AS AQA International Chemistry: CH01 Unit 1 Example Responses | AS AQA 国际化学：CH01 第一单元示例答题

📚 AS AQA International Chemistry: CH01 Unit 1 Example Responses | AS AQA 国际化学：CH01 第一单元示例答题

This guide walks through realistic AQA International AS Chemistry Unit 1 (CH01) questions and shows you the kind of response that earns full marks. Each example is paired with examiner-style commentary, so you can see not just what to write, but why it earns marks.

本指南通过 AQA 国际 AS 化学第一单元（CH01）的真实风格考题，展示如何写出获得满分的答案。每个示例都配有考官式点评，帮助你不仅知道写什么，更明白为什么能得分。

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1. Understanding Command Words | 理解指令词

Before tackling any question, you must decode the command word. In AQA mark schemes, ‘State’ or ‘Write’ requires a single word, phrase, or symbol. ‘Define’ demands a formal statement, usually one that can be applied to any example. ‘Explain’ requires a reason, almost always linked to structure, bonding, or collision theory. ‘Calculate’ tests your working and units. ‘Suggest’ invites you to apply known chemistry to an unfamiliar context, and the examiner will credit any chemically sensible idea.

在作答之前，你首先必须解读指令词。在 AQA 评分方案中，”State / Write（写出）”要求一个单词、短语或符号；”Define（定义）”要求正式表述，通常可适用于任何例子；”Explain（解释）”要求给出理由，几乎总是与结构、键合或碰撞理论相关；”Calculate（计算）”考查你的过程和单位；”Suggest（推测）”则邀请你将所学化学知识应用于陌生情境，考官会对任何化学上合理的想法给分。

For example, if a question says ‘State the meaning of the term relative atomic mass’, a one-line definition is sufficient: the weighted mean mass of an atom of an element relative to 1/12 of the mass of an atom of carbon-12. If the question says ‘Explain’, you must add detail, such as why the abundance of each isotope must be considered.

例如，如果题目要求 “State the meaning of the term relative atomic mass（写出相对原子质量的含义）”，一行定义就足够了：一个元素原子的加权平均质量，相对于碳-12 原子质量的 1/12。如果题目要求 “Explain（解释）”，你必须补充细节，例如为什么需要考虑每种同位素的丰度。

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2. Atomic Structure and Mass Spectrometry | 原子结构与质谱

Question (4 marks): Bromine has two isotopes, ⁷⁹Br and ⁸¹Br, present in approximately equal abundance. Predict the number of molecular ion peaks and their m/z values in the mass spectrum of Br₂. Explain why the peaks appear in a 1 : 2 : 1 ratio.

题目（4 分）：溴有两种同位素 ⁷⁹Br 和 ⁸¹Br，丰度近似相等。预测 Br₂ 质谱中分子离子峰的数量及其 m/z 值，并解释为什么这些峰以 1 : 2 : 1 的比例出现。

Example response that gains full marks:

能获得满分的示例答案：

A Br₂ molecule can contain ⁷⁹Br–⁷⁹Br, ⁷⁹Br–⁸¹Br, or ⁸¹Br–⁸¹Br, giving three molecular ion peaks. Their m/z values are 158, 160 and 162 respectively. The ratio is 1 : 2 : 1 because each bromine atom is 50% ⁷⁹Br and 50% ⁸¹Br; therefore the combination ⁷⁹Br–⁸¹Br has two possible arrangements, making it twice as probable as either the ⁷⁹Br–⁷⁹Br or the ⁸¹Br–⁸¹Br combination.

一个 Br₂ 分子可以含有 ⁷⁹Br–⁷⁹Br、⁷⁹Br–⁸¹Br 或 ⁸¹Br–⁸¹Br，因此产生三个分子离子峰，m/z 值分别为 158、160 和 162。比例为 1 : 2 : 1，因为每个溴原子是 50% 的 ⁷⁹Br 和 50% 的 ⁸¹Br；因此 ⁷⁹Br–⁸¹Br 组合有两种排列方式，其概率是 ⁷⁹Br–⁷⁹Br 或 ⁸¹Br–⁸¹Br 组合的两倍。

Examiner’s comment: Full marks are earned because the response gives the three peaks, links m/z to the sum of the two isotope masses, and uses probability language rather than vague phrases such as ‘there are more combinations’.

考官点评：该答案获得满分，因为给出了三个峰，将 m/z 与两种同位素质量之和联系起来，并使用概率语言而不是”有更多组合”这类模糊表述。

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3. Amount of Substance and Titration Calculations | 物质的量与滴定计算

Question (5 marks): In a titration, 25.0 cm³ of 0.120 mol dm⁻³ NaOH solution exactly neutralised 22.5 cm³ of H₂SO₄ solution. Calculate the concentration of the H₂SO₄ solution in mol dm⁻³.

题目（5 分）：在滴定中，25.0 cm³ 的 0.120 mol dm⁻³ NaOH 溶液恰好中和 22.5 cm³ 的 H₂SO₄ 溶液。计算 H₂SO₄ 溶液的浓度（单位 mol dm⁻³）。

Full-mark worked solution:

满分计算过程：

Step 1 — write the balanced equation. A full equation earns a mark even if the subsequent arithmetic goes wrong.

第一步 — 写出配平的方程式。即使后续计算出错，完整方程式也能获得一分。

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Step 2 — calculate the moles of NaOH used. The units must be shown and the volume converted into dm³.

第二步 — 计算所用 NaOH 的物质的量。必须写出单位，并将体积换算为 dm³。

moles NaOH = 0.120 × 25.0/1000 = 0.00300 mol

Step 3 — use the 1 : 2 mol ratio to find the moles of H₂SO₄.

第三步 — 利用 1 : 2 的摩尔比求出 H₂SO₄ 的物质的量。

moles H₂SO₄ = 0.00300 ÷ 2 = 0.00150 mol

Step 4 — divide by the acid volume to find its concentration.

第四步 — 除以酸的体积，求出其浓度。

concentration = 0.00150 × 1000 ÷ 22.5 = 0.0667 mol dm⁻³ (3 s.f.)

Examiner’s comment: Marks are awarded for the correct equation, the correct conversion, the correct stoichiometry, the correct division, and the correct units. Writing the intermediate value 0.066666… and rounding to 3 significant figures would gain the final mark; leaving an excessive number of decimal places also gains the mark, but a rounded answer with no working may lose the method marks.

考官点评：分数分配给正确的方程式、正确的换算、正确的化学计量关系、正确的除法运算以及正确的单位。写出中间值 0.066666… 并四舍五入到三位有效数字可获得最后的分值；保留过多小数位同样得分，但只写结果而不写过程可能会失去方法分。

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4. Bonding and Structure | 化学键与结构

Question (4 marks): Explain, in terms of structure and bonding, why diamond has a much higher melting point than iodine.

题目（4 分）：从结构和键合的角度解释，为什么金刚石的熔点远高于碘。

Model answer:

模型答案：

Diamond has a giant covalent (macromolecular) structure in which every carbon atom is joined to four others by strong covalent bonds. Melting diamond requires breaking a very large number of these strong covalent bonds, which demands a large amount of energy. Iodine, by contrast, has a simple molecular structure made of discrete I₂ molecules. Within each molecule the covalent bond is strong, but melting iodine only requires overcoming the weak van der Waals’ forces between molecules, not breaking the covalent bonds themselves. Very little energy is therefore needed, giving iodine a low melting point.

金刚石具有巨型共价（大分子）结构，每个碳原子通过强共价键与另外四个碳原子相连。熔化金刚石需要断裂大量强共价键，因此需要巨大的能量。相比之下，碘具有由离散的 I₂ 分子组成的简单分子结构。分子内部的共价键很强，但熔化碘只需要克服分子间微弱的范德华力，而不需要断裂共价键本身。因此所需能量很少，碘的熔点很低。

Examiner’s comment: The answer earns all four marks because it contrasts both structures explicitly, identifies the correct type of force that is overcome in each case, and links energy to melting point. A common weak response is simply ‘diamond has strong bonds and iodine has weak bonds’ — this is too vague because the strong covalent bond within an I₂ molecule is actually comparable to the C–C bond; it is the intermolecular forces in iodine that are weak.

考官点评：该答案获得全部四分，因为它明确对比了两种结构，正确指出了每种情况中被克服的力的类型，并将能量与熔点联系起来。常见的薄弱答案是”金刚石键强而碘键弱”——这太模糊了，因为 I₂ 分子内部的共价键实际上与 C–C 键相当；碘中微弱的是分子间作用力。

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5. Energetics and Hess’s Law | 能量学与赫斯定律

Question (4 marks): Use the standard enthalpy changes of combustion given below to calculate the enthalpy change for the hydrogenation of ethene:

题目（4 分）：利用给出的标准燃烧焓变计算乙烯加氢反应的焓变：

C₂H₄(g) + H₂(g) → C₂H₆(g)

Given: ΔHc(C₂H₄) = −1411 kJ mol⁻¹; ΔHc(H₂) = −286 kJ mol⁻¹; ΔHc(C₂H₆) = −1560 kJ mol⁻¹.

已知：ΔHc(C₂H₄) = −1411 kJ mol⁻¹；ΔHc(H₂) = −286 kJ mol⁻¹；ΔHc(C₂H₆) = −1560 kJ mol⁻¹。

Worked answer with a Hess’s law cycle:

含赫斯循环的计算过程：

ΔH = ΣΔHc(reactants) − ΣΔHc(products)

ΔH = (−1411 + (−286)) − (−1560) = −1697 + 1560 = −137 kJ mol⁻¹

The negative sign shows the reaction is exothermic. In your written answer, draw the Hess’s law cycle with the combustion products (CO₂ and H₂O) at the bottom; the examiner will reward the correct arrows and labels even if the arithmetic is imperfect.

负号表明该反应是放热反应。在书面作答时，画出以燃烧产物（CO₂ 和 H₂O）位于底部的赫斯循环图；即使计算不完全正确，考官也会奖励正确的箭头和标注。

Examiner’s comment: For full marks, you must state the formula used, substitute the values with the correct signs, and give the final answer with units. A very common error is forgetting the minus sign on the combustion of ethane, which gives +169 kJ mol⁻¹ and loses two of the four marks.

考官点评：要获得满分，你必须写出所用公式、以正确符号代入数值并给出带单位的最终答案。一个非常常见的错误是忘记乙烷燃烧焓的负号，这样会得到 +169 kJ mol⁻¹，从而失去四个分值中的两个。

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6. Kinetics and Collision Theory | 动力学与碰撞理论

Question (5 marks): The reaction A + B → C has the rate equation rate = k[A]². State and explain, using collision theory, how the initial rate changes when the concentration of A is doubled while the concentration of B is kept constant.

题目（5 分）：反应 A + B → C 的速率方程为 rate = k[A]²。利用碰撞理论说明并解释：当 A 的浓度加倍而 B 的浓度保持不变时，初始速率如何变化。

Full-mark response:

满分答案：

The rate increases by a factor of 4. Doubling the concentration of A doubles the number of A particles per unit volume, which doubles the collision frequency. Because the reaction is second order with respect to A, the rate depends on [A]², so the rate is multiplied by 2² = 4. The greater collision frequency leads to a greater frequency of successful collisions (collisions with energy greater than or equal to the activation energy), and therefore the initial rate quadruples.

速率增加到原来的 4 倍。A 的浓度加倍使单位体积内 A 的粒子数加倍，从而使碰撞频率加倍。由于该反应对 A 是二级反应，速率取决于 [A]²，因此速率乘以 2² = 4。更高的碰撞频率导致更频繁的有效碰撞（能量大于或等于活化能的碰撞），因此初始速率变为四倍。

Examiner’s comment: The command word ‘explain’ requires a link between concentration, collision frequency, and rate. Simply writing ‘rate = k[A]² so it quadruples’ gains only one or two marks. The phrase ‘successful collisions’ is essential — examiners penalise vague claims such as ‘the molecules react faster’ without reference to collisions or activation energy.

考官点评：指令词 “Explain（解释）”要求在浓度、碰撞频率和速率之间建立联系。只写 “rate = k[A]²，所以变为四倍” 只能得一两分。”有效碰撞”这一表述至关重要——考官会扣掉”分子反应得更快”这类没有提及碰撞或活化能的模糊说法。

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7. Chemical Equilibria and Kc | 化学平衡与 Kc

Question (5 marks): For the following equilibrium, write the Kc expression and state its units:

题目（5 分）：对于以下平衡，写出 Kc 表达式并注明其单位：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Model answer:

模型答案：

Kc = [NH₃]² / ([N₂][H₂]³)

To determine the units, substitute mol dm⁻³ for each concentration term:

为确定单位，将每一项浓度代入 mol dm⁻³：

Units = (mol dm⁻³)² / ((mol dm⁻³)(mol dm⁻³)³) = mol⁻² dm⁶

So the units are mol⁻² dm⁶ (or dm⁶ mol⁻²). Three essential features earn the marks: the product of the products divided by the product of the reactants, the correct stoichiometric indices as powers, and the correct units derived from substituting concentrations in mol dm⁻³.

因此单位为 mol⁻² dm⁶（或 dm⁶ mol⁻²）。三个关键要素可获得分值：产物浓度乘积除以反应物浓度乘积、正确的化学计量数作为幂指数、以及通过代入 mol dm⁻³ 浓度单位导出的正确单位。

Examiner’s comment: A frequent mistake is writing square brackets around the wrong species or forgetting the cubed power on [H₂]. Also, when asked to

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