Chemistry 12 — Unit 2 Learning Guide 详细解答¶
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创建日期: 2026-02-14
最后更新: 2026-02-14
2.1 COLLISION THEORY AND HEAT OF REACTION — 碰撞理论与反应热¶
Understanding Collision Theory — 理解碰撞理论¶
Q1. What two factors determine whether a collision of molecules will result in a successful reaction?
分子碰撞能否导致成功反应取决于哪两个因素？
The two factors are:
两个因素是：
-
Sufficient kinetic energy — Molecules must collide with enough energy to overcome the activation energy barrier (EA).
足够的动能 — 分子必须以足够的能量碰撞，以克服活化能垒 (EA)。
-
Correct orientation — Molecules must be oriented properly so that reactive parts collide.
正确的取向 — 分子必须以正确的方向排列，使反应部位能够碰撞。
ℹ️ 答案页未提供此题答案。
Q2. What is the definition of activation energy (EA)?
活化能 (EA) 的定义是什么？
Activation energy (EA) is the minimum amount of energy required for a collision between molecules to result in a chemical reaction. It is the energy needed to break existing bonds in the reactants so that new bonds can form in the products.
活化能 (EA) 是分子之间碰撞导致化学反应所需的最小能量。它是打破反应物中现有化学键以便形成产物中新化学键所需的能量。
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Q3. When molecules are oriented correctly and collide with sufficient kinetic energy, what is formed?
当分子以正确取向并以足够动能碰撞时，形成什么？
An activated complex (also called transition state) is formed. This is a temporary, high-energy, unstable arrangement of atoms at the peak of the potential energy barrier.
形成活化配合物（也称为过渡态）。这是位于势能垒顶端的临时、高能、不稳定的原子排列。
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Q4a. What type of energy increases as two molecules approach each other?
当两个分子相互接近时，哪种类型的能量增加？
Potential energy (PE) increases as two molecules approach each other. This is because electron clouds begin to repel each other, and energy is required to bring the molecules closer together.
势能 (PE) 增加。这是因为电子云开始相互排斥，将分子拉近需要能量。
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Q4b. What type of energy decreases as two molecules approach each other?
当两个分子相互接近时，哪种类型的能量减少？
Kinetic energy (KE) decreases as two molecules approach each other. As kinetic energy is converted to potential energy during the approach, the molecules slow down.
动能 (KE) 减少。当分子接近时，动能转化为势能，分子速度减慢。
ℹ️ 答案页未提供此题答案。
Q5. What is temperature's role in a reaction?
温度在反应中的作用是什么？
Temperature affects the average kinetic energy of particles. Higher temperature means:
温度影响粒子的平均动能。温度越高意味着：
-
Particles move faster → more frequent collisions
粒子运动更快 → 碰撞更频繁
-
More particles have sufficient energy to overcome EA → higher proportion of successful collisions
更多粒子具有足够能量克服活化能 → 成功碰撞的比例更高
-
Therefore, reaction rate increases with temperature
因此，反应速率随温度升高而增加
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Potential Energy Diagrams — 势能图¶
Q6. Place the correct numbers for each label in the correct circle. Memorize!
将每个标签的正确编号放入正确的圆圈中。要记住！
The diagram shows a potential energy curve with 5 labels to identify:
势能图显示需要标识 5 个位置：
Labels 标签： 1. Activation Energy Forward（正向活化能） 2. Activation Energy Reverse（逆向活化能） 3. ΔH Forward（正向焓变） 4. ΔH Reverse（逆向焓变） 5. Activated Complex（活化配合物）
Placement on diagram 在图上的位置：
PE (kJ)
|
| ⑤ Activated Complex (peak)
| /\
| / \
| ① / \ ②
| / \
|---- ----
| REACTANTS PRODUCTS
|←---- ③ --→|
|←-------- ④ --------→|
|________________________
Progress of Reaction
- ⑤ at the peak = Activated Complex (活化配合物)
- ① vertical arrow from reactants to peak = Activation Energy Forward (EA forward)
- ② vertical arrow from products to peak = Activation Energy Reverse (EA reverse)
- ③ horizontal arrow from reactants to products = ΔH Forward
- ④ horizontal arrow from products to reactants = ΔH Reverse
ℹ️ 答案页未提供此题答案，但这是标准势能图标注法。
Q7. The Enthalpy change (ΔH) for an exothermic reaction will always be (positive/negative)?
放热反应的焓变 (ΔH) 总是（正/负）？
Negative (负)
In an exothermic reaction, energy is released to the surroundings. The products have lower potential energy than the reactants, so ΔH = PE(products) - PE(reactants) < 0.
在放热反应中，能量释放到环境中。产物的势能低于反应物，所以 ΔH = PE(产物) - PE(反应物) < 0。
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Q8. The Enthalpy change (ΔH) for an endothermic reaction will always be (positive/negative)?
吸热反应的焓变 (ΔH) 总是（正/负）？
Positive (正)
In an endothermic reaction, energy is absorbed from the surroundings. The products have higher potential energy than the reactants, so ΔH = PE(products) - PE(reactants) > 0.
在吸热反应中，从环境中吸收能量。产物的势能高于反应物，所以 ΔH = PE(产物) - PE(反应物) > 0。
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Q9. The Activation Energy (EA) for ALL reactions will always be (positive/negative)?
所有反应的活化能 (EA) 总是（正/负）？
Positive (正)
Activation energy is always positive because energy must always be input to break bonds in the reactants before new bonds can form. The activated complex is always at a higher energy than both reactants and products.
活化能总是正的，因为在形成新化学键之前，必须输入能量来打破反应物中的化学键。活化配合物的能量总是高于反应物和产物。
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Q10. Use the graph in question 6 to answer the following questions? Show your work! Be sure to indicate + or - correctly.
使用第 6 题的图表回答以下问题。展示计算过程！确保正确标明正负号。
From the PE diagram in Q6, we can read approximate values: - Reactants: ~50 kJ - Activated Complex (peak): ~85 kJ - Products: ~10 kJ
根据第 6 题的势能图，我们可以读出近似值： - 反应物：~50 kJ - 活化配合物（峰值）：~85 kJ - 产物：~10 kJ
Q10a) What is the value of EA forward?¶
正向活化能的值是多少？
$$\text{EA forward} = \text{PE}{activated\ complex} - \text{PE}$$
$$= 85 - 50 = \boxed{+35 \text{ kJ}}$$
✅ Answer Key 答案: +35 kJ — 一致 ✓
Q10b) What is the value of ΔH forward?¶
正向焓变的值是多少？
$$\Delta H_{\text{forward}} = \text{PE}{products} - \text{PE}$$
$$= 10 - 50 = \boxed{-40 \text{ kJ}}$$
The negative sign indicates this is an exothermic reaction (releases energy).
负号表明这是放热反应（释放能量）。
✅ Answer Key 答案: -40 kJ — 一致 ✓
Q10c) What is the value of EA reverse?¶
逆向活化能的值是多少？
$$\text{EA reverse} = \text{PE}{activated\ complex} - \text{PE}$$
$$= 85 - 10 = \boxed{+75 \text{ kJ}}$$
✅ Answer Key 答案: +75 kJ — 一致 ✓
Q10d) What is the value of ΔH reverse?¶
逆向焓变的值是多少？
$$\Delta H_{\text{reverse}} = \text{PE}{reactants} - \text{PE}$$
$$= 50 - 10 = \boxed{+40 \text{ kJ}}$$
Note: ΔH reverse = -ΔH forward = -(-40) = +40 kJ ✓
注意：ΔH 逆 = -ΔH 正 = -(-40) = +40 kJ ✓
✅ Answer Key 答案: +40 kJ — 一致 ✓
Q10e) What is the energy of the Activated Complex?¶
活化配合物的能量是多少？
From the diagram, the peak of the curve is at:
从图表中，曲线峰值位于：
$$\boxed{85 \text{ kJ}}$$
✅ Answer Key 答案: 85 kJ — 一致 ✓
Q10f) Will the reactants or products have the strongest bonds? Explain how you know.¶
反应物还是产物具有更强的化学键？解释你如何知道。
Products will have the strongest bonds.
产物具有更强的化学键。
Explanation 解释:
Stronger bonds correspond to lower potential energy and greater stability. Since the products (PE = 10 kJ) have much lower potential energy than the reactants (PE = 50 kJ), the products are more stable and have stronger bonds.
更强的化学键对应于更低的势能和更高的稳定性。由于产物（PE = 10 kJ）的势能远低于反应物（PE = 50 kJ），因此产物更稳定，具有更强的化学键。
The energy released in an exothermic reaction (ΔH = -40 kJ) comes from the formation of stronger bonds in the products compared to the bonds broken in the reactants.
放热反应中释放的能量（ΔH = -40 kJ）来自于产物中形成的化学键比反应物中打破的化学键更强。
✅ Answer Key 答案: "products will have the strongest bonds because they have the least PE, therefore the most stable" — 一致 ✓
Drawing Potential Energy Diagrams — 绘制势能图¶
Q11. Use the following information to draw a potential energy diagram. You must first calculate ΔH forward and ΔH reverse. Fully label your diagram similar to question 6.
使用以下信息绘制势能图。你必须首先计算正向和逆向焓变。像第 6 题一样完整标注你的图表。
Given 已知： - EA Forward = +45 kJ - EA Reverse = +30 kJ - Activated Complex Energy = 70 kJ
Step 1: Calculate reactants energy 计算反应物能量
$$\text{EA forward} = \text{PE}{activated\ complex} - \text{PE}$$
$$45 = 70 - \text{PE}_{reactants}$$
$$\text{PE}_{reactants} = 70 - 45 = \boxed{25 \text{ kJ}}$$
Step 2: Calculate products energy 计算产物能量
$$\text{EA reverse} = \text{PE}{activated\ complex} - \text{PE}$$
$$30 = 70 - \text{PE}_{products}$$
$$\text{PE}_{products} = 70 - 30 = \boxed{40 \text{ kJ}}$$
Step 3: Calculate ΔH forward 计算正向焓变
$$\Delta H_{\text{forward}} = \text{PE}{products} - \text{PE}$$
$$= 40 - 25 = \boxed{+15 \text{ kJ}}$$
✅ Answer Key 答案: ΔHforward = +15 — 一致 ✓
Step 4: Calculate ΔH reverse 计算逆向焓变
$$\Delta H_{\text{reverse}} = \text{PE}{reactants} - \text{PE}$$
$$= 25 - 40 = \boxed{-15 \text{ kJ}}$$
Or: ΔH reverse = -ΔH forward = -(+15) = -15 kJ ✓
或：ΔH 逆 = -ΔH 正 = -(+15) = -15 kJ ✓
✅ Answer Key 答案: ΔHreverse = -15 — 一致 ✓
Step 5: Draw and label the diagram 绘制并标注图表
PE (kJ)
100 |
90 |
80 |
70 | ⑤ (70 kJ)
60 | /\
50 | / \
40 | / \____ ② (+30 kJ)
30 | ① (+45 kJ) PRODUCTS (40 kJ)
20 | / ↑
10 | / | ④ (-15 kJ)
0 |___/__________________
REACTANTS (25 kJ)
←------ ③ (+15 kJ) ----→
Progress of Reaction
Labels 标签： 1. EA Forward = +45 kJ (正向活化能) 2. EA Reverse = +30 kJ (逆向活化能) 3. ΔH Forward = +15 kJ (正向焓变) 4. ΔH Reverse = -15 kJ (逆向焓变) 5. Activated Complex = 70 kJ (活化配合物)
Q12. What is the energy of the reactants in question 11?
第 11 题中反应物的能量是多少？
From Step 1 above:
从上面的步骤 1：
$$\boxed{25 \text{ kJ}}$$
✅ Answer Key 答案: 25 kJ — 一致 ✓
Q13. What is the energy of the products in question 11?
第 11 题中产物的能量是多少？
From Step 2 above:
从上面的步骤 2：
$$\boxed{40 \text{ kJ}}$$
✅ Answer Key 答案: 40 kJ — 一致 ✓
Q14. Is the reaction described in question 11 exothermic or endothermic in the forward direction? Explain how you know.
第 11 题描述的反应在正向方向上是放热还是吸热？解释你如何知道。
The reaction is endothermic (吸热) in the forward direction.
Explanation 解释:
We know this because ΔH forward = +15 kJ is positive. A positive ΔH indicates that the products have higher potential energy than the reactants (40 kJ > 25 kJ), meaning energy must be absorbed from the surroundings for the reaction to proceed.
我们知道这一点是因为正向焓变 ΔH = +15 kJ 是正值。正的 ΔH 表明产物的势能高于反应物（40 kJ > 25 kJ），意味着反应进行时必须从环境中吸收能量。
In an endothermic reaction, energy is absorbed, so the system gains energy.
在吸热反应中，能量被吸收，所以体系获得能量。
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Q15. There are a few definitions of Enthalpy. In this course what should we mainly consider Enthalpy to be?
焓有几种定义。在本课程中，我们主要应该将焓视为什么？
In this course, Enthalpy (H) is mainly considered to be the potential energy stored in the chemical bonds of a substance.
在本课程中，焓 (H) 主要被视为储存在物质化学键中的势能。
More precisely: - Enthalpy change (ΔH) represents the energy change associated with breaking and forming chemical bonds - ΔH = H(products) - H(reactants) - It measures whether energy is released (ΔH < 0, exothermic) or absorbed (ΔH > 0, endothermic)
更确切地说： - 焓变 (ΔH) 代表与化学键断裂和形成相关的能量变化 - ΔH = H(产物) - H(反应物) - 它衡量能量是释放（ΔH < 0，放热）还是吸收（ΔH > 0，吸热）
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Identifying Exothermic vs Endothermic Reactions — 识别放热与吸热反应¶
Q16. State whether each of the following reactions is exothermic or endothermic in the forward direction.
判断以下每个反应在正向方向上是放热还是吸热。
| Reaction 反应 | Exo or Endo 放热/吸热 | Explanation 解释 |
|---|---|---|
| Ba(OH)₂ + 2 NH₄Cl + 430 kJ → BaCl₂ + 2 NH₄OH | Endothermic 吸热 | Energy appears on the reactant side (+ 430 kJ), meaning energy must be absorbed for the reaction to proceed. 能量出现在反应物一侧（+ 430 kJ），意味着反应进行需要吸收能量。 |
| 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 16 H₂O + 10110 kJ | Exothermic 放热 | Energy appears on the product side (+ 10110 kJ), meaning energy is released. This is combustion of octane (gasoline). 能量出现在产物一侧（+ 10110 kJ），意味着能量被释放。这是辛烷（汽油）的燃烧反应。 |
| SOCl₂ + CoCl₂•6H₂O → CoCl₂ + HCl + 6 SO₂ ΔH = +360 kJ |
Endothermic 吸热 | ΔH is positive (+360 kJ), meaning products have higher energy than reactants; energy is absorbed. ΔH 为正值（+360 kJ），意味着产物能量高于反应物；能量被吸收。 |
| Mn(s) + 2 HCl(aq) → MnCl₂(aq) + H₂(g) ΔH = -221 kJ |
Exothermic 放热 | ΔH is negative (-221 kJ), meaning products have lower energy than reactants; energy is released. ΔH 为负值（-221 kJ），意味着产物能量低于反应物；能量被释放。 |
Rules to remember 记住的规则: - Energy on reactant side OR positive ΔH → Endothermic (吸热) - Energy on product side OR negative ΔH → Exothermic (放热)
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Q17. If you place calcium in water a spontaneous reaction occurs. Would this be an exothermic or endothermic reaction?
如果将钙放入水中会发生自发反应。这是放热还是吸热反应？
This would be an exothermic (放热) reaction.
Explanation 解释:
The reaction is: Ca(s) + 2 H₂O(l) → Ca(OH)₂(aq) + H₂(g)
Spontaneous reactions that occur rapidly at room temperature are typically exothermic. The reaction of calcium with water is vigorous, produces heat, and can even ignite the hydrogen gas produced. These are all signs of an exothermic reaction.
在室温下快速发生的自发反应通常是放热的。钙与水的反应很剧烈，产生热量，甚至可以点燃产生的氢气。这些都是放热反应的标志。
ℹ️ 答案页未提供此题答案。
Comparing Reaction Rates — 比较反应速率¶
Q18. Determine which of the following reactions is faster and explain your reasoning.
判断以下哪个反应更快并解释你的推理。
a) BaCl₂(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + Ba(NO₃)₂(aq)
b) 2 Ag(s) + Zn(NO₃)₂(aq) → Zn(s) + 2 Ag(NO₃)₂(aq)
Answer 答案: Reaction (a) is faster.
Reasoning 推理:
- Reaction (a) is a homogeneous reaction — all reactants are in the aqueous phase (同相反应 — 所有反应物都在水相中)
- Reaction (b) is a heterogeneous reaction — reactants are in different phases (solid Ag and aqueous Zn(NO₃)₂) (非均相反应 — 反应物处于不同相态，固体 Ag 和水溶液 Zn(NO₃)₂)
Homogeneous reactions are generally faster because: - Reactant particles are evenly distributed throughout the solution - More surface area is available for collisions - Particles can collide freely in all directions
同相反应通常更快，因为： - 反应物粒子均匀分布在溶液中 - 有更多表面积可供碰撞 - 粒子可以在各个方向自由碰撞
In heterogeneous reactions, the reaction can only occur at the interface between phases (solid surface), which limits the reaction rate.
在非均相反应中，反应只能发生在相态之间的界面（固体表面），这限制了反应速率。
ℹ️ 答案页未提供此题答案。
Q19. Determine which of the following reactions is faster and explain your reasoning.
判断以下哪个反应更快并解释你的推理。
a) Cl₂(g) + Br₂(g) → 2 BrCl(g)
b) 3 H₂(g) + N₂(g) → 2 NH₃(g)
Answer 答案: Reaction (a) is faster.
Reasoning 推理:
Both reactions are homogeneous (all gases), but they differ in complexity:
两个反应都是同相的（都是气体），但它们在复杂性上有所不同：
-
Reaction (a): Only 2 molecules need to collide (Cl₂ + Br₂) — this is a simple binary collision.
反应 (a)：只需要 2 个分子碰撞（Cl₂ + Br₂）— 这是一个简单的二元碰撞。
-
Reaction (b): 4 molecules must collide simultaneously (3 H₂ + 1 N₂) — this is a multi-molecular collision.
反应 (b)：4 个分子必须同时碰撞（3 H₂ + 1 N₂）— 这是一个多分子碰撞。
The probability of a successful collision decreases dramatically as the number of molecules increases. A 2-molecule collision is far more likely than a 4-molecule collision occurring with correct orientation and sufficient energy.
随着分子数量的增加，成功碰撞的概率急剧下降。 2 个分子的碰撞比 4 个分子以正确取向和足够能量同时碰撞的可能性大得多。
In reality, reaction (b) proceeds through a multi-step mechanism rather than a single 4-molecule collision, which is why the Haber process requires a catalyst and high temperature/pressure.
实际上，反应 (b) 通过多步反应机理进行，而不是单一的 4 分子碰撞，这就是为什么哈伯法需要催化剂和高温高压。
ℹ️ 答案页未提供此题答案。
2.2 KINETIC ENERGY AND REACTION MECHANISMS — 动能与反应机理¶
Temperature and Reaction Rates — 温度与反应速率¶
Q20. What is the general rule regarding temperature increases and their effect on reaction rates?
关于温度升高及其对反应速率影响的一般规则是什么？
General Rule 一般规则:
For every 10°C increase in temperature, the reaction rate approximately doubles (or increases by a factor of 2).
温度每升高 10°C，反应速率大约翻倍（或增加为原来的 2 倍）。
This is sometimes called the "rule of thumb" or "Van 't Hoff rule."
这有时被称为"经验法则"或"范特霍夫规则"。
Why 为什么: - Higher temperature → particles have greater average kinetic energy - More particles have energy ≥ EA - More frequent collisions - Therefore, reaction rate increases
温度升高 → 粒子具有更大的平均动能 更多粒子的能量 ≥ EA 碰撞更频繁 因此，反应速率增加
ℹ️ 答案页未提供此题答案。
Q21. If a given reaction has a rate of 20 g/s and the temperature is increased by 20°C, what will the new reaction rate be?
如果给定反应的速率为 20 g/s，温度升高 20°C，新的反应速率是多少？
Using the rule: Rate doubles for every 10°C increase.
使用规则：温度每升高 10°C，速率翻倍。
- Temperature increase: 20°C
- Number of 10°C intervals: 20 ÷ 10 = 2
- Rate multiplier: 2² = 4
温度升高：20°C 10°C 间隔数：20 ÷ 10 = 2 速率倍增因子：2² = 4
$$\text{New rate} = \text{Original rate} \times 2^{(\Delta T/10°C)}$$
$$= 20 \text{ g/s} \times 2^{(20/10)}$$
$$= 20 \times 2^2$$
$$= 20 \times 4$$
$$= \boxed{80 \text{ g/s}}$$
✅ Answer Key 答案: 80 g/s — 一致 ✓
Reaction Mechanisms — 反应机理¶
Q22. What is the name given to the slowest step in a reaction mechanism?
反应机理中最慢的步骤叫什么名称？
The slowest step is called the rate-determining step (also called rate-limiting step).
最慢的步骤称为决速步骤（也称为限速步骤）。
Why it's important 为什么重要:
The rate-determining step controls the overall reaction rate because the reaction cannot proceed faster than its slowest step — similar to how the narrowest part of a funnel limits flow rate.
决速步骤控制整体反应速率，因为反应进行的速度不能快于最慢的步骤 — 类似于漏斗最窄的部分限制流速。
ℹ️ 答案页未提供此题答案。
Q23. Consider the following reaction mechanism:
考虑以下反应机理：
- Step 1: CH₄(g) + Cl₂(g) → CH₃(g) + HCl + Cl⁻(g) (slow)
- Step 2: CH₃(g) + Cl₂(g) → CH₃Cl(g) + Cl⁻(g) (fast)
Q23a) Write the equation for the overall reaction.¶
写出总反应方程式。
Method 方法: Add all steps and cancel species that appear on both sides (intermediates).
将所有步骤相加，并消去两侧都出现的物质（中间体）。
Step 1: CH₄(g) + Cl₂(g) → CH₃(g) + HCl(g) + Cl⁻(g)
Step 2: CH₃(g) + Cl₂(g) → CH₃Cl(g) + Cl⁻(g)
Sum: CH₄(g) + 2 Cl₂(g) + CH₃(g) → CH₃(g) + CH₃Cl(g) + HCl(g) + 2 Cl⁻(g)
Cancel CH₃(g) from both sides (it's an intermediate):
从两侧消去 CH₃(g)（它是中间体）：
$$\boxed{\text{CH}_4(g) + 2\text{ Cl}_2(g) \rightarrow \text{CH}_3\text{Cl}(g) + \text{HCl}(g) + 2\text{ Cl}^-(g)}$$
✅ Answer Key 答案: CH₄(g) + 2 Cl₂(g) → CH₃Cl(g) + HCl(g) + 2 Cl⁻(g) — 一致 ✓
Q23b) Identify the reaction intermediate.¶
识别反应中间体。
A reaction intermediate is a species that is produced in one step and consumed in a later step — it does not appear in the overall equation.
反应中间体是在一个步骤中生成并在后续步骤中消耗的物质 — 它不出现在总反应方程式中。
Looking at the mechanism: - CH₃(g) is produced in Step 1 - CH₃(g) is consumed in Step 2 - CH₃(g) does not appear in the overall equation
查看反应机理： - CH₃(g) 在步骤 1 中生成 - CH₃(g) 在步骤 2 中消耗 - CH₃(g) 不出现在总反应方程式中
$$\boxed{\text{CH}_3(g)}$$
✅ Answer Key 答案: CH₃(g) — 一致 ✓
Q23c) Which step is the rate determining step?¶
哪一步是决速步骤？
The rate-determining step is the slowest step.
决速步骤是最慢的步骤。
$$\boxed{\text{Step 1}}$$
✅ Answer Key 答案: Step 1 — 一致 ✓
Q24. Consider the following reaction mechanism:
考虑以下反应机理：
- Step 1: C₃H₆O + H₃O⁺ → C₃H₇O⁺ + H₂O (fast)
- Step 2: C₃H₇O⁺ + H₂O → C₃H₅OH + H₃O⁺ (slowest)
- Step 3: C₃H₅OH + Br₂ → C₃H₅OBr + HBr (slow)
Q24a) Write the equation for the overall reaction.¶
写出总反应方程式。
Add all steps:
相加所有步骤：
Step 1: C₃H₆O + H₃O⁺ → C₃H₇O⁺ + H₂O
Step 2: C₃H₇O⁺ + H₂O → C₃H₅OH + H₃O⁺
Step 3: C₃H₅OH + Br₂ → C₃H₅OBr + HBr
Sum: C₃H₆O + H₃O⁺ + C₃H₇O⁺ + H₂O + C₃H₅OH + Br₂ →
C₃H₇O⁺ + H₂O + C₃H₅OH + H₃O⁺ + C₃H₅OBr + HBr
Cancel intermediates and catalysts: - C₃H₇O⁺ (intermediate) — both sides - C₃H₅OH (intermediate) — both sides - H₂O (appears once on each side) — cancel - H₃O⁺ (catalyst) — both sides
消去中间体和催化剂： - C₃H₇O⁺（中间体）— 两侧都有 - C₃H₅OH（中间体）— 两侧都有 - H₂O（每侧各出现一次）— 消去 - H₃O⁺（催化剂）— 两侧都有
$$\boxed{\text{C}_3\text{H}_6\text{O} + \text{Br}_2 \rightarrow \text{C}_3\text{H}_5\text{OBr} + \text{HBr}}$$
ℹ️ 答案页未提供此题答案。
Q24b) Identify the reaction intermediate(s).¶
识别反应中间体。
Intermediates are produced in one step and consumed in another:
中间体在一个步骤中生成并在另一个步骤中消耗：
- C₃H₇O⁺: produced in Step 1, consumed in Step 2
- C₃H₅OH: produced in Step 2, consumed in Step 3
$$\boxed{\text{C}_3\text{H}_7\text{O}^+ \text{ and } \text{C}_3\text{H}_5\text{OH}}$$
ℹ️ 答案页未提供此题答案。
Q24c) Identify the catalyst(s).¶
识别催化剂。
A catalyst is a species that: - Appears in an early step as a reactant - Is regenerated in a later step as a product - Appears on both sides but is not consumed overall
催化剂是这样的物质： - 在早期步骤中作为反应物出现 - 在后续步骤中作为产物再生 - 出现在两侧但总体上不被消耗
Looking at the mechanism: - H₃O⁺: consumed in Step 1, regenerated in Step 2
查看反应机理： - H₃O⁺：在步骤 1 中消耗，在步骤 2 中再生
$$\boxed{\text{H}_3\text{O}^+}$$
ℹ️ 答案页未提供此题答案。
Q24d) Which step is the rate determining step?¶
哪一步是决速步骤？
The rate-determining step is labeled "slowest" in the mechanism.
决速步骤在机理中标记为"最慢"。
$$\boxed{\text{Step 2}}$$
ℹ️ 答案页未提供此题答案。
Q25. Explain the difference between an activated complex and an intermediate in terms of their stability.
从稳定性角度解释活化配合物和中间体的区别。
| Feature 特征 | Activated Complex 活化配合物 | Intermediate 中间体 |
|---|---|---|
| Stability 稳定性 | Extremely unstable (exists for ~10⁻¹³ seconds) 极其不稳定（存在约 10⁻¹³ 秒） | Relatively stable (can sometimes be isolated) 相对稳定（有时可以分离） |
| Energy 能量 | At the highest point on the PE diagram (peak) 在势能图的最高点（峰值） | At a local minimum (valley) between peaks 在峰值之间的局部最低点（谷底） |
| Location 位置 | At the transition state (top of energy barrier) 在过渡态（能垒顶部） | Between elementary steps in a mechanism 在反应机理的基元步骤之间 |
| Bonds 化学键 | Partially broken and partially formed bonds 部分断裂和部分形成的化学键 | Fully formed bonds (complete molecule) 完全形成的化学键（完整分子） |
| Detection 检测 | Cannot be isolated or detected directly 无法分离或直接检测 | Can sometimes be detected or isolated 有时可以被检测或分离 |
Summary 总结:
-
Activated complex is a fleeting, high-energy, unstable arrangement at the peak of the energy barrier that immediately breaks apart to form products or revert to reactants.
活化配合物是在能垒峰值处的短暂、高能、不稳定的排列，立即分解形成产物或恢复为反应物。
-
Intermediate is a relatively stable species formed between steps of a multi-step reaction that exists long enough to be a distinct chemical entity.
中间体是在多步反应步骤之间形成的相对稳定的物质，存在时间足够长，可以作为一个独特的化学实体。
ℹ️ 答案页未提供此题答案。
Catalysts — 催化剂¶
Q26. When comparing a potential energy diagram for a catalyzed reaction to an uncatalyzed reaction, what value will always decrease and what value will always stay the same?
比较催化反应和非催化反应的势能图时，哪个值总是减小，哪个值总是保持不变？
Will always DECREASE 总是减小:
$$\boxed{\text{Activation Energy (EA)}}$$
A catalyst provides an alternative reaction pathway with a lower activation energy. The catalyst lowers the energy barrier, allowing more molecules to have sufficient energy to react at a given temperature.
催化剂提供一条替代反应路径，具有更低的活化能。催化剂降低能垒，使更多分子在给定温度下具有足够的能量进行反应。
Will always stay the SAME 总是保持不变:
$$\boxed{\text{Enthalpy Change } (\Delta H)}$$
The catalyst does not change the energy difference between reactants and products. It only affects the pathway (activation energy) but not the starting or ending points. ΔH depends only on the identity of reactants and products, not on the reaction pathway.
催化剂不改变反应物和产物之间的能量差。它只影响反应路径（活化能），而不影响起点或终点。ΔH 只取决于反应物和产物的性质，而不取决于反应路径。
Comparison 比较:
PE Uncatalyzed Catalyzed
| /\ /\
| / \ / \
| / ① \ / ② \
| / \ / \
| ___/ \___ ___/ \___
| R P R P
| ←----- ΔH ----→ ←----- ΔH ----→
| (same 相同)
| ① EA (high)
| ② EA (lower)
ℹ️ 答案页未提供此题答案。
Q27. Consider the following reaction where nitrogen monoxide reacts with hydrogen gas to create nitrogen gas and water.
考虑以下一氧化氮与氢气反应生成氮气和水的反应。
Overall reaction 总反应: 2 NO + 2 H₂ → N₂ + 2 H₂O
Given mechanism 给定机理: - Step 1: 2 NO → N₂O₂ (fast) - Step 2: ____ (slow) - Step 3: N₂O + H₂ → N₂ + H₂O (very fast)
Q27a) Determine Step 2 of the reaction mechanism.¶
确定反应机理的步骤 2。
Method 方法: Use the overall equation and known steps to determine what Step 2 must be.
使用总反应方程式和已知步骤来确定步骤 2 必须是什么。
Step 1: 2 NO → N₂O₂
Step 2: ?
Step 3: N₂O + H₂ → N₂ + H₂O
Overall: 2 NO + 2 H₂ → N₂ + 2 H₂O
Analyzing what's needed 分析所需内容:
- Overall needs 2 H₂ consumed, but Step 3 only uses 1 H₂
- Step 2 must consume 1 H₂ (so total = 2 H₂) ✓
- Step 1 produces N₂O₂, which must be consumed
- Step 3 requires N₂O as input
- Therefore, Step 2 must: N₂O₂ + H₂ → ? + ?
总反应需要消耗 2 H₂，但步骤 3 只使用 1 H₂ 步骤 2 必须消耗 1 H₂（总共 = 2 H₂）✓ 步骤 1 产生 N₂O₂，必须被消耗 步骤 3 需要 N₂O 作为输入 因此，步骤 2 必须是：N₂O₂ + H₂ → ? + ?
To produce N₂O for Step 3 要为步骤 3 产生 N₂O:
$$\boxed{\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}}$$
Verification 验证: Add all steps:
Step 1: 2 NO → N₂O₂
Step 2: N₂O₂ + H₂ → N₂O + H₂O
Step 3: N₂O + H₂ → N₂ + H₂O
Sum: 2 NO + 2 H₂ → N₂ + 2 H₂O ✓
(N₂O₂ and N₂O cancel as intermediates)
✅ Answer Key 答案: N₂O₂ + H₂ → N₂O + H₂O — 一致 ✓
Q27b) Identify the reaction intermediate(s).¶
识别反应中间体。
Intermediates are produced in one step and consumed in another:
中间体在一个步骤中生成并在另一个步骤中消耗：
- N₂O₂: produced in Step 1, consumed in Step 2
- N₂O: produced in Step 2, consumed in Step 3
$$\boxed{\text{N}_2\text{O}_2 \text{ and } \text{N}_2\text{O}}$$
✅ Answer Key 答案: N₂O₂ and N₂O — 一致 ✓
Q27c) Which step is the rate determining step?¶
哪一步是决速步骤？
The rate-determining step is the slowest step.
决速步骤是最慢的步骤。
$$\boxed{\text{Step 2}}$$
✅ Answer Key 答案: Step 2 — 一致 ✓
Answer Key Summary — 答案汇总¶
| Q# | Answer 答案 | Match 一致 |
|---|---|---|
| 2.1 Collision Theory and Heat of Reaction | ||
| 10a | +35 kJ | ✅ |
| 10b | -40 kJ | ✅ |
| 10c | +75 kJ | ✅ |
| 10d | +40 kJ | ✅ |
| 10e | 85 kJ | ✅ |
| 10f | products (strongest bonds, least PE, most stable) | ✅ |
| 11 | ΔHforward = +15 kJ; ΔHreverse = -15 kJ | ✅ |
| 12 | 25 kJ | ✅ |
| 13 | 40 kJ | ✅ |
| 2.2 Kinetic Energy and Reaction Mechanisms | ||
| 21 | 80 g/s | ✅ |
| 23a | CH₄(g) + 2 Cl₂(g) → CH₃Cl(g) + HCl(g) + 2 Cl⁻(g) | ✅ |
| 23b | CH₃(g) | ✅ |
| 23c | Step 1 | ✅ |
| 27a | N₂O₂ + H₂ → N₂O + H₂O | ✅ |
| 27b | N₂O₂ and N₂O | ✅ |
| 27c | Step 2 | ✅ |
所有提供答案的题目全部一致！All answers match! ✅
Notes 注释¶
Questions without Answer Key 无答案键的题目¶
The following questions did not have answers provided in the answer key. The solutions above are based on standard Chemistry 12 principles:
以下问题在答案键中未提供答案。上述解答基于标准化学 12 年级原理：
- Q1-Q9: Conceptual questions about collision theory, activation energy, and enthalpy
- Q14-Q20: Conceptual questions about exothermic/endothermic reactions and temperature effects
- Q22: Definition of rate-determining step
- Q24a-d: Reaction mechanism analysis
- Q25-Q26: Conceptual questions about intermediates, activated complexes, and catalysts
Terminology 术语¶
| English 英文 | Chinese 中文 |
|---|---|
| Activation Energy (EA) | 活化能 |
| Activated Complex | 活化配合物 / 过渡态 |
| Enthalpy (H) | 焓 |
| Enthalpy Change (ΔH) | 焓变 |
| Exothermic | 放热 |
| Endothermic | 吸热 |
| Potential Energy (PE) | 势能 |
| Kinetic Energy (KE) | 动能 |
| Reaction Rate | 反应速率 |
| Reaction Mechanism | 反应机理 |
| Rate-Determining Step | 决速步骤 / 限速步骤 |
| Intermediate | 中间体 |
| Catalyst | 催化剂 |
| Collision Theory | 碰撞理论 |
Document completed. All answers verified against answer key with 100% match rate.
文档完成。所有答案已与答案键核对，100% 匹配率。