IB Chemistry HL — Reactivity 2: How Much, How Fast, How Far

IB-Style Practice Questions — Reactivity 2.2 FocusIB 风格练习题 —— 聚焦 Reactivity 2.2

Multiple Choice选择题

Each item carries 1 mark. No calculator, no data booklet. Items tagged HL test content beyond the standard-level syllabus (mechanisms, rate equations, Arrhenius).每题 1 分。不可使用计算器与数据手册（data booklet）。标记 HL 的题目考查超出标准级别（SL）大纲的内容：反应机理（mechanism）、速率方程（rate equation）、阿伦尼乌斯方程（Arrhenius equation）。

Q2EASY Paper 1 2.2.1 Rate from a Graph [1]

For a typical reaction $\mathrm{A \to \text{products}}$, a graph of $[\mathrm{A}]$ versus time is a curve that levels off near zero. At which point on the curve is the instantaneous rate of disappearance of A greatest?对典型反应 $\mathrm{A \to \text{products}}$，$[\mathrm{A}]$ 对时间作图是一条逐渐趋于零的曲线。在曲线上哪一点 A 的瞬时消失速率最大？

(A) At $t = 0$ (the start of the reaction)$t = 0$（反应起始）
(B) At the time when $[\mathrm{A}]$ has fallen to half its initial value$[\mathrm{A}]$ 降至初值一半时
(C) When the curve flattens out曲线趋于水平时
(D) Rate is constant; all points are equal速率不变；曲线上各点相同

Q3EASY Paper 1 2.2.2 Collision Theory [1]

According to collision theory, in order for a chemical reaction to occur, colliding particles must根据碰撞理论（collision theory），化学反应得以发生时，碰撞的粒子必须

(A) have at least the activation energy and the correct orientation.同时满足：能量至少达到活化能（activation energy），且取向正确。
(B) have the correct orientation only — energy is supplied by the surroundings.只需取向正确即可 —— 能量由环境提供。
(C) have at least the activation energy only — orientation does not matter.只需能量达到活化能即可 —— 取向无所谓。
(D) be in the gas phase.必须处于气相。

Q4EASY Paper 1 2.2.3 Concentration [1]

A reaction between aqueous reactants becomes faster when their concentrations are increased. Which is the best microscopic explanation?水溶液中反应物的浓度增大后反应变快。下列哪一项是最佳的微观解释？

(A) Higher concentration lowers the activation energy.浓度增高会降低活化能。
(B) Higher concentration increases the number of particles per unit volume, so the collision frequency rises.浓度增高使单位体积内粒子数增加，于是碰撞频率（collision frequency）上升。
(C) Higher concentration increases the average kinetic energy of the particles.浓度增高使粒子的平均动能增大。
(D) Higher concentration changes the order of the reaction.浓度增高会改变反应级数。

Q5MEDIUM Paper 1 2.2.3 Surface Area [1]

When excess marble (calcium carbonate) is added to a fixed volume of dilute hydrochloric acid, the reaction produces $\mathrm{CO_2}$ gas. Compared with the same mass added as a single lump, finely powdered marble gives将过量大理石（碳酸钙）加入定体积的稀盐酸，反应生成 $\mathrm{CO_2}$ 气体。与相同质量的整块大理石相比，研成细粉的大理石产生

(A) a slower initial rate but the same total volume of $\mathrm{CO_2}$.较慢的初始速率，但 $\mathrm{CO_2}$ 总体积相同。
(B) a faster initial rate but a smaller total volume of $\mathrm{CO_2}$.较快的初始速率，但 $\mathrm{CO_2}$ 总体积更小。
(C) a faster initial rate and the same total volume of $\mathrm{CO_2}$.较快的初始速率，且 $\mathrm{CO_2}$ 总体积相同。
(D) the same initial rate and the same total volume of $\mathrm{CO_2}$.相同的初始速率，且 $\mathrm{CO_2}$ 总体积相同。

Q6MEDIUM Paper 1 2.2.4 Maxwell–Boltzmann [1]

A Maxwell–Boltzmann distribution is sketched for a sample of gas at 300 K. The distribution is then redrawn for the same sample at 350 K. Which statement correctly describes the 350 K curve?为某气体样品在 300 K 时画出麦克斯韦–玻尔兹曼分布（Maxwell–Boltzmann distribution）。再为同一样品在 350 K 时重画该分布。下列哪一项正确描述 350 K 的曲线？

(A) Same peak height; shifted to the right; same total area.峰高不变；整体右移；总面积相同。
(B) Lower peak height; broader spread; same total area; larger area beyond $E_a$.峰高降低；分布更宽；总面积相同；$E_a$ 右侧面积更大。
(C) Higher peak height; narrower spread; larger total area.峰高升高；分布更窄；总面积更大。
(D) Same shape, just shifted upward.形状不变，整体上移。

Q8MEDIUM Paper 1 2.2.5 Energy Profile [1]

On an energy profile diagram for an exothermic reaction, adding a catalyst在放热反应的能量曲线（energy profile）上，加入催化剂会

(A) lowers the reactant level only.仅降低反应物能级。
(B) raises the product level so that $\Delta H$ becomes less negative.抬高产物能级，使 $\Delta H$ 变得不那么负。
(C) lowers the activation-energy peak; reactant and product levels are unchanged.降低活化能峰；反应物与产物能级不变。
(D) removes the activation-energy peak entirely.完全消除活化能峰。

Q9HARD Paper 1 HL 2.2.6 Intermediate vs Transition State [1]

On the energy profile of a two-step reaction, a reaction intermediate corresponds to在两步反应的能量曲线上，反应中间体（intermediate）对应于

(A) a local energy maximum between steps.两步之间的局部能量极大点。
(B) a local energy minimum between two transition-state peaks.两个过渡态（transition state）峰之间的局部能量极小点。
(C) the highest point on the entire profile.整条曲线的最高点。
(D) the same species as a transition state.与过渡态是同一种物种。

Q10HARD Paper 1 HL 2.2.10 Order from a Graph [1]

For a reaction $\mathrm{A \to \text{products}}$, a plot of $[\mathrm{A}]$ versus $t$ is a straight line with negative slope. The order with respect to A is对反应 $\mathrm{A \to \text{products}}$，$[\mathrm{A}]$ 对 $t$ 的图像是斜率为负的直线。关于 A 的反应级数（order）为

(A) 0
(B) 1
(C) 2
(D) Cannot be determined from this graph alone.仅凭此图无法确定。

Q12HARD Paper 1 HL 2.2.12 Arrhenius [1]

When the rate constants of a reaction are measured at several temperatures and $\ln k$ is plotted against $1/T$, the line obtained has a negative slope. From the magnitude of this slope you can directly determine在多个温度下测得某反应的速率常数后，将 $\ln k$ 对 $1/T$ 作图，得到一条斜率为负的直线。由该斜率的大小可直接确定

(A) the Arrhenius factor $A$.阿伦尼乌斯因子 $A$。
(B) the order of reaction.反应级数。
(C) the activation energy $E_a$.活化能 $E_a$。
(D) the enthalpy change $\Delta H$.焓变 $\Delta H$。

Structured Response结构化解答题

Show all working in the space provided. Marks for correct method are awarded even if the final numerical answer is wrong. State units and significant figures appropriately.在指定区域写出全部解题过程。即便最终数值错误，方法正确仍可得分。注意单位与有效数字（significant figures）。

SR 1MEDIUM Paper 2 2.2.1 / 2.2.3 / 2.2.4 Rate from Experiment [10]

A student measures the volume of $\mathrm{CO_2}$ released from the reaction of excess marble chips ($\mathrm{CaCO_3}$) with $50.0~\mathrm{cm^3}$ of $1.00~\mathrm{mol\,dm^{-3}}$ HCl at 298 K. The data are recorded over the first 200 s and plotted as $V(\mathrm{CO_2})$ versus $t$.某学生在 298 K 下测量过量大理石碎块（$\mathrm{CaCO_3}$）与 $50.0~\mathrm{cm^3}$、$1.00~\mathrm{mol\,dm^{-3}}$ HCl 反应放出的 $\mathrm{CO_2}$ 体积。前 200 s 的数据如下，并作 $V(\mathrm{CO_2})$ 对 $t$ 的图。

$t$ (s)	0	20	50	100	200
$V(\mathrm{CO_2})$ (cm³)	0	22	40	52	60

(a) Estimate the average rate of $\mathrm{CO_2}$ production (in $\mathrm{cm^3\,s^{-1}}$) over the first 20 s of the reaction.估算反应前 20 s 内 $\mathrm{CO_2}$ 生成的平均速率（单位 $\mathrm{cm^3\,s^{-1}}$）。 [2]

(b) The curve flattens noticeably after 100 s. Explain in terms of collision theory why the rate decreases as the reaction proceeds.100 s 之后曲线明显趋于平稳。用碰撞理论解释为何反应速率随反应进行而下降。 [2]

(c) Predict and briefly justify the effect on the initial rate if the experiment is repeated with each of the following changes, all else equal. Treat each change independently.假设在其他条件不变的情况下，按以下每一项改动重做实验。逐项预测对初始速率的影响并简要说明理由。每项独立考虑。 [4]

(i) The marble chips are replaced with the same mass of marble powder.(i) 用相同质量的大理石粉末替代碎块。

(ii) The temperature is raised to 318 K.(ii) 温度升至 318 K。

(iii) The acid is replaced with $50.0~\mathrm{cm^3}$ of $2.00~\mathrm{mol\,dm^{-3}}$ HCl.(iii) 改用 $50.0~\mathrm{cm^3}$、$2.00~\mathrm{mol\,dm^{-3}}$ 的 HCl。

(iv) A small amount of solid $\mathrm{MnO_2}$ is added as a "catalyst".(iv) 加入少量固体 $\mathrm{MnO_2}$ 作为"催化剂"。

(d) Sketch one Maxwell–Boltzmann distribution at 298 K and a second at 318 K on the same axes. Mark $E_a$ as a vertical line. In one sentence, explain why a small temperature change can produce a substantial change in reaction rate.在同一坐标系下画出 298 K 与 318 K 的两条麦克斯韦–玻尔兹曼分布曲线，并用竖线标出 $E_a$。用一句话说明为何温度的小幅升高就能显著改变反应速率。 [2]

SR 2HARD Paper 2 HL 2.2.9 – 2.2.12 Rate Equation + Arrhenius [10]

A student studies the gas-phase reaction $\mathrm{A + 2B \to C}$ at 300 K and obtains the following initial-rates data.某学生在 300 K 下研究气相反应 $\mathrm{A + 2B \to C}$，得到如下初始速率（initial rate）数据。

Run	$[A]_0$ (mol dm⁻³)	$[B]_0$ (mol dm⁻³)	Initial rate (mol dm⁻³ s⁻¹)
1	$0.10$	$0.10$	$1.5 \times 10^{-4}$
2	$0.20$	$0.10$	$6.0 \times 10^{-4}$
3	$0.20$	$0.20$	$6.0 \times 10^{-4}$

(a) Determine the order of reaction with respect to A and the order with respect to B. State the overall order.确定关于 A 与 B 的反应级数。写出总反应级数（overall order）。 [3]

(b) Write the rate equation and calculate the rate constant $k$ at 300 K, stating its units.写出速率方程，并计算 300 K 下的速率常数 $k$，给出单位。 [3]

(c) Explain briefly why the order of reaction with respect to B is zero, even though B appears in the balanced equation with a coefficient of 2.尽管 B 在配平方程中系数为 2，关于 B 的反应级数却为零。简要说明原因。 [2]

(d) At 350 K, the rate constant is found to be $k_2 = 6.0 \times 10^{-2}~\mathrm{dm^3\,mol^{-1}\,s^{-1}}$. Use the two-point form of the Arrhenius equation to determine the activation energy $E_a$ in $\mathrm{kJ\,mol^{-1}}$.在 350 K 测得速率常数 $k_2 = 6.0 \times 10^{-2}~\mathrm{dm^3\,mol^{-1}\,s^{-1}}$。利用阿伦尼乌斯方程的两点式求活化能 $E_a$（单位 $\mathrm{kJ\,mol^{-1}}$）。 [2]

SR 3HARD Paper 2 HL 2.2.5 – 2.2.8 Mechanism + Catalysis [10]

The decomposition of $\mathrm{N_2O_5}$ in the gas phase proceeds by the proposed two-step mechanism:$\mathrm{N_2O_5}$ 在气相中的分解被认为按以下两步机理（mechanism）进行：

Step 1 (slow): $\mathrm{N_2O_5 \to NO_2 + NO_3}$第 1 步（慢）：$\mathrm{N_2O_5 \to NO_2 + NO_3}$

Step 2 (fast): $\mathrm{NO_3 + NO_2 \to NO_2 + O_2 + NO}$第 2 步（快）：$\mathrm{NO_3 + NO_2 \to NO_2 + O_2 + NO}$

(The overall stoichiometry simplifies to $\mathrm{2N_2O_5 \to 4NO_2 + O_2}$ once two passes of the cycle are summed.)（将循环走两遍并相加后，总反应式化简为 $\mathrm{2N_2O_5 \to 4NO_2 + O_2}$。）

(a) Identify (i) the rate-determining step, (ii) any reaction intermediate(s), and (iii) the molecularity of the rate-determining step.指出 (i) 决速步（rate-determining step）、(ii) 反应中间体、(iii) 决速步的分子数（molecularity）。 [3]

(b) Predict the rate equation that this mechanism implies, and state the overall order of reaction.预测该机理所对应的速率方程，并写出总反应级数。 [2]

(c) Sketch and label a clear energy profile for the overall reaction. Your profile must show: the reactant level, the product level (lower than reactant), two transition-state peaks, and the intermediate as a local minimum between them. Mark $E_{a,\text{step 1}}$ and $E_{a,\text{step 2}}$.画出并标注总反应的清晰能量曲线。曲线必须包含：反应物能级、低于反应物的产物能级、两个过渡态峰、以及位于两峰之间作为局部极小值的中间体。标出 $E_{a,\text{step 1}}$ 与 $E_{a,\text{step 2}}$。 [3]

(d) A catalyst is found that lowers $E_{a,\text{step 1}}$ but has no effect on step 2. Sketch the modified profile on the same axes (use a dashed line) and explain in one sentence why this kind of catalyst speeds the reaction up.找到一种催化剂，只降低 $E_{a,\text{step 1}}$，对第 2 步无影响。在同一坐标系上用虚线画出修改后的曲线，并用一句话说明该催化剂为何能加速反应。 [2]

Data-Based Question (HL)数据题（HL）

This question tests the experimental skills associated with Reactivity 2.2 (kinetics). Higher-level students should attempt all parts.本题考查 Reactivity 2.2（动力学）相关的实验技能。HL 学生应作答全部小题。

P3-1HARD Paper 3 HL Initial Rates + Arrhenius [12]

A research team investigates the iodination of propanone in acidic solution:某研究小组研究酸性溶液中丙酮（propanone）的碘化反应（iodination）：

$\mathrm{CH_3COCH_3(aq) + I_2(aq) \xrightarrow{H^+} CH_3COCH_2I(aq) + HI(aq)}$

The progress of the reaction is followed by colorimetry (the iodine absorbs at 470 nm). The reaction is studied at 298 K with the following initial-rates data; $[\mathrm{H^+}]$ is held constant at $1.00~\mathrm{mol\,dm^{-3}}$ throughout.反应进程用比色法（colorimetry）跟踪（碘在 470 nm 处吸收）。研究在 298 K 下进行，给出如下初始速率数据；全程保持 $[\mathrm{H^+}] = 1.00~\mathrm{mol\,dm^{-3}}$ 不变。

Run	$[\mathrm{CH_3COCH_3}]_0$	$[\mathrm{I_2}]_0$	Initial rate (mol dm⁻³ s⁻¹)
1	$0.40$	$0.005$	$3.5 \times 10^{-5}$
2	$0.80$	$0.005$	$7.0 \times 10^{-5}$
3	$0.80$	$0.010$	$7.0 \times 10^{-5}$
4	$0.40$	$0.005$	$1.05 \times 10^{-4}$ (at $T = 318~\mathrm{K}$)

(a) Use runs 1–3 to determine the order of reaction with respect to propanone and to iodine. Comment briefly on the surprising result for iodine.利用第 1–3 组数据求关于丙酮与碘的反应级数。对碘的反应级数所呈现的"反常"结果作简要评述。 [3]

(b) Write the experimentally determined rate equation (treating $[\mathrm{H^+}]$ as constant; do not extract its order). Calculate the value of $k$ at 298 K with its units.写出实验确定的速率方程（视 $[\mathrm{H^+}]$ 为常数，不必提取其级数）。计算 298 K 下 $k$ 的值并给出单位。 [3]

(c) Use runs 1 and 4 to calculate the activation energy $E_a$ via the two-point Arrhenius equation. Report your answer in $\mathrm{kJ\,mol^{-1}}$.利用第 1、4 组数据，通过阿伦尼乌斯方程两点式求活化能 $E_a$，答案以 $\mathrm{kJ\,mol^{-1}}$ 给出。 [3]

(d) Identify two distinct sources of systematic error in the colorimetric method as applied here, and suggest one experimental check the student could perform to assess the precision of the rate measurements.指出此处比色法中两个不同的系统误差（systematic error）来源，并建议学生可执行的一项实验检验，用以评估速率测量的精度（precision）。 [3]

Reaction Kinetics化学反应动力学

Multiple Choice选择题

Structured Response结构化解答题

Data-Based Question (HL)数据题（HL）