PROJ-CH-0036: An Exercise in Stress — 勒夏特列原理实验分析 完整解答¶
状态: ✅ 已完成
创建日期: 2026-02-18
最后更新: 2026-03-03
实验背景 | Lab Background¶
Main Reaction 主反应：
$$\text{CO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{COCl}_2(g) + 220\text{ kJ}$$
- 反应类型 | Type：放热反应 Exothermic（正向释放 220 kJ）
- 所有物质均为气态 | All species are gases (g)
- 化学计量比 | Stoichiometry：1 : 1 : 1
- 平衡常数表达式 | Keq expression：
$$K_{eq} = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]}$$
Le Chatelier's Principle 勒夏特列原理： When an equilibrium system is stressed, the reaction will shift to reduce the effect of the stress and try to re-establish equilibrium.
当平衡体系受到干扰时，反应会向减小干扰效果的方向移动，试图重新建立平衡。
图表读数 | Graph Readings¶
⚠️ Note 注意： 以下浓度值为从浓度-时间图表的近似读数，精确到 1 位小数。实际值以你从图表上读取的为准。
图表时间线 | Timeline¶
| 时间段 Period | 事件 Event |
|---|---|
| t = 0 → ~1.5 min | 初始条件 → 第一次平衡 (E1) |
| t ≈ 2 min | 应力 S1 |
| t ≈ 2 → ~3.5 min | 移动 → 第二次平衡 (E2) |
| t ≈ 4 min | 应力 S2 |
| t ≈ 4 → ~5.5 min | 移动 → 第三次平衡 (E3) |
| t ≈ 6 min | 应力 S3 |
| t ≈ 6 → ~7.5 min | 移动 → 第四次平衡 (E4) |
Part 1: First Equilibrium (E1) — 第一次平衡¶
Q1¶
EN (Original): Look at the initial conditions and the changes leading to the first equilibrium. Set up an ICE table using these values. Estimate all concentrations to one decimal place.
中文翻译： 查看初始条件和达到第一次平衡的变化。使用这些数值设置 ICE 表格。将所有浓度估算到小数点后一位。
Answer 答案:
Graph readings 图表读数： - Initial (t=0)：CO = 2.4 M，Cl₂ = 1.4 M，COCl₂ = 0.0 M - E1 (~t=1.5)：CO = 2.0 M，Cl₂ = 1.0 M，COCl₂ = 0.4 M
| CO(g) | + | Cl₂(g) | ⇌ | COCl₂(g) | |
|---|---|---|---|---|---|
| I (Initial) | 2.4 | 1.4 | 0.0 | ||
| C (Change) | −0.4 | −0.4 | +0.4 | ||
| E (Equilibrium) | 2.0 | 1.0 | 0.4 |
Verification 验证： Changes follow 1:1:1 stoichiometric ratio ✓
变化量符合 1:1:1 化学计量比 ✓
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Q2¶
EN (Original): Calculate K_eq at the first equilibrium.
中文翻译： 计算第一次平衡时的 Keq。
Answer 答案:
$$K_{eq} = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} = \frac{0.4}{(2.0)(1.0)} = \boxed{0.20}$$
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Q3¶
EN (Original): Describe the initial rates of the forward and reverse reactions, and how those rates change as the first equilibrium was being established? (No calculations needed)
中文翻译： 描述正向和逆向反应的初始速率，以及在建立第一次平衡过程中这些速率如何变化？（不需要计算）
Answer 答案:
EN: At t = 0, only CO and Cl₂ are present (no COCl₂), so the forward rate is at its maximum and the reverse rate is zero (no product to decompose).
As the reaction proceeds: - Forward rate decreases — as CO and Cl₂ are consumed, their concentrations drop, leading to fewer collisions between reactant molecules - Reverse rate increases — as COCl₂ accumulates, more product molecules can decompose back into reactants
At equilibrium (E1), the forward rate equals the reverse rate. Concentrations no longer change, but both reactions continue at equal rates (dynamic equilibrium).
中文： 在 t = 0 时，仅有 CO 和 Cl₂（无 COCl₂），因此正向速率最大，逆向速率为零（无产物可分解）。
随着反应进行： - 正向速率降低 — 随着 CO 和 Cl₂ 被消耗，浓度下降，反应物分子间碰撞减少 - 逆向速率增加 — 随着 COCl₂ 积累，更多产物分子可以分解回反应物
在平衡 (E1) 时，正向速率等于逆向速率。浓度不再变化，但两个反应以相等的速率持续进行（动态平衡）。
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Part 2: Second Equilibrium (E2) — 第二次平衡¶
Q4¶
EN (Original): Look at the first equilibrium and the changes leading to the second equilibrium. Set up an ICE table using these values.
中文翻译： 查看第一次平衡和达到第二次平衡的变化。设置 ICE 表格。
Answer 答案:
S1 at t = 2 min： CO concentration suddenly jumps from 2.0 to 2.3 M（CO 浓度突然从 2.0 跳升到 2.3 M）→ CO was added（添加了 CO）
System shifts RIGHT to consume excess CO.
体系向右移以消耗多余的 CO。
| CO(g) | + | Cl₂(g) | ⇌ | COCl₂(g) | |
|---|---|---|---|---|---|
| I (After S1) | 2.3 | 1.0 | 0.4 | ||
| C (Change) | −0.3 | −0.3 | +0.3 | ||
| E (E2) | 2.0 | 0.7 | 0.7 |
Verification 验证： 1:1:1 ratio ✓
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Q5¶
EN (Original): Calculate K_eq at the second equilibrium.
中文翻译： 计算第二次平衡时的 Keq。
Answer 答案:
$$K_{eq} = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} = \frac{0.7}{(2.0)(0.7)} = \frac{0.7}{1.4} = \boxed{0.50}$$
⚠️ 注意 Note： Keq 从 0.20 变为 0.50。如果应力仅为添加 CO，Keq 不应改变（温度未变时 Keq 为常数）。差异可能原因：(1) 图表读数的近似误差；(2) 应力可能同时包含温度变化。详见 Q6 和 Q14 的讨论。
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Q6¶
EN (Original): What change led to the stress at 2 minutes? There may be more than one possible answer. Explain your reasoning and be specific.
中文翻译： 什么变化导致了 2 分钟时的应力？可能有不止一个答案。解释你的推理，并具体说明。
Answer 答案:
Most likely answer 最可能的答案：Addition of CO（添加 CO）
EN — Reasoning:
At t = 2 min, we observe that the CO concentration suddenly and sharply increases from ~2.0 to ~2.3 M, while Cl₂ and COCl₂ do not show any sudden jump. A sudden increase in a single species is the hallmark of adding that substance to the system.
According to Le Chatelier's Principle, adding a reactant (CO) disturbs the equilibrium. The system responds by shifting RIGHT (toward products) to consume the excess CO, which: - Decreases [CO] (from 2.3 back toward 2.0) - Decreases [Cl₂] (from 1.0 to 0.7) - Increases [COCl₂] (from 0.4 to 0.7)
This is consistent with the graph observations.
Other possible answers 其他可能的答案：
- Temperature decrease（降低温度） could also shift the reaction right (since the reaction is exothermic, lowering temperature favours the forward reaction). However, a temperature change would cause all three species to change gradually (not one species jumping suddenly). The sudden jump in only CO makes this less likely.
中文 — 推理：
在 t = 2 分钟时，我们观察到 CO 浓度突然急剧上升（从 ~2.0 到 ~2.3 M），而 Cl₂ 和 COCl₂ 没有突然变化。单一物种的突然增加是向体系添加该物质的标志。
根据勒夏特列原理，添加反应物（CO）破坏平衡。体系通过向右移（朝产物方向）来消耗多余的 CO。
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Part 3: Third Equilibrium (E3) — 第三次平衡¶
Q7¶
EN (Original): Look at the second equilibrium and the changes leading to the third equilibrium. Set up an ICE table using these values.
中文翻译： 查看第二次平衡和达到第三次平衡的变化。设置 ICE 表格。
Answer 答案:
S2 at t = 4 min： No single species jumps suddenly. All three concentrations change gradually.（没有单一物种突然跳变，三种物质浓度都逐渐变化。）
This suggests a temperature or pressure/volume change, not a concentration change.
体系向 RIGHT 微移。
| CO(g) | + | Cl₂(g) | ⇌ | COCl₂(g) | |
|---|---|---|---|---|---|
| I (E2) | 2.0 | 0.7 | 0.7 | ||
| C (Change) | −0.1 | −0.1 | +0.1 | ||
| E (E3) | 1.9 | 0.6 | 0.8 |
Note 注意： 由于 S2 不涉及添加/移除物质，初始值就是 E2 的值。
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Q8¶
EN (Original): Calculate K_eq at the third equilibrium.
中文翻译： 计算第三次平衡时的 Keq。
Answer 答案:
$$K_{eq} = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} = \frac{0.8}{(1.9)(0.6)} = \frac{0.8}{1.14} = \boxed{0.70}$$
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Q9¶
EN (Original): What change led to the stress at 4 minutes? There may be more than one possible answer. Explain your reasoning and be specific.
中文翻译： 什么变化导致了 4 分钟时的应力？可能有不止一个答案。具体说明。
Answer 答案:
EN:
Temperature Decrease（降低温度）
Since no single species jumps suddenly (all three change gradually), this is NOT a concentration change. Since the reaction is exothermic (releases 220 kJ):
$$\text{CO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{COCl}_2(g) + 220\text{ kJ}$$
Decreasing temperature shifts the reaction toward the exothermic direction (RIGHT) to produce more heat and counteract the cooling. This matches our observations: - [CO] decreases (consumed) - [Cl₂] decreases (consumed) - [COCl₂] increases (produced)
Furthermore, a temperature change is the only stress that changes Keq. Keq increased from E2 to E3 (0.50 → 0.70), which is only possible through a temperature change — a pressure/volume change would shift the equilibrium RIGHT but would NOT change Keq. Therefore, temperature decrease is the only explanation consistent with all evidence.
📝 Teacher's note: It is a temperature decrease, not a change in pressure.
中文：
降低温度 — 由于反应是放热的，降低温度使反应向右（放热方向）移动以产生更多热量。这与观察一致：[CO] 下降、[Cl₂] 下降、[COCl₂] 上升。
温度变化是唯一能改变 Keq 的因素。Keq 从 0.50 增加到 0.70，只有降温才能解释——增大压力虽然也能使平衡向右移，但不会改变 Keq。因此，降温是唯一与所有证据自洽的答案。
📝 老师批注： 应力是降低温度，不是压力变化。
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Part 4: Fourth Equilibrium (E4) — 第四次平衡¶
Q10¶
EN (Original): Look at the third equilibrium and the changes leading to the fourth equilibrium. Set up an ICE table using these values.
中文翻译： 查看第三次平衡和达到第四次平衡的变化。设置 ICE 表格。
Answer 答案:
S3 at t = 6 min： COCl₂ concentration suddenly spikes UP from ~0.8 to ~1.5 M.（COCl₂ 浓度突然从 ~0.8 跳升到 ~1.5 M。）→ COCl₂ was added（添加了 COCl₂）
System shifts LEFT to consume excess product.
体系向左移以消耗多余的产物。
| CO(g) | + | Cl₂(g) | ⇌ | COCl₂(g) | |
|---|---|---|---|---|---|
| I (After S3) | 1.9 | 0.6 | 1.5 | ||
| C (Change) | +0.5 | +0.5 | −0.5 | ||
| E (E4) | 2.4 | 1.1 | 1.0 |
Verification 验证： 1:1:1 ratio ✓
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Q11¶
EN (Original): Calculate K_eq at the fourth equilibrium.
中文翻译： 计算第四次平衡时的 Keq。
Answer 答案:
$$K_{eq} = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} = \frac{1.0}{(2.4)(1.1)} = \frac{1.0}{2.64} = \boxed{0.38}$$
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Q12¶
EN (Original): What change led to the stress at 6 minutes? There may be more than one possible answer. Explain your reasoning and be specific.
中文翻译： 什么变化导致了 6 分钟时的应力？可能有不止一个答案。具体说明。
Answer 答案:
Most likely answer 最可能的答案：Addition of COCl₂（添加 COCl₂）
EN — Reasoning:
At t = 6 min, COCl₂ concentration suddenly spikes from ~0.8 to ~1.5 M, while CO and Cl₂ show no sudden change. A sudden increase in a single product species indicates COCl₂ was added to the system.
By Le Chatelier's Principle, adding product (COCl₂) shifts the equilibrium LEFT to decompose excess COCl₂: - [CO] increases (produced from decomposition) - [Cl₂] increases (produced from decomposition) - [COCl₂] decreases (consumed by reverse reaction)
Other possible answer 其他可能答案：
- Temperature increase（升高温度） for this exothermic reaction would shift LEFT (endothermic direction). But a temperature change would cause gradual changes in ALL species, not a sudden spike in just one.
中文 — 推理：
在 t = 6 分钟时，COCl₂ 浓度突然急升（从 ~0.8 到 ~1.5 M），而 CO 和 Cl₂ 没有突然变化。单一产物浓度的突然增加表明向体系添加了 COCl₂。
根据勒夏特列原理，添加产物使平衡向左移以分解多余的 COCl₂。
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Additional Questions — 附加问题¶
Q13¶
EN (Original): How would K_eq change in each region if a catalyst was added to the system?
中文翻译： 如果向体系中加入催化剂，每个区域的 Keq 会如何变化？
Answer 答案:
EN: Keq would NOT change in any region.
A catalyst: - Speeds up both the forward and reverse reactions equally - Provides an alternative pathway with lower activation energy - Helps the system reach equilibrium faster - Does NOT change the position of equilibrium - Does NOT change the equilibrium concentrations - Does NOT change Keq
The same equilibrium is reached — just faster.
中文： Keq 在任何区域都不会改变。
催化剂： - 同等加速正向和逆向反应 - 提供活化能更低的替代路径 - 帮助体系更快达到平衡 - 不改变平衡位置 - 不改变平衡浓度 - 不改变 Keq
最终达到相同的平衡状态——只是更快到达。
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Q14¶
EN (Original): K_eq is the equilibrium constant. Does it change in this lab? If so, what could have caused all the K_eq values to be different in this equilibrium system?
中文翻译： Keq 是平衡常数。它在本实验中是否改变？如果是，什么可能导致这个平衡体系中所有 Keq 值不同？
Answer 答案:
EN:
In this lab, we observe different Keq values at each equilibrium:
| Equilibrium | Keq |
|---|---|
| E1 | 0.20 |
| E2 | 0.50 |
| E3 | 0.70 |
| E4 | 0.38 |
Keq is truly constant only at a fixed temperature. The differences we see could be caused by two factors:
1. Temperature change（温度变化） — This is the only legitimate reason for Keq to change. If one or more of the stresses involved a temperature change (e.g., S2 being a temperature decrease), then Keq would genuinely change. For this exothermic reaction: - Temperature ↓ → Keq ↑ (shifts right, favours products) - Temperature ↑ → Keq ↓ (shifts left, favours reactants)
2. Graph reading imprecision（图表读数误差） — Since we estimate concentrations to only 1 decimal place from a graph, small reading errors compound when calculating Keq. For example, reading 2.0 vs 2.1 for CO can change Keq significantly.
In a real lab, if only concentration changes and volume/pressure changes were applied (no temperature change), all Keq values should be identical within experimental error.
中文： 在本实验中，每次平衡的 Keq 值不同。Keq 只在恒温下才是真正的常数。差异可能由两个因素引起：
- 温度变化 — 这是 Keq 改变的唯一合理原因。对于放热反应：温度↓ → Keq↑；温度↑ → Keq↓
- 图表读数误差 — 从图表估算浓度只精确到 1 位小数，小误差在计算 Keq 时会放大
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Taking it Further — 进阶思考¶
EN (Original): Imagine you are setting up a factory to create COCl₂ using the reaction in this lab. What conditions would you create in order to maximize the production of COCl₂?
中文翻译： 假设你正在建立一个使用本实验中的反应生产 COCl₂ 的工厂。你会创造什么条件来最大化 COCl₂ 的产量？
Answer 答案:
EN: To maximize COCl₂ production, apply Le Chatelier's Principle to shift equilibrium RIGHT:
| Strategy 策略 | Reasoning 推理 |
|---|---|
| 1. Use high concentrations of CO and Cl₂ 使用高浓度反应物 | Adding reactants shifts equilibrium RIGHT toward products 添加反应物使平衡向右移 |
| 2. Use LOW temperature 使用低温 | The reaction is exothermic (+220 kJ). Low temperature shifts equilibrium toward the exothermic direction (RIGHT), increasing Keq and producing more COCl₂. However, temperature must not be too low, or the reaction rate becomes impractically slow. 反应是放热的。低温使平衡向放热方向（右）移动。但温度不能太低否则速率过慢。 |
| 3. Use HIGH pressure (small volume) 使用高压（小体积） | Reactant side: 2 moles gas (CO + Cl₂). Product side: 1 mole gas (COCl₂). High pressure shifts toward the side with fewer gas molecules = RIGHT. 反应物侧：2 摩尔气体。产物侧：1 摩尔气体。高压使平衡向气体分子数少的一侧（右）移动。 |
| 4. Continuously remove COCl₂ 持续移除 COCl₂ | Removing product shifts equilibrium RIGHT to make more product 移除产物使平衡向右移以生成更多产物 |
| 5. Add a catalyst 加入催化剂 | A catalyst does NOT change Keq or equilibrium position, but it speeds up the rate at which equilibrium is reached, increasing throughput. 催化剂不改变 Keq，但加速达到平衡的速率，提高产量效率。 |
Optimal factory conditions 最佳工厂条件：
High [CO] + high [Cl₂] + moderately low temperature + high pressure + continuous product removal + catalyst
高浓度 CO + 高浓度 Cl₂ + 适度低温 + 高压 + 持续移除产物 + 催化剂
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Keq Summary — 平衡常数汇总¶
| Equilibrium 平衡 | CO (M) | Cl₂ (M) | COCl₂ (M) | Keq | Stress 应力 |
|---|---|---|---|---|---|
| E1 | 2.0 | 1.0 | 0.4 | 0.20 | (from initial) |
| E2 | 2.0 | 0.7 | 0.7 | 0.50 | S1: Add CO |
| E3 | 1.9 | 0.6 | 0.8 | 0.70 | S2: Decrease T |
| E4 | 2.4 | 1.1 | 1.0 | 0.38 | S3: Add COCl₂ |
ℹ️ 本项目无答案键。 以上解答基于图表近似读数和标准化学平衡原理。实际图表读数可能与上述值略有不同，请以你从图上读取的实际值为准。
术语表 | Terminology¶
| English 英文 | Chinese 中文 |
|---|---|
| Le Chatelier's Principle | 勒夏特列原理 |
| Equilibrium | 化学平衡 |
| Equilibrium Constant (Keq) | 平衡常数 |
| ICE Table | ICE 表格（初始-变化-平衡） |
| Stress / Disturbance | 应力 / 干扰 |
| Exothermic | 放热 |
| Endothermic | 吸热 |
| Forward Reaction | 正向反应 |
| Reverse Reaction | 逆向反应 |
| Shift Right | 向右移动（正向） |
| Shift Left | 向左移动（逆向） |
| Catalyst | 催化剂 |
| Dynamic Equilibrium | 动态平衡 |