← All Units← 返回单元列表 ← Course Hub← 课程主页
I B  C H E M I S T R Y  H L
Reactivity 2.2 · SolutionsReactivity 2.2 · 解析

Reaction Kinetics — Solutions化学反应动力学 —— 解析

Companion to the Reactivity 2.2 Practice SetReactivity 2.2 练习题的解析配套

EASY MEDIUM HARD Paper 1 Paper 2 Paper 3 HL HL

Topics Reactivity 2.2.1 – 2.2.13考点 Reactivity 2.2.1 – 2.2.13HL

PART I  ·  PAPER 1第一部分  ·  第一卷Multiple Choice — Worked Answers选择题 —— 详细解析

Multiple Choice选择题

Q1EASYPaper 12.2.1 Rate Units

Standard units of reaction rate?反应速率的标准单位？

Answer:答案： (C)
Reaction rate = change in concentration ÷ time. Concentration has units $\mathrm{mol\,dm^{-3}}$, divided by seconds → $\mathrm{mol\,dm^{-3}\,s^{-1}}$. Trap (A) is moles per second (used for total amount, not concentration); (B) is concentration without the time dimension.反应速率 = 浓度的变化 ÷ 时间。浓度单位 $\mathrm{mol\,dm^{-3}}$，除以秒 → $\mathrm{mol\,dm^{-3}\,s^{-1}}$。陷阱 (A) 是"摩尔每秒"（用于物质的量而非浓度）；(B) 是浓度本身，缺少时间维度。
Q2EASYPaper 12.2.1 Rate from a Graph

$\mathrm{A \to \text{products}}$; $[\mathrm{A}]$ vs $t$ curve flattens near zero. Where is rate of disappearance of A largest?$\mathrm{A \to \text{products}}$；$[\mathrm{A}]$ 对 $t$ 曲线趋于零。A 的消失速率最大处？

Answer:答案： (A)
Rate is $|d[\mathrm{A}]/dt|$ — the magnitude of the tangent slope. The curve is steepest at $t = 0$ when $[\mathrm{A}]$ is highest (more reactant ⇒ more collisions ⇒ faster). As $[\mathrm{A}]$ falls, the tangent gets shallower, so rate decreases. At the plateau, rate $\to 0$.速率为 $|d[\mathrm{A}]/dt|$ —— 切线斜率的绝对值。$t = 0$ 时 $[\mathrm{A}]$ 最大，曲线最陡（反应物多 ⇒ 碰撞多 ⇒ 速率快）。随 $[\mathrm{A}]$ 下降，切线变缓，速率减小。平台处速率 $\to 0$。
Q3EASYPaper 12.2.2 Collision Theory

Conditions for a successful collision?"有效碰撞"需要满足的条件？

Answer:答案： (A)
Collision theory has two requirements: the colliding particles must have at least the activation energy $E_a$ and they must collide in an orientation that allows the reactive parts to meet. Both are necessary; neither is sufficient on its own. (D) is wrong — kinetics works for reactions in any phase.碰撞理论有两个要求：碰撞粒子的能量至少达到活化能 $E_a$，且取向必须能让反应基团相遇。二者缺一不可。(D) 错误 —— 动力学适用于任何相态的反应。
Q4EASYPaper 12.2.3 Concentration

Why does higher concentration increase rate?为何浓度增大会提高反应速率？

Answer:答案： (B)
Concentration affects how often reactant particles meet, not how energetic each one is. More particles per unit volume → more collisions per second → faster rate. Average kinetic energy depends only on temperature (not concentration), so (C) is wrong. Activation energy is a property of the reaction itself, unchanged by concentration.浓度影响的是反应物粒子相遇的频次，而非单个粒子的能量。单位体积内粒子数增多 → 每秒碰撞次数增多 → 速率加快。平均动能仅与温度有关（与浓度无关），所以 (C) 错误。活化能是反应本身的属性，浓度不能改变它。
Q5MEDIUMPaper 12.2.3 Surface Area

Excess marble + dilute HCl; comparing powder vs lump of same mass.过量大理石 + 稀盐酸；同质量的粉末与整块比较。

Answer:答案： (C)
Powder exposes much more $\mathrm{CaCO_3}$ surface to the acid, so the initial rate is faster. But total $\mathrm{CO_2}$ is fixed by stoichiometry: HCl is the limiting reactant (marble is in excess), so the same amount of HCl produces the same amount of $\mathrm{CO_2}$ regardless of how the marble is presented. The total volume of $\mathrm{CO_2}$ is identical; only the time to reach it differs.粉末暴露出更多 $\mathrm{CaCO_3}$ 表面接触酸，初始速率更快。但 $\mathrm{CO_2}$ 的总量由化学计量决定：HCl 是限量反应物（大理石过量），故等量 HCl 无论与粉末还是整块反应都生成等量 $\mathrm{CO_2}$。总体积相同；只是达到该体积的时间不同。
Q6MEDIUMPaper 12.2.4 Maxwell–Boltzmann

MB curve at 350 K vs 300 K (same sample)?同一样品，麦克斯韦–玻尔兹曼曲线在 350 K 与 300 K 的对比？

Answer:答案： (B)
At higher $T$: particles take a wider spread of energies, so the distribution flattens and broadens. The peak shifts right and down. The total area under the curve is the total number of particles, which is unchanged — so larger spread + larger high-energy tail = lower peak. The area beyond $E_a$ (the reactive fraction) grows substantially.温度升高：粒子能量分布更分散，曲线变得平坦而宽。峰右移并下移。曲线下总面积等于粒子总数，保持不变 —— 故展宽 + 高能尾增大 = 峰高降低。$E_a$ 右侧的面积（即能反应的那部分粒子）大幅增加。
Q7MEDIUMPaper 12.2.5 Catalyst

Which property is not changed by adding a catalyst?加入催化剂后不会改变的是？

Answer:答案： (C)
$\Delta H$ depends only on the energy difference between reactants and products — a thermodynamic property fixed by the chemistry, not the path. A catalyst opens a new pathway (different mechanism, lower $E_a$, faster rate), but the start and end levels of the energy profile stay put. Trap students often pick (D): a catalyst does change the mechanism (by definition — it provides the alternative pathway).$\Delta H$ 仅取决于反应物与产物的能量差 —— 是热力学属性，由化学反应本身决定，与路径无关。催化剂开启新路径（机理不同、$E_a$ 更低、速率更快），但能量曲线的起点与终点不变。常见陷阱选 (D)：催化剂确实会改变机理（按定义 —— 它提供的就是替代路径）。
Q8MEDIUMPaper 12.2.5 Energy Profile

Effect of a catalyst on the energy profile of an exothermic reaction?催化剂对放热反应能量曲线的影响？

Answer:答案： (C)
A catalyst lowers (does not remove) the activation-energy barrier by providing an alternative pathway. Reactant and product energy levels are unchanged, so $\Delta H$ stays the same. Trap (D): no chemical reaction has zero activation energy — there's always some bond reorganisation that costs energy along the way.催化剂通过提供替代路径来降低（而非消除）活化能势垒。反应物与产物能级不变，故 $\Delta H$ 不变。陷阱 (D)：没有任何化学反应的活化能为零 —— 沿反应路径总会有一些键的重组需要消耗能量。
Q9HARDPaper 1HL2.2.6 Intermediate vs Transition State

A reaction intermediate corresponds to反应中间体对应于

Answer:答案： (B)
Intermediates are real, isolable species formed in one step and consumed in a later step — they sit at a local minimum on the energy profile. Transition states are the activated complexes at local maxima between adjacent intermediates (or between reactant and product, in a one-step reaction). They are not the same: transition states cannot be isolated, while intermediates often can.中间体是真实且可分离的物种，由一步生成、被后一步消耗 —— 位于能量曲线的局部极小点。过渡态是位于相邻中间体之间（一步反应中即在反应物与产物之间）的活化络合物，处于局部极大点。二者不同：过渡态无法分离，而中间体通常可以。
Q10HARDPaper 1HL2.2.10 Order from a Graph

$[\mathrm{A}]$ vs $t$ is a straight line with negative slope. Order in A?$[\mathrm{A}]$ 对 $t$ 是斜率为负的直线。关于 A 的级数？

Answer:答案： (A)
A straight line in the $[\mathrm{A}]$-vs-$t$ plot means $d[\mathrm{A}]/dt$ is constant — independent of $[\mathrm{A}]$. That's the signature of a zero-order reaction: rate = $k$ (a constant). First-order gives exponential decay; second-order gives a hyperbola-like decay.$[\mathrm{A}]$ 对 $t$ 是直线，意味着 $d[\mathrm{A}]/dt$ 为常数 —— 与 $[\mathrm{A}]$ 无关。这是零级反应的特征：rate = $k$（常数）。一级是指数衰减；二级是类似双曲线的衰减。
Q11HARDPaper 1HL2.2.11 Units of k

rate $= k[\mathrm{A}][\mathrm{B}]^2$. Units of $k$?rate $= k[\mathrm{A}][\mathrm{B}]^2$。$k$ 的单位？

Answer:答案： (C)
Overall order $= 1 + 2 = 3$. Solving for units:总反应级数 $= 1 + 2 = 3$。求单位：
$$k = \dfrac{\text{rate}}{[\mathrm{A}][\mathrm{B}]^2} \;\Longrightarrow\; \dfrac{\mathrm{mol\,dm^{-3}\,s^{-1}}}{\mathrm{mol^3\,dm^{-9}}} = \mathrm{dm^6\,mol^{-2}\,s^{-1}}$$
Memorise: first-order $\mathrm{s^{-1}}$; second-order $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$; third-order $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$.记住：一级反应 $\mathrm{s^{-1}}$；二级反应 $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$；三级反应 $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$。
Q12HARDPaper 1HL2.2.12 Arrhenius

$\ln k$ vs $1/T$ is linear with negative slope. The slope gives$\ln k$ 对 $1/T$ 为斜率负的直线。该斜率可给出

Answer:答案： (C)
Linear form of Arrhenius: $\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A$. The slope of $\ln k$ vs $1/T$ is $-E_a/R$, so $E_a = -R \times \text{slope}$. The y-intercept gives $\ln A$. Activation energy and $\Delta H$ are different concepts — $E_a$ is about the kinetic barrier, $\Delta H$ is the thermodynamic difference between reactants and products.阿伦尼乌斯方程的线性形式：$\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A$。$\ln k$ 对 $1/T$ 的斜率为 $-E_a/R$，因此 $E_a = -R \times$ 斜率。y 轴截距给出 $\ln A$。$E_a$ 与 $\Delta H$ 是不同概念 —— $E_a$ 是动力学势垒，$\Delta H$ 是反应物与产物之间的热力学能量差。
PART II  ·  PAPER 2第二部分  ·  第二卷Structured Response — Worked Solutions结构化解答题 —— 详细解析

Structured Response结构化解答题

SR 1MEDIUMPaper 2Marble + HCl Experiment

Excess marble chips + 50.0 cm³ of 1.00 mol dm⁻³ HCl at 298 K; $V(\mathrm{CO_2})$ recorded over 200 s.298 K 下过量大理石碎块 + 50.0 cm³ 1.00 mol dm⁻³ HCl；200 s 内记录 $V(\mathrm{CO_2})$。

(a) Average rate over first 20 s:前 20 s 的平均速率：
$$\text{rate}_\text{avg} = \dfrac{\Delta V(\mathrm{CO_2})}{\Delta t} = \dfrac{22 - 0}{20 - 0} = 1.1~\mathrm{cm^3\,s^{-1}}$$
(b) The acid is the limiting reactant; as the reaction proceeds, $[\mathrm{HCl}]$ falls. With fewer acid particles per unit volume, the collision frequency between $\mathrm{H^+}$ and the marble surface drops, so successful-collision rate (and hence reaction rate) falls. When $\mathrm{HCl}$ is essentially exhausted, the curve plateaus.酸是限量反应物；反应进行时 $[\mathrm{HCl}]$ 下降。单位体积内酸粒子减少，$\mathrm{H^+}$ 与大理石表面的碰撞频率降低，有效碰撞率（即反应速率）随之下降。当 $\mathrm{HCl}$ 基本耗尽时，曲线趋于平台。
(c) Effects on the initial rate:对初始速率的影响：
(i) Powdered marble: much larger surface area exposed to the acid; many more solid–liquid collisions per second. Initial rate increases substantially. Total $\mathrm{CO_2}$ unchanged (HCl is still limiting).(i) 大理石粉末：暴露于酸的表面积大大增加，每秒固液碰撞次数大幅增多。初始速率显著增大。$\mathrm{CO_2}$ 总量不变（HCl 仍是限量反应物）。
(ii) 318 K: Maxwell–Boltzmann tail shifts past $E_a$; a much larger fraction of collisions succeed. Initial rate increases substantially (usually $\sim$ 2× per 10 K rise in this regime).(ii) 318 K：麦克斯韦–玻尔兹曼分布的尾部跨过 $E_a$，能反应的碰撞比例大幅上升。初始速率显著增大（在该温区一般每升高 10 K 约翻一倍）。
(iii) 2.00 mol dm⁻³ HCl: $[\mathrm{HCl}]$ doubled → collision frequency between acid and marble surface roughly doubles. Initial rate approximately doubles. Total $\mathrm{CO_2}$ also doubles (more HCl ⇒ more limiting reactant ⇒ more product).(iii) 2.00 mol dm⁻³ HCl：$[\mathrm{HCl}]$ 加倍 → 酸与大理石表面碰撞频率约加倍。初始速率约加倍。$\mathrm{CO_2}$ 总量也加倍（HCl 更多 ⇒ 限量反应物更多 ⇒ 产物更多）。
(iv) $\mathrm{MnO_2}$: $\mathrm{MnO_2}$ is famously a catalyst for the decomposition of $\mathrm{H_2O_2}$, not for the acid–carbonate reaction. Catalysts are reaction-specific. No effect on the initial rate.(iv) $\mathrm{MnO_2}$：$\mathrm{MnO_2}$ 是著名的 $\mathrm{H_2O_2}$ 分解催化剂，不催化酸与碳酸盐的反应。催化剂具有专一性。对初始速率无影响。
(d) Sketch (description): Two MB curves on the same axes. 298 K curve has higher narrower peak; 318 K curve is lower and broader with peak shifted right. Mark $E_a$ as a vertical line at some energy beyond both peaks; shade the area to the right of $E_a$ under each curve. The shaded area is much larger at 318 K. Caption: "A 20 K rise increases the average kinetic energy by only ~7%, but the area beyond $E_a$ — the fraction of particles with enough energy to react — typically more than doubles, accounting for the large change in rate."作图（描述）：在同一坐标系上画出两条 MB 曲线。298 K 曲线峰更高更窄；318 K 曲线更低更宽且峰右移。在两峰右侧某能量处用竖线标出 $E_a$；分别给两条曲线 $E_a$ 右侧的面积上阴影。318 K 时阴影面积明显更大。题注："温度升高 20 K 仅使平均动能上升约 7%，但 $E_a$ 右侧的面积 —— 即能量足以反应的粒子比例 —— 通常会翻倍以上，从而解释速率为何大幅变化。"
SR 2HARDPaper 2HLInitial Rates + Arrhenius

$\mathrm{A + 2B \to C}$ at 300 K. Three runs give rates 1.5e-4, 6.0e-4, 6.0e-4 for ([A], [B]) = (0.10, 0.10), (0.20, 0.10), (0.20, 0.20).300 K 下 $\mathrm{A + 2B \to C}$。三组数据：([A], [B]) = (0.10, 0.10)、(0.20, 0.10)、(0.20, 0.20) 对应速率 1.5e-4、6.0e-4、6.0e-4。

(a) Orders.反应级数。
Runs 1 → 2: [A] doubles, [B] fixed. Rate ×4 (1.5e-4 → 6.0e-4) = $2^2$ ⇒ order in A = 2.第 1 → 2 组：[A] 加倍，[B] 不变。速率 ×4（1.5e-4 → 6.0e-4）= $2^2$ ⇒ 关于 A 的级数 = 2。
Runs 2 → 3: [B] doubles, [A] fixed. Rate unchanged ⇒ order in B = 0.第 2 → 3 组：[B] 加倍，[A] 不变。速率不变 ⇒ 关于 B 的级数 = 0。
Overall order = $2 + 0 = 2$.总反应级数 = $2 + 0 = 2$。
(b) Rate equation and rate constant. rate $= k[A]^2$. From run 1:速率方程与速率常数。rate $= k[A]^2$。由第 1 组：
$$k = \dfrac{\text{rate}}{[A]^2} = \dfrac{1.5 \times 10^{-4}}{(0.10)^2} = 1.5 \times 10^{-2}~\mathrm{dm^3\,mol^{-1}\,s^{-1}}$$
(c) Why order in B is zero despite stoichiometric coefficient 2.为何 B 的化学计量系数为 2 但级数却为零。
The order of reaction with respect to a species is experimentally determined; it depends on the mechanism, not the balanced equation. Here, the rate-determining step (the slowest elementary step) involves only A — for example, A → some intermediate is the slow step, and the intermediate then reacts with B in fast subsequent steps. Because the slow step doesn't involve B, varying $[\mathrm{B}]$ doesn't affect the overall rate. Stoichiometric coefficients become the orders only for elementary single-step reactions; for multi-step mechanisms they generally don't.关于某物种的反应级数是由实验测定的，取决于反应机理，而非配平方程。此处决速步（最慢的基元反应）只涉及 A —— 例如 A → 某中间体为慢步，随后中间体在快步中与 B 反应。慢步不含 B，故改变 $[\mathrm{B}]$ 不影响总速率。化学计量系数仅在单步基元反应中等于反应级数；对多步机理通常不等。
(d) Activation energy. $k_2/k_1 = (6.0\times10^{-2})/(1.5\times10^{-2}) = 4.0$, $\ln 4 = 1.386$. $1/T_2 - 1/T_1 = 1/350 - 1/300 = -4.76\times10^{-4}~\mathrm{K^{-1}}$.活化能。$k_2/k_1 = (6.0\times10^{-2})/(1.5\times10^{-2}) = 4.0$，$\ln 4 = 1.386$。$1/T_2 - 1/T_1 = 1/350 - 1/300 = -4.76\times10^{-4}~\mathrm{K^{-1}}$。
$$E_a = -\dfrac{R\,\ln(k_2/k_1)}{1/T_2 - 1/T_1} = -\dfrac{(8.31)(1.386)}{-4.76\times10^{-4}} \approx 24{,}200~\mathrm{J\,mol^{-1}}$$
$$E_a \approx 24.2~\mathrm{kJ\,mol^{-1}}$$
SR 3HARDPaper 2HLN₂O₅ Mechanism + Catalysis

Two-step mechanism for $\mathrm{N_2O_5}$ decomposition. Step 1 (slow): $\mathrm{N_2O_5 \to NO_2 + NO_3}$. Step 2 (fast): $\mathrm{NO_3 + NO_2 \to NO_2 + O_2 + NO}$.$\mathrm{N_2O_5}$ 分解的两步机理。第 1 步（慢）：$\mathrm{N_2O_5 \to NO_2 + NO_3}$。第 2 步（快）：$\mathrm{NO_3 + NO_2 \to NO_2 + O_2 + NO}$。

(a) Mechanism analysis.机理分析。
  • RDS: Step 1 (explicitly labelled "slow"). The overall rate is set by this step.决速步：第 1 步（题目明确标"慢"）。总反应速率由此步决定。
  • Intermediate: $\mathrm{NO_3}$ — produced in step 1, consumed in step 2 (and does not appear in the overall equation).中间体：$\mathrm{NO_3}$ —— 在第 1 步生成，第 2 步被消耗（不出现在总反应式中）。
  • Molecularity of step 1: unimolecular — only one reactant particle ($\mathrm{N_2O_5}$) is rearranging.第 1 步的分子数：单分子（unimolecular）—— 只有一个反应物粒子（$\mathrm{N_2O_5}$）在重组。
(b) Implied rate equation. The RDS is a unimolecular decomposition of $\mathrm{N_2O_5}$, so机理蕴含的速率方程。决速步是 $\mathrm{N_2O_5}$ 的单分子分解，因此
$$\text{rate} = k\,[\mathrm{N_2O_5}]$$
Overall order = 1.总反应级数 = 1。
(c) Energy profile sketch (description). Y-axis: energy. X-axis: reaction coordinate.能量曲线作图（描述）。y 轴：能量。x 轴：反应坐标。
  • Start at reactant level (label "$\mathrm{N_2O_5}$").起始于反应物能级（标记"$\mathrm{N_2O_5}$"）。
  • Rise sharply to TS1 (label $E_{a,1}$ for the height above the reactant level — this is the large barrier since step 1 is slow).陡升至 TS1（用 $E_{a,1}$ 标注其相对反应物能级的高度 —— 第 1 步慢，故此势垒较大）。
  • Drop to intermediate level (label "$\mathrm{NO_2 + NO_3}$"), still above the product level.下降至中间体能级（标记"$\mathrm{NO_2 + NO_3}$"），仍高于产物能级。
  • Rise to a smaller TS2 (label $E_{a,2}$ for the height above the intermediate — this is small since step 2 is fast).上升至较小的 TS2（用 $E_{a,2}$ 标注其相对中间体的高度 —— 第 2 步快，故此势垒较小）。
  • Drop to product level (label "$\mathrm{NO_2 + O_2 + NO}$"), below the reactant level (the reaction is exothermic overall).下降至产物能级（标记"$\mathrm{NO_2 + O_2 + NO}$"），低于反应物能级（总反应放热）。
Both transition states are local maxima; the intermediate is a local minimum. $E_{a,1} > E_{a,2}$ because step 1 is the slow step.两个过渡态都是局部极大点；中间体是局部极小点。由于第 1 步是慢步，故 $E_{a,1} > E_{a,2}$。
(d) Catalyst lowering $E_{a,1}$ only.只降低 $E_{a,1}$ 的催化剂。
On the same axes, draw a dashed profile that follows the original through the reactant level, then takes a lower path over a reduced TS1, dips back to the same intermediate level, and continues unchanged through TS2 to the same product level.在同一坐标系上用虚线画出修改后的曲线：先沿原曲线穿过反应物能级，然后走较低的路径越过被降低的 TS1，回到同样的中间体能级，再不变地穿过 TS2 到达同样的产物能级。
Explanation: because step 1 is the rate-determining step, its activation energy controls the overall rate. Lowering $E_{a,1}$ shifts the MB tail-fraction that can clear the barrier substantially upward, multiplying the slow-step rate. Lowering $E_{a,2}$ instead would have negligible effect — step 2 is already fast.解释：因为第 1 步是决速步，其活化能控制总反应速率。降低 $E_{a,1}$ 会显著提高能越过该势垒的 MB 尾部比例，从而成倍提升慢步速率。反之降低 $E_{a,2}$ 几乎无影响 —— 第 2 步本就已经很快。
PART III  ·  PAPER 3 HL第三部分  ·  第三卷 HLData-Based — Worked Solution数据题 —— 详细解析

Data-Based Question (HL)数据题（HL）

P3-1HARDPaper 3 HLIodination of Propanone

Iodination of propanone in acid; $[\mathrm{H^+}]$ fixed at $1.00~\mathrm{mol\,dm^{-3}}$. Runs 1–3 at 298 K; run 4 at 318 K with same concentrations as run 1.酸性条件下丙酮的碘化反应；$[\mathrm{H^+}]$ 固定为 $1.00~\mathrm{mol\,dm^{-3}}$。第 1–3 组在 298 K；第 4 组在 318 K，浓度同第 1 组。

(a) Orders.反应级数。
Runs 1 → 2: [propanone] doubles (0.40 → 0.80), [I₂] fixed. Rate doubles (3.5e-5 → 7.0e-5) ⇒ order in propanone = 1.第 1 → 2 组：[丙酮] 加倍（0.40 → 0.80），[I₂] 不变。速率加倍（3.5e-5 → 7.0e-5）⇒ 关于丙酮的级数 = 1。
Runs 2 → 3: [I₂] doubles (0.005 → 0.010), [propanone] fixed. Rate unchanged ⇒ order in I₂ = 0.第 2 → 3 组：[I₂] 加倍（0.005 → 0.010），[丙酮] 不变。速率不变 ⇒ 关于 I₂ 的级数 = 0。
Surprising: I₂ appears in the balanced equation but the reaction is zero-order in I₂ — so I₂ is not involved in the rate-determining step. Consistent with a mechanism in which the slow step is the acid-catalysed enolisation of propanone, with the iodine attacking the enol very fast in a later step.反常之处：I₂ 出现在配平方程中，但反应关于 I₂ 是零级 —— 说明 I₂ 不参与决速步。这与如下机理一致：慢步是酸催化下丙酮的烯醇化（enolisation），随后的快步中碘迅速进攻烯醇。
(b) Rate equation and $k_\text{obs}$ at 298 K.速率方程与 298 K 下的 $k_\text{obs}$。
$$\text{rate} = k_\text{obs}\,[\mathrm{CH_3COCH_3}]$$
Solve for $k_\text{obs}$ from run 1:由第 1 组求 $k_\text{obs}$：
$$k_\text{obs} = \dfrac{\text{rate}}{[\mathrm{CH_3COCH_3}]} = \dfrac{3.5 \times 10^{-5}}{0.40} = 8.75 \times 10^{-5}~\mathrm{s^{-1}}$$
First-order $k$ has units $\mathrm{s^{-1}}$. ($k_\text{obs}$ absorbs the fixed $[\mathrm{H^+}]$ contribution; the full mechanism would unpack $k_\text{obs} = k\,[\mathrm{H^+}]$.)一级反应的 $k$ 单位为 $\mathrm{s^{-1}}$。（$k_\text{obs}$ 把固定的 $[\mathrm{H^+}]$ 贡献吸收进来；完整机理可写为 $k_\text{obs} = k\,[\mathrm{H^+}]$。）
(c) Activation energy from runs 1 (298 K) and 4 (318 K).利用第 1 组（298 K）与第 4 组（318 K）求活化能。
$$k_\text{obs}(318) = \dfrac{1.05 \times 10^{-4}}{0.40} = 2.625 \times 10^{-4}~\mathrm{s^{-1}}$$
$$\dfrac{k_2}{k_1} = \dfrac{2.625 \times 10^{-4}}{8.75 \times 10^{-5}} = 3.0,\quad \ln 3 = 1.099$$
$$\dfrac{1}{T_2} - \dfrac{1}{T_1} = \dfrac{1}{318} - \dfrac{1}{298} = -2.11 \times 10^{-4}~\mathrm{K^{-1}}$$
$$E_a = -\dfrac{R\,\ln(k_2/k_1)}{1/T_2 - 1/T_1} = -\dfrac{(8.31)(1.099)}{-2.11 \times 10^{-4}} \approx 43{,}300~\mathrm{J\,mol^{-1}}$$
$$E_a \approx 43.3~\mathrm{kJ\,mol^{-1}}$$
(d) Systematic errors and precision check.系统误差与精度检验。
Two distinct systematic errors:两个不同的系统误差：
  • Temperature drift — without an actively-thermostatted bath, the temperature rises or falls during a run, biasing the rate (rates roughly double per 10 K).温度漂移 —— 若没有恒温水浴主动控温，运行过程中温度会上升或下降，使速率产生偏差（每升高 10 K 速率约翻倍）。
  • Spectrophotometer calibration — a wavelength offset or zero-baseline error gives a consistent over- or under-read of [I₂], biasing the slope of $A$ vs $t$ and hence the rate.分光光度计校准 —— 波长偏移或零基线误差会导致对 [I₂] 的读数始终偏高或偏低，使 $A$-$t$ 图的斜率（即速率）出现系统偏差。
(Other valid systematic errors: dead-time between mixing and recording; Beer–Lambert breakdown at high [I₂]; trace impurity in the propanone catalysing or inhibiting the reaction.)（其他可接受的系统误差：混合到开始记录之间的死时间；高 [I₂] 下 Beer–Lambert 定律失效；丙酮中痕量杂质催化或抑制反应。）
Precision check: repeat each run independently three times under the same nominal conditions and report the mean ± standard deviation of the initial rate. Large run-to-run spread signals random error (poor precision); a consistent offset between mean and theoretical value signals systematic error (poor accuracy).精度检验：在相同名义条件下独立重复每组实验三次，给出初始速率的均值 ± 标准差。组间差异大说明随机误差大（精度低）；均值与理论值之间存在稳定偏差则说明系统误差（准确度低）。