How Much, How Fast, How Far?反应有多少、有多快、进行到何处？
Three diagnostic questions about any chemical reaction: how much material is produced (stoichiometry & yield), how fast it gets there (kinetics), and how far the reaction proceeds before reaching equilibrium (extent).用三个核心问题剖析任何化学反应：有多少物质生成（化学计量（stoichiometry）与产率（percent yield））、有多快到达终点（反应速率（reaction rate）/ 反应动力学）、以及反应在达到平衡（equilibrium）前进行到何处（反应程度）。
How Much? The Amount of Chemical Change反应有多少？—— 化学变化的量
concentration）推算出其他物质的相应量。
Balanced Equations & State Symbols配平方程式与状态符号
A chemical equation is balanced when every element has equal atom counts on both sides — a statement of conservation of mass. State symbols indicate phase: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution. Use them whenever you write an equation.
当方程式两边每种元素的原子数都相等时，方程式才算配平——这是质量守恒的体现。状态符号（state symbols）标明物质的相态：(s) 固体、(l) 液体、(g) 气体、(aq) 水溶液。只要书写方程式就要标注。
The Mole Ratio摩尔比
The coefficients of a balanced equation are the mole ratio of reactants and products. From the mole ratio you can compute reacting masses, volumes (for gases or solutions), and concentrations using three standard relationships:
配平方程式中的系数就是反应物与产物之间的摩尔比（mole ratio）。由摩尔比出发，可以借助下面三条标准关系式，计算反应中各物质的质量、体积（气体或溶液）以及浓度：
molar mass），单位 g mol⁻¹。理想气体（ideal gas）在 STP（STP）下 $V_m = 22.7~\mathrm{dm^3\,mol^{-1}}$。浓度单位为 $\mathrm{mol\,dm^{-3}}$。Worked Example — Mass from Mole Ratio例题 — 由摩尔比求质量
What mass of CO₂ is produced when 8.0 g of CH₄ burns completely in oxygen?8.0 g 的 CH₄ 在氧气中完全燃烧，生成 CO₂ 的质量是多少？
Limiting Reactant & Theoretical Yield限量试剂与理论产量
When reactants aren't supplied in exact stoichiometric ratio, one runs out first: the limiting reactant. It caps how much product can form (the theoretical yield); the others are in excess.
当反应物不是按精确的化学计量比投料时，总有一种最先耗尽——这就是限量试剂（limiting reagent / limiting reactant）。它决定了产物的最大可能生成量（即理论产量（theoretical yield）），其余物质则属于过量试剂。
Percentage Yield产率（百分产率）
The experimental yield is what you actually obtain in a real reaction; it's almost always less than the theoretical yield due to incomplete reaction, side reactions, losses on transfer, and impure reactants.
实验产量（experimental yield）是真实反应中实际获得的产物量；由于反应不完全、副反应、转移损耗以及反应物纯度不足，它几乎总是小于理论产量。
percent yield）Worked Example — Limiting Reactant & % Yield例题 — 限量试剂与产率
25.0 g of N₂ is reacted with 6.00 g of H₂ to make NH₃ via the Haber process: $\mathrm{N_2 + 3H_2 \to 2NH_3}$. If 22.5 g of NH₃ is obtained, find (i) the limiting reactant and (ii) the percentage yield.将 25.0 g N₂ 与 6.00 g H₂ 按哈伯法（Haber process）合成氨：$\mathrm{N_2 + 3H_2 \to 2NH_3}$。若实际获得 22.5 g NH₃，求 (i) 限量试剂；(ii) 产率。
Atom Economy & Green Chemistry原子经济性与绿色化学
The atom economy measures what fraction of the reactant mass ends up in the desired product (as opposed to wasted by-products). It's a different efficiency metric than % yield: a reaction with a high % yield can still have a low atom economy if it produces a lot of by-product per useful molecule.
原子经济性（atom economy）衡量反应物中有多少质量最终进入目标产物（而不是浪费在副产物上）。它和产率（percent yield）是不同维度的"效率"指标：即使一个反应的产率很高，只要每生成一个目标分子就伴随大量副产物，它的原子经济性仍可能很低。
stoichiometry）来看，反应本身浪费多少？（本征 / 化学层面）工业过程力求两者同时最大化。
How Fast? The Rate of Chemical Change反应有多快？—— 化学反应的速率
reaction rate）？反应速率由碰撞频率、碰撞取向，以及能量达到反应阈值的碰撞所占比例共同决定。
Defining Reaction Rate反应速率的定义
The rate of a reaction is the change in concentration of a reactant or product per unit time. For a generic reaction $aA + bB \to cC + dD$:
反应速率（reaction rate）指单位时间内反应物或产物浓度的变化量。对于通式 $aA + bB \to cC + dD$：
In practice you read the rate off a graph of concentration vs. time: the tangent at any point gives the instantaneous rate at that moment. Concentration is rarely measured directly — typical proxies include gas volume, mass loss (CO₂ released), pH, absorbance, conductivity, or titration of a quenched sample.
实际操作中，速率是从"浓度—时间"曲线上读取的：曲线上任一点的切线斜率就是该时刻的瞬时速率。浓度本身很少直接测量——常用替代量包括气体体积、质量损失（如 CO₂ 释放）、pH、吸光度、电导率，或对淬灭样品进行滴定。
Collision Theory碰撞理论
For two species to react, their particles must collide with two prerequisites:
两种物种要发生反应，其粒子必须发生碰撞（collision theory），且需同时满足两个前提条件：
- Sufficient kinetic energy — at least the activation energy $E_a$.
- Correct orientation — the reactive parts of the molecules must meet (the "steric factor").
- 足够的动能——至少达到活化能（
activation energy）$E_a$。 - 正确的取向——分子上有反应活性的部位必须正好相对（即"空间因子" / steric factor）。
The kinetic energy of particles is determined by temperature: in Kelvin, $\langle KE \rangle \propto T$. Doubling the absolute temperature doubles the average kinetic energy — but, far more importantly, it shifts the tail of the Maxwell–Boltzmann distribution past $E_a$, which is what really drives rate up.
粒子的动能由温度决定：以开尔文为单位时，$\langle KE \rangle \propto T$。绝对温度加倍，平均动能也加倍——但更关键的是，这会把麦克斯韦-玻尔兹曼分布（Maxwell-Boltzmann distribution）的高能尾巴推过 $E_a$，这才是速率显著升高的真正原因。
Factors That Affect Rate影响反应速率的因素
| Factor | Effect on rate | Mechanism |
|---|---|---|
| Concentration ↑ (or pressure ↑ for gases) | ↑ | More particles per volume → more collisions per second |
| Surface area ↑ (for solids) | ↑ | More exposed particles available to collide |
| Temperature ↑ | ↑↑ (large) | More particles exceed $E_a$ (shifts Maxwell–Boltzmann tail) |
| Catalyst added | ↑↑ | Lowers $E_a$ by providing an alternative pathway |
| 因素 | 对速率的影响 | 机理 |
|---|---|---|
| 浓度 ↑（气体则压强 ↑） | ↑ | 单位体积内粒子更多 → 单位时间内碰撞次数更多 |
| 表面积 ↑（固体反应物） | ↑ | 更多粒子暴露在外可发生碰撞 |
| 温度 ↑ | ↑↑（显著） | 更多粒子能量超过 $E_a$（麦克斯韦-玻尔兹曼分布尾部右移） |
加入催化剂（catalyst） | ↑↑ | 提供较低 $E_a$ 的替代路径 |
Activation Energy & the Maxwell–Boltzmann Distribution活化能与麦克斯韦-玻尔兹曼分布
The activation energy $E_a$ is the minimum kinetic energy a colliding pair must have for the collision to be successful. A Maxwell–Boltzmann (MB) distribution plots the number of particles vs. kinetic energy at a given temperature:
活化能（activation energy）$E_a$ 是发生有效碰撞所必须达到的最低动能。麦克斯韦-玻尔兹曼分布（Maxwell-Boltzmann distribution，简称 MB 分布）描绘了在某一温度下粒子数随动能的分布曲线：
- The area to the right of $E_a$ = the fraction of particles with enough energy to react.
- Raising $T$ flattens and broadens the distribution, shifting the peak right and dramatically increasing the high-energy tail.
- A small increase in $T$ produces a large increase in the area beyond $E_a$ — this is why rate is very sensitive to temperature.
- $E_a$ 右侧的面积 = 动能足以反应的粒子所占比例。
- 升高 $T$ 会使曲线变得低而宽、峰值右移，并显著增大高能尾部的面积。
- $T$ 即使只升高一点点，$E_a$ 以上的面积也会大幅增加——这就是反应速率对温度极其敏感的根源。
Catalysts & Energy Profiles催化剂与能量轮廓
A catalyst increases the rate of a reaction by providing an alternative pathway with a lower activation energy. The catalyst is regenerated in a later step, so it isn't consumed. Energy profiles for catalysed vs uncatalysed reactions:
催化剂（catalyst）通过提供一条活化能更低的替代路径来加快反应。它会在后续步骤中再生，所以不被消耗。下面比较催化与未催化反应的能量轮廓（energy profile）：
- Lower $E_a$ for both forward and reverse reactions equally (no shift in equilibrium — see Reactivity 2.3).
- Don't change $\Delta H$ — the energy difference between reactants and products is fixed by the chemistry.
- Enzymes are biological catalysts; they're typically extremely selective.
- Homogeneous vs heterogeneous mechanisms are not assessed at IB.
- 同时等量地降低正反应和逆反应的 $E_a$（因此不影响平衡——见 Reactivity 2.3）。
- 不改变 $\Delta H$——反应物和产物之间的能量差由化学本性决定。
- 酶是生物催化剂，通常具有极高的选择性。
- 均相 vs 非均相催化机理不在 IB 考核范围内。
Worked Example — Rate from a Concentration-Time Tangent例题 — 由"浓度—时间"切线求速率
A reactant's concentration drops from 0.500 to 0.380 mol dm⁻³ over the first 30 s of a reaction. Estimate the initial rate (with respect to this reactant).某反应物在反应前 30 s 内浓度由 0.500 降至 0.380 mol dm⁻³。估算初始速率（以该反应物计）。
Reaction Mechanisms & the Rate-Determining Step反应机理与决速步骤 HL
Most reactions proceed through a series of elementary steps rather than a single concerted event. The overall rate is set by the rate-determining step (RDS) — the slowest step in the mechanism. Reaction intermediates are species formed in one step and consumed in a later step; transition states, by contrast, are unstable maxima on the energy profile and exist only fleetingly.
大多数反应并不是一步到位，而是经过一系列基元反应（elementary step）进行。整个反应的速率由决速步骤（rate-determining step，简称 RDS）决定——也就是机理中最慢的一步。反应中间体（intermediate）是在某一步生成、又在后续步骤中消耗的物种；而过渡态（transition state）则是能量轮廓上不稳定的能量极大值点，仅瞬间存在。
A proposed mechanism must be consistent with both (a) the observed rate equation and (b) the stoichiometry of the overall reaction. The RDS is not necessarily the first step.
所提出的机理必须同时满足两点：(a) 与观测到的速率方程（rate equation）一致；(b) 总和后能给出反应的总化学计量。RDS 不一定是第一步。
Molecularity & Rate Equations分子数与速率方程 HL
The molecularity of an elementary step is the number of particles that react in that step:
一个基元反应（elementary step）的分子数（molecularity）指该步参与反应的粒子数：
- Unimolecular — one particle (e.g. a decomposition).
- Bimolecular — two particles (the most common elementary step).
- Termolecular — three particles (rare; statistically unlikely).
- 单分子反应（unimolecular）——1 个粒子（如分解反应）。
- 双分子反应（bimolecular）——2 个粒子（最常见的基元反应类型）。
- 三分子反应（termolecular）——3 个粒子（极少见；统计上几乎不可能）。
order of reaction），$n$ 为对 $B$ 的级数。反应级数必须通过实验确定——不能直接从配平方程式上读出。The overall reaction order is the sum of the individual orders ($m + n$). At IB HL only integer orders (0, 1, 2) will be assessed.
总反应级数为各组分级数之和（$m + n$）。IB HL 仅考核整数级数（0、1、2）。
Zero, First, and Second Order Reactions零级、一级、二级反应 HL
| Order | Rate law | [A]–t graph | Rate–[A] graph |
|---|---|---|---|
| Zero ($m = 0$) | rate $= k$ | Straight line, negative slope | Horizontal line |
| First ($m = 1$) | rate $= k[A]$ | Exponential decay | Straight line through origin |
| Second ($m = 2$) | rate $= k[A]^2$ | Slower-than-exponential decay | Parabola through origin |
| 反应级数 | 速率方程（rate law） | [A]–t 图 | 速率–[A] 图 |
|---|---|---|---|
| 零级（$m = 0$） | rate $= k$ | 负斜率直线 | 水平直线 |
| 一级（$m = 1$） | rate $= k[A]$ | 指数衰减 | 过原点的直线 |
| 二级（$m = 2$） | rate $= k[A]^2$ | 比指数更缓的衰减 | 过原点的抛物线 |
The Rate Constant k速率常数 k HL
The rate constant $k$ is temperature-dependent (Arrhenius — see below) and has units determined by the overall order so that rate always comes out in $\mathrm{mol\,dm^{-3}\,s^{-1}}$:
速率常数（rate constant）$k$ 随温度变化（Arrhenius——见下文），其单位由总反应级数决定，从而保证 rate 的单位始终是 $\mathrm{mol\,dm^{-3}\,s^{-1}}$：
| Overall order | Units of $k$ |
|---|---|
| 0 | $\mathrm{mol\,dm^{-3}\,s^{-1}}$ |
| 1 | $\mathrm{s^{-1}}$ |
| 2 | $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ |
| 3 | $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ |
| 总反应级数 | $k$ 的单位 |
|---|---|
| 0 | $\mathrm{mol\,dm^{-3}\,s^{-1}}$ |
| 1 | $\mathrm{s^{-1}}$ |
| 2 | $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ |
| 3 | $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ |
Worked Example — Rate Equation from Initial-Rates Data例题 — 由初始速率法导出速率方程
For the reaction $\mathrm{A + 2B \to C}$, initial rates were measured at fixed temperature with different starting concentrations:对于反应 $\mathrm{A + 2B \to C}$，在固定温度下用不同初始浓度测得初始速率（initial rate）如下：
| Run | $[A]_0$ (mol dm⁻³) | $[B]_0$ (mol dm⁻³) | Initial rate (mol dm⁻³ s⁻¹) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | $2.0 \times 10^{-4}$ |
| 2 | 0.20 | 0.10 | $4.0 \times 10^{-4}$ |
| 3 | 0.10 | 0.20 | $8.0 \times 10^{-4}$ |
| 实验 | $[A]_0$ (mol dm⁻³) | $[B]_0$ (mol dm⁻³) | 初始速率 (mol dm⁻³ s⁻¹) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | $2.0 \times 10^{-4}$ |
| 2 | 0.20 | 0.10 | $4.0 \times 10^{-4}$ |
| 3 | 0.10 | 0.20 | $8.0 \times 10^{-4}$ |
Find the order with respect to A, the order with respect to B, the overall order, and $k$ with its units.
求对 A 的级数、对 B 的级数、总反应级数，以及带单位的 $k$。
The Arrhenius Equation阿伦尼乌斯方程 HL
The rate constant's temperature dependence is captured by the Arrhenius equation:
速率常数 $k$ 对温度的依赖关系由阿伦尼乌斯方程（Arrhenius equation）描述：
The linear form — obtained by taking $\ln$ of both sides — is what you actually use in problem-solving:
对两边取 $\ln$ 得到线性形式，这才是做题时常用的形式：
Worked Example — Arrhenius Plot to Find E_a例题 — 用阿伦尼乌斯关系求 $E_a$
A reaction has $k = 2.5 \times 10^{-3}~\mathrm{s^{-1}}$ at 300 K and $k = 4.0 \times 10^{-2}~\mathrm{s^{-1}}$ at 350 K. Estimate $E_a$.某反应在 300 K 时 $k = 2.5 \times 10^{-3}~\mathrm{s^{-1}}$，在 350 K 时 $k = 4.0 \times 10^{-2}~\mathrm{s^{-1}}$。估算 $E_a$。
How Far? The Extent of Chemical Change反应进行到何处？—— 化学变化的程度
equilibrium）常数告诉你反应最终"停"在哪里；勒夏特列原理（Le Chatelier's principle）则告诉你外界扰动会让这个"停留点"如何移动。
Dynamic Equilibrium动态平衡
A reversible reaction in a closed system reaches dynamic equilibrium when the forward and reverse rates become equal. Macroscopic properties (concentrations, colour, pressure) stop changing, but the forward and reverse processes never stop — they merely cancel.
在密闭体系中，可逆反应在正反应和逆反应速率相等时达到动态平衡（dynamic equilibrium）。此时宏观性质（浓度、颜色、压强）保持不变，但正、逆两个方向的过程并未停止——只是互相抵消。
- Forward rate = reverse rate (continuous interchange).
- Closed system (no leakage of reactants or products to the surroundings).
- Constant macroscopic properties (concentration, pressure, colour).
- Reached from either direction (the same $K$ is obtained starting from reactants or starting from products).
- 正反应速率 = 逆反应速率（持续地相互转化）。
- 密闭体系（反应物或产物不与外界交换）。
- 宏观性质保持恒定（浓度、压强、颜色）。
- 从任一方向出发都能达到（从反应物或产物开始都得到相同的 $K$）。
The Equilibrium Constant K平衡常数 K
For a homogeneous reaction $aA + bB \rightleftharpoons cC + dD$ at a given temperature, the equilibrium law says:
对于均相反应 $aA + bB \rightleftharpoons cC + dD$，在给定温度下平衡定律给出平衡常数（equilibrium constant，$K_c$ / $K_p$）的表达式：
For the reverse reaction at the same temperature, $K_\text{rev} = 1/K_\text{fwd}$. If you multiply a reaction's coefficients by $n$, the new equilibrium constant is $K^n$.
在同一温度下，逆反应的平衡常数 $K_\text{rev} = 1/K_\text{fwd}$。若将整条方程式的系数同乘以 $n$，新的平衡常数就变为 $K^n$。
What the Magnitude of K Tells You$K$ 的大小说明什么
| $K$ | Extent of reaction at equilibrium |
|---|---|
| $K \gg 1$ | Products strongly favoured (reaction essentially "goes to completion"). |
| $K > 1$ | Products favoured, but reactants still present. |
| $K = 1$ | Comparable amounts of reactants and products. |
| $K < 1$ | Reactants favoured. |
| $K \ll 1$ | Reactants strongly favoured (reaction barely proceeds). |
| $K$ | 平衡时反应进行程度 |
|---|---|
| $K \gg 1$ | 强烈偏向产物（反应基本"进行到底"）。 |
| $K > 1$ | 产物占优，但仍有反应物存在。 |
| $K = 1$ | 反应物与产物量相当。 |
| $K < 1$ | 反应物占优。 |
| $K \ll 1$ | 强烈偏向反应物（反应几乎不进行）。 |
Le Châtelier's Principle勒夏特列原理
If a system at equilibrium is disturbed, the equilibrium will shift to partially counteract the disturbance. This lets you predict qualitatively how the position changes without doing the full thermodynamics. Important: $K$ itself only changes with temperature; concentration and pressure shifts re-establish the same $K$.
若处于平衡的体系受到外界扰动，平衡会发生移动以部分抵消该扰动。这条原理（Le Chatelier's principle）让你在不做完整热力学计算的情况下定性预测平衡位置如何变化。要点：$K$ 本身只随温度变化；浓度和压强的扰动最终都会回到同一个 $K$ 值。
| Disturbance | Direction of shift | Changes $K$? |
|---|---|---|
| ↑ [reactant] | → products | No |
| ↑ [product] | ← reactants | No |
| ↑ pressure (gas) | Toward side with fewer moles of gas | No |
| ↑ temperature | In endothermic direction | Yes |
| Catalyst added | No shift | No |
| 扰动 | 平衡移动方向 | 是否改变 $K$？ |
|---|---|---|
| ↑ [反应物] | → 正向（产物方向） | 否 |
| ↑ [产物] | ← 逆向（反应物方向） | 否 |
| ↑ 压强（气体） | 向气体摩尔数较小的一侧 | 否 |
| ↑ 温度 | 向吸热方向 | 是 |
加入催化剂（catalyst） | 不移动 | 否 |
Worked Example — Le Châtelier Predictions例题 — 勒夏特列原理预测
For the Haber process: $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$, $\Delta H = -92~\mathrm{kJ\,mol^{-1}}$. Predict the direction of shift and the effect on the yield of NH₃ for each change.哈伯法合成氨：$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$，$\Delta H = -92~\mathrm{kJ\,mol^{-1}}$。预测下列变化下平衡移动方向以及对 NH₃ 产量的影响。
The Reaction Quotient Q反应商 Q HL
The reaction quotient $Q$ has the same form as the equilibrium expression but uses any instantaneous concentrations — not necessarily equilibrium ones. Comparing $Q$ to $K$ tells you which way the reaction will go to reach equilibrium:
反应商（reaction quotient）$Q$ 的表达式形式与平衡常数相同，但代入的是任意瞬时浓度（不必是平衡浓度）。将 $Q$ 与 $K$ 比较即可判断反应朝哪个方向进行才能达到平衡：
| Comparison | Status | Reaction will… |
|---|---|---|
| $Q < K$ | Too few products | Proceed forward (→ products) |
| $Q = K$ | At equilibrium | No net change |
| $Q > K$ | Too many products | Proceed backward (← reactants) |
| 比较 | 当前状态 | 反应将… |
|---|---|---|
| $Q < K$ | 产物太少 | 正向进行（→ 产物方向） |
| $Q = K$ | 处于平衡 | 无净变化 |
| $Q > K$ | 产物太多 | 逆向进行（← 反应物方向） |
Quantitative Equilibrium Calculations (ICE)平衡定量计算（ICE 表） HL
To find equilibrium composition given initial concentrations and $K$, use an ICE table (Initial / Change / Equilibrium). When $K$ is very small, the approximation $[\text{reactant}]_\text{initial} \approx [\text{reactant}]_\text{eqm}$ avoids needing quadratics.
已知初始浓度和 $K$，求平衡组成时采用 ICE 表（Initial 初始 / Change 变化 / Equilibrium 平衡）。当 $K$ 极小时，可用近似 $[\text{反应物}]_\text{initial} \approx [\text{反应物}]_\text{eqm}$，避免解一元二次方程。
Worked Example — Small-K Equilibrium例题 — 小 K 值平衡
For the dissociation $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$, $K_c = 4.6 \times 10^{-3}$ at 298 K. If 0.20 mol of $\mathrm{N_2O_4}$ is placed in a 1.0 dm³ flask, find the equilibrium concentrations.对于离解反应 $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$，298 K 时 $K_c = 4.6 \times 10^{-3}$。若将 0.20 mol $\mathrm{N_2O_4}$ 置于 1.0 dm³ 烧瓶中，求平衡时各组分浓度。
Linking K to Gibbs Energy$K$ 与吉布斯自由能的联系 HL
The equilibrium constant and the standard Gibbs energy change are two views of the same thing — both tell you where a reaction settles:
平衡常数与标准吉布斯自由能（Gibbs free energy）变化是同一件事的两种视角——它们都告诉你反应最终停在哪里：
- $\Delta G^{\ominus} \ll 0 \Rightarrow K \gg 1$: products strongly favoured.
- $\Delta G^{\ominus} = 0 \Rightarrow K = 1$: comparable amounts.
- $\Delta G^{\ominus} \gg 0 \Rightarrow K \ll 1$: reactants strongly favoured.
- $\Delta G^{\ominus} \ll 0 \Rightarrow K \gg 1$：强烈偏向产物。
- $\Delta G^{\ominus} = 0 \Rightarrow K = 1$：反应物与产物量相当。
- $\Delta G^{\ominus} \gg 0 \Rightarrow K \ll 1$：强烈偏向反应物。
Worked Example — Gibbs Energy from K例题 — 由 K 求吉布斯自由能
For the reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$, $K_c = 50.0$ at 700 K. Calculate $\Delta G^{\ominus}$.反应 $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ 在 700 K 时 $K_c = 50.0$。计算 $\Delta G^{\ominus}$。
Exam Strategy考试策略
Reactivity 2 questions here lean on qualitative reasoning: identify the limiting reactant from mole ratios you can do in your head, predict the direction of a Le Châtelier shift, classify orders from a rate–[A] graph, or sketch a Maxwell–Boltzmann shift. Don't burn time on numerical kinetics in Paper 1 — there's no calculator.
本卷的 Reactivity 2 题以定性推理为主：用心算的摩尔比判断限量试剂（limiting reagent）、预测勒夏特列原理（Le Chatelier's principle）下平衡移动方向、从速率–[A] 图判定反应级数（order of reaction），或勾勒麦克斯韦-玻尔兹曼分布的变化。不要在 Paper 1 上死磕动力学数值题——你没有计算器。
This is where the numerical Reactivity 2 work lives. Expect (a) a stoichiometry problem with limiting reactant + % yield (often combined with atom economy), (b) a kinetics question with initial-rates data or an Arrhenius plot, and (c) an equilibrium question — usually $K$ from concentration data, then a Le Châtelier prediction, sometimes plus a $\Delta G^{\ominus}$ calculation at HL.
Reactivity 2 的数值题主要落在这里。常见题型：(a) 化学计量题，含限量试剂 + 产率（往往一并考原子经济性）；(b) 动力学题，给出初始速率数据或阿伦尼乌斯（Arrhenius equation）作图；(c) 平衡题——通常先由浓度数据求 $K$，再做一道勒夏特列预测，HL 还可能加一题 $\Delta G^{\ominus}$ 的计算。
Reactivity 2 commonly drives Paper-3 data analysis: a clock-reaction or gas-collection experiment, where you plot, identify reaction order, and extract $k$ and $E_a$. Practice reading the slope of $\ln k$ vs $1/T$ and connecting it to $E_a = -R \cdot \text{slope}$.
Reactivity 2 是 Paper 3 数据分析题的常客：典型场景是钟摆反应或气体收集实验，要求你作图、判定反应级数、并求出 $k$ 和 $E_a$。务必练熟读取 $\ln k$ 对 $1/T$ 的斜率并据此得到 $E_a = -R \cdot \text{slope}$。
Common Mistakes常见错误
percent yield）和原子经济性混为一谈
% yield is about practical efficiency (you didn't get back as much as you should have); atom economy is about intrinsic efficiency (the reaction by its stoichiometry produces wasteful by-products). A reaction can have 100% yield and only 30% atom economy.产率衡量的是实操层面的效率（你没能拿回应得的全部产物）；原子经济性衡量的是本征层面的效率（反应按化学计量本身就会产生废副产物）。一个反应完全可能 100% 产率但只有 30% 原子经济性。
rate equation）中的反应级数只能由实验确定。rate $= k[A]^m[B]^n$ 中的指数不是化学计量系数（基元反应除外）。