High School Chemistry — Reaction Rates and Kinetics — Solutions

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Reaction rate and stoichiometry反应速率与化学计量 · HS-PS1-5 [3 marks][3 分]

For $\text{A} \to 2\text{B}$: $[\text{A}]$ drops from $1.20$ to $0.60\ \text{mol L}^{-1}$ in $30\ \text{s}$. Average rate of appearance of B?对于 $\text{A} \to 2\text{B}$：$[\text{A}]$ 在 $30\ \text{s}$ 内从 $1.20$ 降至 $0.60\ \text{mol L}^{-1}$。B 的平均生成速率？

Answer:答案： (C) $0.040\ \text{mol L}^{-1}\text{s}^{-1}$

(a) Rate of disappearance of AA 的消失速率 M1

The average rate of disappearance of A over the interval:A 在该时间段内的平均消失速率： $$ \text{rate}_A = \frac{\Delta[\text{A}]}{\Delta t} = \frac{1.20 - 0.60}{30} = \frac{0.60}{30} = 0.020\ \text{mol L}^{-1}\text{s}^{-1}. $$

(b) Apply stoichiometry to find rate of appearance of B利用化学计量换算 B 的生成速率 M1·A1

Since the stoichiometric coefficient of B is 2 relative to A's coefficient of 1, B appears twice as fast as A disappears:由于 B 的化学计量系数为 2，A 的为 1，B 的生成速率是 A 消失速率的两倍： $$ \text{rate}_B = 2 \times \text{rate}_A = 2 \times 0.020 = 0.040\ \text{mol L}^{-1}\text{s}^{-1}. $$ Option (C).选 (C)。

Why the distractors fail.干扰项分析。

(A) $0.010$: divides the concentration change by $\Delta t$ then by 2 again, double-counting the stoichiometry correction.把浓度变化除以 $\Delta t$ 后再除以 2，对化学计量系数重复修正。

(B) $0.020$: this is the rate of disappearance of A, not the rate of appearance of B; it ignores the factor of 2.这是 A 的消失速率，而非 B 的生成速率，忽略了系数 2。

(D) $0.060$: uses $[\text{A}]_0 / \Delta t = 1.20/30 \times \frac{1}{2}$, a mis-application that ignores the $\Delta[\text{A}]$ calculation.误用 $[\text{A}]_0$ 代替 $\Delta[\text{A}]$ 进行计算。

The stoichiometric coefficient ratio links the rates of all species in a reaction.化学计量系数比将反应中所有物种的速率联系起来。 For $a\text{A} \to b\text{B}$, the general relationship is $\text{rate}_B / b = \text{rate}_A / a$, or equivalently $\text{rate}_B = (b/a) \times \text{rate}_A$. Here $b/a = 2/1 = 2$. Always write down the balanced equation first and identify the coefficients before converting between species rates. A common error is to confuse the rate of a reactant disappearing (negative $\Delta[\text{A}]/\Delta t$ in full notation) with a positive rate; by convention, reaction rates are stated as positive quantities.对于 $a\text{A} \to b\text{B}$，通用关系为 $\text{rate}_B / b = \text{rate}_A / a$，即 $\text{rate}_B = (b/a) \times \text{rate}_A$。此题 $b/a = 2/1 = 2$。动手前先写出配平方程式并确认系数，再换算各物种的速率。常见错误是把反应物的消失速率（完整记法中为负的 $\Delta[\text{A}]/\Delta t$）与正速率混淆；按惯例，反应速率均取正值。

Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Collision theory碰撞理论 · HS-PS1-5 [3 marks][3 分]

According to collision theory, which two conditions must be simultaneously satisfied for a collision to produce a reaction?碰撞理论要求哪两个条件同时满足才能发生化学反应？

Answer:答案： (B) Sufficient collision energy and correct molecular orientation足够的碰撞能量和正确的分子方向

(a) Identify the two collision theory criteria确认碰撞理论的两个判据 M1·A1·A1

Collision theory states that for a reaction to occur at a collision, two conditions must be met simultaneously:碰撞理论指出，碰撞发生时必须同时满足以下两个条件：

Sufficient activation energy: the colliding molecules must have kinetic energy at least equal to $E_a$ (the activation energy) so that bonds can be broken and new bonds formed.足够的活化能：碰撞分子必须具有不低于 $E_a$（活化能）的动能，才能断键并形成新键。
Correct orientation: the reactive functional sites on the colliding molecules must be aligned so that the bonds that need to break are exposed to each other.正确的方向：碰撞分子上的活性官能位点必须正确对准，使需要断裂的键相互暴露。

Option (B) captures both conditions exactly.选项 (B) 准确包含了这两个条件。

Why the distractors fail.干扰项分析。

(A): equal masses and equal speeds are irrelevant; molecules react regardless of relative mass as long as (B) holds.质量相等和速度相等与反应无关；只要满足 (B) 的条件，分子无论相对质量如何都能反应。

(C): high temperature and a catalyst both increase the fraction of successful collisions, but they are not conditions for a single collision to succeed. Temperature and catalysts act on populations of molecules, not on individual collisions.高温和催化剂都能增加有效碰撞的比例，但它们不是单次碰撞成功的条件。温度和催化剂作用于分子群体，而非单次碰撞。

(D): high concentration and pressure increase the collision frequency, raising the overall rate, but again these are macroscopic factors, not per-collision criteria.高浓度和高压增加碰撞频率，提高总体速率，但这些是宏观因素，不是单次碰撞的判据。

Collision frequency governs how often molecules meet; collision success governs what fraction of those meetings react.碰撞频率决定分子相遇的频率；碰撞成功率决定这些相遇中有多大比例会发生反应。 Collision theory splits the reaction rate into two factors: rate = (collision frequency) x (fraction of collisions that are effective). A collision is effective only when both the energy threshold ($\geq E_a$) and the orientation requirement are met simultaneously. Increasing temperature raises the fraction of molecules with $\geq E_a$ (the Boltzmann tail), increasing the fraction of effective collisions. A catalyst provides an alternative lower-$E_a$ pathway. Increasing concentration or pressure raises collision frequency. All these levers ultimately work through one or both of the two collision criteria.碰撞理论将反应速率分为两个因素：速率 = （碰撞频率）×（有效碰撞比例）。只有当能量条件（$\geq E_a$）和方向条件同时满足时，碰撞才是有效的。升温增加了具有 $\geq E_a$ 的分子比例（玻尔兹曼分布尾部），从而提高有效碰撞比例。催化剂提供能量要求更低的替代途径。增加浓度或压力提高碰撞频率。所有这些调节手段最终都通过这两个碰撞判据之一起作用。

Q3MEDIUM中 🇨🇦 ON安 🇨🇦 AB阿 ON Provincial-style安大略省考风格 §3 PE diagram势能图 · SCH4U D3.6 · Chem 30 GO2 [6 marks][6 分]

$E_{a,\text{fwd}} = 80\ \text{kJ mol}^{-1}$; $\Delta H = -50\ \text{kJ mol}^{-1}$. (a) Endo or exothermic? (b) $E_{a,\text{rev}}$? (c) Sketch PE diagram.$E_{a,\text{fwd}} = 80\ \text{kJ mol}^{-1}$；$\Delta H = -50\ \text{kJ mol}^{-1}$。(a) 吸热还是放热？(b) $E_{a,\text{rev}}$？(c) 勾画势能图。

Answer:答案： (a) exothermic放热 · (b) $E_{a,\text{rev}} = 130\ \text{kJ mol}^{-1}$ · (c) see diagram description below见下方图形描述

(a) Classify as exothermic or endothermic判断放热或吸热 A1·A1

The reaction is exothermic. $\Delta H = -50\ \text{kJ mol}^{-1}$ is negative, meaning energy is released to the surroundings. Products are at a lower potential energy than reactants.该反应为放热反应。$\Delta H = -50\ \text{kJ mol}^{-1}$ 为负值，意味着能量释放到环境中。生成物的势能低于反应物。

(b) Calculate the reverse activation energy计算逆向活化能 M1·A1

The activated complex sits at the peak of the energy barrier. Starting from the reactant side, the barrier height is $E_{a,\text{fwd}} = 80\ \text{kJ mol}^{-1}$. Since $\Delta H = -50\ \text{kJ mol}^{-1}$, the products are $50\ \text{kJ mol}^{-1}$ lower than reactants. The reverse activation energy is the height of the barrier above the products:活化络合物位于能垒顶点。从反应物一侧出发，能垒高度为 $E_{a,\text{fwd}} = 80\ \text{kJ mol}^{-1}$。由于 $\Delta H = -50\ \text{kJ mol}^{-1}$，生成物比反应物低 $50\ \text{kJ mol}^{-1}$。逆向活化能是能垒高于生成物的部分： $$ E_{a,\text{rev}} = E_{a,\text{fwd}} + |\Delta H| = 80 + 50 = 130\ \text{kJ mol}^{-1}. $$

(c) PE diagram description势能图说明 A1·A1

The diagram should show: reactants at an intermediate potential energy level; a single smooth hill (the activated complex / transition state) that rises $80\ \text{kJ mol}^{-1}$ above the reactants; products $50\ \text{kJ mol}^{-1}$ below the reactants (products at the lowest PE). Label: $E_{a,\text{fwd}} = 80$ kJ/mol (arrow from reactant level to peak); $E_{a,\text{rev}} = 130$ kJ/mol (arrow from product level to peak); $\Delta H = -50$ kJ/mol (arrow from reactant level down to product level, negative).图中应显示：反应物处于中间势能水平；一个平滑的势能山丘（活化络合物/过渡态），其顶点高于反应物 $80\ \text{kJ mol}^{-1}$；生成物低于反应物 $50\ \text{kJ mol}^{-1}$（生成物势能最低）。标注：$E_{a,\text{fwd}} = 80$ kJ/mol（从反应物水平到顶点的箭头）；$E_{a,\text{rev}} = 130$ kJ/mol（从生成物水平到顶点的箭头）；$\Delta H = -50$ kJ/mol（从反应物水平向下到生成物水平的箭头，为负值）。

Key relationship: $E_{a,\text{rev}} = E_{a,\text{fwd}} - \Delta H$ (since $\Delta H$ is negative here, $E_{a,\text{rev}} > E_{a,\text{fwd}}$).关键关系：$E_{a,\text{rev}} = E_{a,\text{fwd}} - \Delta H$（此处 $\Delta H$ 为负，故 $E_{a,\text{rev}} > E_{a,\text{fwd}}$）。 The activated complex (transition state) represents the highest energy species in the reaction coordinate. Both forward and reverse reactions must climb to this same peak. For an exothermic reaction, the products are lower in energy than the reactants, so the reverse reaction faces a larger barrier. This is why exothermic reactions are generally harder to reverse under the same conditions. Catalysts lower the PE diagram's peak symmetrically, reducing $E_{a,\text{fwd}}$ and $E_{a,\text{rev}}$ by the same amount and leaving $\Delta H$ unchanged.活化络合物（过渡态）是反应坐标上能量最高的物种。正向和逆向反应都必须爬升到同一个顶点。对于放热反应，生成物能量低于反应物，因此逆反应面临更大的能垒。这就是为什么在相同条件下放热反应通常更难逆转。催化剂对称地降低势能图的顶点，使 $E_{a,\text{fwd}}$ 和 $E_{a,\text{rev}}$ 减少相同量，同时 $\Delta H$ 不变。

Q4MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Factors affecting rate影响速率的因素 · BC Chem 12 [6 marks][6 分]

CaCO$_3$ + HCl: (I) large chips 20 °C; (II) powder 20 °C; (III) large chips 40 °C. Same acid concentration and total marble mass. (a) Rank slowest to fastest. (b) Control variable for I vs II. (c) Same or different total CO$_2$?CaCO$_3$ 与盐酸反应：（I）大块 20 °C；（II）粉末 20 °C；（III）大块 40 °C。酸浓度和大理石总质量相同。(a) 由慢到快排序。(b) I 与 II 比较的控制变量。(c) 产生 CO$_2$ 总量相同还是不同？

Answer:答案： (a) I < III < III < III < II · (b) temperature (must be identical for I and II)温度（I 和 II 必须相同） · (c) same total CO$_2$总 CO$_2$ 相同

(a) Rank by initial rate using collision theory用碰撞理论按初始速率排序 M1·A1·A1

Condition I (large chips, 20 °C) is slowest: low surface area means fewer acid molecules can contact the marble at any moment, so the collision frequency between H$^+$ ions and CaCO$_3$ is lowest. Condition III (large chips, 40 °C) is intermediate: the surface area is the same as I, but the higher temperature gives more molecules $\geq E_a$, increasing the fraction of effective collisions. Condition II (powder, 20 °C) is fastest: powdering the marble dramatically increases the surface area, producing many more collision sites per unit time even at 20 °C. Ranking from slowest to fastest: I < III < II.条件 I（大块，20 °C）最慢：表面积小，能与大理石接触的酸分子少，H$^+$ 与 CaCO$_3$ 的碰撞频率最低。条件 III（大块，40 °C）居中：表面积与 I 相同，但较高温度使更多分子具有 $\geq E_a$ 的能量，有效碰撞比例增大。条件 II（粉末，20 °C）最快：研磨成粉末显著增大表面积，即使在 20 °C 下，单位时间内的碰撞位点也远多于其他条件。从最慢到最快排序：I < III < II。

(b) Control variable for I vs III 与 II 比较的控制变量 A1·A1

Temperature must be kept constant (both at 20 °C) to make I and II a valid comparison. Only one variable should differ between conditions I and II (particle size / surface area). If temperature were also different, it would be impossible to determine whether the rate change was due to surface area or temperature. A valid comparison isolates the independent variable.温度必须保持不变（均为 20 °C），以使 I 和 II 的比较有效。条件 I 和 II 之间只应有一个变量不同（颗粒大小/表面积）。若温度也不同，则无法判断速率变化是由表面积还是温度造成的。有效比较需要隔离自变量。

(c) Total CO$_2$ produced产生的 CO$_2$ 总量 A1

The total CO$_2$ produced is the same in all three conditions. The total amount of product depends on the amount of limiting reagent consumed, not on how fast the reaction proceeds. Since the total mass of CaCO$_3$ and the acid concentration are identical in all three experiments, the same number of moles of CaCO$_3$ reacts in each, producing the same total moles (and volume) of CO$_2$. Rate affects speed, not yield.三种条件下产生的 CO$_2$ 总量相同。产物总量取决于消耗的限量试剂的量，而非反应速率。由于三组实验中 CaCO$_3$ 总质量和酸浓度完全相同，每组实验中参与反应的 CaCO$_3$ 摩尔数相同，产生的 CO$_2$ 总摩尔数（和体积）也相同。速率影响快慢，不影响产率。

Surface area and temperature both change rate but not yield; only the amounts of reactants determine total product.表面积和温度都改变速率但不改变产率；只有反应物的量决定总产物量。 This distinction is a classic exam trap. Powdering the marble makes the reaction finish faster (high surface area = more collisions per second), but once all the CaCO$_3$ has reacted the total CO$_2$ is fixed by stoichiometry. Likewise, raising temperature speeds up the reaction but does not change the final amount of product for a reaction that goes to completion. In a reversible equilibrium, temperature can shift the equilibrium and change yield, but CaCO$_3$ + HCl is treated as proceeding to completion at high-school level.这种区别是经典考点。研磨大理石使反应更快完成（表面积大 = 每秒碰撞更多），但一旦所有 CaCO$_3$ 反应完，CO$_2$ 的总量由化学计量决定。同样，升温加快反应但不改变完全反应的最终产物量。对于可逆平衡，温度可以改变平衡位置和产率，但高中阶段 CaCO$_3$ + HCl 被视为完全反应。

Q5MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §5 Catalysts催化剂 · Chem 30 GO3 [6 marks][6 分]

$2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$ with MnO$_2$ catalyst. (a) Homogeneous or heterogeneous? (b) How MnO$_2$ increases rate without changing $\Delta H$. (c) One biological catalyst and its reaction type.$2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$，加入 MnO$_2$ 催化剂。(a) 均相还是非均相？(b) MnO$_2$ 如何在不改变 $\Delta H$ 的情况下提高速率。(c) 一种生物催化剂及其催化的反应类型。

Answer:答案： (a) heterogeneous非均相 · (b) lowers $E_a$ via alternative pathway通过替代途径降低 $E_a$ · (c) e.g. catalase / starch hydrolysis如过氧化氢酶 / 淀粉水解

(a) Classify MnO$_2$ as homogeneous or heterogeneous catalyst判断 MnO$_2$ 为均相或非均相催化剂 A1·A1

MnO$_2$ is a heterogeneous catalyst. It is a solid, while H$_2$O$_2$ is a liquid. The catalyst and reactants exist in different phases. A homogeneous catalyst would be dissolved in the same liquid phase as H$_2$O$_2$; here they are not, so it is heterogeneous.MnO$_2$ 是非均相催化剂。它是固体，而 H$_2$O$_2$ 是液体。催化剂与反应物处于不同相中。均相催化剂应溶解在与 H$_2$O$_2$ 相同的液相中；此处并非如此，故为非均相催化剂。

(b) Mechanism of catalysis using activation energy用活化能解释催化机理 M1·A1·A1

MnO$_2$ provides an alternative reaction pathway with a lower activation energy $E_a'$ than the uncatalysed route. Because more molecules in the Boltzmann distribution have kinetic energy $\geq E_a'$, the fraction of effective collisions increases dramatically, and the rate rises. Crucially, the catalyst does not change the energy of reactants or products, so $\Delta H$ is unchanged: the reaction is still exothermic by the same amount. The catalyst is regenerated at the end of each catalytic cycle, which is why MnO$_2$ can be recovered unchanged.MnO$_2$ 提供了一条活化能更低的替代反应途径（$E_a'$ 低于未催化途径）。由于玻尔兹曼分布中具有 $\geq E_a'$ 动能的分子比例大幅增加，有效碰撞比例显著提高，速率随之上升。关键在于催化剂不改变反应物或生成物的能量，因此 $\Delta H$ 不变：反应仍然放热相同的量。催化剂在每次催化循环结束时被再生，这就是 MnO$_2$ 可以不变地回收的原因。

(c) One biological catalyst一种生物催化剂 A1

Catalase is a biological catalyst (enzyme) that catalyses the decomposition of hydrogen peroxide into water and oxygen: $2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$. Other acceptable answers include amylase (catalyses hydrolysis of starch to maltose), lipase (catalyses hydrolysis of fats/lipids), or pepsin (catalyses hydrolysis of proteins).过氧化氢酶是一种生物催化剂（酶），催化过氧化氢分解为水和氧气：$2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$。其他可接受的答案包括淀粉酶（催化淀粉水解为麦芽糖）、脂肪酶（催化脂肪/油脂水解）或胃蛋白酶（催化蛋白质水解）。

A catalyst changes the pathway but not the energy landscape endpoints; $\Delta H$ and the equilibrium position are both unaffected.催化剂改变反应途径但不改变能量图的端点；$\Delta H$ 和平衡位置均不受影响。 The PE diagram for a catalysed reaction has a lower peak (smaller $E_a$) but the same reactant and product energy levels as the uncatalysed diagram. Because both $E_{a,\text{fwd}}$ and $E_{a,\text{rev}}$ decrease by the same amount, the ratio of forward to reverse rate constants is unchanged, which means the equilibrium constant $K$ is unaffected. For heterogeneous catalysts, the reaction happens at the catalyst surface: reactant molecules adsorb onto active sites, bonds weaken, reaction proceeds, products desorb. Poisoning (as in Q8) occurs when impurities block those active sites permanently.催化反应的势能图顶点更低（$E_a$ 更小），但反应物和生成物的能量水平与未催化图相同。由于 $E_{a,\text{fwd}}$ 和 $E_{a,\text{rev}}$ 减少相同量，正逆反应速率常数之比不变，即平衡常数 $K$ 不受影响。对于非均相催化剂，反应在催化剂表面进行：反应物分子吸附到活性位点，键减弱，反应进行，产物脱附。中毒（如 Q8 中所述）发生在杂质永久性堵塞这些活性位点时。

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §1 Rate calculation with stoichiometry含化学计量的速率计算 · HS-PS1-5 [7 marks][7 分]

$2\text{N}_2\text{O}_5(g) \to 4\text{NO}_2(g) + \text{O}_2(g)$. $[\text{N}_2\text{O}_5]$ drops from $0.80$ to $0.20\ \text{mol L}^{-1}$ in $90\ \text{s}$.$2\text{N}_2\text{O}_5(g) \to 4\text{NO}_2(g) + \text{O}_2(g)$。$[\text{N}_2\text{O}_5]$ 在 $90\ \text{s}$ 内从 $0.80$ 降至 $0.20\ \text{mol L}^{-1}$。

Answer:答案： (a) $6.67 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$ · (b) $1.33 \times 10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}$ · (c) $3.33 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$ · (d) concentration is higher at $t = 0$$t = 0$ 时浓度更高

(a) Average rate of disappearance of N$_2$O$_5$N$_2$O$_5$ 消失的平均速率 M1·A1

$$ \text{rate}(\text{N}_2\text{O}_5) = \frac{[\text{N}_2\text{O}_5]_0 - [\text{N}_2\text{O}_5]_{90}}{\Delta t} = \frac{0.80 - 0.20}{90} = \frac{0.60}{90} = 6.67 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}. $$

(b) Average rate of appearance of NO$_2$NO$_2$ 生成的平均速率 M1·A1

The stoichiometric ratio of NO$_2$ to N$_2$O$_5$ is $4:2 = 2:1$, so NO$_2$ appears twice as fast as N$_2$O$_5$ disappears:NO$_2$ 与 N$_2$O$_5$ 的化学计量比为 $4:2 = 2:1$，因此 NO$_2$ 的生成速率是 N$_2$O$_5$ 消失速率的两倍： $$ \text{rate}(\text{NO}_2) = \frac{4}{2} \times 6.67 \times 10^{-3} = 2 \times 6.67 \times 10^{-3} = 1.33 \times 10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}. $$

(c) Average rate of appearance of O$_2$O$_2$ 生成的平均速率 M1·A1

The stoichiometric ratio of O$_2$ to N$_2$O$_5$ is $1:2$, so O$_2$ appears at half the rate N$_2$O$_5$ disappears:O$_2$ 与 N$_2$O$_5$ 的化学计量比为 $1:2$，故 O$_2$ 生成速率是 N$_2$O$_5$ 消失速率的一半： $$ \text{rate}(\text{O}_2) = \frac{1}{2} \times 6.67 \times 10^{-3} = 3.33 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}. $$

(d) Why instantaneous rate at $t = 0$ exceeds the average为何 $t = 0$ 时瞬时速率高于平均速率 A1

At $t = 0$, $[\text{N}_2\text{O}_5]$ is at its maximum ($0.80\ \text{mol L}^{-1}$). The reaction rate depends on the concentration of the reactant(s); as the reaction proceeds, the concentration falls and so does the rate. The average rate ($6.67 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$) spans the entire 90-second interval and includes later, slower moments. The instantaneous rate at $t = 0$ reflects only the initial, high-concentration moment, so it is larger than the average.$t = 0$ 时，$[\text{N}_2\text{O}_5]$ 处于最大值（$0.80\ \text{mol L}^{-1}$）。反应速率取决于反应物浓度；随着反应进行，浓度下降，速率也随之降低。平均速率（$6.67 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$）覆盖整个 90 s 区间，包含了后期较慢的时刻。$t = 0$ 时的瞬时速率只反映初始高浓度时刻，因此大于平均速率。

The average rate is always between the fastest (initial) and slowest (final) instantaneous rates for a reaction where rate decreases over time.对于速率随时间降低的反应，平均速率始终介于最快（初始）和最慢（终态）的瞬时速率之间。 This question tests whether students understand that "average rate over an interval" $\neq$ "rate at the start of the interval." For most simple reactions, [reactant] is highest at $t = 0$, making the instantaneous rate highest there. As [reactant] falls, both the instantaneous rate and the reaction rate fall. The calculated average rate smears out this variation. On a concentration-time curve, the instantaneous rate at any point is the (magnitude of the) slope of the tangent at that point; the average rate over an interval is the slope of the secant line joining the two endpoints.这道题考查学生是否理解"某区间内的平均速率" $\neq$ "该区间起点时的速率"。对于大多数简单反应，$t = 0$ 时 [反应物] 最高，瞬时速率也最高。随着 [反应物] 下降，瞬时速率和反应速率都下降。计算出的平均速率将这种变化"平摊"。在浓度-时间曲线上，任意点的瞬时速率是该点切线的（绝对值）斜率；某区间内的平均速率是连接两端点的割线斜率。

Q7MEDIUM中 🇨🇦 ON安 🇨🇦 BC卑 ON Provincial-style安大略省考风格 §2 + §3 Collision theory + PE diagrams (integrated)碰撞理论 + 势能图（综合） · SCH4U D3.5-D3.6 [8 marks][8 分]

Reactions X ($E_a = 40\ \text{kJ mol}^{-1}$) and Y ($E_a = 120\ \text{kJ mol}^{-1}$) at the same temperature; same negative $\Delta H$. (a) Which is faster and why? (b) PE diagrams on same axes. (c) Two changes to Y's PE diagram when catalyst reduces $E_a$ to $60\ \text{kJ mol}^{-1}$.反应 X（$E_a = 40\ \text{kJ mol}^{-1}$）和 Y（$E_a = 120\ \text{kJ mol}^{-1}$）在相同温度下进行；$\Delta H$ 相同且为负。(a) 哪个更快，为什么？(b) 在同一坐标轴绘制势能图。(c) 催化剂将 Y 的 $E_a$ 降至 $60\ \text{kJ mol}^{-1}$ 后，势能图的两个变化。

Answer:答案： (a) Reaction X is faster反应 X 更快 · (b) two curves sharing reactant/product levels, X with lower peak两条曲线共用反应物/生成物水平，X 顶点更低 · (c) lower peak; unchanged $\Delta H$顶点降低；$\Delta H$ 不变

(a) Which reaction proceeds faster and why哪个反应更快及原因 M1·A1·A1

Reaction X proceeds faster. At the same temperature, both reactions have the same Boltzmann distribution of molecular energies. However, reaction X has a lower activation energy ($40\ \text{kJ mol}^{-1}$ vs $120\ \text{kJ mol}^{-1}$), so a much larger fraction of molecular collisions have kinetic energy $\geq E_a$ for X than for Y. More effective collisions per unit time means a higher rate for X. The fraction of molecules exceeding $E_a$ rises exponentially as $E_a$ decreases (via the Boltzmann factor $e^{-E_a/RT}$), so the difference in rate between X and Y is substantial.反应 X 更快。在相同温度下，两个反应的分子能量玻尔兹曼分布相同。但反应 X 的活化能更低（$40\ \text{kJ mol}^{-1}$ 对比 $120\ \text{kJ mol}^{-1}$），因此碰撞中具有 $\geq E_a$ 动能的比例对 X 远大于 Y。单位时间内有效碰撞更多，意味着 X 的速率更高。超过 $E_a$ 的分子比例随 $E_a$ 降低而指数增大（通过玻尔兹曼因子 $e^{-E_a/RT}$），故 X 和 Y 的速率差异显著。

(b) PE diagrams for X and Y on the same axes在同一坐标轴上绘制 X 和 Y 的势能图 M1·A1·A1

Both curves start at the same reactant PE level and end at the same product PE level (since $\Delta H$ is identical and negative for both). Reaction X has its peak only $40\ \text{kJ mol}^{-1}$ above the reactant level. Reaction Y has its peak $120\ \text{kJ mol}^{-1}$ above the reactant level. Curve X is the "shorter hill"; curve Y is the "taller hill." Label each curve's peak with its $E_a$ value, the shared reactants at the left, the shared (lower) products at the right, and the common $\Delta H$ arrow from reactant level down to product level.两条曲线从相同的反应物势能水平出发，在相同的生成物势能水平结束（因为两者 $\Delta H$ 相同且为负）。反应 X 的顶点仅比反应物水平高 $40\ \text{kJ mol}^{-1}$。反应 Y 的顶点比反应物水平高 $120\ \text{kJ mol}^{-1}$。曲线 X 是"矮山"；曲线 Y 是"高山"。标注每条曲线顶点处的 $E_a$ 值，左侧共用的反应物，右侧共用的（较低）生成物，以及从反应物水平向下到生成物水平的共同 $\Delta H$ 箭头。

(c) Two changes to Y's PE diagram when catalyst is added催化剂加入后 Y 势能图的两个变化 A1·A1

The peak of the curve lowers: from $120\ \text{kJ mol}^{-1}$ above reactants to $60\ \text{kJ mol}^{-1}$ above reactants (the activated complex / transition state is lower in energy).曲线顶点降低：从高于反应物 $120\ \text{kJ mol}^{-1}$ 降至 $60\ \text{kJ mol}^{-1}$（活化络合物/过渡态的能量降低）。
The reactant and product energy levels are unchanged (and so $\Delta H$ is unchanged): the curve's endpoints stay at the same PE values as the uncatalysed reaction. Only the peak shifts.反应物和生成物的能量水平不变（故 $\Delta H$ 不变）：曲线端点的势能值与未催化反应相同，只有顶点位置改变。

A catalyst shifts only the peak of the PE diagram, not its endpoints; this is why it speeds the reaction without altering $\Delta H$ or the equilibrium constant.催化剂只改变势能图的顶点，不改变端点；这正是它在不改变 $\Delta H$ 或平衡常数的情况下加速反应的原因。 The Boltzmann factor $e^{-E_a/RT}$ shows the rate is exponentially sensitive to $E_a$. For Y at 25 °C (298 K), halving $E_a$ from 120 to 60 kJ/mol changes the factor from $e^{-120000/(8.314 \times 298)} \approx e^{-48.4}$ to $e^{-24.2}$ — an increase in the fraction of effective collisions by a factor of $e^{24.2} \approx 3 \times 10^{10}$. This is why catalysts can make otherwise impractically slow reactions proceed at measurable rates.玻尔兹曼因子 $e^{-E_a/RT}$ 表明速率对 $E_a$ 呈指数敏感。对于 Y 在 25 °C（298 K）时，将 $E_a$ 从 120 降至 60 kJ/mol 使该因子从 $e^{-120000/(8.314 \times 298)} \approx e^{-48.4}$ 变为 $e^{-24.2}$，有效碰撞比例提高约 $e^{24.2} \approx 3 \times 10^{10}$ 倍。这就是为什么催化剂能使原本极慢的反应以可测速率进行。

Q8HARD难 🇨🇦 BC卑 🇨🇦 AB阿 BC Provincial-style卑诗省考风格 §4 + §5 Factors and catalysis: Haber process影响因素与催化剂：哈伯法 · BC Chem 12 / Chem 30 GO3 [8 marks][8 分]

Haber process: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g),\ \Delta H = -92\ \text{kJ mol}^{-1}$. Iron catalyst, 150-300 atm, ~450 °C. (a) How pressure increases rate. (b) Homo- or heterogeneous catalysis; where does reaction occur? (c) Why 450 °C is a compromise. (d) Catalyst poisoning by sulfur.哈伯法：$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g),\ \Delta H = -92\ \text{kJ mol}^{-1}$。铁催化剂，150-300 atm，约 450 °C。(a) 增压如何提高速率。(b) 均相还是非均相催化；反应发生的位置。(c) 为何 450 °C 是折中方案。(d) 硫中毒催化剂。

Answer:答案： (a) higher pressure compresses gas molecules into smaller volume, increasing collision frequency高压将气体分子压缩至更小体积，增加碰撞频率 · (b) heterogeneous; at active sites on the iron surface非均相；在铁表面活性位点处 · (c) kinetics vs. equilibrium yield trade-off动力学与平衡产率的权衡 · (d) sulfur atoms block active sites, preventing reactant adsorption硫原子堵塞活性位点，阻止反应物吸附

(a) How increasing pressure increases rate增大压强如何提高速率 M1·A1

Increasing the pressure on a gas mixture compresses the molecules into a smaller volume. This increases the concentration (moles per litre) of both N$_2$ and H$_2$. With more molecules per unit volume, collisions between N$_2$ and H$_2$ molecules occur more frequently per unit time. More frequent collisions means a greater number of effective collisions per second, and therefore a higher reaction rate.增大气体混合物的压强将分子压缩至更小体积，增加了 N$_2$ 和 H$_2$ 的浓度（每升摩尔数）。单位体积内分子更多，N$_2$ 与 H$_2$ 分子之间每单位时间内的碰撞更加频繁。碰撞频率提高意味着每秒有效碰撞次数更多，因此反应速率更高。

(b) Type of catalysis and where reaction occurs催化类型及反应发生的位置 A1·A1

This is heterogeneous catalysis: the catalyst (iron, solid) and the reactants (N$_2$, H$_2$, gases) are in different phases. The reaction occurs at the active sites on the surface of the iron catalyst. Gas molecules of N$_2$ and H$_2$ adsorb onto the iron surface, weaken their strong triple/double bonds as they interact with the iron atoms, react to form NH$_3$, and desorb from the surface as product.这是非均相催化：催化剂（固体铁）与反应物（气体 N$_2$、H$_2$）处于不同相中。反应发生在铁催化剂表面的活性位点。N$_2$ 和 H$_2$ 气体分子吸附到铁表面，与铁原子相互作用使其强三键/双键减弱，反应生成 NH$_3$，再从表面脱附成为产物。

(c) Why 450 °C is a compromise为何 450 °C 是折中方案 A1·A1

Two competing factors operate: (1) Kinetics: higher temperature gives more molecules $\geq E_a$, increasing the reaction rate. (2) Equilibrium (Le Chatelier): the forward reaction is exothermic ($\Delta H = -92\ \text{kJ mol}^{-1}$). Increasing temperature shifts the equilibrium to the left (towards N$_2$ + H$_2$), reducing the equilibrium yield of NH$_3$. At 450 °C, the reaction proceeds at a useful rate while still giving an acceptable yield of NH$_3$. Above this temperature the yield drops too low; below it the rate is too slow to be economically viable. 450 °C is the industrial compromise between rate and yield.两个相互竞争的因素同时作用：(1) 动力学：更高温度使更多分子具有 $\geq E_a$ 的能量，提高反应速率。(2) 平衡（勒夏特列原理）：正反应为放热反应（$\Delta H = -92\ \text{kJ mol}^{-1}$）。升温使平衡向左移动（朝向 N$_2$ + H$_2$），降低 NH$_3$ 的平衡产率。在 450 °C 下，反应以实用速率进行，同时仍能获得可接受的 NH$_3$ 产率。高于此温度产率过低；低于此温度速率太慢，经济上不可行。450 °C 是速率与产率之间的工业折中温度。

(d) Catalyst poisoning by sulfur at the molecular level分子层面解释硫中毒催化剂 A1·A1

At the molecular level, sulfur atoms (from H$_2$S or other sulfur compounds in the feedstock) bind strongly to the active sites on the iron surface, occupying positions where N$_2$ and H$_2$ molecules would normally adsorb. With the active sites blocked, fewer reactant molecules can adsorb and react. This reduces the effective surface area available for catalysis, lowering the number of effective collisions per unit time and therefore reducing the reaction rate. Unlike the reactants, sulfur does not desorb easily, so the poisoning is effectively permanent.在分子层面，硫原子（来自进料中的 H$_2$S 或其他含硫化合物）与铁表面活性位点强力结合，占据了 N$_2$ 和 H$_2$ 分子正常吸附的位置。活性位点被堵塞后，能吸附并反应的反应物分子减少。这降低了可用于催化的有效表面积，减少了单位时间内的有效碰撞次数，从而降低反应速率。与反应物不同，硫不易脱附，因此中毒实际上是永久性的。

The Haber process illustrates all four rate factors in one system: concentration (pressure), temperature, surface area (iron particle size), and catalysis (iron).哈伯法在一个体系中体现了全部四个速率影响因素：浓度（压强）、温度、表面积（铁颗粒大小）和催化（铁）。 High pressure increases collision frequency (factor: concentration). The catalyst lowers $E_a$ (factor: catalyst). Fine iron particles maximize surface area for adsorption. Temperature is the compromise factor: kinetics wants high T; Le Chatelier wants low T for exothermic forward reaction; 450 °C is the industrial sweet spot. The promoters (K$_2$O, Al$_2$O$_3$) further enhance catalytic activity by keeping the iron particles dispersed and avoiding sintering. Understanding this interplay is key to industrial process design.高压增大碰撞频率（浓度因素）。催化剂降低 $E_a$（催化因素）。细铁粉最大化吸附表面积。温度是折中因素：动力学需要高温；勒夏特列原理对放热正反应需要低温；450 °C 是工业最佳点。助催化剂（K$_2$O、Al$_2$O$_3$）通过保持铁颗粒分散、防止烧结进一步提高催化活性。理解这种相互作用是工业过程设计的关键。

Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §6 Rate laws (initial rate method)速率定律（初始速率法） · SCH4U / BC Chem 12 [7 marks][7 分]

$\text{A} + \text{B} \to \text{products}$. Exp 1: [A] = 0.10, [B] = 0.20, rate = $4.0 \times 10^{-3}$; Exp 2: [A] = 0.20, [B] = 0.20, rate = $8.0 \times 10^{-3}$; Exp 3: [A] = 0.10, [B] = 0.40, rate = $1.6 \times 10^{-2}$. All in mol L$^{-1}$ (s$^{-1}$).$\text{A} + \text{B} \to \text{products}$。实验 1：[A] = 0.10，[B] = 0.20，速率 = $4.0 \times 10^{-3}$；实验 2：[A] = 0.20，[B] = 0.20，速率 = $8.0 \times 10^{-3}$；实验 3：[A] = 0.10，[B] = 0.40，速率 = $1.6 \times 10^{-2}$。浓度单位 mol L$^{-1}$，速率单位 mol L$^{-1}$ s$^{-1}$。

Answer:答案： (a) 1st order in A; 2nd order in B对 A 一级；对 B 二级 · (b) $\text{rate} = k[\text{A}][\text{B}]^2;\; k = 1.0\ \text{mol}^{-2}\text{L}^{2}\text{s}^{-1}$ · (c) $1.2 \times 10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}$

(a) Determine orders with respect to A and B确定对 A 和 B 的反应级数 M1·A1·A1

Order with respect to A (compare Experiments 1 and 2, where [B] is held constant at 0.20):对 A 的级数（比较实验 1 和 2，[B] 恒为 0.20）： $$ \frac{\text{rate}_2}{\text{rate}_1} = \frac{8.0 \times 10^{-3}}{4.0 \times 10^{-3}} = 2.0. \quad \frac{[\text{A}]_2}{[\text{A}]_1} = \frac{0.20}{0.10} = 2.0. $$ Rate doubled when [A] doubled; order in A = 1 (first order).[A] 加倍时速率加倍；对 A 的级数 = 1（一级）。

Order with respect to B (compare Experiments 1 and 3, where [A] is held constant at 0.10):对 B 的级数（比较实验 1 和 3，[A] 恒为 0.10）： $$ \frac{\text{rate}_3}{\text{rate}_1} = \frac{1.6 \times 10^{-2}}{4.0 \times 10^{-3}} = 4.0. \quad \frac{[\text{B}]_3}{[\text{B}]_1} = \frac{0.40}{0.20} = 2.0. $$ Rate quadrupled when [B] doubled; $2^n = 4 \Rightarrow n = 2$; order in B = 2 (second order).[B] 加倍时速率变为 4 倍；$2^n = 4 \Rightarrow n = 2$；对 B 的级数 = 2（二级）。

(b) Rate law and rate constant $k$速率定律和速率常数 $k$ M1·A1·A1

Overall rate law:总速率定律： $$ \text{rate} = k[\text{A}]^1[\text{B}]^2. $$ Using Experiment 1 to calculate $k$:用实验 1 计算 $k$： $$ k = \frac{\text{rate}}{[\text{A}][\text{B}]^2} = \frac{4.0 \times 10^{-3}}{(0.10)(0.20)^2} = \frac{4.0 \times 10^{-3}}{(0.10)(0.040)} = \frac{4.0 \times 10^{-3}}{4.0 \times 10^{-3}} = 1.0\ \text{mol}^{-2}\text{L}^{2}\text{s}^{-1}. $$ Overall reaction order = $1 + 2 = 3$ (third order total). Units of $k$ for a third-order reaction are $\text{mol}^{-2}\text{L}^{2}\text{s}^{-1}$.总反应级数 = $1 + 2 = 3$（总体三级）。三级反应 $k$ 的单位为 $\text{mol}^{-2}\text{L}^{2}\text{s}^{-1}$。

(c) Predict initial rate for [A] = 0.30, [B] = 0.20预测 [A] = 0.30，[B] = 0.20 时的初始速率 A1

$$ \text{rate} = k[\text{A}][\text{B}]^2 = (1.0)(0.30)(0.20)^2 = (1.0)(0.30)(0.040) = 1.2 \times 10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}. $$

The initial rate method: always compare two experiments where only one concentration changes, then use the ratio rule to find the exponent.初始速率法：始终比较只有一种浓度改变的两个实验，再用比值规则求指数。 The key steps are: (1) identify a pair of experiments where only one reactant's concentration changes; (2) form the ratio of rates and the ratio of concentrations; (3) solve $(\text{conc ratio})^n = \text{rate ratio}$ for $n$. Common exponents: rate doubles when [X] doubles $\to$ $n = 1$; rate quadruples $\to$ $n = 2$; rate stays same $\to$ $n = 0$. Once the rate law is written, calculate $k$ from any single experiment (cross-check with another to confirm). Units of $k$ must be derived, not guessed: for an overall order $n$, units are $\text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}$.关键步骤：(1) 找出只有一种反应物浓度改变的一对实验；(2) 计算速率比和浓度比；(3) 由 $(\text{浓度比})^n = \text{速率比}$ 求解 $n$。常见指数：[X] 加倍时速率加倍 $\to$ $n = 1$；速率变 4 倍 $\to$ $n = 2$；速率不变 $\to$ $n = 0$。写出速率定律后，从任意单个实验计算 $k$（用另一个实验验证）。$k$ 的单位必须推导而非猜测：对于总级数 $n$，单位为 $\text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}$。

Section C · Worked SolutionsC 部分 · 详细解答

Q10MEDIUM中Honors荣誉级 🇨🇦 ON安 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §7 Reaction mechanisms and rate-determining step反应机理与速率决定步骤 · SCH4U / BC Chem 12 [9 marks][9 分]

Two-step mechanism for $2\text{NO}(g) + \text{Cl}_2(g) \to 2\text{NOCl}(g)$: Step 1 (fast eq): $\text{NO} + \text{Cl}_2 \rightleftharpoons \text{NOCl}_2$. Step 2 (slow): $\text{NOCl}_2 + \text{NO} \to 2\text{NOCl}$.$2\text{NO}(g) + \text{Cl}_2(g) \to 2\text{NOCl}(g)$ 的两步机理：步骤 1（快速平衡）：$\text{NO} + \text{Cl}_2 \rightleftharpoons \text{NOCl}_2$。步骤 2（慢）：$\text{NOCl}_2 + \text{NO} \to 2\text{NOCl}$。

Answer:答案： (a) Step 2 (slow step)步骤 2（慢步骤） · (b) NOCl$_2$ is the intermediateNOCl$_2$ 是中间体 · (c) $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ · (d) consistent一致

(a) Identify the rate-determining step确定速率决定步骤 A1·A1

The rate-determining step (RDS) is Step 2, the slow step. The overall reaction can only proceed as fast as its slowest elementary step; a fast preceding step can reach equilibrium rapidly, but the slow step is the bottleneck that controls the overall rate. Analogy: the slowest lane on a highway determines traffic throughput regardless of how fast other lanes move.速率决定步骤（RDS）是步骤 2，即慢步骤。整个反应的速率不能超过其最慢的基元步骤；前面的快速步骤可以迅速达到平衡，但慢步骤是控制总体速率的瓶颈。类比：高速公路最慢的车道决定通行量，无论其他车道多快。

(b) Identify the reaction intermediate; compare to a catalyst确定反应中间体；与催化剂比较 A1·A1

NOCl$_2$ is the reaction intermediate. It is produced in Step 1 and consumed in Step 2; it does not appear in the overall balanced equation. A reaction intermediate differs from a catalyst in two ways: (1) A catalyst is present at the start of the reaction, is not consumed, and is regenerated at the end. An intermediate is produced during the reaction and fully consumed before the reaction is complete. (2) A catalyst is added externally; an intermediate is a transient species generated from reactants within the mechanism.NOCl$_2$ 是反应中间体。它在步骤 1 中生成，在步骤 2 中被消耗；它不出现在总配平方程式中。反应中间体与催化剂有两点区别：(1) 催化剂在反应开始时存在，不被消耗，反应结束后再生。中间体在反应过程中生成，在反应完成前被完全消耗。(2) 催化剂是外加的；中间体是在机理中由反应物生成的瞬态物种。

(c) Derive the overall rate law推导总速率定律 M1·A1·A1·A1

Step 2 (the RDS) involves NOCl$_2$ and NO, so the rate law from the RDS is:步骤 2（RDS）涉及 NOCl$_2$ 和 NO，故 RDS 的速率定律为： $$ \text{rate} = k_2[\text{NOCl}_2][\text{NO}]. $$ NOCl$_2$ is an intermediate (not measurable directly). Step 1 is a fast equilibrium, so:NOCl$_2$ 是中间体（无法直接测量）。步骤 1 为快速平衡，故： $$ K_{eq,1} = \frac{[\text{NOCl}_2]}{[\text{NO}][\text{Cl}_2]} \;\Longrightarrow\; [\text{NOCl}_2] = K_{eq,1}[\text{NO}][\text{Cl}_2]. $$ Substituting to eliminate the intermediate:代入消去中间体： $$ \text{rate} = k_2 \cdot K_{eq,1}[\text{NO}][\text{Cl}_2] \cdot [\text{NO}] = k_{\text{obs}}[\text{NO}]^2[\text{Cl}_2], $$ where $k_{\text{obs}} = k_2 K_{eq,1}$ is the observed rate constant.其中 $k_{\text{obs}} = k_2 K_{eq,1}$ 是观测速率常数。

(d) Consistency with experimental rate law与实验速率定律的一致性 A1

The mechanism predicts $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$, which is consistent with the experimentally observed rate law $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$. The orders match: first order in Cl$_2$ and second order in NO. This agreement supports (but does not prove) the proposed mechanism.该机理预测 $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$，与实验观察到的速率定律 $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ 一致。级数吻合：对 Cl$_2$ 一级，对 NO 二级。这种一致性支持（但不能证明）所提出的机理。

A mechanism consistent with the rate law is supported, not proven; the rate law alone cannot uniquely identify the mechanism.与速率定律一致的机理只是被支持，而非被证明；速率定律本身不能唯一确定机理。 The substitution trick for eliminating intermediates is a key AP/honors skill. The equilibrium approximation for a fast pre-equilibrium step is valid whenever the fast step can be considered to reach its equilibrium position before the slow step significantly depletes its products. The combined rate constant $k_{\text{obs}} = k_2 K_{eq,1}$ explains why the overall rate constant changes with temperature (both $k_2$ and $K_{eq,1}$ are temperature-dependent). Multiple mechanisms can often predict the same rate law, so experimental evidence beyond rate data (e.g., isotope labelling, detection of intermediates) is needed to distinguish them.消去中间体的代换技巧是 AP/荣誉课程的关键技能。快速预平衡步骤的平衡近似在快速步骤能在慢步骤显著消耗其产物之前达到平衡位置时成立。总速率常数 $k_{\text{obs}} = k_2 K_{eq,1}$ 解释了为何总体速率常数随温度变化（$k_2$ 和 $K_{eq,1}$ 均与温度有关）。多种机理常常能预测相同的速率定律，因此需要速率数据以外的实验证据（如同位素标记、中间体检测）来区分它们。

Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §1 + §4 Rate calculation + factors (applied)速率计算 + 影响因素（应用） · Chem 30 GO2-GO3 [9 marks][9 分]

$\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \to \text{ZnSO}_4(aq) + \text{H}_2(g)$. Exp A: 2.0 g Zn granules, 50 mL of 1.0 mol L$^{-1}$ H$_2$SO$_4$, 25 °C; after 60 s, $[\text{H}_2\text{SO}_4]$ falls to 0.40 mol L$^{-1}$. Exp B: Zn powder (same mass). Exp C: 2.0 mol L$^{-1}$ acid.$\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \to \text{ZnSO}_4(aq) + \text{H}_2(g)$。实验 A：2.0 g 锌粒，50 mL，1.0 mol L$^{-1}$ H$_2$SO$_4$，25 °C；60 s 后 $[\text{H}_2\text{SO}_4]$ 降至 0.40 mol L$^{-1}$。实验 B：等质量锌粉。实验 C：2.0 mol L$^{-1}$ 酸。

Answer:答案： (a) $1.0 \times 10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}$ · (b) B is faster (more surface area)B 更快（表面积更大） · (c) C has higher initial rate (higher [H$_2$SO$_4$])C 初始速率更高（[H$_2$SO$_4$] 更高） · (d) A and B produce same total H$_2$; C produces same total H$_2$ as A and B (Zn is limiting in all)A 和 B 产生相同总量 H$_2$；C 与 A、B 产生相同总量 H$_2$（所有实验中 Zn 均为限量试剂）

(a) Average rate of disappearance of H$_2$SO$_4$ in Experiment A实验 A 中 H$_2$SO$_4$ 消失的平均速率 M1·A1

$$ \text{rate} = \frac{[\text{H}_2\text{SO}_4]_0 - [\text{H}_2\text{SO}_4]_{60}}{\Delta t} = \frac{1.0 - 0.40}{60} = \frac{0.60}{60} = 1.0 \times 10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}. $$

(b) Rate comparison for Experiment B (Zn powder)实验 B（锌粉）的速率比较 M1·A1·A1

Experiment B has a greater initial rate than Experiment A. Replacing granules with powder dramatically increases the surface area of Zn exposed to the acid. With more Zn surface sites available, H$^+$ ions collide with Zn atoms more frequently per unit time, increasing the collision frequency between acid and metal. This raises the number of effective collisions per second and therefore the reaction rate. All other variables (temperature, acid concentration, total mass of Zn) are unchanged, so surface area is the only factor changing.实验 B 的初始速率大于实验 A。将锌粒换成锌粉显著增大了 Zn 与酸接触的表面积。可利用的 Zn 表面位点更多，H$^+$ 离子每单位时间与 Zn 原子碰撞更频繁，增大了酸与金属之间的碰撞频率。这提高了每秒有效碰撞次数，从而提高反应速率。其他所有变量（温度、酸浓度、Zn 总质量）不变，故表面积是唯一改变的因素。

(c) Effect of doubling acid concentration in Experiment C实验 C 酸浓度加倍的影响 A1·A1

Doubling the acid concentration to $2.0\ \text{mol L}^{-1}$ increases the initial reaction rate. With more H$^+$ ions per unit volume, acid molecules collide with the Zn surface more frequently. This increases the number of effective collisions per second, raising the initial rate. The rate approximately doubles compared to Experiment A (assuming the reaction is first order in [H$^+$] under these conditions).将酸浓度加倍至 $2.0\ \text{mol L}^{-1}$ 提高了初始反应速率。单位体积内 H$^+$ 离子更多，酸分子与 Zn 表面的碰撞更频繁。这增加了每秒有效碰撞次数，提高初始速率。与实验 A 相比，速率约为两倍（假设在这些条件下反应对 [H$^+$] 为一级）。

(d) Total H$_2$ produced in Experiments A, B, and C实验 A、B、C 中产生的 H$_2$ 总量 M1·A1

First identify the limiting reagent in each experiment. Moles of Zn (molar mass $\approx 65.4\ \text{g mol}^{-1}$): $n(\text{Zn}) = 2.0 / 65.4 \approx 0.0306\ \text{mol}$. Experiments A and B: moles of H$_2$SO$_4$ = $0.050\ \text{L} \times 1.0\ \text{mol L}^{-1} = 0.050\ \text{mol}$. Since $0.0306 < 0.050$, Zn is the limiting reagent. Experiment C: moles of H$_2$SO$_4$ = $0.050\ \text{L} \times 2.0\ \text{mol L}^{-1} = 0.10\ \text{mol}$. Zn ($0.0306\ \text{mol}$) is still limiting. Since 1 mol Zn produces 1 mol H$_2$ (1:1 stoichiometry), all three experiments produce the same moles of H$_2$ ($\approx 0.0306\ \text{mol}$), and hence the same total mass. A and B produce the same total H$_2$; C also produces the same total H$_2$.首先确定各实验的限量试剂。Zn 的摩尔数（摩尔质量 $\approx 65.4\ \text{g mol}^{-1}$）：$n(\text{Zn}) = 2.0 / 65.4 \approx 0.0306\ \text{mol}$。实验 A 和 B：$n(\text{H}_2\text{SO}_4) = 0.050\ \text{L} \times 1.0\ \text{mol L}^{-1} = 0.050\ \text{mol}$。因为 $0.0306 < 0.050$，Zn 为限量试剂。实验 C：$n(\text{H}_2\text{SO}_4) = 0.050\ \text{L} \times 2.0\ \text{mol L}^{-1} = 0.10\ \text{mol}$。Zn（$0.0306\ \text{mol}$）仍为限量试剂。由于 1 mol Zn 生成 1 mol H$_2$（1:1 化学计量比），三组实验产生的 H$_2$ 摩尔数相同（$\approx 0.0306\ \text{mol}$），总质量也相同。A 和 B 产生相同总量 H$_2$；C 也产生相同总量 H$_2$。

Rate affects how quickly a reaction reaches completion; the limiting reagent determines total yield regardless of rate.速率影响反应达到完全的快慢；限量试剂决定总产率，与速率无关。 A classic trap in AB/BC diploma and AP exams: students often assume that higher concentration or more surface area produces more product. It does not. The total product is fixed by stoichiometry and the amount of the limiting reagent. Rate only controls the time taken to consume that limiting reagent. In Experiment C, doubling the acid concentration makes the reaction faster (reaches completion sooner) but does not change how much Zn reacts (still all 2.0 g) or how much H$_2$ is produced. Always identify the limiting reagent first before making claims about total yield.AB/BC 毕业考和 AP 考试中的经典陷阱：学生常以为更高浓度或更大表面积会产生更多产物。事实并非如此。总产物由化学计量关系和限量试剂的量决定。速率只控制消耗限量试剂所需的时间。在实验 C 中，酸浓度加倍使反应更快完成（更快消耗完），但不改变参与反应的 Zn 量（仍然是全部 2.0 g）或产生的 H$_2$ 量。在对总产率作出任何判断之前，始终先确定限量试剂。

Q12HARD难 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §1-§5 Integrated kinetics: H$_2$O$_2$ decomposition data analysis动力学综合：H$_2$O$_2$ 分解数据分析 · HS-PS1-5 / SCH4U [10 marks][10 分]

H$_2$O$_2$ decomposition at 25 °C (no catalyst). Data: $t = 0$ s: $[\text{H}_2\text{O}_2] = 0.900$; $t = 60$ s: $0.720$; $t = 120$ s: $0.576$; $t = 180$ s: $0.461$ (mol L$^{-1}$).25 °C 下 H$_2$O$_2$ 分解（无催化剂）。数据：$t = 0$ s：$0.900$；$t = 60$ s：$0.720$；$t = 120$ s：$0.576$；$t = 180$ s：$0.461$（mol L$^{-1}$）。

Answer:答案： (a) $3.00 \times 10^{-3}$ and $1.92 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$ · (b) fewer H$_2$O$_2$ molecules, fewer collisionsH$_2$O$_2$ 分子减少，碰撞减少 · (c) ratios both 0.800; implies first-order reaction两个比值均为 0.800；意味着一级反应 · (d) faster rate of O$_2$ production; shorter time to completion; both due to lower $E_a$O$_2$ 产生速率更快；完成时间更短；均因 $E_a$ 更低

(a) Average rates over two intervals两段时间内的平均速率 M1·A1·A1

Interval $t = 0$ to $60\ \text{s}$:区间 $t = 0$ 至 $60\ \text{s}$： $$ \text{rate}_{0-60} = \frac{0.900 - 0.720}{60} = \frac{0.180}{60} = 3.00 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}. $$ Interval $t = 120\ \text{s}$ to $180\ \text{s}$:区间 $t = 120\ \text{s}$ 至 $180\ \text{s}$： $$ \text{rate}_{120-180} = \frac{0.576 - 0.461}{60} = \frac{0.115}{60} = 1.92 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}. $$

(b) Why rate decreases over time (collision theory)速率随时间降低的原因（碰撞理论） A1·A1

As the reaction proceeds, the concentration of H$_2$O$_2$ decreases. With fewer H$_2$O$_2$ molecules per unit volume, there are fewer molecule-molecule collisions per unit time. The collision frequency therefore falls, reducing the number of effective collisions per second, and so the reaction rate slows. This is the essence of why concentration affects rate: fewer reactant molecules mean fewer opportunities for productive collisions.随着反应进行，H$_2$O$_2$ 浓度下降。单位体积内 H$_2$O$_2$ 分子减少，单位时间内分子间碰撞次数减少。碰撞频率因此降低，每秒有效碰撞次数减少，反应速率随之减慢。这正是浓度影响速率的本质：反应物分子越少，发生有效碰撞的机会越少。

(c) Constant ratio test for reaction order等时间比值检验法判断反应级数 M1·A1·A1

$$ \frac{[\text{H}_2\text{O}_2]_{60}}{[\text{H}_2\text{O}_2]_0} = \frac{0.720}{0.900} = 0.800. $$ $$ \frac{[\text{H}_2\text{O}_2]_{120}}{[\text{H}_2\text{O}_2]_{60}} = \frac{0.576}{0.720} = 0.800. $$ Both ratios are equal ($0.800$). A constant fractional decrease in concentration over equal time intervals is the signature of a first-order reaction. For a first-order process, a fixed fraction of the remaining reactant decomposes in each equal time period, regardless of the starting concentration. This is mathematically equivalent to saying the half-life is constant: $[\text{H}_2\text{O}_2]$ falls to 80% of its previous value every 60 s.两个比值相等（均为 $0.800$）。在等时间间隔内浓度的恒定分数下降是一级反应的标志。对于一级过程，每个等时间段内消耗的是剩余反应物的固定比例，与初始浓度无关。这在数学上等价于说半衰期是恒定的：每 60 s，$[\text{H}_2\text{O}_2]$ 降为前值的 80%。

(d) Two observable differences with catalase enzyme加入过氧化氢酶后的两个可观察差异 A1·A1

Faster rate of O$_2$ production (observable as more rapid bubbling): catalase provides an alternative pathway with a much lower activation energy than the uncatalysed reaction. More H$_2$O$_2$ molecules have sufficient energy to react per unit time, increasing the rate dramatically.O$_2$ 产生速率更快（表现为气泡更剧烈）：过氧化氢酶提供了一条活化能远低于未催化反应的替代途径。每单位时间内有足够能量参与反应的 H$_2$O$_2$ 分子更多，速率大幅提高。
Shorter time to complete the reaction: because the rate is much higher with catalase, the same initial amount of H$_2$O$_2$ is consumed in a much shorter time. The total amount of O$_2$ produced at the end is the same (same initial [H$_2$O$_2$] and stoichiometry), but it is produced faster. Again this follows from the lower $E_a$: more effective collisions per second means the reactant is consumed sooner.完成反应所需时间更短：由于加入过氧化氢酶后速率大幅提高，相同初始量的 H$_2$O$_2$ 在更短时间内被消耗完。最终产生的 O$_2$ 总量相同（初始 [H$_2$O$_2$] 和化学计量比相同），但生成更快。这同样源于更低的 $E_a$：每秒有效碰撞更多意味着反应物更快被消耗。

The constant ratio 0.800 over 60-second intervals reveals a first-order process with a "half-life" of about 270 s (time to fall to 0.500 of initial).每 60 s 的恒定比值 0.800 揭示了一个一级过程，其"半衰期"约为 270 s（降至初始值 0.500 所需时间）。 For a first-order reaction with rate constant $k$: $[\text{A}] = [\text{A}]_0 e^{-kt}$. After one interval of 60 s, $[\text{A}]/[\text{A}]_0 = e^{-k \cdot 60} = 0.800$, giving $k = -\ln(0.800)/60 = 0.223/60 = 3.72 \times 10^{-3}\ \text{s}^{-1}$. Half-life: $t_{1/2} = \ln 2 / k = 0.693 / (3.72 \times 10^{-3}) \approx 186\ \text{s}$. The data-analysis approach used in this question (constant ratio test) is a core AP exam skill for identifying reaction order from experimental data without needing to fit the integrated rate law.对于速率常数为 $k$ 的一级反应：$[\text{A}] = [\text{A}]_0 e^{-kt}$。60 s 后，$[\text{A}]/[\text{A}]_0 = e^{-k \cdot 60} = 0.800$，得 $k = -\ln(0.800)/60 = 0.223/60 = 3.72 \times 10^{-3}\ \text{s}^{-1}$。半衰期：$t_{1/2} = \ln 2 / k = 0.693 / (3.72 \times 10^{-3}) \approx 186\ \text{s}$。本题使用的数据分析方法（恒定比值检验）是 AP 考试中从实验数据确定反应级数的核心技能，无需拟合积分速率方程。