Chemistry 12 — Unit 3 Learning Guide 详细解答¶
状态: ✅ 已完成
创建日期: 2026-02-18
最后更新: 2026-02-28
3.1 EQUILIBRIUM — 化学平衡¶
Q1¶
EN: When does equilibrium occur?
中文翻译： 化学平衡何时发生？
Answer 答案:
EN: Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a closed system. At this point, the concentrations of all reactants and products remain constant over time (though both reactions continue — dynamic equilibrium).
中文： 当封闭体系中正向反应速率等于逆向反应速率时，达到化学平衡。此时，所有反应物和产物的浓度随时间保持恒定（但两个反应仍在持续进行——动态平衡）。
ℹ️ 答案页未提供此题答案。
Q2¶
EN: Give an example of two physical processes that oppose each other during equilibrium?
中文翻译： 举例说明平衡过程中两个相互对立的物理过程。
Answer 答案:
EN: Evaporation and condensation of water in a sealed container.
- In a sealed bottle half-filled with water, water molecules at the surface evaporate (liquid → gas)
- At the same time, water vapour molecules hit the surface and condense (gas → liquid)
- At equilibrium, the rate of evaporation = rate of condensation
- The water level and vapour pressure remain constant
Other examples: dissolving and crystallization of a solute in saturated solution.
中文： 密封容器中水的蒸发与凝结。
半满水的密封瓶中，表面水分子蒸发（液→气），同时水蒸气分子撞击表面凝结（气→液）。平衡时，蒸发速率 = 凝结速率，水面高度和蒸气压保持恒定。
ℹ️ 答案页未提供此题答案。
Q3¶
EN: Complete the ICE chart: 5.0 moles of N₂ and 10.0 moles of H₂ are placed into a 1.00 L flask. At equilibrium, 3.6 moles of NH₃ are present.
中文翻译： 完成 ICE 表格：5.0 摩尔 N₂ 和 10.0 摩尔 H₂ 放入 1.00 L 烧瓶中。平衡时有 3.6 摩尔 NH₃。
Reaction 反应： N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
Step 1: Convert to molarity（Volume = 1.00 L, so M = mol）
Step 2: Find change. NH₃ goes from 0 → 3.6 M, so C(NH₃) = +3.6. By ratio: C(N₂) = −3.6/2 = −1.8; C(H₂) = −3(3.6/2) = −5.4
| N₂(g) | + | 3H₂(g) | ⇌ | 2 NH₃(g) | |
|---|---|---|---|---|---|
| [I] | 5.0 | 10.0 | 0 | ||
| [C] | −1.8 | −5.4 | +3.6 | ||
| [E] | 3.2 | 4.6 | 3.6 |
✅ Answer Key 答案: N₂=3.2 M, H₂=4.6 M, NH₃=3.6 M — 一致 ✓
Q4¶
EN: Complete the ICE chart: Unknown amounts of CO and O₂ are placed into a 2.00 L flask. At equilibrium: 3.2 mol CO, 4.6 mol O₂, 1.2 mol CO₂.
中文翻译： 完成 ICE 表格：未知量的 CO 和 O₂ 放入 2.00 L 烧瓶。平衡时：3.2 mol CO、4.6 mol O₂、1.2 mol CO₂。
Reaction 反应： 2 CO(g) + O₂(g) ⇌ 2 CO₂(g)
Step 1: Convert to molarity (÷ 2.00 L)：CO = 1.6 M, O₂ = 2.3 M, CO₂ = 0.6 M
Step 2: CO₂ went from 0 → 0.6 M, so C(CO₂) = +0.6. By ratio: C(CO) = −0.6; C(O₂) = −0.6/2 = −0.3
| 2 CO(g) | + | O₂(g) | ⇌ | 2 CO₂(g) | |
|---|---|---|---|---|---|
| [I] | 2.2 | 2.6 | 0 | ||
| [C] | −0.6 | −0.3 | +0.6 | ||
| [E] | 1.6 | 2.3 | 0.6 |
✅ Answer Key 答案: CO=1.6 M, O₂=2.3 M, CO₂=0.6 M — 一致 ✓
Q5¶
EN: Fill in the Le Chatelier's Principle chart. Memorize!
中文翻译： 填写勒夏特列原理表格。记住！
| Disturbance 干扰 | Counteraction 反应 | Shift 移动方向 |
|---|---|---|
| Add Reactant 添加反应物 | Consume that Reactant 消耗该反应物 | Right 向右 |
| Remove Reactant 移除反应物 | Replace that Reactant 补充该反应物 | Left 向左 |
| Add Product 添加产物 | Consume that Product 消耗该产物 | Left 向左 |
| Remove Product 移除产物 | Make more Product 生成更多产物 | Right 向右 |
| Increase Temperature 升高温度 | Absorb Heat 吸收热量 | Endothermic Direction 吸热方向 |
| Decrease Temperature 降低温度 | Release Heat 释放热量 | Exothermic Direction 放热方向 |
| Increase Pressure (↓V) 增大压力 | Decrease Pressure 减小压力 | To side with fewest gas molecules 气体分子数少的一侧 |
| Decrease Pressure (↑V) 减小压力 | Increase Pressure 增大压力 | To side with most gas molecules 气体分子数多的一侧 |
✅ Answer Key — 一致 ✓
Q6¶
EN: Given: C₃H₈(l) + 5 O₂(g) ⇌ 3 CO₂(g) + 4 H₂O(g) + energy. Determine the direction of shift for each stress and give a reason.
中文翻译： 判断每种应力的平衡移动方向并给出理由。
注意：左侧 5 mol 气体，右侧 7 mol 气体（C₃H₈ 是液态不计）
| Stress 应力 | Shift 移动 | Reason 理由 |
|---|---|---|
| Increase temperature 升温 | Left 向左 | Energy is a product (exothermic). Increasing T → system absorbs heat → shift LEFT (endothermic direction). 能量在产物侧，升温使系统吸热向左移。 |
| Removal of CO₂ 移除CO₂ | Right 向右 | Product removed → system shifts RIGHT to replace CO₂. 产物被移除 → 系统向右移补充 CO₂。 |
| Decrease Pressure (↑V) 降压 | Right 向右 | Products side has MORE gas molecules (7 vs 5) → system shifts to side with more moles = RIGHT. 产物侧气体分子更多（7 vs 5）→ 向右移。 |
| Add a catalyst 加催化剂 | None 无移动 | Catalyst speeds up both forward and reverse reactions equally. No shift. 催化剂同等加速正逆反应，不移动平衡。 |
| Addition of O₂ 添加O₂ | Right 向右 | Reactant added → system shifts RIGHT to consume excess O₂. 添加反应物 → 向右移消耗多余 O₂。 |
✅ Answer Key 答案: Left, Right, Right, None, Right — 一致 ✓
Q7¶
EN: Cl₂(g) + F₂(g) ⇌ 2ClF(g), ΔH = −56.5 kJ. Colors: Cl₂=green, F₂=brown, ClF=colorless.
中文翻译： 考虑反应（Cl₂绿色，F₂棕色，ClF无色）。
a) Heat it up — will the test tube become darker, lighter, or not change?
加热——试管会变深、变浅还是不变？
Answer: Darker 变深。 Reaction is exothermic (ΔH = −56.5 kJ). Heating shifts LEFT (endothermic direction). More Cl₂ (green) and F₂ (brown) are produced → colour becomes darker.
反应放热，升温使平衡向左（吸热方向）移动。更多 Cl₂（绿色）和 F₂（棕色）生成 → 颜色变深。
b) Add a catalyst?
加入催化剂？
Answer: No change 不变。 Catalyst does not change equilibrium position, only speeds up attainment. 催化剂不改变平衡位置。
c) Increase the volume?
增大体积？
Answer: No change 不变。 Left side: 2 moles gas (Cl₂ + F₂). Right side: 2 moles gas (2 ClF). Equal moles on both sides → volume/pressure change has no effect. 两侧气体分子数相等（2 = 2）→ 体积变化无影响。
ℹ️ 答案页未提供此题答案。
Q8¶
EN: SOCl₂(aq) + CoCl₂•6H₂O(aq) ⇌ CoCl₂(aq) + HCl(aq) + 6 SO₂(g), ΔH = +360 kJ
中文翻译： 起始浓度 SOCl₂=10M, CoCl₂•6H₂O=8M, CoCl₂=6M, HCl=4M, SO₂=2M
a) Add HCl → ?
Answer: Shift LEFT 向左. HCl is a product; adding product shifts equilibrium LEFT to consume it. HCl 是产物，添加产物使平衡向左移。
b) Heat up → ?
Answer: Shift RIGHT 向右. ΔH = +360 kJ (endothermic). Heat acts as a "reactant." Adding heat shifts toward the endothermic direction = RIGHT. 反应吸热，加热使平衡向右（吸热方向）移动。
c) Decrease pressure by increasing volume → ?
Answer: Shift RIGHT 向右. Product side has 6 moles of gas (SO₂), reactant side has 0 moles of gas (all aqueous). Decreasing pressure shifts to side with MORE gas molecules = RIGHT. 产物侧 6 摩尔气体 vs 反应物侧 0 摩尔气体。减压向气体分子数多的一侧（右）移动。
d) Sketch graph — 草图说明：
三次应力后浓度变化趋势： - 加 HCl 后：[HCl]↑ then ↓, [SOCl₂]↑, [CoCl₂•6H₂O]↑, [CoCl₂]↓, [SO₂]↓ - 加热后：[SOCl₂]↓, [CoCl₂•6H₂O]↓, [CoCl₂]↑, [HCl]↑, [SO₂]↑ - 增大体积后：所有浓度先因稀释而下降，然后 [SO₂]↑ 方向调整，最终产物侧增大
ℹ️ 答案页未提供此题答案。
Q9¶
EN: 2 CO(g) + O₂(g) ⇌ 2CO₂(g) + 210 kJ. State the disturbance and shift at each time point on the graph.
中文翻译： 说明图表上每个时间点的干扰和移动方向。
| Time 时间 | Disturbance 干扰 | Shift 移动 |
|---|---|---|
| Time 1 | CO₂ is added 添加 CO₂ | Left 向左 |
| Time 2 | Pressure increase or Temperature decrease 增大压力或降低温度 | Right 向右 |
| Time 3 | CO is removed 移除 CO | Left 向左 |
Explanation for Time 2 解释： At Time 2, CO and O₂ decrease while CO₂ increases (shift right). Since all species change gradually (no sudden jump), this is either a pressure increase (3 moles gas on left → 2 on right, shifts toward fewer moles = RIGHT) or a temperature decrease (exothermic, shifts toward products = RIGHT).
✅ Answer Key 答案: Time 1=CO₂ added/Left, Time 2=Pressure increase or Temperature decrease/Right, Time 3=CO removed/Left — 一致 ✓
3.2 THE EQUILIBRIUM LAW — 平衡定律¶
Q10¶
EN: Write the equilibrium constant equation for: aA + bB ⇌ xX + yY
中文翻译： 写出平衡常数表达式。
$$K_{eq} = \frac{[X]^x[Y]^y}{[A]^a[B]^b}$$
Products over reactants, each raised to the power of its coefficient. 产物在分子，反应物在分母，各取系数次幂。
ℹ️ 答案页未提供此题答案。
Q11¶
EN: What number do we use to represent the concentration of a solid, liquid or solvent?
中文翻译： 固体、液体或溶剂的浓度用什么数字表示？
Answer: 1. Pure solids, liquids, and solvents are NOT included in the Keq expression (their concentrations are effectively constant and incorporated into Keq). 纯固体、液体和溶剂不写入 Keq 表达式中（浓度恒定，视为 1）。
ℹ️ 答案页未提供此题答案。
Q12¶
EN: If you have a small Keq, do you have more reactants or more products?
中文翻译： 如果 Keq 很小，反应物还是产物更多？
Answer: More reactants 反应物. A small Keq means the denominator (reactants) is much larger than the numerator (products), so equilibrium lies to the LEFT. 小 Keq 意味着分母（反应物）远大于分子（产物），平衡偏左。
ℹ️ 答案页未提供此题答案。
Q13¶
EN: What are the three cases in which Keq will change?
中文翻译： Keq 会改变的三种情况是什么？
- Temperature change 温度变化
- Reverse the reaction 反转反应（Keq 变为倒数）
- Change the stoichiometric coefficients 改变化学计量系数（Keq 取相应次幂）
ℹ️ 答案页未提供此题答案。
Q14¶
EN: If the Keq of a given forward reaction is m/n, what is the Keq of the reverse reaction?
中文翻译： 如果正向反应的 Keq 为 m/n，逆向反应的 Keq 是多少？
$$K_{eq(\text{reverse})} = \frac{1}{K_{eq(\text{forward})}} = \boxed{\frac{n}{m}}$$
ℹ️ 答案页未提供此题答案。
Q15¶
EN: Mg + 2HCl ⇌ MgCl₂ + H₂, Keq = 5.0×10⁻². Find Keq for: 3MgCl₂ + 3H₂ ⇌ 3Mg + 6HCl
中文翻译： 已知正向 Keq = 5.0×10⁻²，求逆向乘以 3 后的 Keq。
Step 1: The new reaction is the REVERSE of the original multiplied by 3.
$$\text{Reverse: } K' = \frac{1}{5.0 \times 10^{-2}} = 20$$
Step 2: Multiply coefficients by 3 → raise Keq to the power of 3:
$$K_{new} = (K')^3 = 20^3 = \boxed{8000}$$
✅ Answer Key 答案: 8000 — 一致 ✓
Q16¶
EN: What temperature scale must be used in thermodynamic and equilibrium calculations?
中文翻译： 热力学和平衡计算中必须使用什么温度标度？
Answer 答案:
EN: Kelvin (K). Always convert °C to K (K = °C + 273) for thermodynamic calculations.
中文： 开尔文 (K)。热力学计算中需将摄氏度转换为开尔文（K = °C + 273）。
ℹ️ 答案页未提供此题答案。
Q17¶
EN: A reaction occurs in which the entropy of the system decreases (ΔS < 0). Why is this still possible?
中文翻译： 某反应使体系熵减少（ΔS < 0），为什么该反应仍然可能发生？
Answer 答案:
EN: The entropy of the surroundings must increase to compensate. The Second Law of Thermodynamics states that the total entropy of the universe must always increase (ΔS_universe ≥ 0). If ΔS_system < 0, then ΔS_surroundings must be large enough so that ΔS_universe ≥ 0.
中文： 环境的熵必须增加以补偿。热力学第二定律要求宇宙总熵始终增加（ΔS_宇宙 ≥ 0）。若 ΔS_体系 < 0，则 ΔS_环境 必须足够大以保证 ΔS_宇宙 ≥ 0。
ℹ️ 答案页未提供此题答案。
Q18¶
EN: State three events (changes) that would increase the entropy of a system.
中文翻译： 列举三种能增加体系熵的变化。
Answer 答案:
- Increase temperature 升高温度
- Increase number of moles of gas 增加气体摩尔数
- Phase change toward more disorder 向更无序的相态转变 (solid → liquid → gas)
ℹ️ 答案页未提供此题答案。
Q19¶
EN: Reactions naturally proceed towards _ (minimum/maximum) enthalpy and _ (minimum/maximum) entropy.
中文翻译： 反应自然趋向于__（最低/最高）焓和____（最低/最高）熵。
Answer 答案:
EN: Reactions go towards lower (minimum) enthalpy and higher (maximum) entropy.
中文： 反应趋向最低焓和最高熵。
ℹ️ 答案页未提供此题答案。
Q20¶
EN: Based on the drive toward minimum enthalpy: does an exothermic reaction favour the reactants or the products? What about an endothermic reaction?
中文翻译： 基于趋向最低焓的驱动力：放热反应有利于反应物还是产物？吸热反应呢？
Answer 答案:
EN: Exothermic: minimum enthalpy favours the products (energy is released → products are lower in enthalpy → more stable). Endothermic: minimum enthalpy favours the reactants (energy is absorbed → products are higher in enthalpy → less stable).
中文： 放热反应：最低焓有利于产物（能量释放 → 产物焓更低 → 更稳定）。 吸热反应：最低焓有利于反应物（能量吸收 → 产物焓更高 → 不稳定）。
ℹ️ 答案页未提供此题答案。
Q21¶
EN: For each reaction, determine if ΔS and ΔH favour R or P, and the result (C/E/NR).
中文翻译： 判断每个反应的 ΔS 和 ΔH 是否有利于反应物 (R) 或产物 (P)，以及结果。
Rules 规则: - Both favour P → Completion (C) 完成 - Both favour R → No Reaction (NR) 不反应 - One each → Equilibrium (E) 平衡
| Reaction 反应 | ΔS | ΔH | Result 结果 | Reasoning 推理 |
|---|---|---|---|---|
| 2C₈H₁₈(l) + 25O₂(g) ⇌ 16CO₂(g) + 16H₂O(g) + 10110 kJ | P | P | C | Gas moles: 25→32 (↑entropy→P). Exothermic (↓enthalpy→P). Both favour P → Completion. 气体增多→P，放热→P。 |
| 6SOCl₂(s) + CoCl₂•6H₂O(aq) ⇌ CoCl₂(aq) + HCl(aq) + 6SO₂(g), ΔH=+360 kJ | P | R | E | Gas produced (0→6 mol)→P. Endothermic→R. Mixed → Equilibrium. 产生气体→P，吸热→R。 |
| 2BiCl₃(aq) + 3H₂S(g) + 510 kJ ⇌ Bi₂S₃(s) + 6HCl(aq) | R | R | NR | Gas consumed (3→0), solid formed→R. Endothermic→R. Both R → No Reaction. 气体消耗→R，吸热→R。 |
| AgNO₃(aq) + NaCl(aq) ⇌ AgCl(s) + NaNO₃(aq) + 315 kJ | R | P | E | Solid formed→R. Exothermic→P. Mixed → Equilibrium. 固体形成→R，放热→P。 |
✅ Answer Key 答案 (Row 1): ΔS=P, ΔH=P, C — 一致 ✓
ℹ️ 答案页仅提供第一行答案，其余为标准解法。
Q22¶
EN: Ca(OH)₂(aq) + 2CO₂(g) ⇌ Ca(HCO₃)₂(aq). Find Keq at 0°C.
中文翻译： 已知 0°C 时平衡浓度如下，求 Keq。
[Ca(OH)₂] = 1.25 M, [CO₂] = 0.563 M, [Ca(HCO₃)₂] = 0.0578 M
$$K_{eq} = \frac{[\text{Ca(HCO}_3)_2]}{[\text{Ca(OH)}_2][\text{CO}_2]^2} = \frac{0.0578}{(1.25)(0.563)^2}$$
$$= \frac{0.0578}{(1.25)(0.317)} = \frac{0.0578}{0.396} = \boxed{0.146}$$
✅ Answer Key 答案: 0.146 — 一致 ✓
Q23¶
EN: 2H₂O₂(l) ⇌ 2H₂O(l) + O₂(g). [O₂] = 0.200 M. Find Keq.
中文翻译： 已知平衡时 [O₂] = 0.200 M，求 Keq。
The important fact: H₂O₂ and H₂O are liquids → their concentrations are NOT included in the Keq expression (= 1).
重要事实：H₂O₂ 和 H₂O 是液体 → 不写入 Keq 表达式。
$$K_{eq} = [\text{O}_2] = \boxed{0.200}$$
ℹ️ 答案页未提供此题答案。
Q24¶
EN: H₂(g) + Br₂(g) ⇌ 2HBr(g), Keq = 0.0204. 3.00 L vessel: 4.78×10⁻³ mol H₂, 0.0643 mol HBr. Find mol Br₂.
中文翻译： 已知 Keq = 0.0204，在 3.00 L 容器中有 4.78×10⁻³ mol H₂ 和 0.0643 mol HBr，求平衡时 Br₂ 的摩尔数。
Step 1: Convert to molarity:
$$[\text{H}_2] = \frac{4.78 \times 10^{-3}}{3.00} = 1.593 \times 10^{-3} \text{ M}$$
$$[\text{HBr}] = \frac{0.0643}{3.00} = 0.02143 \text{ M}$$
Step 2: Solve for [Br₂]:
$$K_{eq} = \frac{[\text{HBr}]^2}{[\text{H}_2][\text{Br}_2]}$$
$$0.0204 = \frac{(0.02143)^2}{(1.593 \times 10^{-3})[\text{Br}_2]}$$
$$[\text{Br}_2] = \frac{(0.02143)^2}{(0.0204)(1.593 \times 10^{-3})} = \frac{4.593 \times 10^{-4}}{3.250 \times 10^{-5}} = 14.1 \text{ M}$$
Step 3: Convert back to moles:
$$n_{\text{Br}_2} = 14.1 \times 3.00 = \boxed{42.3 \text{ mol}}$$
✅ Answer Key 答案: 42.3 moles Br₂ — 一致 ✓
Q25¶
EN: 3NO₂(g) ⇌ N₂O₅(g) + NO(g), Keq = 7.31×10⁻⁴. 2.00 L vessel: 0.0415 mol N₂O₅, 0.0273 mol NO. Find mol NO₂.
中文翻译： 已知 Keq = 7.31×10⁻⁴，在 2.00 L 容器中有 0.0415 mol N₂O₅ 和 0.0273 mol NO，求平衡时 NO₂ 的摩尔数。
Step 1: Convert:
$$[\text{N}_2\text{O}_5] = \frac{0.0415}{2.00} = 0.02075 \text{ M}; \quad [\text{NO}] = \frac{0.0273}{2.00} = 0.01365 \text{ M}$$
Step 2: Solve:
$$K_{eq} = \frac{[\text{N}_2\text{O}_5][\text{NO}]}{[\text{NO}_2]^3}$$
$$[\text{NO}_2]^3 = \frac{(0.02075)(0.01365)}{7.31 \times 10^{-4}} = \frac{2.832 \times 10^{-4}}{7.31 \times 10^{-4}} = 0.3874$$
$$[\text{NO}_2] = \sqrt[3]{0.3874} = 0.729 \text{ M}$$
$$n_{\text{NO}_2} = 0.729 \times 2.00 = \boxed{1.46 \text{ mol}}$$
✅ Answer Key 答案: 1.46 mols NO₂ — 一致 ✓
Q26¶
EN: 2O₂(g) + N₂(g) ⇌ N₂O₄(g). 6.0 mol O₂, 9.0 mol N₂ in 10.0 L. At equilibrium [N₂O₄] = 0.15 M.
中文翻译： 将 6.0 mol O₂ 和 9.0 mol N₂ 置于 10.0 L 容器中，平衡时 [N₂O₄] = 0.15 M，完成 ICE 表格并求 Keq。
| 2O₂(g) | + | N₂(g) | ⇌ | N₂O₄(g) | |
|---|---|---|---|---|---|
| [I] | 0.60 | 0.90 | 0 | ||
| [C] | −2(0.15) = −0.30 | −0.15 | +0.15 | ||
| [E] | 0.30 | 0.75 | 0.15 |
$$K_{eq} = \frac{[\text{N}_2\text{O}_4]}{[\text{O}_2]^2[\text{N}_2]} = \frac{0.15}{(0.30)^2(0.75)} = \frac{0.15}{0.0675} = \boxed{2.2}$$
✅ Answer Key 答案: 2.2 — 一致 ✓
Q27¶
EN: CO(g) + 2H₂(g) ⇌ CH₃OH(g). 1.35 mol CO, 2.50 mol H₂ in 2.00 L. At equilibrium [CH₃OH] = 0.560 M.
中文翻译： 将 1.35 mol CO 和 2.50 mol H₂ 置于 2.00 L 容器中，平衡时 [CH₃OH] = 0.560 M，完成 ICE 表格并求 Keq。
| CO(g) | + | 2H₂(g) | ⇌ | CH₃OH(g) | |
|---|---|---|---|---|---|
| [I] | 0.675 | 1.25 | 0 | ||
| [C] | −0.560 | −1.120 | +0.560 | ||
| [E] | 0.115 | 0.130 | 0.560 |
$$K_{eq} = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2} = \frac{0.560}{(0.115)(0.130)^2} = \frac{0.560}{(0.115)(0.0169)} = \frac{0.560}{1.944 \times 10^{-3}} = \boxed{2.88 \times 10^2}$$
✅ Answer Key 答案: 2.88 × 10² — 一致 ✓
Q28¶
EN: 2COF₂(g) ⇌ CO₂(g) + CF₄(g), Keq = 53.6. 0.500 mol COF₂, 3.60 mol CO₂, 4.00 mol CF₄ in 10.0 L.
中文翻译： 已知 Keq = 53.6，在 10.0 L 容器中有 0.500 mol COF₂、3.60 mol CO₂ 和 4.00 mol CF₄。(a) 反应向哪个方向移动？(b) [COF₂] 增大还是减小？(c) 总压增大、减小还是不变？
Concentrations: [COF₂] = 0.0500 M, [CO₂] = 0.360 M, [CF₄] = 0.400 M
a) Direction?
$$Q = \frac{[\text{CO}_2][\text{CF}_4]}{[\text{COF}_2]^2} = \frac{(0.360)(0.400)}{(0.0500)^2} = \frac{0.144}{0.00250} = 57.6$$
Q = 57.6 > Keq = 53.6 → too many products → shift LEFT 向左
✅ Answer Key 答案: Left — 一致 ✓
b) Will [COF₂] increase or decrease?
Shift LEFT produces more reactant → [COF₂] increases 增加。
c) Will total pressure increase or decrease?
2COF₂ ⇌ CO₂ + CF₄: Left side 2 moles, Right side 2 moles. Equal moles → pressure stays the SAME 不变。
ℹ️ Q28b, Q28c 答案页未提供。
Q29¶
EN: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Keq = 33.3. 8.0 mol PCl₅, 6.0 mol PCl₃, 4.00 mol Cl₂ in 2.0 L.
中文翻译： 已知 Keq = 33.3，在 2.0 L 容器中有 8.0 mol PCl₅、6.0 mol PCl₃ 和 4.00 mol Cl₂。(a) 反应向哪个方向移动？(b) [PCl₅] 增大还是减小？(c) 总压增大、减小还是不变？
Concentrations: [PCl₅] = 4.0 M, [PCl₃] = 3.0 M, [Cl₂] = 2.0 M
a) Direction?
$$Q = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{(3.0)(2.0)}{4.0} = 1.5$$
Q = 1.5 < Keq = 33.3 → too few products → shift RIGHT 向右
✅ Answer Key 答案: Right — 一致 ✓
b) [PCl₅] will decrease 减少 (consumed as reaction shifts right).
c) 1 mole reactant → 2 moles products. Shift right → more total moles → pressure increases 增大。
ℹ️ Q29b, Q29c 答案页未提供。
Q30¶
EN: N₂(g) + O₂(g) ⇌ 2NO(g), Keq = 0.00100. 7.50 mol NO in 250. mL flask.
中文翻译： 已知 Keq = 0.00100，在 250. mL 烧瓶中初始仅有 7.50 mol NO，求平衡时各物质浓度。
$$[\text{NO}]_i = \frac{7.50}{0.250} = 30.0 \text{ M}$$
| N₂(g) | + | O₂(g) | ⇌ | 2NO(g) | |
|---|---|---|---|---|---|
| [I] | 0 | 0 | 30.0 | ||
| [C] | +x | +x | −2x | ||
| [E] | x | x | 30.0−2x |
$$K_{eq} = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} = \frac{(30.0 - 2x)^2}{x^2} = 0.00100$$
Take square root of both sides:
$$\frac{30.0 - 2x}{x} = \sqrt{0.00100} = 0.03162$$
$$30.0 - 2x = 0.03162x$$
$$30.0 = 2.03162x$$
$$x = \frac{30.0}{2.03162} = 14.77$$
$$\boxed{[\text{N}_2] = 14.8 \text{ M}, \quad [\text{O}_2] = 14.8 \text{ M}, \quad [\text{NO}] = 30.0 - 2(14.77) = 0.46 \approx 0.467 \text{ M}}$$
✅ Answer Key 答案: [N₂]=14.8 M, [O₂]=14.8 M, [NO]=0.467 M — 一致 ✓
Q31¶
EN: HBr(g) + ClF(g) ⇌ HF(g) + BrCl(g), Keq = 17.8. 3.00 mol each reactant in 1.00 L.
中文翻译： 已知 Keq = 17.8，将 3.00 mol HBr 和 3.00 mol ClF 置于 1.00 L 容器中，求平衡时各物质浓度。
| HBr(g) | + | ClF(g) | ⇌ | HF(g) | + | BrCl(g) | |
|---|---|---|---|---|---|---|---|
| [I] | 3.00 | 3.00 | 0 | 0 | |||
| [C] | −x | −x | +x | +x | |||
| [E] | 3.00−x | 3.00−x | x | x |
$$K_{eq} = \frac{x^2}{(3.00 - x)^2} = 17.8$$
$$\frac{x}{3.00 - x} = \sqrt{17.8} = 4.219$$
$$x = 4.219(3.00 - x) = 12.66 - 4.219x$$
$$5.219x = 12.66$$
$$x = 2.43$$
$$\boxed{[\text{HBr}] = 0.57\text{ M},\ [\text{ClF}] = 0.57\text{ M},\ [\text{HF}] = 2.43\text{ M},\ [\text{BrCl}] = 2.43\text{ M}}$$
✅ Answer Key 答案: [HBr]=0.57, [ClF]=0.57, [HF]=2.43, [BrCl]=2.43 — 一致 ✓
Q32¶
EN: Cl₂(g) + O₂(g) ⇌ 2OCl(g), Keq = 0.0500. Equal moles of Cl₂ and O₂ in 4.00 L. [OCl]eq = 0.568 M. Find mass of Cl₂ initially added.
中文翻译： 已知 Keq = 0.0500，将等摩尔 Cl₂ 和 O₂ 置于 4.00 L 容器中，平衡时 [OCl] = 0.568 M，求最初加入的 Cl₂ 的质量（克）。
Let initial concentration of each = y M.
| Cl₂(g) | + | O₂(g) | ⇌ | 2OCl(g) | |
|---|---|---|---|---|---|
| [I] | y | y | 0 | ||
| [C] | −0.284 | −0.284 | +0.568 | ||
| [E] | y−0.284 | y−0.284 | 0.568 |
$$K_{eq} = \frac{[\text{OCl}]^2}{[\text{Cl}_2][\text{O}_2]} = \frac{(0.568)^2}{(y - 0.284)^2} = 0.0500$$
$$(y - 0.284)^2 = \frac{0.3226}{0.0500} = 6.453$$
$$y - 0.284 = \sqrt{6.453} = 2.540$$
$$y = 2.824 \text{ M}$$
$$n_{\text{Cl}_2} = 2.824 \times 4.00 = 11.30 \text{ mol}$$
$$m_{\text{Cl}_2} = 11.30 \times 70.90 \text{ g/mol} = \boxed{801 \text{ g}}$$
✅ Answer Key 答案: 801 g Cl₂ — 一致 ✓
Answer Key Summary — 答案汇总¶
| Q# | Answer 答案 | Match 一致 |
|---|---|---|
| 3.1 Equilibrium | ||
| 3 [E] | N₂=3.2, H₂=4.6, NH₃=3.6 M | ✅ |
| 4 [E] | CO=1.6, O₂=2.3, CO₂=0.6 M | ✅ |
| 5 | Le Chatelier table (8 entries) | ✅ |
| 6 | Left, Right, Right, None, Right | ✅ |
| 9 | Time 1: CO₂ added/Left; Time 2: ↑P or ↓T/Right; Time 3: CO removed/Left | ✅ |
| 3.2 The Equilibrium Law | ||
| 15 | 8000 | ✅ |
| 21 (row 1) | ΔS=P, ΔH=P, C | ✅ |
| 22 | 0.146 | ✅ |
| 24 | 42.3 mol Br₂ | ✅ |
| 25 | 1.46 mol NO₂ | ✅ |
| 26 | 2.2 | ✅ |
| 27 | 2.88 × 10² | ✅ |
| 28a | Left | ✅ |
| 29a | Right | ✅ |
| 30 | [N₂]=14.8, [O₂]=14.8, [NO]=0.467 M | ✅ |
| 31 | [HBr]=0.57, [ClF]=0.57, [HF]=2.43, [BrCl]=2.43 | ✅ |
| 32 | 801 g Cl₂ | ✅ |
所有提供答案的题目全部一致！All answers match! ✅
术语表 | Terminology¶
| English 英文 | Chinese 中文 |
|---|---|
| Equilibrium | 化学平衡 |
| Equilibrium Constant (Keq) | 平衡常数 |
| Le Chatelier's Principle | 勒夏特列原理 |
| ICE Table | ICE 表格（Initial-Change-Equilibrium） |
| Reaction Quotient (Q) | 反应商 |
| Exothermic | 放热 |
| Endothermic | 吸热 |
| Entropy (S) | 熵 |
| Enthalpy (H) | 焓 |
| Completion (C) | 完全反应 |
| No Reaction (NR) | 不反应 |
| Dynamic Equilibrium | 动态平衡 |
| Stoichiometric Ratio | 化学计量比 |