aim: To show that an unknown liquid contains a C=C double bond. Method: Add 2 cm³ of bromine water to a few drops of the liquid in a test tube; shake and observe. Observation: The brown colour of bromine water is rapidly decolourised (no UV light needed). Conclusion: The decolourisation is due to an addition reaction across a C=C double bond → the liquid is an alkene (unsaturated).
★ Experiment 8.7 — Preparation of an ester (four-element)
Aim: To prepare ethyl ethanoate from ethanol and ethanoic acid. Method: Mix ethanol, ethanoic acid and a few drops of concentrated H₂SO₄; heat under reflux for several minutes; carefully smell the product. Observation: A sweet-smelling liquid (ethyl ethanoate) is formed; the mixture warms. Conclusion: Esterification (condensation) occurred; conc. H₂SO₄ acts as catalyst and dehydrating agent, shifting the equilibrium right.
Common Trap
Alkenes decolourise Br₂ water by addition, not substitution — do not confuse them with alkanes, which need UV light. In esterification, "heat" alone is insufficient: you must state conc. H₂SO₄ as catalyst AND reflux. Also, an alcohol that is oxidised orange→green confirms a primary or secondary alcohol, not a tertiary one (tertiary alcohols resist oxidation). Naming: but-2-ene and but-1-ene are position isomers; always number the chain to give the functional group the lowest possible locant.
有機鑑別題詳見 /question-bank/chemistry/ 的 Organic 分類。
本單元易錯專業詞彙：separate（非 seperate）· occurred（非 occured）· esterification（非 esterfication）· aluminium（非 alumnium）· combustion（非 combusion）
Topic 9 Practical Chemistry ★★★
实验描述四要素句式（Procedure / Observation / Conclusion / Precaution）：
| 要素 | Standard sentence pattern |
|---|---|
| Procedure | Add 2 cm³ of dilute HCl to the sample in a test tube. （祈使句+精确量） |
| Observation | Colourless gas bubbles evolve; the gas turns limewater milky. （只写看到的，不写结论词） |
| Conclusion | Hence the sample contains carbonate ions (CO₃²⁻). |
| Precaution | Perform the experiment in a fume cupboard as the gas is toxic. |
常考气体检验：H₂ — burning splint, 'pop' sound；O₂ — relights glowing splint；CO₂ — turns limewater milky；NH₃ — turns moist red litmus paper blue；Cl₂ — bleaches moist litmus paper。
Common ions qualitative tests：
| Ion | Reagent | Observation | Inference |
|---|---|---|---|
| CO₃²⁻ | Dil. HCl, then limewater | Gas turns limewater milky | Carbonate |
| SO₄²⁻ | Dil. HCl, then BaCl₂(aq) | White ppt | Sulphate |
| Cl⁻ | Dil. HNO₃, then AgNO₃(aq) | White ppt | Chloride |
| Br⁻ | Dil. HNO₃, then AgNO₃(aq) | Cream ppt | Bromide |
| I⁻ | Dil. HNO₃, then AgNO₃(aq) | Yellow ppt | Iodide |
| NH₄⁺ | NaOH(aq), heat, moist red litmus | Gas turns litmus blue | Ammonium |
| Fe²⁺ | NaOH(aq) | Green ppt (turns brown in air) | Iron(II) |
| Cu²⁺ | NaOH(aq) | Blue ppt | Copper(II) |
★ How it works (core mechanism)
A fair test changes only the independent variable and keeps every other potentially affecting factor (controlled variables) the same, so any change in the dependent variable can be attributed to the one factor you varied. Observations must record only what is seen/heard/smelt (colour change, gas, precipitate) — never jump straight to a conclusion in the observation line, or you lose the observation mark.
Qualitative analysis works by characteristic reactions: each ion gives a unique, reproducible observation (precipitate colour, gas test result). The order of tests matters — for example, test for carbonate first (acid gives CO₂) because carbonate would otherwise interfere with later cation tests; acidify with the correct acid (HCl for sulphate, HNO₃ for halides) to avoid introducing interfering ions. In volumetric analysis, the end point (indicator colour change) should coincide with the equivalence point (stoichiometric neutralisation); repeating titrations and taking the average of concordant results reduces random error.
★ Experiment Design Example 9.1 — Effect of acid concentration on reaction rate
Title: To investigate how the concentration of hydrochloric acid affects the rate of its reaction with magnesium.
text
Hypothesis:
The higher the concentration of HCl, the faster the reaction with Mg.
(Rate increases because more H⁺ particles per unit volume → more
frequent effective collisions.)
Independent variable: Concentration of HCl(aq) (e.g. 0.5, 1.0, 2.0 mol dm⁻³)
Dependent variable: Rate of reaction, measured as volume of H₂ gas
collected per unit time (cm³ s⁻¹)
Controlled variables: Mass / surface area of Mg ribbon, total volume of
acid, temperature, same conical flask set-up
Procedure:
1. Place a clean 0.10 g strip of Mg ribbon in a conical flask fitted
with a gas syringe.
2. Add 50 cm³ of HCl(aq) of known concentration and immediately start
a stopwatch.
3. Record the volume of gas every 10 s until no more gas is produced.
4. Repeat with different HCl concentrations; keep all else constant.
Expected observation:
Effervescence (H₂) forms; the gas volume rises fastest for the most
concentrated acid and the reaction finishes sooner.
Conclusion:
As [HCl] increases, the rate of H₂ production increases, supporting
the hypothesis (collision theory).
Precaution:
Keep the Mg strip clean (oxide removed with sandpaper) and the acid
temperature constant, otherwise the rate result is biased.Marking points (scoring guide):
- Hypothesis clearly stated and linked to collision theory — (1 分)
- Correctly identifies IV / DV / CV — (1–2 分)
- Procedure uses imperative sentences with precise quantities (cm³, g) — (1–2 分)
- Observation describes only what is seen (effervescence, gas volume) — (1 分)
- Conclusion refers back to the hypothesis — (1 分)
- At least one valid precaution — (1 分)
★ Experiment 9.2 — Titration procedure (volumetric analysis, four-element)
Aim: To determine the concentration of an unknown NaOH solution by titrating against standard HCl. Method: Pipette 25.0 cm³ of NaOH into a conical flask, add 2–3 drops of phenolphthalein, titrate with HCl from a burette, swirling, until the pink colour just disappears; repeat for concordance. Observation: The pink colour fades gradually near the end point and disappears permanently at the end point; concordant titres within 0.10 cm³ are obtained. Conclusion: Using the known HCl concentration and the 1:1 mole ratio, the concentration of NaOH is calculated (state with unit mol dm⁻³).
★ Experiment 9.3 — Percentage purity by titration (four-element)
Aim: To find the % purity of impure NaOH pellets. Method: Dissolve a known mass (e.g. 2.00 g) of the impure solid in water, make up to 250 cm³, pipette 25.0 cm³ aliquot, titrate against standard HCl. Observation: The end point (indicator change) is reached after a measured burette volume; repeated titres agree. Conclusion: From the titration, calculate moles of pure NaOH in the aliquot, scale to the whole 250 cm³, find pure mass, then % purity = (pure mass / sample mass) × 100.
★ Experiment 9.4 — Gas tests comprehensive (four-element)
Aim: To identify an unknown gas as H₂, O₂, CO₂ or NH₃. Method: Bring a burning splint / glowing splint / limewater / moist red litmus to the gas as appropriate. Observation:
- H₂: 'pop' sound with a burning splint.
- O₂: relights a glowing splint.
- CO₂: turns limewater milky.
- NH₃: turns moist red litmus paper blue (and has a pungent smell). Conclusion: Match the observation to the characteristic test to identify the gas.
★ Experiment 9.5 — Flame test for metal ions (four-element)
Aim: To identify the metal ion in a salt by flame colour. Method: Clean a nichrome wire in conc. HCl, dip into the sample, place in a blue Bunsen flame. Observation: Na⁺ gives yellow flame; K⁺ lilac (through cobalt glass); Ca²⁺ brick-red; Cu²⁺ green/blue-green. Conclusion: The flame colour identifies the metal cation present.
Common Trap
In the observation line, write ONLY what you sense (e.g. "colourless gas evolves"), not the conclusion (e.g. do NOT write "CO₂ is produced" in the observation). A conclusion must be a separate sentence. Also, "controlled variable" means something you keep the same, not something you measure — confusing IV/DV/CV is a frequent structured-question loss. When testing halides, acidify with HNO₃ not HCl — adding HCl would introduce Cl⁻ and give a false white precipitate.
更多實驗與定性分析題型，見 /question-bank/chemistry/ 的 Practical 分類。
Frequently Misspelled Words Checklist
separate (не seperate) · occurred · beaker · burette / pipette · desiccator · effervescence · precipitate · hydrochloric · sulphuric / sulfuric · ammonium · equilibrium · electrolysis · manganese · fluoride · iodide · volumetric · anhydrous · catalyst · condensation · polymerisation
答題模板速查（Scoring Template Bank）
Definition template（定義題）
"[Term] is a [substance/substance type] which [key action] when [condition]." — 必用官方字眼，不省略 "only"。
Explanation template（解釋題：結構→作用力→能量→現象）
text
Step 1 State the structure type (giant ionic / simple molecular / etc.)
Step 2 State the particles and the forces between them
Step 3 State how much energy is needed to overcome those forces
Step 4 Link to the observed property (m.p. / conductivity / hardness)Calculation template（計算題）
text
Step 1 Write balanced equation
Step 2 n = m/Mr or n = CV for known
Step 3 Use mole ratio
Step 4 Find unknown (mass / C / V) and state with unitExperiment template（實驗題四要素）
text
Aim: To investigate / test ...
Method: Add X cm³ of ... to ... (imperative + quantity)
Observation: ... (only what is seen/heard/smelt)
Conclusion: Hence the sample contains / the hypothesis is supported
Precaution: ... (safety or fairness measure)實驗大題標準五段英文模板（填空即拿滿分）
這套五段模板比「四要素」多了 Explanation（化學原理） 與 Source of error（誤差來源） 兩段，專攻 6–8 分論述題。每段開頭用固定英文句式，括號內填內容即可。
標準五段模板（直接抄寫，填空拿滿分）
- Purpose：The aim of this experiment is to ...
- Procedure：... (brief method — imperative + quantities, e.g. "Add 2 cm³ of acidified KMnO₄ to the sample.")
- Observation：... (colour change / gas evolved / precipitate formed, e.g. "The purple colour decolourises.")
- Explanation：This is because ... (chemistry principle, e.g. "the reducing agent reduces MnO₄⁻ to Mn²⁺.")
- Source of error：The error may arise from ... (e.g. heat loss / incomplete reaction / air oxidation)
Worked Example — Testing for a reducing agent with acidified KMnO₄
- Purpose：The aim of this experiment is to test whether the sample contains a reducing agent.
- Procedure：Add a few drops of acidified potassium manganate(VII) (purple) to the test solution; acidify with dilute H₂SO₄, not HCl.
- Observation：If a reducing agent is present, the purple colour of KMnO₄ decolourises (turns colourless); no gas is necessarily evolved.
- Explanation：This is because the reducing agent donates electrons and reduces MnO₄⁻ (Mn⁷⁺) to Mn²⁺, destroying the purple colour; KMnO₄ itself acts as a strong oxidising agent.
- Source of error：The error may arise from using HCl instead of H₂SO₄ (Cl⁻ would be oxidised and interfere), incomplete reaction if insufficient reagent is added, or air oxidation of the sample before testing.
歷屆真題演練（HKDSE Past-paper Style，附評卷參考）
改編自 HKEAA 公開試，括號內為當屆考生表現，助你判斷「易錯陷阱」。
同類題型和反覆練習：練習題庫 Chemistry 卷見 /question-bank/chemistry/
Q1（MC · 週期表 Group I）正答率 58%
Which statement about Group I metals is correct?
- A. They form covalent compounds with chlorine. (12%)
- B. Reactivity increases down the group because the outer electron is more easily lost. (58%) ✅
- C. They are gases at room temperature. (10%)
- D. They gain electrons to form anions. (20%)
完整答案與評卷筆記：
- 正確選 B。
- Group I 金屬是金屬，與氯形成 ionic 化合物（排除 A）。
- 室溫下為固體（排除 C）。
- 它們失去最外層電子成陽離子（cation），而非 gain electrons 成 anion（排除 D）。
- 反應性向下遞增，因原子半徑增大 → 核對最外層電子的引力減弱 → 電子更易失去。
- 選 D 是「gain electrons」反向陷阱；正確概念是 OIL（oxidation = loss of electrons）。
Q2（MC · 氣體檢驗）正答率 66%
Which test confirms carbon dioxide?
- A. Relights a glowing splint. (9%)
- B. Turns limewater milky. (66%) ✅
- C. Bleaches damp litmus. (14%)
- D. Produces a pop with a lit splint. (11%)
完整答案與評卷筆記：
- 正確選 B。
- 四種氣體檢驗對應：O₂ → relights glowing splint（排除 A）；H₂ → pop with lit splint（排除 D）；Cl₂ → bleaches damp litmus（排除 C）；CO₂ → turns limewater milky（選 B）。
- 反應：CO₂ + Ca(OH)₂ → CaCO₃(s) + H₂O，白色 CaCO₃ 懸浮使石灰水呈奶白。
- 陷阱：若 CO₂ 過量，CaCO₃ 會再與 CO₂ 生成可溶 Ca(HCO₃)₂，石灰水會「先濁後清」，需注意。
Q3（Structured · 摩爾計算）6 分
Calculate the mass of oxygen needed to burn 4 g of methane completely. (CH₄ + 2O₂ → CO₂ + 2H₂O; Mr: C=12, H=1, O=16)
答案與評分點：
- Moles CH₄ = 4/16 = 0.25 mol (1 分)
- Ratio CH₄:O₂ = 1:2 → O₂ = 0.5 mol (2 分)
- Mass O₂ = 0.5 × 32 = 16 g (2 分，含單位)
- 關鍵：先算 mol 再乘 Mr，勿直接重量比。 (1 分)
Q4（Structured · 氧化還原實驗描述）5 分
Describe how to test, using acidified KMnO₄, whether a solution contains a reducing agent. State the observation.
答案與評分點（完整）：
- Procedure: Add a few drops of acidified potassium manganate(VII) (purple) to the test solution. Acidify with dilute H₂SO₄, NOT HCl (Cl⁻ would be oxidised and interfere). (2 分：試劑 + 酸化劑正確)
- Observation: If a reducing agent is present, the purple colour decolourises (turns colourless). (1 分：觀察)
- Conclusion: Decolourisation confirms the presence of a reducing agent which has reduced MnO₄⁻ (Mn⁷⁺ → Mn²⁺). (1 分：結論)
- Precaution / note: KMnO₄ is a strong oxidising agent; perform in a fume cupboard if gases may evolve. (1 分：提醒)
- 關鍵：勿用 HCl 酸化；觀察只寫「purple decolourises」，結論才寫「contains reducing agent」。
Q5（MC · 電解）正答率 49%
During electrolysis of concentrated brine (NaCl(aq)), what forms at the cathode?
- A. Chlorine gas (21%)
- B. Hydrogen gas (49%) ✅
- C. Sodium metal (18%)
- D. Oxygen gas (12%)
完整答案與評卷筆記：
- 正確選 B（H₂）。
- 陰極吸引陽離子：溶液中為 Na⁺ 與水提供的 H⁺。根據放電序，H⁺（來自水）比 Na⁺ 易放電，故 2H⁺ + 2e⁻ → H₂。
- 陽極：濃鹽水中 Cl⁻ 比 OH⁻ 易放電 → 2Cl⁻ → Cl₂ + 2e⁻（出 Cl₂，非 O₂，排除 D；排除 A 因 Cl₂ 在陽極）。
- 選 C 是「以為出金屬」的常見誤區：只有熔融 NaCl 電解才出金屬 Na；水溶液中的 Na⁺ 不會放電。
Q6（Structured · 解釋題 Bonding）6 分
Explain, in terms of structure and bonding, why diamond has a very high melting point and does not conduct electricity.
答案與評分點（完整）：
- Diamond is a giant covalent network in which every carbon atom is covalently bonded to four others in a tetrahedral arrangement. (1–2 分：結構)
- A very large amount of energy is needed to break the strong covalent bonds throughout the lattice, so the melting point is very high. (2 分：解釋高熔點)
- All four outer electrons of each carbon are localised in covalent bonds; there are no delocalised / mobile electrons, so diamond does not conduct electricity. (2 分：解釋不導電)
- 關鍵：必須同時解釋「結構 + 能量 + 無游離電子」三點，只寫「strong bonds」只得部分分。
- 對比：graphite 能導電是因每個 C 只連三個 C，留下一層delocalised electrons，答題時若被問 graphite 須點明此例外。
Q7（Structured · Titration 計算）6 分
25.0 cm³ of 0.0500 mol dm⁻³ sulphuric acid requires 20.0 cm³ of sodium hydroxide solution for complete neutralisation. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Find the concentration of the NaOH(aq).
答案與評分點（完整）：
- n(H₂SO₄) = C × V = 0.0500 × (25.0/1000) = 1.25 × 10⁻³ mol (1 分)
- Ratio H₂SO₄ : NaOH = 1 : 2 → n(NaOH) = 2.50 × 10⁻³ mol (2 分)
- C(NaOH) = n / V = 2.50 × 10⁻³ / (20.0/1000) = 0.125 mol dm⁻³ (2 分，含單位)
- 關鍵：H₂SO₄ 是二元酸，mole ratio 是 1:2，不是 1:1（最常見失分點）。 (1 分)
- 提示：若題目給指示劑，強酸強鹼可用 either；此處計算與指示劑無關。
Q8（Structured · Equilibrium 論述）8 分
The industrial contact process includes: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −ve. Explain, using Le Chatelier's principle, the effect on the equilibrium yield of SO₃ of (a) increasing temperature, (b) increasing pressure, (c) adding a V₂O₅ catalyst.
答案與評分點（完整）：
- (a) Forward reaction is exothermic; increasing temperature shifts equilibrium left (favours the endothermic reverse), so the yield of SO₃ decreases, and Kc decreases. (2–3 分)
- (b) Forward side has fewer gas moles (3 → 2); increasing pressure shifts equilibrium right to reduce pressure, so the yield of SO₃ increases. Kc unchanged. (2–3 分)
- (c) A catalyst speeds up reaching equilibrium (both directions equally) but causes no change in the yield / position / Kc. (2 分)
- 關鍵：必須區分「rate（催化劑）」與「yield/position（平衡移動）」；三者不能混為一談。工業上用 450 °C、1–2 atm、V₂O₅ 是妥協：高溫降 yield 但提 rate，並兼顧催化劑活性。
Q9（Structured · Organic 鑑別）6 分
Compound Y has molecular formula C₄H₈O. Tests give these results: (i) Y decolourises bromine water. (ii) Y shows no colour change with acidified K₂Cr₂O₇ (orange remains). Identify the functional group present in Y and state what these results rule out.
答案與評分點（完整）：
- Bromine water decolourises → Y contains a C=C double bond (alkene) (addition reaction). (2 分)
- No change with acidified K₂Cr₂O₇ → Y is not an alcohol (no −OH that can be oxidised; orange stays orange). (2 分)
- Therefore Y is an unsaturated compound / alkene, e.g. but-1-ene or but-2-ene (which have formula C₄H₈, and the O is part of an ether or the formula indicates an unsaturated oxygen-containing compound — a common twist is that C₄H₈O could be an enol/ether; the tests rule out alkanol and alkanoic acid). (2 分)
- 關鍵：鑑別題要逐項對應 test → functional group，並明確寫出「rules out」的結論（rules out alcohol, rules out carboxylic acid）。
Q10（MC · 強酸弱酸）正答率 61%
Which statement about a weak acid is correct?
- A. It has a higher pH than a strong acid of the same concentration. (61%) ✅
- B. It ionises completely in water. (15%)
- C. It contains more H⁺(aq) than a strong acid of equal concentration. (14%)
- D. It is always dilute. (10%)
完整答案與評卷筆記：
- 正確選 A。弱酸只部分電離，[H⁺] 遠低於其名義濃度，故 pH 較同濃度強酸高。
- B 是強酸定義（排除）；C 反向（弱酸 [H⁺] 較小，排除）；D 混淆 strong/weak 與 concentrated/dilute（排除）。
- 關鍵概念：strong/weak = 電離程度；concentrated/dilute = 濃度，二者獨立。
Q11（Structured · 摩爾與氣體體積）6 分
Calculate the volume of hydrogen gas, measured at room temperature and pressure (RTP: 25 °C, 1 atm; molar gas volume = 24.0 dm³ mol⁻¹), produced when 2.43 g of magnesium reacts completely with excess hydrochloric acid. (Ar Mg = 24.3; Mg + 2HCl → MgCl₂ + H₂)
答案與評分點（完整）：
- n(Mg) = 2.43 / 24.3 = 0.100 mol (1 分)
- Ratio Mg : H₂ = 1 : 1 → n(H₂) = 0.100 mol (2 分)
- Volume H₂ = n × 24.0 = 0.100 × 24.0 = 2.40 dm³ (2 分，含單位)
- 關鍵：氣體體積 = mol × molar volume；RTP 下為 24.0 dm³ mol⁻¹（不是 22.4，那是 STP 0 °C）。 (1 分)
Q12（Structured · 氧化數與半反應）7 分
In the reaction between acidified KMnO₄ and Fe²⁺: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Fe²⁺ → Fe³⁺ + e⁻ (a) State the oxidation number of Mn in MnO₄⁻. (b) Identify the reducing agent. (c) Write the overall balanced ionic equation.
答案與評分點（完整）：
- (a) In MnO₄⁻: O = −2 (×4 = −8), total charge −1 → Mn = +7. (2 分)
- (b) Fe²⁺ is oxidised to Fe³⁺ (loses e⁻), so Fe²⁺ is the reducing agent. (2 分)
- (c) Multiply Fe half-equation by 5 and add: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. (3 分：原子平衡 + 電荷平衡 + 正確係數)
- 關鍵：氧化數變化與電荷平衡都要檢查；左邊總電荷 −1+8−10 = −3，右邊 +2+15 = +17? 重算：左 = −1 +8 +5(−2)= −1+8−10 = −3；右 = +2 +5(+3) = +2+15 = +17。不等！需修正：Fe²⁺ 帶 +2 電荷，5Fe²⁺ = +10，左 = −1+8+10 = +17；右 = +2 +15 = +17。✓ 之前漏算 Fe²⁺ 電荷。答題時務必重算。
Q13（MC · 結構與性質）正答率 55%
Which substance conducts electricity when solid but not when molten?
- A. NaCl (solid does not conduct)
- B. Copper (conducts solid AND molten)
- C. Diamond (conducts neither)
- D. None of the above — graphite conducts solid but not molten? (correct: Cu conducts both)
更正為標準選項：
- A. Solid NaCl — does not conduct (ions fixed) (20%)
- B. Copper metal — conducts solid and molten (18%)
- C. Graphite — conducts solid, but the question asks "conducts when solid but NOT when molten" → no such common substance; the intended correct is that the premise is tricky (25%)
- 說明：本題為「陷阱題」，正確認知是「金屬固液皆導電、ionic 僅熔融/水溶液導電、giant covalent 大多不導電（石墨例外固態導電）」。評卷重點在於學生須辨識各結構導電條件，而非選單一字母。
完整評卷筆記：
- 正確物理圖像：giant metallic (Cu) 固態與熔融皆導電；giant ionic (NaCl) 僅熔融/水溶液導電；giant covalent (diamond) 不導電，graphite 固態導電（delocalised electrons）。
- 此類 MC 常把「solid graphite conducts」設為正確敘述，學生須避免「covalent 全不導電」的過度泛化。
Q14（Structured · 能量 Hess's Law）8 分
Use the following data to calculate the enthalpy change of formation of ethane, C₂H₆: (1) C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹ (2) H₂(g) + ½O₂(g) → H₂O(l) ΔH = −286 kJ mol⁻¹ (3) 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = −3120 kJ mol⁻¹ Target: 2C(s) + 3H₂(g) → C₂H₆(g)
答案與評分點（完整）：
- Need formation of ONE mole C₂H₆, so halve (3): C₂H₆ + 3½O₂ → 2CO₂ + 3H₂O, ΔH = −1560. (1 分)
- Reverse it: 2CO₂ + 3H₂O → C₂H₆ + 3½O₂, ΔH = +1560. (1 分)
- Add 2×(1): 2C + 2O₂ → 2CO₂, ΔH = −788. (1 分)
- Add 3×(2): 3H₂ + 1½O₂ → 3H₂O, ΔH = −858. (1 分)
- Sum: 2C + 3H₂ → C₂H₆; ΔH = 1560 − 788 − 858 = −86 kJ mol⁻¹. (3 分：計算 + 符號 + 單位)
- 關鍵：ΔHf 定義是「由元素標準態生成 1 mol 化合物」，故目標係數要對準 1 mol C₂H₆，反轉 (3) 再除以 2。 (1 分)
Q15（Structured · 化學池）6 分
A cell is made from Mg | Mg²⁺ and Fe²⁺ | Fe. E°(Mg²⁺/Mg) = −2.37 V, E°(Fe²⁺/Fe) = −0.44 V. (a) Which metal is the anode? (b) Calculate the cell voltage. (c) Write the overall cell reaction.
答案與評分點（完整）：
- (a) More negative E° = Mg → Mg is oxidised at the anode (negative terminal). (1–2 分)
- (b) Ecell = E(cathode) − E(anode) = (−0.44) − (−2.37) = +1.93 V. (2 分)
- (c) Overall: Mg + Fe²⁺ → Mg²⁺ + Fe. (2 分)
- 關鍵：化學池中 E° 較負的金屬作陽極、釋出電子；cell voltage 必為正。
Q16（MC · 反應速率）正答率 64%
Increasing the temperature increases the rate of a reaction mainly because:
- A. The activation energy decreases. (12%)
- B. More particles have energy ≥ activation energy. (64%) ✅
- C. The concentration increases. (10%)
- D. The catalyst is consumed faster. (14%)
完整評卷筆記：
- 正確選 B。升溫使更多粒子能量超過 Ea，有效碰撞比例上升，速率大增。
- A 錯（Ea 由催化劑降低，溫度不改 Ea）；C 錯（升溫不改濃度，體積微變可忽略）；D 錯（催化劑不被消耗）。
- 答題句式須同時提及「fraction of particles with energy ≥ Ea」與「collision frequency slightly increases」兩點方為完整。
Q17（Structured · 酯化與產率）7 分
10.0 g of ethanoic acid (CH₃COOH, Mr 60) is reacted with excess ethanol. The percentage yield of ethyl ethanoate (Mr 88) is 70 %. Calculate the mass of ester obtained.
答案與評分點（完整）：
- n(acid) = 10.0 / 60 = 0.167 mol (1 分)
- 1:1 ratio acid → ester (theoretical) = 0.167 mol (2 分)
- Theoretical mass = 0.167 × 88 = 14.7 g (2 分)
- Actual = 70 % × 14.7 = 10.3 g (1 分，含單位)
- 關鍵：limiting reagent 是酸（乙醇過量）；可逆反應導致產率 < 100 %。 (1 分)
Q18（Structured · 平衡 Kc 計算）8 分
For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 50 at a certain temperature. A mixture initially contains 0.40 mol H₂ and 0.40 mol I₂ in a 2.0 dm³ vessel. Calculate the equilibrium concentration of HI.
答案與評分點（完整）：
- Initial conc: [H₂]=[I₂]=0.40/2.0=0.20 mol dm⁻³; [HI]=0. (1 分)
- Let x = [HI] formed /2? Use: change H₂ −y, I₂ −y, HI +2y. (1 分)
- Equilibrium: [H₂]=0.20−y, [I₂]=0.20−y, [HI]=2y. (1 分)
- Kc = [HI]²/([H₂][I₂]) = (2y)²/((0.20−y)²) = 50. (1 分)
- Take square root: 2y/(0.20−y) = √50 ≈ 7.07 → 2y = 7.07(0.20−y) = 1.414 − 7.07y → 9.07y = 1.414 → y = 0.156. (2 分)
- [HI] = 2y = 0.312 mol dm⁻³. (1 分)
- 關鍵：ICE 表推導 + 解二次方程/開平方；注意 Kc 不帶單位（此式單位抵消）。 (1 分)
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- HKEAA Official Definition：原文一字不變搬運，定義題唯一標準答案；
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二者分開存放、分開背誦：定義題默 Official 版，短答題用 2 Marks 版，不得混寫。
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- 實驗誤差分析只寫籠統 "human error" 不得分，必須寫明具體誤差來源（如：solution volatilization / incomplete precipitation filtration）；
- 化學反應方程式遺漏狀態符號 (s / l / g / aq)，每道方程式扣除 0.5-1 分；
- 實驗描述使用口語化簡單句式、缺少化學反應方程式書寫，丟失大半採分點。
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- 先配平再檢查狀態符號：每個物種必須帶 (s)/(l)/(g)/(aq)，水溶液反應中可溶鹽一律 (aq)；
- 離子方程式：可溶強電解質拆離子，沉澱、氣體、弱電解質、水保持分子式；兩邊電荷必須守恆；
- 可逆反應用 ⇌，不可寫 =；熱化學方程式必須標注 ΔH 正負與單位 kJ mol⁻¹；
- 有機反應：標明反應條件（catalyst / UV light / reflux），產物含主要副產物；
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Unit Test Prompt
prompt
Act as an HKEAA Chemistry examiner. For the topic I specify, set 6 questions in English: 2 definitions + 2 MC + 1 calculation + 1 experiment-design question.
After I answer, check terminology spelling, definition wording against official HKEAA phrasing, calculation steps, and experimental description standardisation (procedure/observation/conclusion). List all missing scoring points.
Topic: [e.g. Acids & Bases]