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A-Level 化学 Paper 1 结构题模拟卷（100 分・2 小时・附评分要点）｜TutorZone
TUTORZONE 原创模拟卷 · A-Level 化学 9701 Paper 1（结构题）
本卷为 TutorZone 原创 A-Level 化学 Paper 1 结构题模拟卷，覆盖计量、酸碱滴定、焓变、平衡、速率、原子结构、化学键、有机化学与实验计算九大范畴。题型与深浅按 A-Level 化学 Paper 1 的课程范围编写，卷制（10 题 30 部分、100 分、2 小时）为本平台自订练习安排，并非任何考试局官方卷。全卷附逐部分评分要点，可用作操卷与自我批改。
卷制速览
| 项目 | 内容 |
|---|---|
| 试卷 | Chemistry Paper 1 (Structured) |
| 题数 | 10 题（30 部分） |
| 分数 | 100 分 |
| 时限 | 2 小时 |
| 占比 | 本平台自订练习卷（非考试局官方格式） |
| 计算机 | 准用计算机 |
题目
全卷 10 题、30 部分，答案须写出步骤；除特别指明外，非准确值取 3 位有效数字。
1. （10 分）摩尔计算与结晶水
(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）酸碱滴定与浓度
(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）量热法与焓变
(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）化学平衡与 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）反应速率
(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）原子结构与游离能
(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）化学键与分子形状
(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）有机命名与异构
(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）有机反应机制
(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）气体体积、产率与限量试剂
(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）
答案与评分要点
以下为每部分参考答案与评分要点，M 为方法分、A 为答案分、B 为独立分。
第 1 题 · 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 题 · 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 题 · 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 题 · 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 题 · 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 题 · 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 题 · 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 题 · 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 题 · 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 题 · 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.
常见失分位
- 摩尔计算忘记乘结晶水的 5，或把 Mr 与 M 混淆，令水质量减半。
- 量热法用溶液的体积当质量，或忘记 ΔT 要用 K（数值与 °C 相同但单位要写对）。
- Kc 表达式把生成物与反应物的次方倒转，或忘记平方 [HI]。
- 电子排布写成 3d⁴ 4s²，忽略半满 3d 的额外稳定性。
- 有机机制只写产物，没有画出由孤电子对指向碳的箭嘴，拿不到 M 分。
时间分配
| 题目 | 分数 | 建议时间 |
|---|---|---|
| 第 1–2 题 | 20 分 | 约 24 分钟 |
| 第 3–4 题 | 20 分 | 约 24 分钟 |
| 第 5–6 题 | 20 分 | 约 24 分钟 |
| 第 7–8 题 | 20 分 | 约 24 分钟 |
| 第 9–10 题 | 20 分 | 约 24 分钟 |
下载与打印
题目卷与答案卷均为 A4 版本，可直接打印：
