PART I · PAPER 1 — CONCEPTUAL STRETCHES第一部分 · 第一卷 —— 概念延伸No calculator · multiple choice · 8 marks不可使用计算器 · 选择题 · 8 分
Multiple Choice (Hard)选择题（困难）
Each item carries 1 mark. No calculator, no data booklet. Every item requires interpretation or short reasoning — there are no factual lookups here. HL content (mechanism, rate equation, Arrhenius) is assumed throughout.每题 1 分。不可使用计算器与数据手册（data booklet）。每题都要求推理或解释 —— 不含纯记忆性内容。本卷默认掌握 HL 内容（反应机理、速率方程、阿伦尼乌斯方程）。
C1HARDPaper 12.2.1 Tangent Rate[1]
A student records the volume $V$ of $\mathrm{H_2}$ collected over a gas burette during the reaction $\mathrm{Zn(s) + 2HCl(aq) \to ZnCl_2(aq) + H_2(g)}$. The burette already contained 5.0 cm³ of trapped air at $t = 0$, so the recorded curve does not pass through the origin. To obtain the initial rate of reaction, the student should学生在 $\mathrm{Zn(s) + 2HCl(aq) \to ZnCl_2(aq) + H_2(g)}$ 反应中用气体量筒（gas burette）记录 $\mathrm{H_2}$ 的体积 $V$。$t = 0$ 时筒内已封存 5.0 cm³ 空气，因此记录曲线不过原点。要得到初始反应速率，学生应
(A)read $V$ at $t = 30~\mathrm{s}$, divide by $30~\mathrm{s}$, and call it the initial rate.读取 $t = 30~\mathrm{s}$ 时的 $V$，除以 $30~\mathrm{s}$，作为初始速率。
(B)draw a tangent at $t = 0$ to the curve after subtracting the 5.0 cm³ offset, then report its slope.先扣除 5.0 cm³ 偏移后在曲线上 $t = 0$ 处作切线，给出其斜率。
(C)average the slope over the entire curve, since instantaneous rates are unreliable.对整条曲线取平均斜率，因为瞬时速率不可靠。
(D)ignore the offset because it cancels when computing $\Delta V / \Delta t$.忽略偏移，因为计算 $\Delta V / \Delta t$ 时会自然抵消。
C2HARDPaper 1HL2.2.10 Half-Life ↔ Order[1]
For the reaction $\mathrm{A \to \text{products}}$, a student runs two experiments under identical conditions except for $[\mathrm{A}]_0$. When $[\mathrm{A}]_0$ is halved from $0.20$ to $0.10~\mathrm{mol\,dm^{-3}}$, the measured half-life $t_{1/2}$ doubles. What is the order of reaction with respect to A?在反应 $\mathrm{A \to \text{products}}$ 中，学生在除 $[\mathrm{A}]_0$ 之外的相同条件下做两组实验。将 $[\mathrm{A}]_0$ 由 $0.20$ 减半至 $0.10~\mathrm{mol\,dm^{-3}}$ 时，测得半衰期（half-life）$t_{1/2}$ 加倍。关于 A 的反应级数是？
(A) 0
(B) 1
(C) 2
(D)cannot be determined无法确定
C3HARDPaper 1HL2.2.9 Pseudo-First-Order[1]
The hydrolysis of a small amount of an ester is studied in pure water as solvent. The true rate equation is $\text{rate} = k\,[\mathrm{ester}]\,[\mathrm{H_2O}]$, but the observed rate equation reduces to $\text{rate} = k_\text{obs}\,[\mathrm{ester}]$. The best explanation is that在以纯水为溶剂时研究少量酯（ester）的水解（hydrolysis）。真实速率方程为 $\text{rate} = k\,[\mathrm{ester}]\,[\mathrm{H_2O}]$，但实验得到 $\text{rate} = k_\text{obs}\,[\mathrm{ester}]$。最佳解释是
(A)water is the solvent and never appears in any rate equation.水作为溶剂，永远不出现在速率方程中。
(B)water is in such large excess that $[\mathrm{H_2O}]$ is effectively constant; it folds into $k_\text{obs}$, giving pseudo-first-order behaviour.水大幅过量，$[\mathrm{H_2O}]$ 实际上视为常数；它被并入 $k_\text{obs}$，呈现准一级（pseudo-first-order）行为。
(C)the order of reaction with respect to water is genuinely zero in dilute solution.在稀溶液中关于水的反应级数实际上就是零。
(D)the experimental method cannot measure the order in water, so it is set to zero by convention.实验方法无法测定关于水的级数，按惯例取零。
C4HARDPaper 1HL2.2.7 Mechanism ↔ Rate Law[1]
A reaction $\mathrm{2A + B \to P}$ obeys the experimental rate equation $\text{rate} = k\,\dfrac{[\mathrm{A}]^2}{[\mathrm{B}]}$ (B in the denominator). Which mechanism is consistent with this rate law?反应 $\mathrm{2A + B \to P}$ 实验测得速率方程为 $\text{rate} = k\,\dfrac{[\mathrm{A}]^2}{[\mathrm{B}]}$（B 在分母）。下列哪一机理与该速率方程一致？
(A)A single elementary step: $\mathrm{2A + B \to P}$.单步基元反应：$\mathrm{2A + B \to P}$。
(B)A fast pre-equilibrium $\mathrm{A \rightleftharpoons X + B}$ followed by slow $\mathrm{X + A \to P}$, where X is an intermediate.快速预平衡 $\mathrm{A \rightleftharpoons X + B}$，随后慢步 $\mathrm{X + A \to P}$，X 为中间体。
(C)A slow step $\mathrm{A + B \to Y}$ followed by fast $\mathrm{Y + A \to P}$.慢步 $\mathrm{A + B \to Y}$，随后快步 $\mathrm{Y + A \to P}$。
(D)No mechanism can give a negative order in a reactant.没有机理可以给出对反应物的负级数。
C5HARDPaper 12.2.5 Catalyst Poisoning[1]
A car's three-way catalytic converter uses Pt nanoparticles dispersed on a ceramic honeycomb to oxidise CO to $\mathrm{CO_2}$. When the car is run on leaded fuel, Pb atoms bind irreversibly to Pt surface sites. The rate of CO oxidation falls, even though the bulk amount of Pt in the converter is unchanged. The best explanation is that lead汽车的三元催化转化器（three-way catalytic converter）用 Pt 纳米颗粒分散在陶瓷蜂窝（ceramic honeycomb）上，将 CO 氧化为 $\mathrm{CO_2}$。当使用含铅汽油时，Pb 原子不可逆地结合在 Pt 表面位点上。即便转化器内 Pt 总量不变，CO 氧化速率仍下降。最佳解释是铅
(A)raises the activation energy of every elementary step uniformly.把每一个基元反应的活化能均匀地提高。
(B)reduces the number of accessible active sites, so collisions with the Pt surface become less frequent.减少了可用的活性位点（active site）数量，使粒子与 Pt 表面碰撞的频次下降。
(C)lowers the temperature of the exhaust gas, decreasing the kinetic energy of CO molecules.降低尾气温度，使 CO 分子的动能下降。
(D)changes the enthalpy change $\Delta H$ of the CO oxidation reaction.改变 CO 氧化反应的焓变 $\Delta H$。
C6MEDIUMPaper 1HL2.2.10 Enzyme Saturation[1]
An enzyme-catalysed reaction is studied by varying the substrate concentration $[\mathrm{S}]$ at constant total enzyme. A plot of initial rate $r_0$ versus $[\mathrm{S}]$ rises linearly at low $[\mathrm{S}]$, then curves over to a plateau at high $[\mathrm{S}]$. In the linear region, the reaction is — with respect to $[\mathrm{S}]$ — and in the plateau region it is在保持总酶量不变的情况下改变底物浓度 $[\mathrm{S}]$，研究某酶催化反应。初始速率 $r_0$ 对 $[\mathrm{S}]$ 的图在低 $[\mathrm{S}]$ 时呈线性上升，在高 $[\mathrm{S}]$ 时趋于平台。该反应对 $[\mathrm{S}]$ 在线性区为——，在平台区为
A reaction proceeds by the two-step mechanism反应按以下两步机理进行
Step 1 (fast equilibrium): $\mathrm{A + B \rightleftharpoons X}$, equilibrium constant $K_\text{eq}$第 1 步（快平衡）：$\mathrm{A + B \rightleftharpoons X}$，平衡常数 $K_\text{eq}$
Step 2 (slow): $\mathrm{X + C \to P}$, rate constant $k$第 2 步（慢）：$\mathrm{X + C \to P}$，速率常数 $k$
The experimental rate equation, written in terms of measured species concentrations $[\mathrm{A}]$, $[\mathrm{B}]$, $[\mathrm{C}]$ (the intermediate X cannot be measured), is用可测物种浓度 $[\mathrm{A}]$、$[\mathrm{B}]$、$[\mathrm{C}]$ 表达的实验速率方程为（中间体 X 不可直接测得）
A first-order decomposition $\mathrm{A \to 2\,B}$ has half-life $60~\mathrm{s}$. Starting from $[\mathrm{A}]_0 = 0.40~\mathrm{mol\,dm^{-3}}$ in a closed flask of fixed volume (with no B initially present), the concentration of B at $t = 180~\mathrm{s}$ is一级分解反应 $\mathrm{A \to 2\,B}$ 的半衰期为 $60~\mathrm{s}$。在定容封闭烧瓶中，初始 $[\mathrm{A}]_0 = 0.40~\mathrm{mol\,dm^{-3}}$、不含 B。$t = 180~\mathrm{s}$ 时 B 的浓度为
(A) $0.050~\mathrm{mol\,dm^{-3}}$
(B) $0.35~\mathrm{mol\,dm^{-3}}$
(C) $0.70~\mathrm{mol\,dm^{-3}}$
(D) $0.80~\mathrm{mol\,dm^{-3}}$
PART II · PAPER 2 — STRUCTURED RESPONSE第二部分 · 第二卷 —— 结构化解答Calculator + data booklet · 3 multi-step problems · 36 marks可使用计算器与数据手册 · 三道多步题 · 36 分
Structured Response (Hard)结构化解答题（困难）
Show all working in the space provided. Method marks are awarded even if a final number is wrong. State units; give answers to 2–3 significant figures unless the problem indicates otherwise. Where the problem specifies a chemical context (atmospheric ozone, persulfate iodide clock), the chemistry described is real.在指定区域写出全部解题过程。即便最终数值有误，方法正确仍可得分。注意单位；除题目另有要求外，保留 2–3 位有效数字。题目设定的化学情境（如平流层臭氧、过硫酸根–碘离子时钟）均取自真实化学。
SR 1HARDPaper 2HLCHALLENGE2.2.5–8 + 2.1 — Stratospheric Ozone Catalysis[12]
In the lower stratosphere ($T \approx 220~\mathrm{K}$), ozone ($\mathrm{O_3}$) is destroyed via two competing pathways:在下平流层（lower stratosphere，$T \approx 220~\mathrm{K}$），臭氧（$\mathrm{O_3}$）通过两个相互竞争的途径被破坏：
(Atomic O in the stratosphere is produced by UV photolysis of $\mathrm{O_2}$ and is present at low but steady concentration.)（平流层中原子 O 由 $\mathrm{O_2}$ 经紫外光解（UV photolysis）产生，浓度低但稳定。）
(a)For the catalysed pathway, identify (i) the catalyst, (ii) the reaction intermediate, and (iii) the net overall reaction obtained by summing steps 1 and 2.对氯催化途径：指出 (i) 催化剂、(ii) 反应中间体、(iii) 将第 1、2 步相加所得的净总反应。[3]
(b)Write the rate equation implied by the catalysed mechanism (rate of $\mathrm{O_3}$ destruction in terms of the species in step 1).写出氯催化机理蕴含的速率方程（用第 1 步中物种表达 $\mathrm{O_3}$ 被消耗的速率）。[1]
(c)On the same set of axes, sketch a single Maxwell–Boltzmann distribution for collision kinetic energies at $T = 220~\mathrm{K}$, and mark on it two vertical lines: one at $E_{a,\text{cat}} = 2.2~\mathrm{kJ\,mol^{-1}}$ and one at $E_{a,\text{uncat}} = 17.1~\mathrm{kJ\,mol^{-1}}$. Shade the area to the right of each line. Using these shaded areas, explain in one or two sentences why the catalysed channel dominates ozone destruction at stratospheric $T$.在同一坐标系上画出 $T = 220~\mathrm{K}$ 下碰撞动能的麦克斯韦–玻尔兹曼分布，并在曲线上标出两条竖线：一条在 $E_{a,\text{cat}} = 2.2~\mathrm{kJ\,mol^{-1}}$，另一条在 $E_{a,\text{uncat}} = 17.1~\mathrm{kJ\,mol^{-1}}$。给每条竖线右侧面积上阴影。利用这两块阴影面积，用一两句话解释为何氯催化途径在平流层温度下主导臭氧的破坏。[4]
(d)Stratospheric measurements suggest each $\mathrm{Cl}$ atom turns over the cycle $\sim 10^{5}$ times before being terminated (e.g. by reaction with $\mathrm{CH_4}$ to form HCl). Estimate the mass of stratospheric $\mathrm{O_3}$ ($M = 48.00~\mathrm{g\,mol^{-1}}$) that could be destroyed by $1.00~\mathrm{mg}$ of chlorine atoms ($M_\mathrm{Cl} = 35.45~\mathrm{g\,mol^{-1}}$) before deactivation. (This part links 2.2 catalysis with 2.1 stoichiometry.)平流层观测表明，每个 $\mathrm{Cl}$ 原子在被终止前（例如与 $\mathrm{CH_4}$ 反应生成 HCl）平均完成约 $10^{5}$ 次催化循环。请估算 $1.00~\mathrm{mg}$ 氯原子（$M_\mathrm{Cl} = 35.45~\mathrm{g\,mol^{-1}}$）在被终止前可破坏的平流层 $\mathrm{O_3}$（$M = 48.00~\mathrm{g\,mol^{-1}}$）质量。（本小题将 2.2 的催化与 2.1 的化学计量联系起来。）[4]
SR 2HARDPaper 2HLCHALLENGE2.2.10–11 + 2.1 — Pressure-Rise Kinetics of N₂O₅[12]
The gas-phase decomposition of dinitrogen pentoxide,五氧化二氮的气相分解反应：
$\mathrm{2\,N_2O_5(g) \to 4\,NO_2(g) + O_2(g)}$
is followed in a sealed, rigid 1.00 dm³ vessel held at 338 K. Only $\mathrm{N_2O_5}$ is present initially ($P_0 = 50.0~\mathrm{kPa}$). The total pressure $P_\text{tot}$ is recorded at four times:在 338 K 的密闭刚性容器（体积 1.00 dm³）内进行；初始仅含 $\mathrm{N_2O_5}$（$P_0 = 50.0~\mathrm{kPa}$）。在四个时刻记录总压 $P_\text{tot}$：
$t$ (s)
0
600
1200
1800
$P_\text{tot}$ (kPa)
50.0
65.0
77.0
86.6
(a)Let $x$ be the fraction of $\mathrm{N_2O_5}$ decomposed at time $t$. Derive an expression for $P_\text{tot}$ in terms of $P_0$ and $x$, and hence show that设 $t$ 时刻 $\mathrm{N_2O_5}$ 的分解分数为 $x$。用 $P_0$ 与 $x$ 写出 $P_\text{tot}$ 的表达式，从而证明
(b)Use the four data points to demonstrate that the reaction is first order in $\mathrm{N_2O_5}$, and determine the rate constant $k$ at 338 K. State the units of $k$.利用四组数据证明反应关于 $\mathrm{N_2O_5}$ 为一级，并求 338 K 下的速率常数 $k$，写出 $k$ 的单位。[4]
(c)Calculate the half-life $t_{1/2}$, and predict $P_\text{tot}$ at $t = 5\,t_{1/2}$.计算半衰期 $t_{1/2}$，并预测 $t = 5\,t_{1/2}$ 时的 $P_\text{tot}$。[3]
(d)At very long $t$, the experimenter measures $P_\text{tot} \approx 119~\mathrm{kPa}$, slightly lower than the value predicted by complete decomposition. Suggest one chemical and one experimental reason for this small deficit. (Hint for the chemical reason: $\mathrm{NO_2}$ in cold gas establishes its own equilibrium.)长时间后，实验者测得 $P_\text{tot} \approx 119~\mathrm{kPa}$，略低于完全分解所预测的值。请各给出一个化学原因与一个实验原因解释此微小亏损。（化学方面提示：低温下 $\mathrm{NO_2}$ 会建立自身的平衡。）[2]
SR 3HARDPaper 2HLCHALLENGE2.2.9–12 — Persulfate–Iodide Clock + Catalysis[12]
The peroxodisulfate–iodide reaction过二硫酸根–碘离子（peroxodisulfate–iodide）反应：
is followed by an iodine-clock method (a small fixed amount of thiosulfate plus starch indicator gives a sharp endpoint when $\mathrm{S_2O_3^{2-}}$ is exhausted, defining the initial-rate window). All runs are at constant ionic strength.采用碘时钟法（iodine-clock method）跟踪反应：加入少量定量的硫代硫酸根（thiosulfate）与淀粉指示剂，当 $\mathrm{S_2O_3^{2-}}$ 被消耗完时给出显著终点，从而确定初始速率窗口。所有实验在恒定离子强度（ionic strength）下进行。
All four runs are at 298 K.四组实验均在 298 K 下进行。
Run
$[\mathrm{S_2O_8^{2-}}]_0$ (mol dm⁻³)
$[\mathrm{I^-}]_0$ (mol dm⁻³)
Initial rate (mol dm⁻³ s⁻¹)
1
0.010
0.010
$2.0 \times 10^{-6}$
2
0.020
0.010
$4.0 \times 10^{-6}$
3
0.020
0.020
$8.0 \times 10^{-6}$
(a)Use runs 1–3 to determine the order of reaction with respect to $\mathrm{S_2O_8^{2-}}$ and to $\mathrm{I^-}$. Write the rate equation, calculate $k$ at 298 K, and give its units.利用第 1–3 组数据确定关于 $\mathrm{S_2O_8^{2-}}$ 与 $\mathrm{I^-}$ 的反应级数。写出速率方程，计算 298 K 下的 $k$，并给出其单位。[4]
(b)Propose a two-step mechanism consistent with this rate law. Label the rate-determining step and identify any intermediate.提出与该速率方程一致的两步机理。标出决速步，并指出中间体。[3]
(c)In a new experiment at 298 K with $[\mathrm{S_2O_8^{2-}}]_0 = 0.050~\mathrm{mol\,dm^{-3}}$ and $[\mathrm{I^-}]_0 = 0.030~\mathrm{mol\,dm^{-3}}$, the iodine-clock endpoint (blue-black starch–$\mathrm{I_2}$ colour) is observed when $1.0 \times 10^{-3}~\mathrm{mol\,dm^{-3}}$ of $\mathrm{I_2}$ has been produced (set by the small amount of thiosulfate added). Predict the time at which the endpoint will appear, assuming initial-rate conditions hold throughout. State one assumption that limits the validity of your answer.在 298 K 下的新实验中，初始浓度为 $[\mathrm{S_2O_8^{2-}}]_0 = 0.050~\mathrm{mol\,dm^{-3}}$、$[\mathrm{I^-}]_0 = 0.030~\mathrm{mol\,dm^{-3}}$；当产生 $1.0 \times 10^{-3}~\mathrm{mol\,dm^{-3}}$ 的 $\mathrm{I_2}$（由所加少量硫代硫酸根决定）时，碘时钟终点（蓝黑色淀粉–$\mathrm{I_2}$）出现。在全程保持初始速率条件的假设下，预测终点出现的时间。并指出一项限制该答案有效性的假设。[3]
(d)A trace of $\mathrm{Fe^{3+}(aq)}$ is added. The measured rate increases sharply, but a final-state analysis shows the products and stoichiometry are identical to the uncatalysed reaction. Explain in one or two sentences (i) how $\mathrm{Fe^{3+}}$ catalyses the reaction (the catalytic cycle uses $\mathrm{Fe^{3+}/Fe^{2+}}$), and (ii) why this constitutes homogeneous catalysis rather than heterogeneous.向体系中加入痕量 $\mathrm{Fe^{3+}(aq)}$。测得速率显著上升，但产物与化学计量关系与无催化反应完全一致。用一两句话解释 (i) $\mathrm{Fe^{3+}}$ 如何催化反应（催化循环利用 $\mathrm{Fe^{3+}/Fe^{2+}}$）、(ii) 为何这是均相催化（homogeneous catalysis）而非异相催化。[2]
PART III · PAPER 3 HL — RESEARCH DATA第三部分 · 第三卷 HL —— 研究数据Calculator + data booklet · one extended data-based problem · 18 marks可使用计算器与数据手册 · 一道延伸数据题 · 18 分
Data-Based Question — Iodide-Catalysed Decomposition of $\mathrm{H_2O_2}$数据题 —— $\mathrm{H_2O_2}$ 的碘离子催化分解
This problem couples 2.2 method-of-initial-rates with integrated-rate-law analysis, catalysis interrogation, and a 2.1 stoichiometric end-state calculation. The iodide-catalysed decomposition of hydrogen peroxide is a classic HL kinetics study system.本题将 2.2 的初始速率法（method of initial rates）与积分速率方程分析、催化机理的考查以及 2.1 的化学计量末态计算结合。$\mathrm{H_2O_2}$ 的碘离子催化分解是 HL 动力学研究的经典体系。
In aqueous solution, hydrogen peroxide decomposes slowly:在水溶液中，过氧化氢缓慢分解：
$\mathrm{2\,H_2O_2(aq) \to 2\,H_2O(l) + O_2(g)}$
Adding a small amount of potassium iodide accelerates the reaction dramatically without altering the stoichiometry. A student measures the initial rate at 298 K for three runs:加入少量碘化钾后反应大大加速，且总化学计量不变。学生在 298 K 下测得三组初始速率：
Run
$[\mathrm{H_2O_2}]_0$ (mol dm⁻³)
$[\mathrm{I^-}]_0$ (mol dm⁻³)
Initial rate (mol dm⁻³ s⁻¹)
1
0.10
0.010
$5.0 \times 10^{-5}$
2
0.20
0.010
$1.0 \times 10^{-4}$
3
0.10
0.020
$1.0 \times 10^{-4}$
For Run 1 (with $[\mathrm{I^-}]_0 = 0.010~\mathrm{mol\,dm^{-3}}$ held effectively constant by a large excess of KI), the student follows $[\mathrm{H_2O_2}]$ over time:在 Run 1 中（KI 大幅过量，$[\mathrm{I^-}]_0 = 0.010~\mathrm{mol\,dm^{-3}}$ 视为常数），学生跟踪 $[\mathrm{H_2O_2}]$ 随时间的变化：
$t$ (s)
0
500
1000
1500
2000
$[\mathrm{H_2O_2}]$ (mol dm⁻³)
0.100
0.0779
0.0607
0.0472
0.0368
(a)Use runs 1–3 to determine the order of reaction with respect to $\mathrm{H_2O_2}$ and to $\mathrm{I^-}$. Write the rate equation, and calculate the rate constant $k$ at 298 K (state its units).利用第 1–3 组数据确定关于 $\mathrm{H_2O_2}$ 与 $\mathrm{I^-}$ 的反应级数。写出速率方程，并计算 298 K 下的速率常数 $k$（给出单位）。[5]
(b)Propose a two-step mechanism for the catalysed reaction in which $\mathrm{I^-}$ participates. Label the rate-determining step and identify any intermediate. Verify that summing your two steps reproduces the net stoichiometric equation.为该催化反应提出一个使 $\mathrm{I^-}$ 参与的两步机理。标出决速步并指出中间体。验证将两步相加后能再现净化学计量方程。[3]
(c)Use the time-resolved data from Run 1 to verify that the reaction is first order in $\mathrm{H_2O_2}$. (Compute $\ln[\mathrm{H_2O_2}]$ at each $t$; show the differences over equal time intervals are constant.) Hence determine the pseudo-first-order rate constant $k'$ and the half-life $t_{1/2}$ for Run 1. Confirm that $k' \approx k\,[\mathrm{I^-}]_0$.利用 Run 1 的时间数据验证反应关于 $\mathrm{H_2O_2}$ 为一级。（计算每个 $t$ 下的 $\ln[\mathrm{H_2O_2}]$；说明等时间间隔下差值相等。）由此求 Run 1 的准一级速率常数 $k'$ 与半衰期 $t_{1/2}$。验证 $k' \approx k\,[\mathrm{I^-}]_0$。[4]
(d)Explain in one or two sentences why $\mathrm{I^-}$ appears in the rate equation even though it does not appear in the net stoichiometric equation. (Hint: which step does it participate in?)用一两句话解释：为何 $\mathrm{I^-}$ 出现在速率方程中，却不出现在净化学计量方程中？（提示：它参与了哪一步？）[2]
(e)Suggest two distinct experimental observations that, if made, would confirm that $\mathrm{I^-}$ is acting as a catalyst (rather than as a stoichiometric reactant that just happens to be in small amount).提出两项不同的实验观测，若得到该观测结果，可证实 $\mathrm{I^-}$ 是催化剂（而非恰好用量较小的化学计量反应物）。[2]
(f)A separate sample consists of $100~\mathrm{cm^3}$ of $1.00~\mathrm{mol\,dm^{-3}}$ $\mathrm{H_2O_2}$. A trace of KI is added and the mixture is left until decomposition is complete. Calculate the mass of $\mathrm{O_2(g)}$ liberated. (Hint: the catalyst doesn't affect the stoichiometry — this is a 2.1 cross-over.) Take $M(\mathrm{O_2}) = 32.00~\mathrm{g\,mol^{-1}}$.另取 $100~\mathrm{cm^3}$、$1.00~\mathrm{mol\,dm^{-3}}$ 的 $\mathrm{H_2O_2}$ 溶液。加入少量 KI，静置至完全分解。计算释出的 $\mathrm{O_2(g)}$ 质量。（提示：催化剂不影响化学计量 —— 此为 2.1 交叉。）取 $M(\mathrm{O_2}) = 32.00~\mathrm{g\,mol^{-1}}$。[2]