📚 Year 13 WJEC Chemistry: Mock Unit Test Analysis | Year 13 WJEC 化学：单元测试模拟卷解析
Welcome to this detailed walkthrough of a mock unit test for WJEC Year 13 Chemistry. This resource covers key topics from A2 Physical, Inorganic and Organic Chemistry, providing fully worked answers and examiner-style commentary. Use it to identify gaps in your knowledge and sharpen your exam technique under timed conditions.
欢迎阅读这份 WJEC Year 13 化学单元模拟卷的详细解析。本资源涵盖 A2 物理化学、无机化学和有机化学的核心主题，提供完整的解题过程和考官风格的点评。用它来诊断知识漏洞，并在限时条件下磨炼你的应试技巧。
1. Rate Equation Determination | 速率方程确定
The reaction 2NO(g) + O₂(g) → 2NO₂(g) was studied at 298 K. Initial rates are given below.
在 298 K 下研究了反应 2NO(g) + O₂(g) → 2NO₂(g)。初始速率数据如下。
| Experiment | [NO] / mol dm⁻³ | [O₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
| 1 | 1.0 × 10⁻³ | 1.0 × 10⁻³ | 2.0 × 10⁻⁴ |
| 2 | 2.0 × 10⁻³ | 1.0 × 10⁻³ | 8.0 × 10⁻⁴ |
| 3 | 1.0 × 10⁻³ | 4.0 × 10⁻³ | 8.0 × 10⁻⁴ |
he rate (8.0×10⁻⁴ ÷ 2.0×10⁻⁴ = 4). Hence, the reaction is second order with respect to NO.
比较实验 1 和 2，在 [O₂] 恒定时将 [NO] 加倍，速率变为原来的四倍 (8.0×10⁻⁴ ÷ 2.0×10⁻⁴ = 4)。因此，反应对 NO 为二级。
Comparing experiments 1 and 3, quadrupling [O₂] while [NO] is constant also quadruples the rate. Thus, the order with respect to O₂ is 1.
比较实验 1 和 3，在 [NO] 恒定时将 [O₂] 增至四倍，速率也增至四倍。因此，对 O₂ 为一级。
The rate equation is: rate = k[NO]²[O₂]. Using experiment 1, k = (2.0×10⁻⁴) ÷ [(1.0×10⁻³)² × (1.0×10⁻³)] = 2.0×10⁵ mol⁻² dm⁶ s⁻¹.
速率方程为：rate = k[NO]²[O₂]。利用实验 1 数据，k = (2.0×10⁻⁴) ÷ [(1.0×10⁻³)² × (1.0×10⁻³)] = 2.0×10⁵ mol⁻² dm⁶ s⁻¹。
A plausible two-step mechanism: (1) 2NO ⇌ he rate law derived from step (2) using the equilibrium assumption matches the experimental orders.
可能的双步机理：(1) 2NO ⇌ N₂O₂（快速平衡），(2) N₂O₂ + O₂ → 2NO₂（慢，速率控制步）。由步骤 (2) 结合平衡假设推导出的速率定律与实验级数吻合。
2. Equilibrium Constant Kp | 平衡常数 Kp
Consider the equilibrium 2SO₂(g) + O₂(g) ⇌ The total pressure is 200 kPa. Calculate the partial pressures and Kp, stating its units.
考虑 700 K 下的平衡 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)。平衡时摩尔分数为：x(SO₂) = 0.200，x(O₂) = 0.300，x(SO₃) = 0.500。总压为 200 kPa。计算分压和 Kp，并注明单位。
SO₂) = 0.200 × 200 = 40 kPa; p(O₂) = 0.300 × 200 = 60 kPa; p(SO₃) = 0.500 × 200 = 100 kPa.
分压 = 摩尔分数 × 总压：p(SO₂) = 0.200 × 200 = 40 kPa；p(O₂) = 0.300 × 200 = 60 kPa；p(SO₃) = 0.500 × 200 = 100 kPa。
significant figures). The units: (kPa)² / [(kPa)² × kPa] = kPa⁻¹. Hence Kp ≈ 0.104 kPa⁻¹.
)² / [(pSO₂)² × pO₂] = (100)² / [(40)² × 60] = 10000 / (1600 × 60) = 10000 / 96000 = 0.104（保留三位有效数字）。单位：(kPa)² / [(kPa)² × kPa] = kPa⁻¹。因此 Kp ≈ 0.104 kPa⁻¹。
3. Weak Acid pH and Buffer Calculation | 弱酸 pH 与缓冲溶液计算
⁻³. Calculate its pH. Then, 25.0 cm³ of 0.10 mol dm⁻³ NaOH is added to 50.0 cm³ of the acid. Determine the pH of the resulting buffer solution.
浓度为 0.10 mol dm⁻³ 的乙酸 (CH₃COOH) 溶液的 Ka = 1.8 × 10⁻⁵ mol dm⁻³。计算其 pH。然后将 25.0 cm³ 的 0.10 mol dm⁻³ NaOH 加入 50.0 cm³ 的该酸中。确定所得缓冲溶液的 pH。
For a weak acid, [H⁺] ≈ √Ka × C = √1.8×10⁻⁵ × 0.10 = √1.8×10⁻⁶ = 1.34×10⁻³ mol dm⁻³. pH = –log(1.34×10⁻³) ≈ 2.87.
对于弱酸，[H⁺] ≈ √Ka × C = √1.8×10⁻⁵ × 0.10 = √1.8×10⁻⁶ = 1.34×10⁻³ mol dm⁻³。pH = –log(1.34×10⁻³) ≈ 2.87。
5 × 0.10 = 2.5×10⁻³ mol. The neutralisation produces 2.5×10⁻³ mol of CH₃COO⁻, leaving (5.0 – H. Total volume = 75.0 cm³, so [CH₃COOH] = [CH₃COO⁻] = 2.5×10⁻³ / 0.075 = 0.0333 mol dm⁻³.
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