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
Guiding Question导引问题How are chemical equations used to calculate reacting ratios? A balanced equation is a recipe in molar units — once you know the ratio, you can find any mass, volume, or concentration from any other.如何用化学方程式计算反应物之间的比例？配平后的方程式（balanced equation）就是一份以摩尔为单位的"配方"——只要知道这个比例，就能由任一物质的质量、体积或浓度（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.
Atom check: 1 C, 4 H, 4 O on each side. ✓原子核对：两边各 1 C、4 H、4 O。✓
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:
$$n = \dfrac{m}{M_r} \qquad n = c\,V \qquad n = \dfrac{V_{\text{gas}}}{V_m}$$
$M_r$ is the molar mass (g mol⁻¹). For ideal gases at STP, $V_m = 22.7~\mathrm{dm^3\,mol^{-1}}$. Concentrations in $\mathrm{mol\,dm^{-3}}$.$M_r$ 是摩尔质量（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？
From the 1:1 ratio: $n(\mathrm{CO_2}) = 0.498~\mathrm{mol}$.由 1:1 比例：$n(\mathrm{CO_2}) = 0.498~\mathrm{mol}$。
$$m(\mathrm{CO_2}) = n M_r = 0.498 \times 44.01 = 21.9~\mathrm{g}$$
Evaluate检验
Roughly tripled mass (16 → 44 g mol⁻¹), consistent with the C atom picking up two extra O atoms.质量大约变为三倍（16 → 44 g mol⁻¹），与 C 原子额外结合两个 O 原子相符。
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.
How to Identify the Limiting Reactant如何判断限量试剂Convert each reactant's mass (or volume) to moles, then divide each by its coefficient in the balanced equation. The reactant with the smallest ratio is the limiting reactant.把每种反应物的质量（或体积）换算为摩尔数，再除以它在配平方程式中的系数。所得比值最小的那一种即为限量试剂。
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.
Always between 0 and 100% (any higher means impure product or weighing error).数值始终介于 0 与 100% 之间（超出则说明产物不纯或称量出错）。
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。
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.
Higher = less wasted mass = greener. The ideal addition reaction (one product) has 100% atom economy.数值越高，浪费的质量越少，越"绿色"。理想的加成反应（只有一个产物）原子经济性为 100%。
Yield vs Atom Economy产率 vs 原子经济性Both report "efficiency" but answer different questions. % yield: did the reaction actually run to completion? (kinetic / practical) Atom economy: is the reaction inherently wasteful by stoichiometry? (intrinsic / chemical) Industrial processes aim to maximise both.两者都是"效率"指标，但回答的是不同问题。产率：反应是否真的进行到底？（动力学 / 实操层面）原子经济性：从化学计量（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。
Iron(III) oxide is reduced by carbon monoxide: $\mathrm{Fe_2O_3 + 3\,CO \to 2\,Fe + 3\,CO_2}$. If 32.0 g of $\mathrm{Fe_2O_3}$ ($M_r = 159.7$) reacts with excess CO, the maximum mass of iron ($M_r = 55.85$) that can be produced is closest to用一氧化碳还原氧化铁(III)：$\mathrm{Fe_2O_3 + 3\,CO \to 2\,Fe + 3\,CO_2}$。若 32.0 g $\mathrm{Fe_2O_3}$（$M_r = 159.7$）与过量 CO 反应，能生成铁（$M_r = 55.85$）的最大质量最接近
Two ways to make ethanol: (i) hydration of ethene $\mathrm{C_2H_4 + H_2O \to C_2H_5OH}$, and (ii) fermentation $\mathrm{C_6H_{12}O_6 \to 2\,C_2H_5OH + 2\,CO_2}$. Which statement is correct?两条制乙醇的路线：(i) 乙烯水合 $\mathrm{C_2H_4 + H_2O \to C_2H_5OH}$；(ii) 发酵 $\mathrm{C_6H_{12}O_6 \to 2\,C_2H_5OH + 2\,CO_2}$。下列说法正确的是？
Route (i) has 100% atom economy; route (ii) has ~51% atom economy.路线 (i) 原子经济性 100%；路线 (ii) 约 51%。
Both routes have the same atom economy.两条路线的原子经济性相同。
Route (i) has lower atom economy because it uses water.路线 (i) 因为用了水所以原子经济性更低。
Atom economy can't be calculated for either route.两条路线都无法计算原子经济性。
Correct! Route (i) is a single-product addition reaction (100% atom economy). Route (ii) produces $2 \times M_r(\mathrm{C_2H_5OH}) = 92.16$ g of ethanol per $92.16 + 2 \times 44.01 = 180.18$ g of total products, giving $\approx 51.1\%$.正确！路线 (i) 是只生成一种产物的加成反应（原子经济性 100%）。路线 (ii) 每生成 $2 \times M_r(\mathrm{C_2H_5OH}) = 92.16$ g 乙醇就同时产生 $92.16 + 2 \times 44.01 = 180.18$ g 总产物，故原子经济性约为 $51.1\%$。
Atom economy depends on whether by-products form. Addition reactions (no by-products) have 100% atom economy by construction. Route (ii) loses 2 CO₂ molecules per glucose. Answer: (A).原子经济性取决于是否生成副产物。加成反应（无副产物）天然就是 100%。路线 (ii) 每分子葡萄糖损失 2 个 CO₂。答案：(A)。
Worked Example — Mass of Excess Reactant Left Over例题 — 求过量反应物的剩余质量
4.00 g of magnesium ($M_r = 24.31$) is added to 100.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid: $\mathrm{Mg(s) + 2HCl(aq) \to MgCl_2(aq) + H_2(g)}$. Identify the limiting reactant and find the mass of magnesium left unreacted.将 4.00 g 镁（$M_r = 24.31$）加入 100.0 cm³、2.00 mol dm⁻³ 的盐酸中：$\mathrm{Mg(s) + 2HCl(aq) \to MgCl_2(aq) + H_2(g)}$。判断限量试剂（limiting reactant），并求剩余未反应镁的质量。
About 39% of the magnesium is unreacted — sensible, since the acid ran out first. Always base the product amount on the limiting reactant, never the excess one.约 39% 的镁未反应——合理，因为酸先耗尽。产物的量必须以限量试剂为准，绝不能用过量试剂来算。
Worked Example — Reactant Mass Needed for a Target Yield例题 — 由目标产量反推所需反应物质量
Aspirin ($\mathrm{C_9H_8O_4}$, $M_r = 180.17$) is made from salicylic acid ($\mathrm{C_7H_6O_3}$, $M_r = 138.13$): $\mathrm{C_7H_6O_3 + C_4H_6O_3 \to C_9H_8O_4 + C_2H_4O_2}$. If the process runs at 78.0% yield, what mass of salicylic acid is required to make 50.0 g of aspirin?阿司匹林（$\mathrm{C_9H_8O_4}$，$M_r = 180.17$）由水杨酸（$\mathrm{C_7H_6O_3}$，$M_r = 138.13$）制得：$\mathrm{C_7H_6O_3 + C_4H_6O_3 \to C_9H_8O_4 + C_2H_4O_2}$。若产率（percent yield）为 78.0%，制备 50.0 g 阿司匹林需要多少质量的水杨酸？
The key move is dividing by the yield (not multiplying): a sub-100% yield always means you need more starting material than a perfect reaction would.关键一步是除以产率（而非乘以）：产率不足 100% 时，所需起始原料总是比理想反应更多。
Two Independent Axes of "Efficiency""效率"的两个独立维度Green chemistry treats % yield and atom economy as orthogonal levers. A process chemist maximises both, then also tracks the E-factor (mass of waste per mass of product) — solvents and work-up reagents count too, even though they never appear in the balanced equation. A reaction with 95% yield and 90% atom economy can still be "dirty" if it needs 50 kg of solvent per kg of product. IB only assesses % yield and atom economy numerically, but knowing they answer different questions — did it work? vs is it inherently wasteful? — is exactly the framing examiners reward in extended-response marks.绿色化学把产率与原子经济性（atom economy）视为相互独立的两个调节杠杆。工艺化学家会同时最大化二者，还会追踪 E 因子（E-factor，每单位产物质量对应的废物质量）——溶剂和后处理试剂即便从不出现在配平方程式中，也计入其中。一个产率 95%、原子经济性 90% 的反应，若每千克产物需要 50 kg 溶剂，仍可能"不环保"。IB 只在数值上考核产率与原子经济性，但理解二者回答的是不同问题——反应成功了吗？ 与 反应本身是否浪费？——正是长答题中阅卷人青睐的分析框架。
Ammonia is oxidised by oxygen: $\mathrm{4NH_3 + 5O_2 \to 4NO + 6H_2O}$. A mixture of 4.0 mol $\mathrm{NH_3}$ and 4.0 mol $\mathrm{O_2}$ is reacted. Which reactant is limiting, and how many moles of NO can form?氨被氧气氧化：$\mathrm{4NH_3 + 5O_2 \to 4NO + 6H_2O}$。将 4.0 mol $\mathrm{NH_3}$ 与 4.0 mol $\mathrm{O_2}$ 混合反应。哪种反应物是限量试剂？能生成多少摩尔 NO？
$\mathrm{NH_3}$ limiting; 4.0 mol NO$\mathrm{NH_3}$ 限量；4.0 mol NO
$\mathrm{O_2}$ limiting; 4.0 mol NO$\mathrm{O_2}$ 限量；4.0 mol NO
$\mathrm{O_2}$ limiting; 3.2 mol NO$\mathrm{O_2}$ 限量；3.2 mol NO
Neither is limiting; 4.0 mol NO两者都不限量；4.0 mol NO
Correct! Divide by coefficients: $\mathrm{NH_3} \to 4.0/4 = 1.0$; $\mathrm{O_2} \to 4.0/5 = 0.80$. $\mathrm{O_2}$ has the smaller ratio, so it is limiting. NO : $\mathrm{O_2}$ = 4 : 5, so $n(\mathrm{NO}) = \tfrac{4}{5}(4.0) = 3.2$ mol.正确！除以系数：$\mathrm{NH_3} \to 4.0/4 = 1.0$；$\mathrm{O_2} \to 4.0/5 = 0.80$。$\mathrm{O_2}$ 的比值更小，故为限量试剂。NO : $\mathrm{O_2}$ = 4 : 5，所以 $n(\mathrm{NO}) = \tfrac{4}{5}(4.0) = 3.2$ mol。
Equal moles does not mean neither is limiting — divide each by its coefficient. $\mathrm{O_2}$ (ratio 0.80) runs out before $\mathrm{NH_3}$ (ratio 1.0), and NO scales off the limiting $\mathrm{O_2}$ via the 4:5 ratio, giving 3.2 mol. Answer: (C).摩尔数相等并不代表都不限量——要各自除以系数。$\mathrm{O_2}$（比值 0.80）先于 $\mathrm{NH_3}$（比值 1.0）耗尽，NO 依据 4:5 的比例由限量的 $\mathrm{O_2}$ 决定，得 3.2 mol。答案：(C)。
Reactivity 2.2
How Fast? The Rate of Chemical Change反应有多快？—— 化学反应的速率
Guiding Question导引问题How can the rate of a reaction be controlled? Reaction rate is set by collision frequency, collision geometry, and the fraction of collisions with enough energy to react.如何控制反应速率（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 $aA + bB \to cC + dD$：
Negative signs for reactants (their concentration falls); positive for products. Common units: $\mathrm{mol\,dm^{-3}\,s^{-1}}$.反应物取负号（浓度下降）；产物取正号。常用单位：$\mathrm{mol\,dm^{-3}\,s^{-1}}$。
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.
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:
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:
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）。
Set Up列式
Average rate over the interval (a coarse proxy for the initial rate):用该时段的平均速率近似估算初始速率：
This is the average rate over the first 30 s; the true initial rate (the tangent to the curve at $t=0$) is typically a bit higher because rate falls as the reactant depletes.这是前 30 s 的平均速率；真正的初始速率（$t=0$ 处的切线斜率）通常会更高一些，因为反应物消耗后速率会下降。
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.
Intermediate vs Transition State中间体 vs 过渡态An intermediate appears as a local minimum on the energy profile — it has a finite (if short) lifetime and a real chemical formula. A transition state is at the energy maximum between steps and cannot be isolated; partial bonds are forming or breaking.中间体在能量轮廓上是一个局部极小值——它有真实的化学式和（虽然很短的）有限寿命。过渡态则位于两步之间的能量极大值，无法分离；其中的化学键正在形成或断裂，处于半键状态。
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.
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）。
General Rate Equation速率方程（通式）
$$\text{rate} = k\,[A]^m\,[B]^n$$
$m$ is the order with respect to $A$; $n$ is the order with respect to $B$. Orders must be determined experimentally — you cannot read them off the balanced equation.$m$ 为对 $A$ 的反应级数（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 kHL
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 $\mathrm{mol\,dm^{-3}\,s^{-1}}$:
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 $k$。
Order in A对 A 的级数
Compare runs 1 and 2: [B] held constant, [A] doubled, rate doubled. Order in A = 1.比较实验 1 和 2：[B] 不变，[A] 加倍，速率也加倍。对 A 的级数 = 1。
Order in B对 B 的级数
Compare runs 1 and 3: [A] held constant, [B] doubled, rate quadrupled ($2^2$). Order in B = 2.比较实验 1 和 3：[A] 不变，[B] 加倍，速率变为四倍（$2^2$）。对 B 的级数 = 2。
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：
Arrhenius (data booklet)阿伦尼乌斯方程（数据手册）
$$k = A\,e^{-E_a / RT}$$
$A$ is the Arrhenius factor (collision frequency × steric factor); $R = 8.31~\mathrm{J\,K^{-1}\,mol^{-1}}$; $T$ in K. As $T \uparrow$, the exponent gets less negative, so $k$ rises sharply.$A$ 为阿伦尼乌斯因子（碰撞频率 × 空间因子）；$R = 8.31~\mathrm{J\,K^{-1}\,mol^{-1}}$；$T$ 取开尔文。$T$ 升高时指数的负值绝对值减小，$k$ 急剧增大。
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：
Linear Arrhenius Plot线性阿伦尼乌斯作图
$$\ln k = -\dfrac{E_a}{R}\cdot\dfrac{1}{T} + \ln A$$
Plot $\ln k$ vs $1/T$ to get a straight line with slope $-E_a/R$ and y-intercept $\ln A$.作 $\ln k$ 对 $1/T$ 的图，得到斜率为 $-E_a/R$、截距为 $\ln A$ 的直线。
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 $E_a$。
$E_a \approx 48~\mathrm{kJ\,mol^{-1}}$ — a typical magnitude for a moderately fast organic reaction.$E_a \approx 48~\mathrm{kJ\,mol^{-1}}$，与中等速率有机反应的典型量级一致。
A reaction's rate doubles when the temperature is raised from 300 K to 310 K. Which best explains this?当温度从 300 K 升到 310 K 时，反应速率加倍。下面哪种解释最佳？
The average kinetic energy of particles doubles.粒子的平均动能加倍。
The fraction of particles with energy ≥ $E_a$ increases substantially.能量大于等于 $E_a$ 的粒子比例显著增加。
Collision frequency between reactants doubles.反应物之间的碰撞频率加倍。
The activation energy decreases.活化能降低。
Correct! Raising $T$ from 300 to 310 K only increases the mean $KE$ by ~3% but shifts the Maxwell–Boltzmann tail beyond $E_a$ enough to roughly double the reactive fraction.正确！$T$ 从 300 K 升至 310 K 仅使平均动能增加约 3%，但麦克斯韦-玻尔兹曼分布尾部超过 $E_a$ 的部分扩大，足以使能发生反应的粒子比例约加倍。
Mean $KE$ goes up by only $\sim T_2/T_1 = 1.03$ (a few %), so (A) and (C) overestimate. $E_a$ is a property of the reaction itself (unless a catalyst is added). The big change is the high-energy tail. Answer: (B).平均动能仅增大约 $T_2/T_1 = 1.03$（即几个百分点），故 (A) 和 (C) 都高估。$E_a$ 是反应本身的属性（除非加入催化剂）。真正的大变化在高能尾部。答案：(B)。
(HL) The rate law for a reaction is rate $= k[A]^2[B]$. Which statement is correct?（HL）某反应的速率方程为 rate $= k[A]^2[B]$。下列说法正确的是？
The reaction is third order with respect to A.该反应对 A 是三级。
Doubling [B] halves the rate.[B] 加倍则速率减半。
Doubling [A] quadruples the rate; doubling [B] doubles the rate.[A] 加倍速率变为四倍；[B] 加倍速率加倍。
$k$ has units of $\mathrm{s^{-1}}$.$k$ 的单位为 $\mathrm{s^{-1}}$。
Correct! Order in A is 2 (so [A]² scaling means doubling [A] gives $2^2 = 4\times$ rate). Order in B is 1 (linear scaling). Overall order = 3, so $k$ has units of $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$.正确！对 A 是 2 级（[A]² 表示 [A] 加倍则速率变为 $2^2 = 4$ 倍）。对 B 是 1 级（线性）。总级数 = 3，因此 $k$ 的单位为 $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$。
Read the exponents carefully: [A]² means order 2 in A, [B]¹ means order 1 in B. The overall third-order $k$ has units $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$. Answer: (C).仔细看指数：[A]² 表示对 A 是 2 级，[B]¹ 表示对 B 是 1 级。总级数为 3 时 $k$ 的单位是 $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$。答案：(C)。
Worked Example — Testing a Mechanism Against the Rate Law HL例题 — 用速率方程检验反应机理 HL
The reaction $\mathrm{NO_2 + CO \to NO + CO_2}$ is found experimentally to follow rate $= k[\mathrm{NO_2}]^2$, independent of $[\mathrm{CO}]$. Show that the two-step mechanism below is consistent, and identify the intermediate. Step 1 (slow): $\mathrm{NO_2 + NO_2 \to NO_3 + NO}$; Step 2 (fast): $\mathrm{NO_3 + CO \to NO_2 + CO_2}$.实验测得反应 $\mathrm{NO_2 + CO \to NO + CO_2}$ 遵循 rate $= k[\mathrm{NO_2}]^2$，与 $[\mathrm{CO}]$ 无关。证明下面的两步机理与之相符，并指出中间体。第 1 步（慢）：$\mathrm{NO_2 + NO_2 \to NO_3 + NO}$；第 2 步（快）：$\mathrm{NO_3 + CO \to NO_2 + CO_2}$。
Rate from the RDS由决速步写速率
The rate-determining step (rate-determining step) is the slow Step 1, a bimolecular collision of two $\mathrm{NO_2}$ molecules. Its rate law comes straight from its molecularity:决速步骤（rate-determining step）是较慢的第 1 步，即两个 $\mathrm{NO_2}$ 分子的双分子碰撞。其速率方程直接由分子数写出：
CO reacts only in the fast Step 2, which occurs after the RDS. Species consumed only after the slow step never appear in the rate law — matching the observed independence from $[\mathrm{CO}]$.CO 只在快的第 2 步中反应，而这一步位于决速步之后。仅在慢步之后才消耗的物种不会出现在速率方程中——这与观测到的"速率与 $[\mathrm{CO}]$ 无关"一致。
Add the steps两步相加
$$\mathrm{2NO_2 + NO_3 + CO \to NO_3 + NO + NO_2 + CO_2}$$
Cancel $\mathrm{NO_3}$ (made in Step 1, used in Step 2) and one $\mathrm{NO_2}$: overall $\mathrm{NO_2 + CO \to NO + CO_2}$. ✓消去 $\mathrm{NO_3}$（第 1 步生成、第 2 步消耗）和一个 $\mathrm{NO_2}$：总反应为 $\mathrm{NO_2 + CO \to NO + CO_2}$。✓
Intermediate中间体
$\mathrm{NO_3}$ is the intermediate: produced in one step and consumed in a later one, so it cancels from the overall equation. The mechanism reproduces both the stoichiometry and the rate law, so it is consistent.$\mathrm{NO_3}$ 是中间体（intermediate）：在某一步生成、又在后续步骤消耗，因而从总方程式中抵消。该机理同时再现了化学计量和速率方程，故相符。
Worked Example — Half-Life of a First-Order Reaction HL例题 — 一级反应的半衰期 HL
A first-order decomposition has rate constant $k = 0.0231~\mathrm{s^{-1}}$ at a fixed temperature. Find (i) the half-life, and (ii) the time for the reactant concentration to fall to one-eighth of its initial value.某一级分解反应在固定温度下速率常数 $k = 0.0231~\mathrm{s^{-1}}$。求 (i) 半衰期（half-life）；(ii) 反应物浓度降至初始值 1/8 所需的时间。
First-order half-life一级反应半衰期
For a first-order reaction the half-life is constant (independent of concentration):一级反应的半衰期为常数（与浓度无关）：
A constant half-life is the diagnostic fingerprint of first-order kinetics — if successive half-lives from a graph are equal, the reaction is first order in that reactant.恒定的半衰期正是一级动力学的判别特征——若从图上读出的连续半衰期相等，则反应对该反应物为一级。
Molecularity vs Order — Don't Conflate Them分子数 vs 反应级数——切勿混淆Molecularity is a theoretical count of particles in a single elementary step, and is always a small whole number you read off the mechanism. Order is an experimental quantity for the overall reaction, extracted from rate data, and can even be zero or fractional. They coincide only for a one-step reaction. The bridge between them is the rate-determining step: because the RDS is elementary, its molecularity becomes the overall order — which is why a proposed mechanism is judged valid only if its slow-step molecularity reproduces the measured order. This is exactly the reasoning IB Paper 2 tests when it hands you a mechanism and a rate law and asks "are these consistent?"分子数（molecularity）是对单个基元步骤中粒子数的理论计数，总是从机理中读出的小整数。反应级数（order of reaction）则是整个反应的实验量，由速率数据得出，甚至可以为零或分数。二者只有在一步反应中才一致。连接它们的桥梁正是决速步：由于 RDS 是基元步骤，它的分子数就成为总反应级数——这正是"只有当慢步分子数能再现实测级数时，所提机理才成立"的原因。IB Paper 2 给出机理和速率方程、问"两者是否相符"时，考的恰恰是这一推理。
(HL) For $\mathrm{2NO + Cl_2 \to 2NOCl}$, doubling $[\mathrm{NO}]$ (with $[\mathrm{Cl_2}]$ fixed) quadruples the rate, and doubling $[\mathrm{Cl_2}]$ (with $[\mathrm{NO}]$ fixed) doubles the rate. What are the overall order and the units of $k$?（HL）对于 $\mathrm{2NO + Cl_2 \to 2NOCl}$：固定 $[\mathrm{Cl_2}]$、$[\mathrm{NO}]$ 加倍时速率变为四倍；固定 $[\mathrm{NO}]$、$[\mathrm{Cl_2}]$ 加倍时速率加倍。总反应级数与 $k$ 的单位各是多少？
Second order; $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$二级；$\mathrm{dm^3\,mol^{-1}\,s^{-1}}$
Third order; $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$三级；$\mathrm{dm^6\,mol^{-2}\,s^{-1}}$
Third order; $\mathrm{s^{-1}}$三级；$\mathrm{s^{-1}}$
Second order; $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$二级；$\mathrm{dm^6\,mol^{-2}\,s^{-1}}$
Correct! Rate quadruples when $[\mathrm{NO}]$ doubles → order 2 in NO ($2^2 = 4$); rate doubles when $[\mathrm{Cl_2}]$ doubles → order 1 in $\mathrm{Cl_2}$. Overall order $= 2 + 1 = 3$, so rate $= k[\mathrm{NO}]^2[\mathrm{Cl_2}]$ and $k$ has third-order units $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$.正确！$[\mathrm{NO}]$ 加倍速率变四倍 → 对 NO 为 2 级（$2^2 = 4$）；$[\mathrm{Cl_2}]$ 加倍速率加倍 → 对 $\mathrm{Cl_2}$ 为 1 级。总级数 $= 2 + 1 = 3$，故 rate $= k[\mathrm{NO}]^2[\mathrm{Cl_2}]$，$k$ 取三级单位 $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$。
"Quadruples" signals order 2 (not 1); "doubles" signals order 1. Summing gives overall order 3, whose $k$ units are $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ — derive units from the overall order every time. Answer: (B)."变四倍"对应 2 级（不是 1 级）；"加倍"对应 1 级。相加得总级数 3，其 $k$ 的单位为 $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$——务必每次都由总级数推导单位。答案：(B)。
Reactivity 2.3
How Far? The Extent of Chemical Change反应进行到何处？—— 化学变化的程度
Guiding Question导引问题How can the extent of a reversible reaction be influenced? The equilibrium constant tells you where the reaction settles; Le Châtelier's principle tells you how disturbances shift that settling point.如何影响可逆反应进行的程度？平衡（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.
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 $K$）。
Common Mistake — "Equilibrium" Does Not Mean "Equal Amounts"常见错误 — "平衡"不等于"等量"Students often assume equilibrium means $[\text{reactants}] = [\text{products}]$. It doesn't. Equilibrium means the rates are equal, not the concentrations — $K$ can be $10^{-10}$ or $10^{10}$ and the system is still perfectly at equilibrium. Only $K = 1$ (roughly) gives comparable amounts of each.学生常误以为平衡意味着 $[\text{反应物}] = [\text{产物}]$。并非如此。平衡指的是速率相等，而不是浓度相等——$K$ 可以是 $10^{-10}$ 或 $10^{10}$，体系依然完全处于平衡。只有当 $K \approx 1$ 时，两者的量才大致相当。
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:
Only equilibrium concentrations enter; coefficients become exponents. $K$ is temperature-dependent.仅代入平衡时刻的浓度；化学计量数变为指数。$K$ 随温度变化。
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 $K^n$.
$K_c \approx 146 \gg 1$: at equilibrium, products (HI) are strongly favoured over reactants — consistent with the reaction going nearly, but not entirely, to completion.$K_c \approx 146 \gg 1$：平衡时产物（HI）远多于反应物——说明反应接近（但并非完全）进行到底。
Practice: for $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ at equilibrium, $[\mathrm{NH_3}] = 0.4$, $[\mathrm{N_2}] = 0.2$, $[\mathrm{H_2}] = 0.2$ (mol dm⁻³). What is $K_c$?练习：反应 $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ 平衡时 $[\mathrm{NH_3}] = 0.4$，$[\mathrm{N_2}] = 0.2$，$[\mathrm{H_2}] = 0.2$（mol dm⁻³）。$K_c$ 是多少？
Going Deeper — Heterogeneous Equilibria拓展 — 多相平衡When a reaction includes a pure solid or pure liquid, its "concentration" is constant and is folded into $K$ itself — so solids and pure liquids are omitted from the equilibrium expression. For $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$, only the gas matters: $K_c = [\mathrm{CO_2}]$. This is why decomposition equilibria like this one depend only on temperature (through $K$), not on how much solid is present.当反应中含有纯固体或纯液体时，它们的"浓度"是常数，已经并入 $K$ 本身——因此固体和纯液体不出现在平衡表达式中。对于 $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$，只有气体起作用：$K_c = [\mathrm{CO_2}]$。这也是为什么这类分解平衡只取决于温度（通过 $K$），而与固体的量无关。
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").
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 $K$.
Why Catalysts Don't Shift Equilibrium为什么催化剂不会使平衡移动A catalyst lowers $E_a$ for the forward and reverse reactions equally — so both rates rise by the same factor and the equilibrium position is unchanged. The system just gets there faster.催化剂对正、逆反应的 $E_a$ 等量降低——因此两个方向的速率以相同倍数增大，平衡位置不变。体系只是更快地达到同一个平衡。
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。
(a) Increase pressure(a) 升高压强
4 mol of gas on the reactant side, 2 mol on the product side. Shift → toward products. Yield of NH₃ increases.反应物侧 4 mol 气体，产物侧 2 mol。平衡 → 向产物侧移动。NH₃ 产量增加。
(b) Increase temperature(b) 升高温度
Forward reaction is exothermic ($\Delta H < 0$). The system absorbs the added heat by shifting in the endothermic direction → toward reactants. Yield of NH₃ decreases, and $K$ also decreases.正反应放热（$\Delta H < 0$）。体系通过向吸热方向移动来吸收加入的热量 → 向反应物侧移动。NH₃ 产量降低，$K$ 也减小。
(c) Remove NH₃ as it forms(c) 边生成边移走 NH₃
Lowering [NH₃] forces the system to make more. Shift → products. Total yield over time increases (this is how industrial Haber plants operate — continuous NH₃ removal by condensation).降低 [NH₃] 会迫使体系继续生成 NH₃。平衡 → 向产物侧移动。累计产量随时间增加（这正是工业哈伯法的运作方式——通过冷凝持续移走 NH₃）。
(d) Add an iron catalyst(d) 加入铁催化剂
No shift in equilibrium position. Yield unchanged — but the time to reach equilibrium is much shorter, which is what makes the process economic.平衡位置不移动。产量不变——但达到平衡所需时间大大缩短，这才是该工艺具有经济效益的关键。
A gas-phase equilibrium has fewer moles of gas on the product side. Increasing the pressure will:某气相平衡反应中产物侧气体摩尔数较少。增大压强将：
2.3c
Shift toward reactants and change K向反应物侧移动并改变 K
Shift toward products; K unchanged向产物侧移动；K 不变
Have no effect on the position对平衡位置没有影响
Shift toward reactants; K unchanged向反应物侧移动；K 不变
Correct! Higher pressure shifts equilibrium toward the side with fewer moles of gas — here, the products. $K$ only changes with temperature, so it stays the same.正确！增大压强会使平衡向气体摩尔数较少的一侧移动——此处即产物侧。$K$ 只随温度变化，故保持不变。
Le Châtelier: the system relieves added pressure by favouring the side with fewer gas moles (the product side here). $K$ itself is temperature-dependent only.根据勒夏特列原理：体系通过偏向气体摩尔数较少的一侧（此处为产物侧）来缓解压强增大。$K$ 本身只随温度变化。
The Reaction Quotient Q反应商 QHL
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:
Worked Example — Predicting Shift Direction from Q vs. K例题 — 由 Q 与 K 的比较预测移动方向
For $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$, $K_c = 4.6 \times 10^{-3}$ at 298 K. A flask contains $[\mathrm{N_2O_4}] = 0.10~\mathrm{mol\,dm^{-3}}$ and $[\mathrm{NO_2}] = 0.050~\mathrm{mol\,dm^{-3}}$. Which way will the reaction proceed?对于 $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$，298 K 时 $K_c = 4.6 \times 10^{-3}$。某烧瓶中 $[\mathrm{N_2O_4}] = 0.10~\mathrm{mol\,dm^{-3}}$，$[\mathrm{NO_2}] = 0.050~\mathrm{mol\,dm^{-3}}$。反应将朝哪个方向进行？
$Q = 0.025 > K_c = 4.6 \times 10^{-3}$: there are too many products for equilibrium.$Q = 0.025 > K_c = 4.6 \times 10^{-3}$：产物相对偏多，尚未达到平衡。
Conclusion结论
The reaction proceeds in reverse (toward $\mathrm{N_2O_4}$) until $Q$ falls to meet $K_c$.反应将逆向进行（生成更多 $\mathrm{N_2O_4}$），直到 $Q$ 降低至等于 $K_c$ 为止。
If $Q < K$ for a reaction, the reaction will:若某反应 $Q < K$，该反应将：
2.3b
Proceed in reverse, toward reactants逆向进行，趋向反应物
Already be at equilibrium已经处于平衡
Proceed forward, toward products正向进行，趋向产物
Stop reacting entirely完全停止反应
Correct! $Q < K$ means there are too few products relative to equilibrium, so the reaction shifts forward to make more — increasing $Q$ until it equals $K$.正确！$Q < K$ 说明相对于平衡而言产物太少，反应会正向移动以生成更多产物——使 $Q$ 增大直至等于 $K$。
$Q < K$ means the products-to-reactants ratio is below its equilibrium value, so the reaction proceeds forward (making more products) until $Q$ rises to $K$.$Q < K$ 说明产物与反应物的比值低于平衡值，反应会正向进行（生成更多产物），直到 $Q$ 升高至 $K$。
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.
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。
Sanity check: $x / 0.20 = 7.6\%$, comfortably below the usual 5–10% threshold for the small-$K$ approximation.合理性检验：$x / 0.20 = 7.6\%$，在小 $K$ 近似常用的 5–10% 阈值之下，可放心使用。
Practice: for $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$, $K_c$ is very large. If you start with only $\mathrm{SO_2}$ and $\mathrm{O_2}$ (no $\mathrm{SO_3}$), is the small-$K$ or the "goes nearly to completion" approximation more appropriate?练习：对于 $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$，$K_c$ 非常大。若起始只有 $\mathrm{SO_2}$ 和 $\mathrm{O_2}$（无 $\mathrm{SO_3}$），应采用"小 K 近似"还是"接近完全反应"的近似？
Worked Example — Equilibrium Calculation Needing the Quadratic Formula例题 — 需用求根公式求解的平衡计算
For $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$, $K_c = 54.0$ at a given temperature. If 1.00 mol $\mathrm{H_2}$ and 1.00 mol $\mathrm{I_2}$ are placed in a 1.00 dm³ container, find the equilibrium concentration of $\mathrm{HI}$.对于 $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$，某温度下 $K_c = 54.0$。若将 1.00 mol $\mathrm{H_2}$ 与 1.00 mol $\mathrm{I_2}$ 置于 1.00 dm³ 容器中，求平衡时 $\mathrm{HI}$ 的浓度。
Why the small-K trick fails here为何小 K 近似在此不适用
$K_c = 54.0$ is not $\ll 1$, so $x$ is not negligible compared to 1.00 — the approximation from the earlier example would give a wrong answer here.$K_c = 54.0$ 并非 $\ll 1$，故 $x$ 相对于 1.00 不可忽略——前一例题中的近似法用在这里会得出错误结果。
ICE table and expressionICE 表与表达式
$$K_c = \dfrac{(2x)^2}{(1.00-x)^2} = 54.0$$
Take the square root of both sides两边同时开平方
Because both sides are perfect squares, take $\sqrt{\,\cdot\,}$ directly instead of expanding into a full quadratic:因两边均为完全平方，可直接开平方，而不必展开成完整的一元二次方程：
General lesson: when both sides of the $K$ expression are perfect squares, take the square root before reaching for the quadratic formula — it is faster and avoids extraneous roots.通用经验：当 $K$ 表达式两边都是完全平方时，先开平方再考虑一元二次求根公式——这样更快，也能避免产生多余的根。
Which signal tells you the small-$K$ approximation is not safe to use?哪个信号说明小 K 近似不适用？
2.3d
$K$ is much less than 1$K$ 远小于 1
$K$ is close to or greater than 1$K$ 接近或大于 1
The initial concentration is large初始浓度较大
The reaction is exothermic该反应为放热反应
Correct! The approximation $[\text{initial}] \approx [\text{eqm}]$ relies on $x$ being negligibly small, which only holds when $K$ is very small. Once $K \gtrsim 1$, $x$ becomes comparable to the initial concentration and you must solve exactly.正确！近似 $[\text{initial}] \approx [\text{eqm}]$ 依赖于 $x$ 可忽略不计，而这只有在 $K$ 很小时才成立。一旦 $K \gtrsim 1$，$x$ 就与初始浓度相当，必须精确求解。
The approximation's validity depends on $K$ being small (so $x \ll$ initial concentration) — check the size of $K$, not the reaction's enthalpy or initial amount alone.该近似是否成立取决于 $K$ 是否够小（使 $x \ll$ 初始浓度）——应检查 $K$ 的大小，而非仅看反应焓变或初始量。
Going Deeper — $K_c$ vs. $K_p$拓展 — $K_c$ 与 $K_p$For gas-phase equilibria, $K_p$ uses partial pressures instead of concentrations. The two are related by $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas (products − reactants). If $\Delta n = 0$ (equal moles each side), $K_p = K_c$ regardless of temperature or units — a quick check worth remembering.对于气相平衡，$K_p$ 使用分压而非浓度。两者关系为 $K_p = K_c(RT)^{\Delta n}$，其中 $\Delta n$ 为气体摩尔数变化（产物 − 反应物）。若 $\Delta n = 0$（两侧摩尔数相等），则无论温度或单位如何，$K_p = K_c$——这是一个值得记住的快速检验。
Worked Example — Converting Kc to Kp例题 — 由 Kc 换算 Kp
A negative $\Delta n$ makes $K_p$ much smaller than $K_c$ here — a reminder that $K_p$ and $K_c$ are numerically equal only when $\Delta n = 0$.此处 $\Delta n$ 为负，使 $K_p$ 远小于 $K_c$——提醒我们只有当 $\Delta n = 0$ 时，$K_p$ 与 $K_c$ 在数值上才相等。
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:
$R = 8.31~\mathrm{J\,K^{-1}\,mol^{-1}}$. Sign of $\Delta G^{\ominus}$ matches $K$: $\Delta G^{\ominus} < 0 \Leftrightarrow K > 1$ (products favoured at equilibrium).$R = 8.31~\mathrm{J\,K^{-1}\,mol^{-1}}$。$\Delta G^{\ominus}$ 的符号与 $K$ 对应：$\Delta G^{\ominus} < 0 \Leftrightarrow K > 1$（平衡时产物占优）。
Worked Example — From K to ΔG° and Back例题 — 由 K 求 ΔG°
At 298 K, a reaction has $K = 2.5 \times 10^{3}$. Calculate $\Delta G^{\ominus}$, and predict its sign before you compute.298 K 时，某反应的 $K = 2.5 \times 10^{3}$。计算 $\Delta G^{\ominus}$，并在计算前先预测其符号。
Predict the sign预测符号
$K > 1$, so products are favoured at equilibrium — expect $\Delta G^{\ominus} < 0$.$K > 1$，说明平衡时产物占优——预期 $\Delta G^{\ominus} < 0$。
Apply the formula代入公式
$$\Delta G^{\ominus} = -RT\ln K = -(8.31)(298)\ln(2.5 \times 10^3)$$
Negative, as predicted, and only moderately so — a large $K$ does not require an enormous $|\Delta G^{\ominus}|$ because of the logarithm.结果为负，与预测一致，且数值并不算很大——由于对数关系，较大的 $K$ 并不需要非常大的 $|\Delta G^{\ominus}|$。
Going Deeper — Non-Standard ΔG and the Direction of Change拓展 — 非标准 ΔG 与变化方向$\Delta G^{\ominus}$ only describes standard conditions. At any other composition, $\Delta G = \Delta G^{\ominus} + RT\ln Q$. This is why a reaction with $\Delta G^{\ominus} > 0$ can still run forward spontaneously away from standard conditions (e.g. very low product concentration makes $\ln Q$ very negative) — and why every reaction, whatever its $\Delta G^{\ominus}$, drives toward the composition where $\Delta G = 0$, which is precisely $Q = K$: equilibrium.$\Delta G^{\ominus}$ 只描述标准状态。在其他任意组成下，$\Delta G = \Delta G^{\ominus} + RT\ln Q$。这也是为什么某个 $\Delta G^{\ominus} > 0$ 的反应，在偏离标准状态时仍可能自发正向进行（例如产物浓度极低会使 $\ln Q$ 变得非常负）——并且说明无论 $\Delta G^{\ominus}$ 为何值，任何反应都会驱使体系趋向 $\Delta G = 0$ 的组成，而这恰好对应 $Q = K$，即平衡。
Going Deeper — Why K Changes with Temperature拓展 — K 为何随温度变化Combine $\Delta G^{\ominus} = -RT\ln K$ with $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$ to get $\ln K = -\dfrac{\Delta H^{\ominus}}{RT} + \dfrac{\Delta S^{\ominus}}{R}$. For an exothermic reaction ($\Delta H^{\ominus} < 0$), raising $T$ makes $-\Delta H^{\ominus}/RT$ smaller, so $\ln K$ (and $K$) falls — exactly the Le Châtelier prediction that heating an exothermic equilibrium shifts it toward reactants, now derived rather than just stated.将 $\Delta G^{\ominus} = -RT\ln K$ 与 $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$ 联立，得到 $\ln K = -\dfrac{\Delta H^{\ominus}}{RT} + \dfrac{\Delta S^{\ominus}}{R}$。对放热反应（$\Delta H^{\ominus} < 0$），升高 $T$ 会使 $-\Delta H^{\ominus}/RT$ 变小，故 $\ln K$（及 $K$）减小——这正是勒夏特列原理所说的"加热放热平衡会使其向反应物方向移动"，只是这里是推导出来的，而非直接给出的结论。
Negative $\Delta G^{\ominus}$ and $K > 1$ agree — products are favoured under standard conditions.$\Delta G^{\ominus}$ 为负且 $K > 1$ 相互一致——在标准条件下产物占优。
For the equilibrium $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ with $\Delta H < 0$, which change would increase both the yield of $\mathrm{SO_3}$ and the value of $K$?对于平衡 $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$，$\Delta H < 0$。下列哪种变化能同时提高 $\mathrm{SO_3}$ 的产量和 $K$ 值？
Increase the temperature升高温度
Add a catalyst加入催化剂
Increase the pressure at constant temperature恒温下升高压强
Decrease the temperature降低温度
Correct! Forward reaction is exothermic, so lowering $T$ shifts toward products AND increases $K$. Pressure shifts the position but leaves $K$ unchanged; catalysts change neither.正确！正反应放热，因此降低 $T$ 既会向产物方向移动，又会增大 $K$。压强变化只移动平衡位置但不改变 $K$；催化剂两者都不改变。
Only temperature changes $K$. For an exothermic forward reaction, lowering $T$ shifts toward products and raises $K$. Answer: (D).只有温度能改变 $K$。对于放热的正反应，降低 $T$ 会向产物方向移动并提高 $K$。答案：(D)。
(HL) For a reaction with $K = 1$, what is $\Delta G^{\ominus}$ at any temperature?（HL）当反应的 $K = 1$ 时，任意温度下 $\Delta G^{\ominus}$ 等于多少？
$\Delta G^{\ominus} > 0$
$\Delta G^{\ominus} = 0$
$\Delta G^{\ominus} < 0$
$\Delta G^{\ominus}$ depends on temperature; cannot be determined$\Delta G^{\ominus}$ 取决于温度，无法确定
Correct! $\Delta G^{\ominus} = -RT\ln K = -RT \cdot 0 = 0$. The standard Gibbs energy is zero precisely when reactants and products are equally favoured at equilibrium.正确！$\Delta G^{\ominus} = -RT\ln K = -RT \cdot 0 = 0$。当反应物和产物在平衡时同等存在时，标准吉布斯自由能恰为 0。
Paper 1 (MC, no calc, no data booklet)Paper 1（选择题，无计算器、无数据手册）
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。
Paper 2 (short + long, calculator + data booklet)Paper 2（短题 + 长题，可用计算器和数据手册）
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 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 $E_a = -R \cdot \text{slope}$.
Data Booklet — Useful Tables数据手册中的实用表格Section 1 (Avogadro, $V_m$), Section 2 (constants — $R$, $k_B$), Section 5 ($A_r$ to 2 d.p.), and the formula sheet (atom economy, $\Delta G^{\ominus} = -RT\ln K$, Arrhenius equation and its linear form). Bookmark these before the exam.第 1 部分（阿伏伽德罗常数、$V_m$）、第 2 部分（常数 $R$、$k_B$）、第 5 部分（保留两位小数的 $A_r$）以及公式页（原子经济性、$\Delta G^{\ominus} = -RT\ln K$、阿伦尼乌斯方程及其线性形式）。考前先把这些页标好。
Watch Out小心陷阱
Common Mistakes常见错误
Forgetting to balance before mole-ratio work用摩尔比之前忘了配平The mole ratio comes from the balanced equation. Always verify atom counts before extracting the ratio.摩尔比来自配平后的方程式。提取比例之前一定先核对各元素的原子数。
Confusing % yield with atom economy把产率（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。
Reading orders off the balanced equation直接从配平方程式上"读出"反应级数Orders in the rate equation are only determined experimentally. The exponents in rate $= k[A]^m[B]^n$ are not the stoichiometric coefficients (except for elementary steps).速率方程（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）。
"Catalysts shift equilibrium""催化剂会让平衡移动"Never. A catalyst speeds up forward and reverse equally — same equilibrium position, same $K$. It only changes how fast you get there.绝不会。催化剂等量加速正反应和逆反应——平衡位置不变，$K$ 也不变。它改变的只是达到平衡的速度。
Treating $K$ as a constant across temperatures把 $K$ 当作跨温度的常数$K$ is "constant" in the sense that for a given temperature it has a fixed value — but it does change with $T$. Don't reuse a $K$ from one $T$ at a different $T$.$K$ "恒定"的含义是：在给定温度下取固定值——但它确实随 $T$ 变化。不要把某一温度下的 $K$ 套用到另一温度。
Forgetting to convert temperature to Kelvin忘记把温度换算为开尔文Every kinetics and equilibrium equation that involves $T$ requires absolute temperature. $0~^\circ\mathrm{C} = 273.15~\mathrm{K}$ — not 273.凡是涉及 $T$ 的动力学和平衡方程都必须使用绝对温度。$0~^\circ\mathrm{C} = 273.15~\mathrm{K}$——不是 273。
Skipping units of $k$漏写 $k$ 的单位Examiners check whether you derived $k$'s units from the overall order. A second-order $k$ in $\mathrm{s^{-1}}$ is wrong even if the numerical value is right.阅卷人会检查你是否依据总反应级数推出了 $k$ 的单位。二级反应的 $k$ 即使数值正确，单位若写成 $\mathrm{s^{-1}}$ 也算错。
Review复习
Flashcards闪卡
Identify the limiting reactant?如何判定限量试剂？
Divide each reactant's moles by its coefficient in the balanced equation; the smallest ratio = limiting.各反应物摩尔数除以方程式中的系数；比值最小者 = 限量。
Two conditions for a successful collision?有效碰撞的两个条件？
Sufficient energy ($\geq E_a$) and correct orientation.能量足够（$\geq E_a$）且取向正确。
How does a catalyst increase rate?催化剂如何加快反应速率？
Provides an alternative pathway with a lower activation energy. Doesn't affect $\Delta H$ or the equilibrium position.提供活化能更低的替代路径。不改变 $\Delta H$，也不改变平衡位置。
(HL) Linear Arrhenius equation?（HL）线性形式的阿伦尼乌斯方程？
$\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$ — plot $\ln k$ vs $1/T$ to get slope $= -E_a/R$.作 $\ln k$ 对 $1/T$ 的图，斜率 $= -E_a/R$。
Le Châtelier — what changes $K$?勒夏特列——什么会改变 $K$？
Only temperature. Pressure and concentration changes shift the position but $K$ stays the same.只有温度。压强和浓度的变化会移动平衡位置，但 $K$ 不变。
$K$ expression for $aA + bB \rightleftharpoons cC + dD$?$aA + bB \rightleftharpoons cC + dD$ 的 $K$ 表达式？
$K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$ — equilibrium concentrations only, coefficients become exponents.只用平衡浓度，系数变指数。