A-Level Chemistry: Insert 5 Jan 21 Calculation Questions | A-Level 化学：2021年1月数据插页5计算题型

📚 A-Level Chemistry: Insert 5 Jan 21 Calculation Questions | A-Level 化学：2021年1月数据插页5计算题型

The A-Level Chemistry exam papers often provide an Insert booklet that includes essential constants, key equations, and a periodic table. For the January 2021 series, Insert 5 is a critical resource that candidates must use efficiently to solve calculation-based questions. This article will guide you through the typical calculation question types that rely on the data and formulas given in Insert 5, ensuring you can approach them with confidence and accuracy.

A-Level 化学试卷通常会提供包含基本常数、关键方程式和周期表的数据插页。在 2021 年 1 月的考试系列中，插入页 5 是考生必须高效利用的关键资源，用来解答计算类题目。本文将带你梳理依赖插入页 5 所给数据和公式的典型计算题型，让你能够自信、准确地应对它们。

1. What Does Insert 5 Include? | 插入页5包含哪些内容？

Insert 5 typically supplies the gas constant R (8.31 J K⁻¹ mol⁻¹), the Avogadro constant L (6.022 × 10²³ mol⁻¹), the Faraday constant F (9.65 × 10⁴ C mol⁻¹), and standard temperature 298 K. It also lists important equations: pV = nRT, ΔG = ΔH – TΔS, ΔG° = –RT ln K, the Nernst equation E = E° – (RT/nF) ln Q, and the Arrhenius equation ln k = –Ea/RT + ln A. Familiarity with exactly where to find each formula and constant saves precious time in the exam.

插入页 5 通常会提供气体常数 R（8.31 J K⁻¹ mol⁻¹）、阿伏伽德罗常数 L（6.022 × 10²³ mol⁻¹）、法拉第常数 F（9.65 × 10⁴ C mol⁻¹）以及标准温度 298 K。它还列出了重要的方程式：pV = nRT、ΔG = ΔH – TΔS、ΔG° = –RT ln K、能斯特方程 E = E° – (RT/nF) ln Q 和阿伦尼乌斯方程 ln k = –Ea/RT + ln A。熟悉每个公式和常数在何处可以节省考试中宝贵的时间。

2. Ideal Gas Equation pV = nRT | 理想气体状态方程 pV = nRT

The most frequently tested equation in Insert 5 is the ideal gas equation. You may be asked to calculate the number of moles n, the volume V, the pressure p, or the temperature T. Remember that temperature must be in kelvin (T(K) = T(°C) + 273) and that the value of R used depends on the units of p and V. If pressure is given in kPa and volume in dm³, convert both to Pa and m³ before inserting them: 1 kPa = 1 × 10³ Pa, 1 dm³ = 1 × 10⁻³ m³. Then R = 8.31 J K⁻¹ mol⁻¹ correctly applies because J ≡ Pa m³.

插入页 5 中最常考的方程就是理想气体状态方程。你可能需要计算物质的量 n、体积 V、压强 p 或温度 T。须记温度必须使用开尔文（T(K) = T(°C) + 273），且 R 的取值取决于 p 与 V 的单位。若压强以 kPa 给出、体积以 dm³ 给出，应先转换为 Pa 和 m³ 再代入：1 kPa = 1 × 10³ Pa，1 dm³ = 1 × 10⁻³ m³。这样 R = 8.31 J K⁻¹ mol⁻¹ 才能正确使用，因为 J ≡ Pa m³。

For example, to find the molar mass of a gas when 2.00 dm³ has a mass of 3.56 g at 100 kPa and 298 K, first calculate n = pV/RT. Convert: p = 100 × 10³ Pa, V = 2.00 × 10⁻³ m³, T = 298 K. n = (100 × 10³ × 2.00 × 10⁻³) / (8.31 × 298) = 0.0808 mol. Then molar mass M = mass / n = 3.56 g / 0.0808 mol = 44.1 g mol⁻¹. Always show clear unit conversions to avoid losing marks.

例如，某气体在 100 kPa、298 K 下体积为 2.00 dm³，质量为 3.56 g，求其摩尔质量：先求 n = pV/RT。单位转换：p = 100 × 10³ Pa，V = 2.00 × 10⁻³ m³，T = 298 K。n = (100 × 10³ × 2.00 × 10⁻³) / (8.31 × 298) = 0.0808 mol。然后摩尔质量 M = 质量 / n = 3.56 g / 0.0808 mol = 44.1 g mol⁻¹。一定要清晰地展示单位转换以避免失分。

3. Manipulating the Equation for Different Unknowns | 变换公式求不同的未知量

Rearrange pV = nRT to solve for any variable. For pressure: p = nRT/V; for temperature: T = pV/(nR). A common pitfall is forgetting to use kelvin or using the wrong volume unit. When the question gives the gas volume as 24.5 dm³ at room temperature and pressure (RTP), you can directly use the molar gas volume 24.5 dm³ mol⁻¹ (at 101 kPa and 298 K) to find moles if conditions match precisely. Otherwise, use pV = nRT with the actual pressure and temperature provided.

变换 pV = nRT 可求解任一变量。求压强：p = nRT/V；求温度：T = pV/(nR)。常见错误是忘记使用开尔文温度或使用了错误的体积单位。当题目给出在室温和常压（RTP）下气体体积为 24.5 dm³ 时，若条件完全吻合，可以直接使用摩尔气体体积 24.5 dm³ mol⁻¹（101 kPa 和 298 K）求物质的量；否则，请用 pV = nRT 带入实际的压强和温度计算。

You may also encounter questions that ask for the density of a gas. Density ρ = mass / V = pM / RT, where M is molar mass. This rearranged form is not given in Insert 5, but you can derive it from pV = nRT by substituting n = m/M. Practice combining equations to handle such situations confidently.

你还可能遇到求气体密度的题目。密度 ρ = 质量 / V = pM / RT，其中 M 是摩尔质量。这一变形公式并未在插入页 5 中给出，但你可以通过将 n = m/M 代入 pV = nRT 推导出来。练习组合方程以自信应对此类情形。

4. Gibbs Free Energy ΔG = ΔH – TΔS | 吉布斯自由能 ΔG = ΔH – TΔS

Insert 5 provides this equation to link thermodynamic quantities. You must be comfortable working with any missing term. Watch the units: ΔH is usually given in kJ mol⁻¹, while ΔS is in J K⁻¹ mol⁻¹. Convert ΔS to kJ K⁻¹ mol⁻¹ by dividing by 1000, or convert ΔH to J mol⁻¹ to keep everything consistent. For example, for a reaction with ΔH = –196 kJ mol⁻¹ and ΔS = –223 J K⁻¹ mol⁻¹ at 298 K, calculate ΔG = –196 – 298 × (–0.223) = –196 + 66.5 = –129.5 kJ mol⁻¹. The negative ΔG indicates the reaction is feasible under these conditions.

插入页 5 提供了这一联系热力学量的方程。你必须能熟练地计算任一缺失项。注意单位：ΔH 通常以 kJ mol⁻¹ 给出，而 ΔS 则以 J K⁻¹ mol⁻¹ 给出。需要将 ΔS 除以 1000 转换为 kJ K⁻¹ mol⁻¹，或者将 ΔH 转换为 J mol⁻¹ 以保持单位一致。例如，某反应的 ΔH = –196 kJ mol⁻¹，ΔS = –223 J K⁻¹ mol⁻¹，温度为 298 K，计算 ΔG = –196 – 298 × (–0.223) = –196 + 66.5 = –129.5 kJ mol⁻¹。ΔG 为负表明该反应在此条件下是可行的。

Sometimes the question asks for the temperature at which the reaction becomes feasible, i.e. ΔG = 0. Set ΔH – TΔS = 0 and solve for T = ΔH / ΔS. Remember to use consistent units: if ΔH = 110 kJ mol⁻¹ and ΔS = 200 J K⁻¹ mol⁻¹, then T = (110 × 10³ J mol⁻¹) / (200 J K⁻¹ mol⁻¹) = 550 K. Below this temperature, ΔG is positive and the reaction is not spontaneous.

有时题目会问反应在什么温度下变得可行，即 ΔG = 0。设 ΔH – TΔS = 0，解得 T = ΔH / ΔS。记得单位要统一：如果 ΔH = 110 kJ mol⁻¹，ΔS = 200 J K⁻¹ mol⁻¹，则 T = (110 × 10³ J mol⁻¹) / (200 J K⁻¹ mol⁻¹) = 550 K。低于此温度时 ΔG 为正，反应不自发。

5. Relating ΔG° to the Equilibrium Constant K | ΔG° 与平衡常数 K 的关系

The equation ΔG° = –RT ln K is directly given in Insert 5 and is fundamental for linking thermodynamics to equilibrium. If you know ΔG°, you can find K, and vice versa. At 298 K, the product RT = 8.31 × 298 ≈ 2.48 kJ mol⁻¹ (when expressed in kJ). Convert ΔG° to J mol⁻¹ before plugging into the formula if you use R = 8.31 J K⁻¹ mol⁻¹. For example, if ΔG° = –32.0 kJ mol⁻¹ = –32000 J mol⁻¹, then ln K = –(–32000) / (8.31 × 298) = 32000 / 2476 = 12.9. Therefore K = e¹²·⁹ ≈ 4.0 × 10⁵. A large K indicates that the equilibrium lies far to the right.

方程 ΔG° = –RT ln K 直接列于插入页 5 中，是将热力学与平衡联系起来的基础。若已知 ΔG°，可求出 K，反之亦然。在 298 K 时，RT = 8.31 × 298 ≈ 2.48 kJ mol⁻¹（以 kJ 表示时）。若使用 R = 8.31 J K⁻¹ mol⁻¹，则在代入前应将 ΔG° 转换为 J mol⁻¹。例如，若 ΔG° = –32.0 kJ mol⁻¹ = –32000 J mol⁻¹，则 ln K = –(–32000) / (8.31 × 298) = 32000 / 2476 = 12.9。因此 K = e¹²·⁹ ≈ 4.0 × 10⁵。大的 K 值说明平衡位置远向右移。

Conversely, if the question gives the equilibrium constant, you can determine ΔG°. If K = 2.0 × 10⁻³ at 298 K, then ln K = –6.21, ΔG° = –8.31 × 298 × (–6.21) = +15.4 kJ mol⁻¹ (positive, so the equilibrium favours the reactants). Always check the sign: negative ΔG° gives K > 1, positive ΔG° gives K < 1.

反过来，如果题目给出平衡常数，可以求 ΔG°。若 K = 2.0 × 10⁻³（298 K），则 ln K = –6.21，ΔG° = –8.31 × 298 × (–6.21) = +15.4 kJ mol⁻¹（正值，因此平衡偏向反应物）。务必检查符号：ΔG° 为负时 K > 1，ΔG° 为正时 K < 1。

6. The Nernst Equation for Electrochemical Cells | 电化学电池的能斯特方程

Insert 5 includes the Nernst equation in the form E = E° – (RT/nF) ln Q. At 298 K, this simplifies to E = E° – (0.0592/n) log₁₀ Q, a convenient version for calculating cell potentials under non-standard conditions. Here n is the number of electrons transferred in the redox reaction, and Q is the reaction quotient. For a cell Zn|Zn²⁺ (0.10 M) || Cu²⁺ (0.010 M)|Cu with E°(cell) = +1.10 V, the reaction is Zn + Cu²⁺ → Zn²⁺ + Cu, so n = 2 and Q = [Zn²⁺]/[Cu²⁺] = 0.10 / 0.010 = 10. Then E = 1.10 – (0.0592/2) log₁₀ 10 = 1.10 – 0.0296 × 1 = 1.0704 V. The cell potential decreases slightly because the product ion concentration is higher than the reactant ion concentration.

插入页 5 给出了能斯特方程 E = E° – (RT/nF) ln Q。在 298 K 时，可简化为 E = E° – (0.0592/n) log₁₀ Q，这是计算非标准状态下电池电势的便捷版本。其中 n 是氧化还原反应中转移的电子数，Q 是反应商。例如电池 Zn|Zn²⁺ (0.10 M) || Cu²⁺ (0.010 M)|Cu，标准电池电势 E°(cell) = +1.10 V，反应为 Zn + Cu²⁺ → Zn²⁺ + Cu，故 n = 2，Q = [Zn²⁺]/[Cu²⁺] = 0.10 / 0.010 = 10。则 E = 1.10 – (0.0592/2) log₁₀ 10 = 1.10 – 0.0296 × 1 = 1.0704 V。由于产物离子浓度高于反应物离子浓度，电池电势略微下降。

If you need to find the concentration of an ion from a measured cell potential, rearrange the equation. For example, if E = 1.12 V, E° = 1.10 V, n = 2, then log Q = (E° – E) × n / 0.0592 = (1.10 – 1.12) × 2 / 0.0592 = –0.675. So Q = 10

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