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A-Level 化學 Paper 1 結構題模擬卷（100 分・2 小時・附評分要點）｜TutorZone
TUTORZONE 原創模擬卷 · A-Level 化學 9701 Paper 1（結構題）
本卷為 TutorZone 原創 A-Level 化學 Paper 1 結構題模擬卷，覆蓋計量、酸鹼滴定、焓變、平衡、速率、原子結構、化學鍵、有機化學與實驗計算九大範疇。題型與深淺按 A-Level 化學 Paper 1 的課程範圍編寫，卷制（10 題 30 部分、100 分、2 小時）為本平台自訂練習安排，並非任何考試局官方卷。全卷附逐部分評分要點，可用作操卷與自我批改。
卷制速覽
| 項目 | 內容 |
|---|---|
| 試卷 | Chemistry Paper 1 (Structured) |
| 題數 | 10 題（30 部分） |
| 分數 | 100 分 |
| 時限 | 2 小時 |
| 佔比 | 本平台自訂練習卷（非考試局官方格式） |
| 計算機 | 准用計算機 |
題目
全卷 10 題、30 部分，答案須寫出步驟；除特別指明外，非準確值取 3 位有效數字。
1. （10 分）摩爾計算與結晶水
(a) Calculate the relative molecular mass of hydrated copper(II) sulfate, CuSO4·5H2O. (Ar: Cu = 63.5, S = 32.0, O = 16.0, H = 1.0) （3 分）
(b) A sample of the hydrated salt has mass 12.475 g. Calculate the amount, in mol, of the hydrated salt and the amount, in mol, of water of crystallisation it contains. （3 分）
(c) The sample is heated gently until the mass is constant. Calculate the mass of anhydrous copper(II) sulfate obtained and the mass of water lost, and show that the two masses add up to the original mass. （4 分）
2. （10 分）酸鹼滴定與濃度
(a) In a titration, 25.0 cm³ of sodium hydroxide solution of unknown concentration required 23.60 cm³ of 0.100 mol dm−³ hydrochloric acid to reach the end-point. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l). Calculate the amount, in mol, of hydrochloric acid used. （3 分）
(b) Calculate the concentration, in mol dm−³, of the sodium hydroxide solution. （4 分）
(c) Calculate the mass of sodium hydroxide dissolved in 250 cm³ of this solution. (M = 40.0 g mol−¹) （3 分）
3. （10 分）量熱法與焓變
(a) 25.0 cm³ of 0.500 mol dm−³ copper(II) sulfate solution was added to excess zinc powder. The temperature rose from 19.0 °C to 22.4 °C. The solution has mass 50.0 g and specific heat capacity 4.18 J g−¹ K−¹. Calculate the heat energy released, in J. （4 分）
(b) Calculate the amount, in mol, of copper(II) sulfate used. （3 分）
(c) Calculate the enthalpy change of the reaction, in kJ mol−¹, and state whether the reaction is exothermic or endothermic. （3 分）
4. （10 分）化學平衡與 Kc
(a) 1.00 mol of hydrogen and 1.00 mol of iodine are mixed in a 1.00 dm³ vessel and left to reach equilibrium: H2(g) + I2(g) ⇌ 2HI(g). At equilibrium 1.58 mol of hydrogen iodide is present. Calculate the amount of hydrogen and of iodine at equilibrium. （3 分）
(b) Calculate the value of Kc for this equilibrium, and state its units. （4 分）
(c) The forward reaction is exothermic. State and explain the effect of increasing the temperature on the value of Kc. （3 分）
5. （10 分）反應速率
(a) In a reaction the concentration of a reactant fell from 0.80 mol dm−³ to 0.20 mol dm−³ in 120 s. Calculate the average rate of reaction in mol dm−³ s−¹. （4 分）
(b) Explain what is meant by the rate-determining step of a reaction mechanism. （3 分）
(c) State and explain the effect of adding a catalyst on the rate of reaction and on the value of the equilibrium constant. （3 分）
6. （10 分）原子結構與游離能
(a) Write the electron configuration of a chromium atom (Z = 24) using the 3d and 4s subshells, and explain why it is not what the simple filling order predicts. （3 分）
(b) The first ionisation energy of magnesium is 738 kJ mol−¹ and that of aluminium is 578 kJ mol−¹. Explain why the value for aluminium is lower, even though it has more protons. （4 分）
(c) Explain why the first ionisation energy generally increases across Period 3 from sodium to argon. （3 分）
7. （10 分）化學鍵與分子形狀
(a) The ammonium ion, NH4+, has four bonds and no lone pair on the nitrogen. State the shape of the ion and the bond angle, and explain your answer. （4 分）
(b) Ammonia, NH3, is a polar molecule but carbon dioxide, CO2, is not. Explain this difference in terms of shape and bond polarity. （3 分）
(c) Hydrogen fluoride boils at 20 °C while hydrogen chloride boils at −85 °C, although HCl has the larger Mr. Explain this difference. （3 分）
8. （10 分）有機命名與異構
(a) Give the IUPAC name of CH3CH(CH3)CH2OH. （2 分）
(b) There are four structural isomers with the molecular formula C4H9Br. Name or describe all four. （4 分）
(c) Explain why 2-bromobutane has optical isomers but 1-bromobutane does not. （4 分）
9. （10 分）有機反應機制
(a) Bromoethane reacts with hydroxide ions in aqueous ethanol. Describe the mechanism, naming the type of substitution, the role of the hydroxide ion and the species formed. （5 分）
(b) Iodoethane reacts faster than bromoethane with hydroxide ions. Explain this using bond enthalpies. （3 分）
(c) Ethanoic acid reacts with ethanol to form ethyl ethanoate. State the equation, the catalyst and one condition needed. （2 分）
10. （10 分）氣體體積、產率與限量試劑
(a) 0.486 g of magnesium reacts completely with excess dilute hydrochloric acid. Calculate the amount of magnesium used and the volume of hydrogen produced, in cm³, at room temperature and pressure (molar volume = 24.0 dm³ mol−¹). (Ar: Mg = 24.3) （4 分）
(b) In a repeat experiment only 0.0175 mol of hydrogen was collected. Calculate the percentage yield. （3 分）
(c) In another experiment 0.0200 mol of magnesium is added to hydrochloric acid containing 0.050 mol of HCl. Show by calculation which reactant is in excess. （3 分）
答案與評分要點
以下為每部分參考答案與評分要點，M 為方法分、A 為答案分、B 為獨立分。
第 1 題 · 10 分
(a) Mr = 63.5 + 32.0 + (4 × 16.0) + 5 × 18.0 = 249.5
B1 correct Ar sum for CuSO4 (159.5); B1 5 × 18.0 = 90.0 for the water; B1 total 249.5.
(b) 0.0500 mol of CuSO4·5H2O and 0.250 mol of H2O
M1 n = m / Mr; A1 0.0500 mol; M1 multiplies by 5; A1 0.250 mol (allow ecf).
(c) anhydrous CuSO4 = 7.97 g; water lost = 4.50 g; 7.97 + 4.50 = 12.475 g
M1 0.0500 × 159.5; A1 7.98 g; M1 0.250 × 18.0; A1 4.50 g and the check 7.975 + 4.5 = 12.475.
第 2 題 · 10 分
(a) 0.00236 mol
M1 n = cV with V in dm³ (0.02360); A1 0.00236 mol.
(b) 0.0944 mol dm−³
M1 1:1 mole ratio; M1 c = n / 0.0250; A1 0.0944 mol dm−³ (allow 0.094).
(c) 0.944 g
M1 n = 0.0944 × 0.250 = 0.0236 mol; A1 0.0236 × 40.0; A1 0.944 g.
第 3 題 · 10 分
(a) q = 50.0 × 4.18 × 3.4 = 711 J
M1 q = mcΔT with ΔT = 3.4 K; A1 50.0 × 4.18; A1 209; A1 711 J (allow 710-711).
(b) 0.0125 mol
M1 n = cV = 0.500 × 0.0250; A1 0.0125 mol.
(c) −56.8 kJ mol−¹; exothermic
M1 divides 711 J by 0.0125 mol; A1 56.8 kJ mol−¹; B1 negative sign with exothermic.
第 4 題 · 10 分
(a) 0.21 mol H2 and 0.21 mol I2
M1 1.58 / 2 = 0.79 mol reacted; A1 0.21 mol of each remaining; B1 states both are equal by symmetry.
(b) Kc = 56.6; no units
M1 Kc = [HI]² / ([H2][I2]); M1 substitutes 1.58² / (0.21 × 0.21); A1 56.6; B1 no units because the powers cancel.
(c) Kc decreases
B1 Kc decreases; M1 equilibrium shifts in the endothermic (reverse) direction; A1 explained as Le Chatelier opposing the rise in temperature.
第 5 題 · 10 分
(a) 0.0050 mol dm−³ s−¹
M1 Δ[reactant] = 0.60; M1 divides by 120 s; A1 5.0 × 10⁻³ mol dm−³ s−¹.
(b) the slowest step of the mechanism, which limits the overall rate
B1 slowest step; B1 it controls the overall rate; B1 only species up to and including this step appear in the rate equation.
(c) rate increases; Kc is unchanged
B1 rate increases (alternative route of lower activation energy); B1 Kc unchanged; B1 forward and reverse rates are increased equally.
第 6 題 · 10 分
(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹, i.e. [Ar] 3d⁵ 4s¹
B1 [Ar] 3d⁵ 4s¹ (24 electrons in total); B1 one electron occupies 4s and five occupy 3d; B1 a half-filled 3d subshell has extra stability from exchange.
(b) the outer electron of Al is in a 3p orbital, which is higher in energy and better shielded
M1 states the electron removed from Al is in 3p and from Mg in 3s; M1 3p is at a higher energy level (further from the nucleus); A1 more shielding by the 3s electrons; A1 the shielding and distance effect outweighs the extra proton.
(c) nuclear charge increases while shielding stays similar, so the outer electrons are held more strongly
B1 the number of protons increases; B1 the outer electrons are in the same shell so shielding is similar; B1 the atomic radius decreases, so attraction increases.
第 7 題 · 10 分
(a) tetrahedral; 109.5°
B1 tetrahedral; B1 109.5° (allow 109-110°); M1 four bonding pairs; A1 they repel equally and arrange as far apart as possible.
(b) NH3 is trigonal pyramidal so the bond dipoles do not cancel; CO2 is linear so the two C=O dipoles cancel
B1 NH3 has a lone pair giving a pyramidal shape; B1 the dipoles add up; B1 CO2 is linear and the equal and opposite dipoles cancel.
(c) HF molecules form hydrogen bonds; HCl has only permanent dipole-dipole forces and van der Waals forces
B1 HF forms hydrogen bonds (H bonded to F); B1 hydrogen bonds are much stronger than dipole-dipole forces; B1 more energy is needed to separate HF molecules.
第 8 題 · 10 分
(a) 2-methylpropan-1-ol
B1 correct stem (propan-1-ol); B1 correct methyl position (2-).
(b) 1-bromobutane; 2-bromobutane; 1-bromo-2-methylpropane; 2-bromo-2-methylpropane
B1 two isomers with a straight butane chain; B1 the two positions on that chain distinguished; B1 two isomers with a branched (2-methylpropane) skeleton; B1 all four correctly named.
(c) C2 of 2-bromobutane carries four different groups, so it is chiral; C1 of 1-bromobutane carries two identical hydrogen atoms
B1 2-bromobutane has a carbon with four different groups; B1 the two enantiomers are non-superimposable mirror images; B1 1-bromobutane has two H atoms on C1; B1 so no chiral centre exists.
第 9 題 · 10 分
(a) nucleophilic substitution; OH− attacks the δ+ carbon of C–Br and Br− leaves; ethanol and bromide ions are formed
B1 nucleophilic substitution; M1 curly arrow from the lone pair on OH− to the carbon; M1 curly arrow from the C–Br bond to Br; A1 the transition state with a partial C–O and partial C–Br bond; A1 products ethanol + Br− and the name SN2.
(b) the C–I bond is weaker (lower bond enthalpy) than the C–Br bond, so it breaks more easily
B1 C–I bond enthalpy is lower than C–Br; B1 less energy needed to break the bond; B1 so the activation energy is lower.
(c) CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O; concentrated H2SO4; heat under reflux
B1 correct equation; B1 concentrated sulfuric acid and warming (reflux).
第 10 題 · 10 分
(a) 0.0200 mol Mg; 480 cm³ H2 (0.48 dm³)
M1 n = 0.486 / 24.3; A1 0.0200 mol; M1 V = n × 24.0; A1 0.480 dm³ = 480 cm³.
(b) 87.5 %
M1 percentage yield = actual / theoretical × 100 with 0.0200 mol theoretical; A1 87.5 %; B1 the answer is quoted to 3 significant figures.
(c) Mg + 2HCl → MgCl2 + H2; 0.0200 mol Mg needs 0.0400 mol HCl, so HCl (0.050 mol) is in excess and magnesium is the limiting reactant
M1 states the 1:2 ratio; M1 calculates 0.0400 mol HCl needed; A1 correct conclusion that HCl is in excess.
常見失分位
- 摩爾計算忘記乘結晶水的 5，或把 Mr 與 M 混淆，令水質量減半。
- 量熱法用溶液的體積當質量，或忘記 ΔT 要用 K（數值與 °C 相同但單位要寫對）。
- Kc 表達式把生成物與反應物的次方倒轉，或忘記平方 [HI]。
- 電子排布寫成 3d⁴ 4s²，忽略半滿 3d 的額外穩定性。
- 有機機制只寫產物，沒有畫出由孤電子對指向碳的箭嘴，拿不到 M 分。
時間分配
| 題目 | 分數 | 建議時間 |
|---|---|---|
| 第 1–2 題 | 20 分 | 約 24 分鐘 |
| 第 3–4 題 | 20 分 | 約 24 分鐘 |
| 第 5–6 題 | 20 分 | 約 24 分鐘 |
| 第 7–8 題 | 20 分 | 約 24 分鐘 |
| 第 9–10 題 | 20 分 | 約 24 分鐘 |
下載與列印
題目卷與答案卷均為 A4 版本，可直接列印：
