IB Chemistry HL — Reactivity 2: How Much, How Fast, How Far — Challenge Solutions (2.2 Focus, 2.1 Cross-overs)

C1HARDPaper 12.2.1 Tangent Rate

Gas-burette method with 5.0 cm³ trapped air offset at $t = 0$; how to extract the initial rate?气体量筒法，$t = 0$ 时已有 5.0 cm³ 空气偏移；如何获得初始速率？

(A) $V(30)/30$ as initial rate.$V(30)/30$ 作为初始速率。
(B) Subtract 5.0 cm³ offset, then tangent at $t = 0$.先扣除 5.0 cm³ 偏移，再在 $t = 0$ 处作切线。
(C) Average slope over the entire curve.对整条曲线取平均斜率。
(D) Ignore the offset — it cancels.忽略偏移 —— 会自然抵消。

Answer:答案： (B)

"Initial rate" means the instantaneous rate at $t = 0$ — the tangent slope at the start, not an average. The 5.0 cm³ offset shifts every reading uniformly; you must subtract it to get $V(\mathrm{H_2})$. The offset does cancel in $\Delta V$, but it doesn't cancel in the tangent at $t = 0$, because the curve doesn't start at the origin. (A) gives an average over 0–30 s, not an instantaneous rate. (C) averages out the fast initial behaviour we care about. (D) is the seductive wrong answer — true for $\Delta V$ between two times but false at the boundary $t = 0$."初始速率"指的是 $t = 0$ 时的瞬时速率 —— 起始处的切线斜率，而非平均值。5.0 cm³ 偏移把每个读数都均匀抬高，必须先扣除才能得到 $V(\mathrm{H_2})$。偏移在 $\Delta V$ 中可以抵消，但在 $t = 0$ 处作切线时不能抵消，因为曲线不过原点。(A) 给出的是 0–30 s 平均速率，不是瞬时速率。(C) 把我们关心的快速初始行为平均掉了。(D) 是看似合理的错误答案 —— 两时刻之差的 $\Delta V$ 确实可以抵消偏移，但边界处 $t = 0$ 的切线不行。

C2HARDPaper 1HL2.2.10 Half-Life ↔ Order

$t_{1/2}$ doubles when $[\mathrm{A}]_0$ is halved. Order in A?$[\mathrm{A}]_0$ 减半时 $t_{1/2}$ 加倍。关于 A 的级数？

(A) 0
(B) 1
(C) 2
(D) cannot tell无法确定

Answer:答案： (C)

General rule for an $n$-th order reaction: $t_{1/2} \propto [\mathrm{A}]_0^{\,1-n}$.对 $n$ 级反应的通用规律：$t_{1/2} \propto [\mathrm{A}]_0^{\,1-n}$。

$n = 0$: $t_{1/2} \propto [\mathrm{A}]_0$ → halving $[\mathrm{A}]_0$ halves $t_{1/2}$. ✗$n = 0$：$t_{1/2} \propto [\mathrm{A}]_0$ → $[\mathrm{A}]_0$ 减半使 $t_{1/2}$ 减半。✗
$n = 1$: $t_{1/2}$ independent of $[\mathrm{A}]_0$. ✗$n = 1$：$t_{1/2}$ 与 $[\mathrm{A}]_0$ 无关。✗
$n = 2$: $t_{1/2} \propto 1/[\mathrm{A}]_0$ → halving $[\mathrm{A}]_0$ doubles $t_{1/2}$. ✓$n = 2$：$t_{1/2} \propto 1/[\mathrm{A}]_0$ → $[\mathrm{A}]_0$ 减半使 $t_{1/2}$ 加倍。✓

Memorise: zero-order $t_{1/2}$ scales with $[\mathrm{A}]_0$; first-order is independent; second-order scales inversely. Knowing this lets you read off the order from a single concentration-doubling/halving experiment.记住：零级 $t_{1/2}$ 与 $[\mathrm{A}]_0$ 同向；一级无关；二级反比。掌握这一规律即可从一次浓度加倍/减半实验直接读出反应级数。

C3HARDPaper 1HL2.2.9 Pseudo-First-Order

Ester hydrolysis in water: true rate $= k[\text{ester}][\mathrm{H_2O}]$; observed $= k_\text{obs}[\text{ester}]$.酯在水中水解：真实速率 $= k[\text{酯}][\mathrm{H_2O}]$；观察 $= k_\text{obs}[\text{酯}]$。

(A) Water as solvent never appears in any rate equation.水作溶剂时不出现在速率方程中。
(B) Large excess of water → $[\mathrm{H_2O}]$ effectively constant; pseudo-first-order.水大幅过量 → $[\mathrm{H_2O}]$ 视为常数；准一级。
(C) Order in water is genuinely zero.关于水的级数确实是零。
(D) Experimental method can't measure it; set to zero by convention.实验无法测定，按惯例取零。

Answer:答案： (B)

Pure water is $\sim 55.5~\mathrm{mol\,dm^{-3}}$. A few mmol of ester reacting changes $[\mathrm{H_2O}]$ by less than $0.01\%$, so it stays effectively constant. The kinetic equation absorbs the constant into a new rate constant: $k_\text{obs} = k\,[\mathrm{H_2O}]_0$. The observed first-order behaviour is the experimental signature of this approximation — hence "pseudo"-first-order. Genuine zero-order in water (option C) would mean water plays no role at all, which is wrong: water is the actual nucleophile.纯水浓度约 $55.5~\mathrm{mol\,dm^{-3}}$。几毫摩尔酯反应对 $[\mathrm{H_2O}]$ 的影响小于 $0.01\%$，实际上保持不变。动力学方程把这个常数并入新的速率常数：$k_\text{obs} = k\,[\mathrm{H_2O}]_0$。观察到的一级行为正是这一近似的实验印记 —— 故称"准一级"。选项 (C) 即"关于水的级数确实为零"则错了：水正是反应中的真正亲核试剂（nucleophile）。

C4HARDPaper 1HL2.2.7 Mechanism ↔ Rate Law

$\mathrm{2A + B \to P}$, rate $= k[\mathrm{A}]^2/[\mathrm{B}]$. Consistent mechanism?$\mathrm{2A + B \to P}$，rate $= k[\mathrm{A}]^2/[\mathrm{B}]$。一致的机理？

(A) Single elementary $\mathrm{2A + B \to P}$.单步基元 $\mathrm{2A + B \to P}$。
(B) Fast $\mathrm{A \rightleftharpoons X + B}$ then slow $\mathrm{X + A \to P}$.快平衡 $\mathrm{A \rightleftharpoons X + B}$，慢步 $\mathrm{X + A \to P}$。
(C) Slow $\mathrm{A + B \to Y}$ then fast $\mathrm{Y + A \to P}$.慢 $\mathrm{A + B \to Y}$，快 $\mathrm{Y + A \to P}$。
(D) No mechanism gives a negative order.没有机理能给出负级数。

Answer:答案： (B)

A negative order in B is the classic signature of B being a product of a fast pre-equilibrium that occurs before the rate-determining step. For mechanism (B):对 B 呈负级数，是 B 作为决速步之前的快速预平衡产物的经典特征。机理 (B)：

$$K_\text{eq} = \dfrac{[\mathrm{X}][\mathrm{B}]}{[\mathrm{A}]} \;\Longrightarrow\; [\mathrm{X}] = \dfrac{K_\text{eq}[\mathrm{A}]}{[\mathrm{B}]}$$

$$\text{rate} = k_\text{slow}\,[\mathrm{X}]\,[\mathrm{A}] = k_\text{slow}\,K_\text{eq}\,\dfrac{[\mathrm{A}]^2}{[\mathrm{B}]}$$

Matching: $k = k_\text{slow}\,K_\text{eq}$. Order in A = +2, order in B = –1, overall +1.对照可知 $k = k_\text{slow}\,K_\text{eq}$。关于 A 级数 = +2，关于 B 级数 = –1，总级数 +1。 (A) would give rate $= k[\mathrm{A}]^2[\mathrm{B}]$ (positive order in B). (C)'s slow step is bimolecular A + B, giving rate $\propto [\mathrm{A}][\mathrm{B}]$. (D) is just false: when species are consumed in faster steps before the RDS, the rate law can show negative orders.(A) 给出 rate $= k[\mathrm{A}]^2[\mathrm{B}]$（B 为正级数）。(C) 的慢步是 A + B 双分子，给出 rate $\propto [\mathrm{A}][\mathrm{B}]$。(D) 错误：当物种在决速步前的快步中被消耗时，速率方程可呈现负级数。

C5HARDPaper 12.2.5 Catalyst Poisoning

Pb on Pt converter; rate falls.Pt 转化器被 Pb 毒化；速率下降。

(A) Raises $E_a$ of every step.提高每步活化能。
(B) Reduces accessible active sites.减少可用的活性位点。
(C) Lowers exhaust temperature.降低尾气温度。
(D) Changes $\Delta H$.改变 $\Delta H$。

Answer:答案： (B)

A heterogeneous catalyst works at active sites on its surface — specific atomic arrangements where reactant molecules adsorb and react. Pb binds irreversibly to those sites, occupying them so CO molecules can no longer adsorb there. The bulk amount of Pt is unchanged, but the active surface area is reduced, so the rate of CO+O₂ collisions on Pt drops. (A) is wrong because the surviving Pt sites still operate at their normal $E_a$ — it's not the energetics that change, it's the available area. (D) is wrong because $\Delta H$ is a thermodynamic property of the reaction, independent of catalyst.异相催化剂的工作发生在表面的活性位点 —— 反应物分子在此吸附（adsorb）并反应的特定原子排列。Pb 不可逆地结合在这些位点上，使 CO 分子无法再在此吸附。Pt 总量不变，但有效表面积减小，CO 与 $\mathrm{O_2}$ 在 Pt 上的碰撞速率下降。(A) 错误：未被毒化的 Pt 位点仍以原有 $E_a$ 工作 —— 改变的不是能量学，而是可用面积。(D) 错误：$\Delta H$ 是反应的热力学属性，与催化剂无关。

C6MEDIUMPaper 1HL2.2.10 Enzyme Saturation

Enzyme $r_0$ vs $[\mathrm{S}]$: linear at low, plateau at high. Orders?酶的 $r_0$ 对 $[\mathrm{S}]$ 图：低 $[\mathrm{S}]$ 线性，高 $[\mathrm{S}]$ 平台。级数？

(A) 0; 10；1
(B) 1; 01；0
(C) 1; 21；2
(D) 2; 12；1

Answer:答案： (B)

"$r_0$ rises linearly with $[\mathrm{S}]$" is $r_0 \propto [\mathrm{S}]^1$ — first-order in substrate. "$r_0$ plateaus, independent of $[\mathrm{S}]$" is $r_0 \propto [\mathrm{S}]^0$ — zero-order in substrate. Mechanistic story (Michaelis–Menten): at low $[\mathrm{S}]$ enzyme sites are mostly vacant, so adding substrate raises the encounter rate linearly; at high $[\mathrm{S}]$ every enzyme is bound (saturated), so adding more substrate doesn't speed anything up — the rate is set by enzyme turnover. This is the same shape as a Langmuir adsorption isotherm and the same idea as "catalyst surface saturation" in heterogeneous catalysis."$r_0$ 与 $[\mathrm{S}]$ 线性增大"即 $r_0 \propto [\mathrm{S}]^1$ —— 对底物为一级。"$r_0$ 趋于平台，不再依赖 $[\mathrm{S}]$"即 $r_0 \propto [\mathrm{S}]^0$ —— 对底物为零级。机理（Michaelis–Menten）：低 $[\mathrm{S}]$ 时大部分酶位点空闲，增加底物使相遇速率线性升高；高 $[\mathrm{S}]$ 时每个酶都已结合（饱和），再加底物无济于事 —— 速率由酶的周转速率决定。其形状与朗缪尔吸附等温线（Langmuir adsorption isotherm）相同，思想与异相催化中"催化剂表面饱和"完全一致。

C7HARDPaper 1HL2.2.7 Pre-Equilibrium Derivation

Step 1 (fast eq): $\mathrm{A + B \rightleftharpoons X}$, $K_\text{eq}$; Step 2 (slow): $\mathrm{X + C \to P}$, $k$. Rate law in measured species?第 1 步（快平衡）：$\mathrm{A + B \rightleftharpoons X}$，$K_\text{eq}$；第 2 步（慢）：$\mathrm{X + C \to P}$，$k$。以可测物种表达的速率方程？

(A) $\text{rate} = k\,[\mathrm{A}]\,[\mathrm{B}]\,[\mathrm{C}]$
(B) $\text{rate} = k\,K_\text{eq}\,[\mathrm{A}]\,[\mathrm{B}]\,[\mathrm{C}]$
(C) $\text{rate} = k\,[\mathrm{A}]\,[\mathrm{B}]$
(D) $\text{rate} = k\,[\mathrm{X}]\,[\mathrm{C}]$

Answer:答案： (B)

The rate-determining step (step 2) gives the formal rate $= k\,[\mathrm{X}]\,[\mathrm{C}]$ — option (D). But $[\mathrm{X}]$ is the concentration of an unmeasurable intermediate, so we must eliminate it. Applying the equilibrium condition for step 1:决速步（第 2 步）给出形式速率 $= k\,[\mathrm{X}]\,[\mathrm{C}]$ —— 即选项 (D)。但 $[\mathrm{X}]$ 是不可测的中间体浓度，必须消去。由第 1 步的平衡条件：

$$K_\text{eq} = \dfrac{[\mathrm{X}]}{[\mathrm{A}]\,[\mathrm{B}]} \;\Longrightarrow\; [\mathrm{X}] = K_\text{eq}\,[\mathrm{A}]\,[\mathrm{B}]$$

$$\therefore \text{rate} = k\,[\mathrm{X}]\,[\mathrm{C}] = k\,K_\text{eq}\,[\mathrm{A}]\,[\mathrm{B}]\,[\mathrm{C}]$$

So the observed rate constant is the product $k_\text{obs} = k\,K_\text{eq}$. Option (A) drops the equilibrium constant — a common error. Option (C) ignores [C] from the slow step. Option (D) is formally true but useless experimentally because [X] is not measurable.故观察速率常数是乘积 $k_\text{obs} = k\,K_\text{eq}$。选项 (A) 漏掉了平衡常数 —— 常见错误。选项 (C) 漏掉了来自慢步的 [C]。选项 (D) 形式上正确，但因 [X] 不可测，实验上无用。

C8HARDPaper 1HL2.2.10 Half-Life + Stoichiometry

First-order $\mathrm{A \to 2\,B}$; $t_{1/2} = 60~\mathrm{s}$; $[\mathrm{A}]_0 = 0.40$. $[\mathrm{B}]$ at $t = 180~\mathrm{s}$?一级 $\mathrm{A \to 2\,B}$；$t_{1/2} = 60~\mathrm{s}$；$[\mathrm{A}]_0 = 0.40$。$t = 180~\mathrm{s}$ 时 $[\mathrm{B}]$？

(A) $0.050~\mathrm{mol\,dm^{-3}}$
(B) $0.35~\mathrm{mol\,dm^{-3}}$
(C) $0.70~\mathrm{mol\,dm^{-3}}$
(D) $0.80~\mathrm{mol\,dm^{-3}}$

Answer:答案： (C)

Step 1 — find $[\mathrm{A}]$ at $t = 180~\mathrm{s}$. For a first-order reaction, $t_{1/2}$ is independent of concentration, so each successive 60 s halves $[\mathrm{A}]$:第 1 步 —— 求 $t = 180~\mathrm{s}$ 时的 $[\mathrm{A}]$。一级反应的 $t_{1/2}$ 与浓度无关，故每过 60 s $[\mathrm{A}]$ 减半：

$$180~\mathrm{s} = 3\,t_{1/2} \;\Longrightarrow\; [\mathrm{A}](180) = (1/2)^3 \times 0.40 = 0.050~\mathrm{mol\,dm^{-3}}$$

Step 2 — use stoichiometry to get $[\mathrm{B}]$. Amount of A consumed:第 2 步 —— 用化学计量求 $[\mathrm{B}]$。A 被消耗的量：

$$\Delta[\mathrm{A}] = 0.40 - 0.050 = 0.35~\mathrm{mol\,dm^{-3}}$$

From $\mathrm{A \to 2\,B}$: each mole of A consumed produces 2 mol B, so由 $\mathrm{A \to 2\,B}$：每消耗 1 mol A 生成 2 mol B，故

$$[\mathrm{B}](180) = 2 \times 0.35 = 0.70~\mathrm{mol\,dm^{-3}}$$

Trap (A): the residual $[\mathrm{A}]$, not $[\mathrm{B}]$. Trap (B): forgetting the factor of 2 from stoichiometry. Trap (D): assuming complete reaction (which would give $2 \times 0.40 = 0.80$). The two-step structure — half-life arithmetic then stoichiometric scaling — is the whole skill.陷阱 (A)：写成残余的 $[\mathrm{A}]$，不是 $[\mathrm{B}]$。陷阱 (B)：忘记化学计量的 2 倍因子。陷阱 (D)：假设反应完全（即 $2 \times 0.40 = 0.80$）。两步结构 —— 先做半衰期算术，再做化学计量换算 —— 正是本题考查的全部技能。

SR 1HARDPaper 2HLCHALLENGEStratospheric Ozone Catalysis

Cl-catalysed vs uncatalysed $\mathrm{O_3}$ destruction in the stratosphere at 220 K, with $E_{a,\text{cat}} = 2.2~\mathrm{kJ\,mol^{-1}}$ and $E_{a,\text{uncat}} = 17.1~\mathrm{kJ\,mol^{-1}}$.平流层 220 K 下，氯催化与无催化 $\mathrm{O_3}$ 销毁，$E_{a,\text{cat}} = 2.2~\mathrm{kJ\,mol^{-1}}$、$E_{a,\text{uncat}} = 17.1~\mathrm{kJ\,mol^{-1}}$。

(a) Identifying species.物种的识别。

(i) Catalyst: $\mathrm{Cl}$ — consumed in step 1, regenerated in step 2, so its concentration is unchanged across one full cycle.(i) 催化剂：$\mathrm{Cl}$ —— 第 1 步被消耗，第 2 步重新生成，单次循环后浓度不变。
(ii) Intermediate: $\mathrm{ClO}$ — formed in step 1, consumed in step 2; does not appear in the net equation.(ii) 中间体：$\mathrm{ClO}$ —— 第 1 步生成，第 2 步被消耗；不出现在净反应中。
(iii) Net reaction (add steps 1 and 2; cancel $\mathrm{Cl}$ and $\mathrm{ClO}$ on both sides):(iii) 净反应（将第 1、2 步相加，两边消去 $\mathrm{Cl}$ 与 $\mathrm{ClO}$）：
$$\mathrm{O_3 + O \to 2\,O_2}$$
— identical to the uncatalysed reaction, as it must be (a catalyst cannot change the overall stoichiometry).—— 与无催化反应完全一致，必然如此（催化剂不能改变总化学计量）。

(b) Implied rate equation. The slow step is bimolecular in $\mathrm{Cl}$ and $\mathrm{O_3}$:机理蕴含的速率方程。慢步对 $\mathrm{Cl}$ 与 $\mathrm{O_3}$ 双分子：

$$-\dfrac{d[\mathrm{O_3}]}{dt} = k_\text{cat}\,[\mathrm{Cl}]\,[\mathrm{O_3}]$$

(c) Maxwell–Boltzmann comparison at 220 K.220 K 下的麦克斯韦–玻尔兹曼分布比较。

Sketch (description): Draw a single MB curve for collision kinetic energy at $T = 220~\mathrm{K}$ — the usual asymmetric distribution rising from zero, peaking, then decaying with a long high-energy tail. On the energy axis, mark two vertical lines:作图（描述）：画出 $T = 220~\mathrm{K}$ 下碰撞动能的单条 MB 曲线 —— 由零升起、达峰后衰减并带长高能尾的不对称分布。在能量轴上画两条竖线：

$E_{a,\text{cat}} = 2.2~\mathrm{kJ\,mol^{-1}}$ — far to the left, well below the peak of the distribution at 220 K (since $RT \approx 1.83~\mathrm{kJ\,mol^{-1}}$, this barrier sits just past the peak).$E_{a,\text{cat}} = 2.2~\mathrm{kJ\,mol^{-1}}$ —— 远在左侧，远低于 220 K 分布的峰位（$RT \approx 1.83~\mathrm{kJ\,mol^{-1}}$，故此势垒刚越过峰）。
$E_{a,\text{uncat}} = 17.1~\mathrm{kJ\,mol^{-1}}$ — far to the right, deep in the high-energy tail.$E_{a,\text{uncat}} = 17.1~\mathrm{kJ\,mol^{-1}}$ —— 远在右侧，处于高能尾深处。

Shaded areas (= fraction of collisions with enough energy to react):阴影面积（= 能量足以反应的碰撞所占比例）：

Area to the right of $E_{a,\text{cat}}$: nearly the entire distribution — almost every $\mathrm{Cl}$–$\mathrm{O_3}$ collision has enough energy.$E_{a,\text{cat}}$ 右侧面积：几乎涵盖整条分布 —— 几乎每一次 $\mathrm{Cl}$–$\mathrm{O_3}$ 碰撞都具有足够能量。
Area to the right of $E_{a,\text{uncat}}$: a tiny sliver in the tail — only the rare high-energy $\mathrm{O}$–$\mathrm{O_3}$ collisions clear the barrier.$E_{a,\text{uncat}}$ 右侧面积：仅是尾部一小片 —— 只有罕见的高能 $\mathrm{O}$–$\mathrm{O_3}$ 碰撞能越过势垒。

One-sentence explanation: At 220 K the high-energy tail of the MB distribution is depleted, so the fraction of collisions clearing the high 17.1 kJ mol⁻¹ uncatalysed barrier is vanishingly small — meanwhile virtually every Cl–O₃ collision clears the tiny 2.2 kJ mol⁻¹ catalysed barrier, so the catalysed channel dominates ozone destruction even when $[\mathrm{Cl}]$ is parts-per-trillion.一句话解释：220 K 时 MB 分布的高能尾被显著抑制，能越过 17.1 kJ mol⁻¹ 无催化势垒的碰撞比例可忽略不计 —— 与此同时几乎每次 Cl–O₃ 碰撞都能轻易越过 2.2 kJ mol⁻¹ 的催化势垒，故即便 $[\mathrm{Cl}]$ 只有万亿分之几，氯催化通道仍主导臭氧的破坏。

(d) Turnover stoichiometry (2.2 catalysis ↔ 2.1 stoichiometry).周转化学计量（2.2 催化 ↔ 2.1 化学计量）。

$$n(\mathrm{Cl}) = \dfrac{1.00 \times 10^{-3}~\mathrm{g}}{35.45~\mathrm{g\,mol^{-1}}} = 2.82 \times 10^{-5}~\mathrm{mol}$$

Each Cl atom turns over the cycle $10^5$ times before being deactivated, so the number of $\mathrm{O_3}$ molecules destroyed per Cl atom is $10^5$ (each cycle destroys one $\mathrm{O_3}$, via step 1).每个 Cl 原子在被终止前完成 $10^5$ 次循环，故每个 Cl 原子可破坏 $10^5$ 个 $\mathrm{O_3}$（每次循环在第 1 步破坏一个 $\mathrm{O_3}$）。

$$n(\mathrm{O_3})\;\text{destroyed} = 2.82 \times 10^{-5} \times 10^{5} = 2.82~\mathrm{mol}$$

$$m(\mathrm{O_3}) = 2.82~\mathrm{mol} \times 48.00~\mathrm{g\,mol^{-1}} \approx 135~\mathrm{g}$$

$\boxed{\approx 1.4 \times 10^{2}~\mathrm{g}}$ of stratospheric ozone destroyed per milligram of $\mathrm{Cl}$ atoms — a $\sim 10^{5}\times$ mass amplification. This is the kinetic origin of why CFCs were banned: a trickle of Cl atoms (released by photolysis of $\mathrm{CFCl_3}$ etc.) destroys vast amounts of ozone before the chain terminates.$\boxed{\approx 1.4 \times 10^{2}~\mathrm{g}}$ 平流层臭氧被破坏 / 每毫克 $\mathrm{Cl}$ 原子 —— 质量放大约 $10^{5}$ 倍。这正是 CFC 被禁用的动力学根源：少量氯原子（由 $\mathrm{CFCl_3}$ 等光解释放）在链反应被终止前可销毁巨量臭氧。

SR 2HARDPaper 2HLCHALLENGEN₂O₅ Pressure-Rise

$\mathrm{2\,N_2O_5(g) \to 4\,NO_2(g) + O_2(g)}$ at 338 K; $P_0 = 50.0~\mathrm{kPa}$; $P_\text{tot}$ at 0, 600, 1200, 1800 s = 50.0, 65.0, 77.0, 86.6 kPa.$\mathrm{2\,N_2O_5(g) \to 4\,NO_2(g) + O_2(g)}$，338 K；$P_0 = 50.0~\mathrm{kPa}$；$t$ = 0、600、1200、1800 s 时 $P_\text{tot}$ = 50.0、65.0、77.0、86.6 kPa。

(a) Pressure–extent relationship.压强–反应程度的关系。

If $x$ = fraction of $\mathrm{N_2O_5}$ decomposed, then per mole of original $\mathrm{N_2O_5}$ the partial pressures are设 $x$ = $\mathrm{N_2O_5}$ 已分解的分数，则按每摩尔初始 $\mathrm{N_2O_5}$，各分压为

$$P_{\mathrm{N_2O_5}} = (1 - x)\,P_0,\quad P_{\mathrm{NO_2}} = 2x\,P_0,\quad P_{\mathrm{O_2}} = \tfrac{1}{2}x\,P_0$$

(Coefficient 2 in $\mathrm{NO_2}$ comes from $\mathrm{2\,N_2O_5 \to 4\,NO_2}$: each mole of $\mathrm{N_2O_5}$ produces 2 mol $\mathrm{NO_2}$; similarly $\tfrac{1}{2}$ mol $\mathrm{O_2}$.) Summing:（$\mathrm{NO_2}$ 的系数 2 来自 $\mathrm{2\,N_2O_5 \to 4\,NO_2}$：每摩尔 $\mathrm{N_2O_5}$ 生成 2 mol $\mathrm{NO_2}$；类似地，$\tfrac{1}{2}$ mol $\mathrm{O_2}$。）求和：

$$P_\text{tot} = \bigl(1 - x + 2x + \tfrac{1}{2}x\bigr)\,P_0 = \bigl(1 + \tfrac{3}{2}\,x\bigr)\,P_0$$

Solving for $x$ and substituting back:解出 $x$ 并代回：

$$x = \dfrac{2\,(P_\text{tot} - P_0)}{3\,P_0} \;\Longrightarrow\; P_{\mathrm{N_2O_5}} = P_0\,(1 - x) = \dfrac{5P_0 - 2P_\text{tot}}{3}\quad\checkmark$$

(b) First-order test.一级动力学检验。

$t$ (s)	$P_\text{tot}$ (kPa)	$P_{\mathrm{N_2O_5}}$ (kPa)	$\ln P_{\mathrm{N_2O_5}}$
0	50.0	$\tfrac{1}{3}(250-100)=50.0$	$3.912$
600	65.0	$\tfrac{1}{3}(250-130)=40.0$	$3.689$
1200	77.0	$\tfrac{1}{3}(250-154)=32.0$	$3.466$
1800	86.6	$\tfrac{1}{3}(250-173.2)=25.6$	$3.243$

Successive differences in $\ln P_{\mathrm{N_2O_5}}$ over 600 s intervals are $-0.223, -0.223, -0.223$ — constant. So $\ln P_{\mathrm{N_2O_5}}$ is linear in $t$, confirming the integrated rate law of a first-order reaction $\ln P = \ln P_0 - kt$. The slope gives$\ln P_{\mathrm{N_2O_5}}$ 在 600 s 区间上的相邻差值均为 $-0.223$ —— 恒定。说明 $\ln P_{\mathrm{N_2O_5}}$ 对 $t$ 线性，符合一级反应积分速率方程 $\ln P = \ln P_0 - kt$。由斜率得

$$k = \dfrac{0.223}{600~\mathrm{s}} \approx 3.72 \times 10^{-4}~\mathrm{s^{-1}}$$

Units of a first-order $k$: $\mathrm{s^{-1}}$.一级反应 $k$ 的单位：$\mathrm{s^{-1}}$。

(c) Half-life and prediction at $5\,t_{1/2}$.半衰期与 $5\,t_{1/2}$ 时的预测。

$$t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{3.72 \times 10^{-4}} \approx 1.86 \times 10^{3}~\mathrm{s} \approx 31~\mathrm{min}$$

After $5\,t_{1/2}$, fraction decomposed $x = 1 - (1/2)^5 = 31/32 = 0.969$.$5\,t_{1/2}$ 时分解分数 $x = 1 - (1/2)^5 = 31/32 = 0.969$。

$$P_\text{tot} = (1 + 1.5 \times 0.969)(50.0) = 2.453 \times 50.0 \approx 123~\mathrm{kPa}$$

Quick cross-check via partial pressures: $P_{\mathrm{N_2O_5}} = 0.031 \times 50 = 1.55$, $P_{\mathrm{NO_2}} = 2 \times 0.969 \times 50 = 96.9$, $P_{\mathrm{O_2}} = 0.5 \times 0.969 \times 50 = 24.2$; sum $= 122.6~\mathrm{kPa}$. ✓用分压交叉验证：$P_{\mathrm{N_2O_5}} = 0.031 \times 50 = 1.55$，$P_{\mathrm{NO_2}} = 2 \times 0.969 \times 50 = 96.9$，$P_{\mathrm{O_2}} = 0.5 \times 0.969 \times 50 = 24.2$；和 $= 122.6~\mathrm{kPa}$。✓

(d) Why observed $P_\text{tot} \approx 119~\mathrm{kPa}$ < predicted $125~\mathrm{kPa}$.为何观察 $P_\text{tot} \approx 119~\mathrm{kPa}$ < 预测的 $125~\mathrm{kPa}$。

Chemical reason: $\mathrm{NO_2}$ is in equilibrium with its dimer: $\mathrm{2\,NO_2(g) \rightleftharpoons N_2O_4(g)}$. The equilibrium converts 2 mol of gas to 1 mol, reducing total moles and hence $P_\text{tot}$. At 338 K the equilibrium is mostly shifted toward $\mathrm{NO_2}$ but a small fraction of $\mathrm{N_2O_4}$ persists, producing the small deficit observed.化学原因：$\mathrm{NO_2}$ 与其二聚体存在平衡 $\mathrm{2\,NO_2(g) \rightleftharpoons N_2O_4(g)}$。该平衡把 2 mol 气体合并为 1 mol，使气体总摩尔数减少，进而使 $P_\text{tot}$ 下降。338 K 时平衡主要偏向 $\mathrm{NO_2}$，但仍有少量 $\mathrm{N_2O_4}$ 存在，造成观察到的小亏损。

Experimental reason (any one of):实验原因（任举一项）：

slow gas leak through fittings over the long observation time;长时间观察过程中接口处气体缓慢泄漏；
drift of the thermostat away from 338 K — at lower $T$, the $\mathrm{N_2O_4}$ equilibrium shifts further toward dimer;恒温装置温度偏离 338 K —— 温度降低时 $\mathrm{N_2O_4}$ 平衡进一步偏向二聚体；
pressure-gauge zero offset / calibration error.压力计零点偏移或校准误差。

SR 3HARDPaper 2HLCHALLENGEPersulfate–Iodide Clock

$\mathrm{S_2O_8^{2-} + 2\,I^- \to 2\,SO_4^{2-} + I_2}$ at 298 K; orders, rate constant, mechanism, iodine-clock prediction, Fe³⁺ catalysis.298 K 下 $\mathrm{S_2O_8^{2-} + 2\,I^- \to 2\,SO_4^{2-} + I_2}$；级数、速率常数、机理、碘时钟预测、$\mathrm{Fe^{3+}}$ 催化。

(a) Orders, rate equation, and rate constant at 298 K.298 K 下的反应级数、速率方程与速率常数。

Runs 1 → 2: $[\mathrm{S_2O_8^{2-}}]$ doubles, $[\mathrm{I^-}]$ fixed. Rate doubles ($2.0\times10^{-6} \to 4.0\times10^{-6}$) ⇒ order in $\mathrm{S_2O_8^{2-}}$ = 1.第 1 → 2 组：$[\mathrm{S_2O_8^{2-}}]$ 加倍，$[\mathrm{I^-}]$ 不变。速率加倍（$2.0\times10^{-6} \to 4.0\times10^{-6}$）⇒ 关于 $\mathrm{S_2O_8^{2-}}$ 的级数 = 1。

Runs 2 → 3: $[\mathrm{I^-}]$ doubles, $[\mathrm{S_2O_8^{2-}}]$ fixed. Rate doubles ($4.0\times10^{-6} \to 8.0\times10^{-6}$) ⇒ order in $\mathrm{I^-}$ = 1.第 2 → 3 组：$[\mathrm{I^-}]$ 加倍，$[\mathrm{S_2O_8^{2-}}]$ 不变。速率加倍（$4.0\times10^{-6} \to 8.0\times10^{-6}$）⇒ 关于 $\mathrm{I^-}$ 的级数 = 1。

$$\text{rate} = k\,[\mathrm{S_2O_8^{2-}}]\,[\mathrm{I^-}],\quad\text{overall order} = 2$$

From run 1:由第 1 组：

$$k = \dfrac{2.0 \times 10^{-6}}{(0.010)(0.010)} = 2.0 \times 10^{-2}~\mathrm{dm^3\,mol^{-1}\,s^{-1}}$$

(b) Two-step mechanism. The rate-determining step must involve one $\mathrm{S_2O_8^{2-}}$ and one $\mathrm{I^-}$ (to reproduce the observed first-order dependence on each), so a bimolecular slow step followed by a fast step that consumes the second $\mathrm{I^-}$:两步机理。决速步必须涉及一个 $\mathrm{S_2O_8^{2-}}$ 与一个 $\mathrm{I^-}$（以再现观察到的对二者均一级的依赖关系），故双分子慢步 + 快步消耗第二个 $\mathrm{I^-}$：

Step 1 (slow, RDS): $\mathrm{S_2O_8^{2-} + I^- \to SO_4^{2-} + SO_4I^{-}}$ (or $\mathrm{S_2O_8^{2-} + I^- \to 2\,SO_4^{2-} + I^\bullet}$ via radicals)第 1 步（慢，决速步）：$\mathrm{S_2O_8^{2-} + I^- \to SO_4^{2-} + SO_4I^{-}}$ （或经自由基 $\mathrm{S_2O_8^{2-} + I^- \to 2\,SO_4^{2-} + I^\bullet}$）
Step 2 (fast): $\mathrm{SO_4I^- + I^- \to SO_4^{2-} + I_2}$ (or $\mathrm{2\,I^\bullet \to I_2}$ in the radical variant)第 2 步（快）：$\mathrm{SO_4I^- + I^- \to SO_4^{2-} + I_2}$ （自由基版本中为 $\mathrm{2\,I^\bullet \to I_2}$）
Intermediate: $\mathrm{SO_4I^-}$ (or $\mathrm{I^\bullet}$) — formed in step 1, consumed in step 2; absent from the net equation.中间体：$\mathrm{SO_4I^-}$（或 $\mathrm{I^\bullet}$）—— 第 1 步生成，第 2 步消耗；不出现在净反应中。

Either mechanism gives $\text{rate of RDS} = k_1\,[\mathrm{S_2O_8^{2-}}]\,[\mathrm{I^-}]$, which matches the experimental rate law.两种机理的决速步速率都给出 $\text{rate} = k_1\,[\mathrm{S_2O_8^{2-}}]\,[\mathrm{I^-}]$，与实验速率方程相符。

(c) Iodine-clock prediction.碘时钟预测。

Step 1 — initial rate at the new concentrations. Using the rate law and $k = 2.0 \times 10^{-2}~\mathrm{dm^3\,mol^{-1}\,s^{-1}}$:第 1 步 —— 新浓度下的初始速率。用速率方程与 $k = 2.0 \times 10^{-2}~\mathrm{dm^3\,mol^{-1}\,s^{-1}}$：

$$\text{rate} = k\,[\mathrm{S_2O_8^{2-}}]_0\,[\mathrm{I^-}]_0 = (0.020)(0.050)(0.030) = 3.0 \times 10^{-5}~\mathrm{mol\,dm^{-3}\,s^{-1}}$$

Step 2 — relate the rate to $\mathrm{I_2}$ production. From the stoichiometric equation $\mathrm{S_2O_8^{2-} + 2\,I^- \to 2\,SO_4^{2-} + I_2}$, every mole of $\mathrm{S_2O_8^{2-}}$ consumed produces one mole of $\mathrm{I_2}$, so第 2 步 —— 将速率与 $\mathrm{I_2}$ 的生成联系起来。由化学计量方程 $\mathrm{S_2O_8^{2-} + 2\,I^- \to 2\,SO_4^{2-} + I_2}$，每消耗 1 mol $\mathrm{S_2O_8^{2-}}$ 生成 1 mol $\mathrm{I_2}$：

$$\dfrac{d[\mathrm{I_2}]}{dt} \approx 3.0 \times 10^{-5}~\mathrm{mol\,dm^{-3}\,s^{-1}}$$

Step 3 — time to reach the endpoint. The thiosulfate keeps $[\mathrm{I_2}] \approx 0$ until exhausted, defining the clock window. Endpoint appears once $1.0 \times 10^{-3}~\mathrm{mol\,dm^{-3}}$ of $\mathrm{I_2}$ has been produced:第 3 步 —— 终点出现时间。硫代硫酸根使 $[\mathrm{I_2}] \approx 0$ 直至自身耗尽，从而界定时钟窗口。当生成 $1.0 \times 10^{-3}~\mathrm{mol\,dm^{-3}}$ 的 $\mathrm{I_2}$ 时出现终点：

$$t_\text{endpoint} \approx \dfrac{1.0 \times 10^{-3}}{3.0 \times 10^{-5}} \approx 33~\mathrm{s}$$

Limiting assumption (any one): the rate is taken to be constant over the 33 s window, but in reality $[\mathrm{S_2O_8^{2-}}]$ and $[\mathrm{I^-}]$ both fall as the reaction proceeds, so the true rate is slightly lower and $t_\text{endpoint}$ slightly longer. (Other valid answers: that the thiosulfate truly keeps $[\mathrm{I_2}]$ at zero; that no other oxidant interferes; that the temperature is constant.)限制性假设（任举一项）：上述计算把速率视为在 33 s 内恒定，但实际上 $[\mathrm{S_2O_8^{2-}}]$ 与 $[\mathrm{I^-}]$ 都会随反应进行而下降，真实速率略低，故 $t_\text{endpoint}$ 略长。（其他可接受的答案：硫代硫酸根确实能将 $[\mathrm{I_2}]$ 维持在 0；体系中无其他氧化剂干扰；温度保持不变。）

(d) Fe³⁺ catalysis.$\mathrm{Fe^{3+}}$ 催化。

(i) The slow step in the uncatalysed reaction is collision between two anions ($\mathrm{S_2O_8^{2-}}$ and $\mathrm{I^-}$) — electrostatically unfavourable. $\mathrm{Fe^{3+}}$ opens a faster two-step cycle through cation–anion collisions:(i) 无催化反应的慢步是两个阴离子之间的碰撞（$\mathrm{S_2O_8^{2-}}$ 与 $\mathrm{I^-}$） —— 静电上不利。$\mathrm{Fe^{3+}}$ 通过阳离子–阴离子碰撞开启更快的两步循环：

$\mathrm{2\,Fe^{3+} + 2\,I^- \to 2\,Fe^{2+} + I_2}$ (fast, favourable: $+3$ meeting $-1$);$\mathrm{2\,Fe^{3+} + 2\,I^- \to 2\,Fe^{2+} + I_2}$（快、有利：$+3$ 与 $-1$ 相吸）；
$\mathrm{2\,Fe^{2+} + S_2O_8^{2-} \to 2\,Fe^{3+} + 2\,SO_4^{2-}}$ (fast, regenerates $\mathrm{Fe^{3+}}$).$\mathrm{2\,Fe^{2+} + S_2O_8^{2-} \to 2\,Fe^{3+} + 2\,SO_4^{2-}}$（快，再生 $\mathrm{Fe^{3+}}$）。

Sum: identical to the uncatalysed overall equation. Each step has a lower $E_a$ than the direct anion–anion path, so the rate rises.两步相加：与无催化总反应一致。每步的 $E_a$ 都比直接的阴离子–阴离子碰撞低，故速率上升。

(ii) Homogeneous because $\mathrm{Fe^{3+}/Fe^{2+}}$ are in the same (aqueous) phase as the reactants — no separate solid surface or different phase boundary. Contrast with Pt in C5, which sits as a solid in a gas stream (heterogeneous).(ii) 均相，因为 $\mathrm{Fe^{3+}/Fe^{2+}}$ 与反应物处于同一（水溶液）相 —— 没有独立的固体表面或不同相界。与 C5 中作为固体置于气流的 Pt 形成对比（异相催化）。

P3-1HARDPaper 3 HLHLCHALLENGEH₂O₂ + I⁻

$\mathrm{2\,H_2O_2 \to 2\,H_2O + O_2}$ catalysed by $\mathrm{I^-}$ at 298 K; three initial-rate runs varying $[\mathrm{H_2O_2}]$ and $[\mathrm{I^-}]$, plus a time-resolved $[\mathrm{H_2O_2}]$ table for Run 1.298 K 下 $\mathrm{I^-}$ 催化 $\mathrm{2\,H_2O_2 \to 2\,H_2O + O_2}$；三组改变 $[\mathrm{H_2O_2}]$ 与 $[\mathrm{I^-}]$ 的初始速率数据，外加 Run 1 的 $[\mathrm{H_2O_2}]$ 随时间变化数据。

(a) Orders, rate equation, and rate constant.反应级数、速率方程与速率常数。

Runs 1 → 2: $[\mathrm{H_2O_2}]$ doubles (0.10 → 0.20), $[\mathrm{I^-}]$ fixed. Rate doubles ($5.0\times10^{-5} \to 1.0\times10^{-4}$) ⇒ order in $\mathrm{H_2O_2}$ = 1.Run 1 → 2：$[\mathrm{H_2O_2}]$ 加倍（0.10 → 0.20），$[\mathrm{I^-}]$ 不变。速率加倍（$5.0\times10^{-5} \to 1.0\times10^{-4}$）⇒ 关于 $\mathrm{H_2O_2}$ 的级数 = 1。

Runs 1 → 3: $[\mathrm{I^-}]$ doubles (0.010 → 0.020), $[\mathrm{H_2O_2}]$ fixed. Rate doubles ($5.0\times10^{-5} \to 1.0\times10^{-4}$) ⇒ order in $\mathrm{I^-}$ = 1.Run 1 → 3：$[\mathrm{I^-}]$ 加倍（0.010 → 0.020），$[\mathrm{H_2O_2}]$ 不变。速率加倍（$5.0\times10^{-5} \to 1.0\times10^{-4}$）⇒ 关于 $\mathrm{I^-}$ 的级数 = 1。

$$\text{rate} = k\,[\mathrm{H_2O_2}]\,[\mathrm{I^-}],\quad\text{overall order} = 2$$

From Run 1:由 Run 1：

$$k = \dfrac{5.0 \times 10^{-5}}{(0.10)(0.010)} = 5.0 \times 10^{-2}~\mathrm{dm^3\,mol^{-1}\,s^{-1}}$$

Units are those of a second-order rate constant — exactly what's expected for a bimolecular rate-determining step.单位为二级反应速率常数的单位 —— 与双分子决速步的预期一致。

(b) Mechanism. The slow step must be bimolecular in $\mathrm{H_2O_2}$ and $\mathrm{I^-}$ (to reproduce the observed first-order dependence on each). A standard two-step mechanism with $\mathrm{IO^-}$ as the intermediate:机理。决速步必须对 $\mathrm{H_2O_2}$ 与 $\mathrm{I^-}$ 双分子（以再现对二者均一级的依赖）。以 $\mathrm{IO^-}$ 为中间体的标准两步机理：

Step 1 (slow, RDS): $\mathrm{H_2O_2 + I^- \to H_2O + IO^-}$第 1 步（慢、决速步）：$\mathrm{H_2O_2 + I^- \to H_2O + IO^-}$
Step 2 (fast): $\mathrm{IO^- + H_2O_2 \to H_2O + O_2 + I^-}$第 2 步（快）：$\mathrm{IO^- + H_2O_2 \to H_2O + O_2 + I^-}$
Intermediate: $\mathrm{IO^-}$ (hypoiodite) — formed in step 1, consumed in step 2.中间体：$\mathrm{IO^-}$（次碘酸根）—— 第 1 步生成，第 2 步消耗。

Sum check: add the two steps and cancel intermediates ($\mathrm{IO^-}$) and the catalyst ($\mathrm{I^-}$ appears on both sides):加和验证：两步相加，消去中间体（$\mathrm{IO^-}$）与催化剂（$\mathrm{I^-}$ 两边各出现一次）：

$$\mathrm{2\,H_2O_2 \to 2\,H_2O + O_2}\quad\checkmark$$

Matches the net stoichiometric equation.与净化学计量方程一致。

(c) Integrated first-order verification + half-life.积分一级动力学验证 + 半衰期。

In Run 1, the large excess of $\mathrm{I^-}$ keeps $[\mathrm{I^-}] \approx 0.010$ constant, so the rate law reduces to $\text{rate} = k'\,[\mathrm{H_2O_2}]$ with $k' = k\,[\mathrm{I^-}]_0$. Computing $\ln[\mathrm{H_2O_2}]$:Run 1 中 $\mathrm{I^-}$ 大幅过量，$[\mathrm{I^-}] \approx 0.010$ 视为常数，速率方程化为 $\text{rate} = k'\,[\mathrm{H_2O_2}]$，其中 $k' = k\,[\mathrm{I^-}]_0$。计算 $\ln[\mathrm{H_2O_2}]$：

$t$ (s)	$[\mathrm{H_2O_2}]$	$\ln[\mathrm{H_2O_2}]$	$\Delta \ln$ over 500 s
0	0.100	$-2.303$	—
500	0.0779	$-2.552$	$-0.249$
1000	0.0607	$-2.801$	$-0.249$
1500	0.0472	$-3.053$	$-0.252$
2000	0.0368	$-3.303$	$-0.250$

Successive differences in $\ln[\mathrm{H_2O_2}]$ over equal 500 s intervals are essentially constant ($\approx -0.250$). $\ln[\mathrm{H_2O_2}]$ vs $t$ is therefore linear — confirming first-order behaviour in $\mathrm{H_2O_2}$. Slope:$\ln[\mathrm{H_2O_2}]$ 在相等 500 s 区间上的相邻差值基本相同（约 $-0.250$）。故 $\ln[\mathrm{H_2O_2}]$ 对 $t$ 线性 —— 验证关于 $\mathrm{H_2O_2}$ 为一级。斜率：

$$\text{slope} = \dfrac{-0.250}{500~\mathrm{s}} = -5.0 \times 10^{-4}~\mathrm{s^{-1}} \;\Longrightarrow\; k' = 5.0 \times 10^{-4}~\mathrm{s^{-1}}$$

$$t_{1/2} = \dfrac{\ln 2}{k'} = \dfrac{0.693}{5.0 \times 10^{-4}} \approx 1.4 \times 10^{3}~\mathrm{s} \approx 23~\mathrm{min}$$

Cross-check with (a): $k\,[\mathrm{I^-}]_0 = (5.0 \times 10^{-2})(0.010) = 5.0 \times 10^{-4}~\mathrm{s^{-1}}$ — agrees exactly with $k'$. ✓ The two independent measurements (initial rates and integrated kinetics) are mutually consistent.与 (a) 的交叉验证：$k\,[\mathrm{I^-}]_0 = (5.0 \times 10^{-2})(0.010) = 5.0 \times 10^{-4}~\mathrm{s^{-1}}$ —— 与 $k'$ 完全一致。✓ 两种独立测量（初始速率法与积分动力学）互相印证。

(d) Why $\mathrm{I^-}$ appears in the rate equation.为何 $\mathrm{I^-}$ 出现在速率方程中。

$\mathrm{I^-}$ participates in the rate-determining step (step 1), so its concentration directly controls how fast that step proceeds — hence it appears in the rate law with order 1. It does not appear in the net stoichiometric equation because step 2 regenerates it: every $\mathrm{I^-}$ consumed in step 1 is recovered in step 2, so its concentration is unchanged over a full catalytic cycle. Kinetic role (in rate law) and stoichiometric role (in net equation) are distinct.$\mathrm{I^-}$ 参与决速步（第 1 步），其浓度直接决定该步速率 —— 故以一级出现在速率方程中。它不出现在净化学计量方程中，因为第 2 步将其再生：第 1 步消耗的每个 $\mathrm{I^-}$ 都在第 2 步被回收，故完成一次催化循环后浓度不变。动力学角色（出现在速率方程）与化学计量角色（出现在净方程）截然不同。

(e) Two experimental confirmations that $\mathrm{I^-}$ is the catalyst (not a consumed reactant).证实 $\mathrm{I^-}$ 是催化剂（非被消耗反应物）的两项实验观测。

Titrate $[\mathrm{I^-}]$ before and after reaction (e.g. with $\mathrm{AgNO_3}$ to give $\mathrm{AgI}$): if $\mathrm{I^-}$ is a catalyst, the measured $[\mathrm{I^-}]$ should be unchanged within experimental error. A stoichiometric reactant would show a measurable decrease.反应前后滴定 $[\mathrm{I^-}]$（例如以 $\mathrm{AgNO_3}$ 生成 $\mathrm{AgI}$）：若 $\mathrm{I^-}$ 为催化剂，所测 $[\mathrm{I^-}]$ 应在实验误差范围内不变。化学计量反应物则会显示可测量的下降。
Run the reaction with no $\mathrm{I^-}$ present: $\mathrm{H_2O_2}$ should still decompose by the same overall stoichiometry, just very much more slowly. Net mol of $\mathrm{O_2}$ produced (predicted from $\mathrm{H_2O_2}$ alone) is the same as the catalysed run — confirming $\mathrm{I^-}$ doesn't appear in the stoichiometry.不加 $\mathrm{I^-}$ 直接运行反应：$\mathrm{H_2O_2}$ 仍按相同的总化学计量分解，只是慢得多。所得 $\mathrm{O_2}$ 的摩尔数（仅由 $\mathrm{H_2O_2}$ 预测）与催化反应相同 —— 说明 $\mathrm{I^-}$ 不出现在化学计量中。

(Other valid checks: a single small portion of $\mathrm{I^-}$ catalyses repeated additions of fresh $\mathrm{H_2O_2}$ without depletion; detection of transient $\mathrm{IO^-}$ at short times via UV-vis absorption.)（其他可接受的检验：一小份 $\mathrm{I^-}$ 可催化多次新加入的 $\mathrm{H_2O_2}$ 而不被耗尽；通过紫外–可见吸收在短时间内检测瞬时 $\mathrm{IO^-}$。）

(f) Stoichiometric end-state (2.1 cross-over).化学计量末态（2.1 交叉）。

$$n(\mathrm{H_2O_2}) = c\,V = (1.00~\mathrm{mol\,dm^{-3}})(0.100~\mathrm{dm^3}) = 0.100~\mathrm{mol}$$

From $\mathrm{2\,H_2O_2 \to 2\,H_2O + O_2}$, the mol ratio is $\mathrm{H_2O_2 : O_2 = 2 : 1}$:由 $\mathrm{2\,H_2O_2 \to 2\,H_2O + O_2}$，摩尔比 $\mathrm{H_2O_2 : O_2 = 2 : 1}$：

$$n(\mathrm{O_2}) = \tfrac{1}{2}\,n(\mathrm{H_2O_2}) = 0.0500~\mathrm{mol}$$

$$m(\mathrm{O_2}) = 0.0500 \times 32.00 = 1.60~\mathrm{g}$$

$\boxed{m(\mathrm{O_2}) = 1.60~\mathrm{g}}$. The catalyst speeds the reaction up but plays no role in this calculation — proof of (d) in numerical form.$\boxed{m(\mathrm{O_2}) = 1.60~\mathrm{g}}$。催化剂加快反应却在本计算中不出现 —— 这是 (d) 的数值化证明。

Reaction Kinetics — Challenge Solutions化学反应动力学 —— 挑战题详解

Multiple Choice选择题

Structured Response结构化解答题

Data-Based — H₂O₂ + I⁻ Catalysis数据题 —— H₂O₂ + I⁻ 催化