| 核心 | 补充 |
|---|---|
| 1 识别物理变化和化学变化,并描述它们之间的区别 |
化学反应
IGCSE 化学 · 知识点 6
15:53
速率与平衡
牛奶在柜台上一天就会变质。在冰箱里却能放一周。同样的牛奶——但反应速度不同。首先快速检查:一种化学…
英语旁白 · 英中内嵌字幕
6.1
物理变化与化学变化
教学大纲
来源:Cambridge International 教学大纲

A physical change 物理变化 does not make a new substance. The substance only changes its state or shape, and the change can usually be reversed. Melting ice and dissolving sugar are physical changes.
A chemical change 化学变化 (a chemical reaction) makes one or more new substances and is usually hard to reverse. Signs of a chemical change include a colour change, a gas being given off, an energy change, or a precipitate 沉淀 (a solid) forming.
| Physical change | Chemical change |
|---|---|
| no new substance made | new substance(s) made |
| easy to reverse | usually hard to reverse |
| e.g. melting, boiling, dissolving | e.g. burning, rusting |
物理变化对化学变化
逐步看区别。一个物理变化不产生新物质且通常容易逆转;一个化学变化产生一种新物质且难以撤销。
| English | 中文 | 拼音 |
|---|---|---|
| physical change/ˈfɪzɪkl tʃeɪndʒ/ | 物理变化 | wù lǐ biàn huà |
| chemical change/ˈkemɪkl tʃeɪndʒ/ | 化学变化 | huà xué biàn huà |
| precipitate/prɪˈsɪpɪteɪt/ | 沉淀 | chén diàn |
6.2
反应速率
教学大纲
| 核心内容 | 补充内容 |
|---|---|
| 5 从以下方面描述碰撞理论:(a) 单位体积内的粒子数 (b) 粒子间碰撞的频率 (c) 粒子的动能 (d) 活化能,$E_a$ | |
| 1 描述以下因素对反应速率的影响:(a) 改变溶液的浓度 (b) 改变气体的压强 (c) 改变固体的表面积 (d) 改变温度 (e) 添加或移除催化剂,包括酶 | 6 利用碰撞理论描述并解释以下因素对反应速率的影响:(a) 改变溶液的浓度 (b) 改变气体的压强 (c) 改变固体的表面积 (d) 改变温度 (e) 添加或移除催化剂,包括酶 |
| 2 陈述催化剂能增加反应速率,且在反应结束时保持不变 | 7 陈述催化剂能降低反应的活化能,$E_a$ |
| 3 描述探究反应速率的实验方法,包括反应物或生成物的质量变化以及气体的生成 | 8 评估探究反应速率的实验方法,包括反应物或生成物的质量变化以及气体的生成 |
| 4 解读来自反应速率实验的数据,包括图表 |
来源:Cambridge International 教学大纲
The rate of reaction 反应速率 tells you how fast the reactants 反应物 change into products 生成物.
Collision theory
Collision theory 碰撞理论 explains what controls the rate. For a reaction to happen, the particles 粒子 must collide 碰撞, and they must collide with enough energy. The least energy they need is the activation energy 活化能 ($E_a$). A faster rate happens when the particles have successful collisions more often.
What changes the rate
| Change | Effect | Reason (collision theory) |
|---|---|---|
| increase concentration 浓度 of a solution | faster | more particles in the same volume, so collisions happen more often |
| increase pressure 压强 of gases | faster | particles are pushed closer, so collisions happen more often |
| increase surface area 表面积 of a solid | faster | more particles are exposed, so collisions happen more often |
| increase temperature | faster | particles gain kinetic energy 动能 and move faster, so collisions are more frequent and more of them have enough energy |
| add a catalyst 催化剂 | faster | the catalyst lowers the activation energy, so more collisions are successful |


Catalysts
A catalyst speeds up a reaction but is not used up — it is unchanged at the end. It works by lowering the activation energy. Enzymes 酶 are biological catalysts (catalysts that work in living things).

Measuring the rate
You can follow a reaction over time in these ways:
- Change in mass: stand the flask on a balance. If a gas escapes, the mass falls. Record the mass at regular times.
- Volume of gas: collect the gas in a gas syringe 注射器 and read its volume at regular times.
- Formation of a precipitate: in a reaction that turns cloudy, time how long it takes for a mark under the flask to disappear.

On a graph of product against time, the line is steepest at the start (fastest rate), becomes less steep as reactants are used up, and goes flat when the reaction has finished.

从气体体积图求速率
体积–时间曲线的梯度就是速率;开始时最陡,随着反应物耗尽而变平。
反应速率
浓度 = c₀·bᵗ
浓度随时间下降——起初快,然后更慢。
碰撞理论在行动
观察粒子碰撞。热加速它们,所以它们更频繁、更猛地碰撞;更多粒子或一个催化剂给出更多成功的碰撞——速率攀升。
| English | 中文 | 拼音 |
|---|---|---|
| rate of reaction/reɪt ɒv rɪˈækʃn/ | 反应速率 | fǎn yìng sù lǜ |
| reactants/rɪˈæktənts/ | 反应物 | fǎn yìng wù |
| products/ˈprɒdʌkts/ | 生成物 | shēng chéng wù |
| collision theory/kəˈlɪʒn ˈθɪəri/ | 碰撞理论 | pèng zhuàng lǐ lùn |
| particles/ˈpɑːtɪklz/ | 粒子 | lì zǐ |
| collide/kəˈlaɪd/ | 碰撞 | pèng zhuàng |
| activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ | 活化能 | huó huà néng |
| concentration/ˌkɒnsənˈtreɪʃn/ | 浓度 | nóng dù |
| pressure/ˈpreʃə/ | 压强 | yā qiáng |
| surface area/ˈsɜːfɪs ˈeərɪə/ | 表面积 | biǎo miàn jī |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | 动能 | dòng néng |
| catalyst/ˈkætəlɪst/ | 催化剂 | cuī huà jì |
| enzymes/ˈenzaɪmz/ | 酶 | méi |
| syringe/sɪˈrɪndʒ/ | 注射器 | zhù shè qì |
6.3
可逆反应与平衡
教学大纲
| 核心 | 补充 |
|---|---|
| 1 说明某些化学反应是可逆的,如符号 $\rightleftharpoons$ 所示 | 3 说明在封闭系统中,当满足以下条件时,可逆反应处于平衡状态:(a) 正反应速率等于逆反应速率 (b) 反应物和产物的浓度不再变化 |
| 2 描述改变条件如何改变可逆反应的方向,包括:(a) 热量对水合化合物的影响 (b) 向无水化合物中加水(仅限于硫酸铜和氯化钴) | 4 根据提供的信息,预测并解释对于可逆反应,平衡位置受以下因素的影响:(a) 温度变化 (b) 压力变化 (c) 浓度变化 (d) 使用催化剂 |
| 5 写出哈伯法生产氨气的符号方程式 $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ | |
| 6 说明哈伯法中氢气(甲烷)和氮气(空气)的来源 | |
| 7 说明哈伯法的典型条件为 $450\text{ }^{\circ}\text{C}$、$20\,000\text{ kPa}/200\text{ atm}$ 以及铁催化剂 | |
| 8 写出接触法中将二氧化硫转化为三氧化硫的符号方程式 $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$ | |
| 9 说明接触法中二氧化硫(燃烧硫或焙烧硫化矿石)和氧气(空气)的来源 | |
| 10 说明接触法中将二氧化硫转化为三氧化硫的典型条件为 $450\text{ }^{\circ}\text{C}$、$200\text{ kPa}/2\text{ atm}$ 以及五氧化二钒催化剂 | |
| 11 从反应速率和平衡位置的角度,解释为何在哈伯法和接触法中使用所述的典型条件,包括安全考虑和经济因素 |
来源:Cambridge International 教学大纲
Some reactions are reversible 可逆反应: the products can react to form the reactants again. We show this with the symbol $\rightleftharpoons$.

Changing the direction
A clear example uses hydrated 水合 and anhydrous 无水 compounds:
- Blue hydrated copper(II) sulfate, when heated, loses its water to become white anhydrous copper(II) sulfate. Adding water turns it blue again.
- Pink hydrated cobalt(II) chloride loses water when heated to become blue anhydrous cobalt(II) chloride. Adding water turns it pink again.
Adding water to the anhydrous solid (and seeing the colour return) is used as a test for water.
Equilibrium
In a closed system 密闭系统 (where no matter enters or leaves, although energy may be transferred), a reversible reaction reaches equilibrium 平衡 when:
- the rate of the forward reaction 正反应 equals the rate of the reverse reaction 逆反应, and
- the concentrations of the reactants and products are no longer changing.
Changing the position of equilibrium
The position of equilibrium 平衡位置 tells you whether there are more reactants or more products. You can move it:
- Temperature: heating moves the equilibrium in the direction that takes in heat (the endothermic 吸热反应 direction); cooling moves it in the direction that gives out heat (the exothermic 放热反应 direction).
- Pressure (gases): increasing pressure by compressing the gases at constant temperature favours the side with fewer gas molecules. If both sides have the same number, this does not shift the equilibrium.
- Concentration: adding more of a substance moves the equilibrium to the other side, to use it up.
- A catalyst does not move the position of equilibrium. It only helps the reaction reach equilibrium faster.
The Haber process
The Haber process 哈伯法 makes ammonia:
- The hydrogen 氢气 comes from methane 甲烷 (natural gas); the nitrogen 氮气 comes from the air.
- Typical conditions: a temperature of $450\,{}^{\circ}\text{C}$, a pressure of about $200$ atm ($20\,000$ kPa), and an iron 铁 catalyst.
The Contact process
The Contact process 接触法 turns sulfur dioxide into sulfur trioxide:
- The sulfur dioxide comes from burning sulfur 硫 (or roasting sulfide ores); the oxygen 氧气 comes from the air.
- Typical conditions: a temperature of $450\,{}^{\circ}\text{C}$, a pressure of about $2$ atm ($200$ kPa), and a vanadium(V) oxide 五氧化二钒 catalyst.
Why these conditions are chosen
The conditions are a compromise 折中:
- Higher pressure favours products in both equations. Haber uses high pressure as a compromise with cost and risk; Contact uses only low pressure because its conversion is already high.
- A lower temperature would give more product (both forward reactions are exothermic), but the reaction would be too slow, so a fairly high temperature is used to keep a good rate.
- The catalyst speeds up the reaction without changing the position of equilibrium, which lowers cost.

(Supplement) Worked example. Explain the effect of raising the pressure on the Haber process. Count the gas molecules on each side of the equation: on the left $1 + 3 = 4$, on the right $2$. A higher pressure moves the equilibrium to the side with fewer gas molecules, which is the right. So more ammonia is made. The same count for the Contact process gives $2 + 1 = 3$ on the left and $2$ on the right, so a higher pressure would also help there. But the yield at $2$ atm is already very high, so a higher pressure is not worth its cost.
Common mistakes.
- Saying a higher temperature gives more ammonia because "the reaction goes faster". A higher temperature makes it reach equilibrium faster, but the equilibrium mixture contains less ammonia, because the forward reaction is exothermic.
- Saying a catalyst gives a higher yield. It gives the same yield, only sooner.
- Counting atoms instead of gas molecules when you predict the effect of pressure.
- Saying the conditions are chosen "to give the highest yield". They are a compromise between yield, rate, cost and safety.
平衡的移动
改变温度、压力或浓度,观察可逆反应移动至新的平衡状态。
| English | 中文 | 拼音 |
|---|---|---|
| reversible reaction/rɪˈvɜːsɪbl rɪˈækʃn/ | 可逆反应 | kě nì fǎn yìng |
| hydrated/haɪˈdreɪtɪd/ | 水合 | shuǐ hé |
| anhydrous/ænˈhaɪdrəs/ | 无水 | wú shuǐ |
| closed system/kləʊzd ˈsɪstəm/ | 密闭系统 | mì bì xì tǒng |
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | 平衡 | píng héng |
| forward reaction/ˈfɔːwəd rɪˈækʃn/ | 正反应 | zhèng fǎn yìng |
| reverse reaction/rɪˈvɜːs rɪˈækʃn/ | 逆反应 | nì fǎn yìng |
| position of equilibrium/pəˈzɪʃn ɒv ˌiːkwɪˈlɪbrɪəm/ | 平衡位置 | píng héng wèi zhì |
| endothermic/ˌendəʊˈθɜːmɪk/ | 吸热反应 | xī rè fǎn yìng |
| exothermic/eɡzəˈðɜːmɪk/ | 放热反应 | fàng rè fǎn yìng |
| Haber process/ˈheɪbə ˈprəʊses/ | 哈伯法 | hā bó fǎ |
| hydrogen/ˈhaɪdrədʒn/ | 氢气 | qīng qì |
| methane/ˈmiːθeɪn/ | 甲烷 | jiǎ wán |
| nitrogen/ˈnaɪtrədʒn/ | 氮气 | dàn qì |
| iron/ˈaɪən/ | 铁 | tiě |
| Contact process/ˈkɒntækt ˈprəʊses/ | 接触法 | jiē chù fǎ |
| sulfur/ˈsʌlfɜː/ | 硫 | liú |
| oxygen/ˈɒksɪdʒn/ | 氧气 | yǎng qì |
| vanadium(V) oxide/væˈneɪdɪəm ˈɒksaɪd/ | 五氧化二钒 | wǔ yǎng huà èr fán |
| compromise/ˈkɒmprəmaɪz/ | 折中 | zhé zhōng |
6.4
氧化还原
教学大纲
| 核心内容 | 补充内容 |
|---|---|
| 1 使用罗马数字表示化合物中某元素的氧化数 | |
| 2 将氧化还原反应定义为同时发生氧化和还原的反应 | |
| 3 将氧化定义为得氧,将还原定义为失氧 | 6 从以下方面定义氧化:(a) 失去电子 (b) 氧化数升高 |
| 7 从以下方面定义还原:(a) 得到电子 (b) 氧化数降低 | |
| 4 识别涉及得氧和失氧的氧化还原反应 | 8 识别涉及电子得失的氧化还原反应 |
| 5 在氧化还原反应中识别氧化过程和还原过程 | 9 通过氧化数的变化来识别氧化还原反应,依据包括:(a) 游离态元素的氧化数为零 (b) 单原子离子的氧化数等于该离子所带电荷 (c) 化合物中各元素氧化数之和为零 (d) 离子中各元素氧化数之和等于该离子所带电荷 |
| 10 利用酸化的酸性高锰酸钾(VII)溶液或碘化钾水溶液反应时的颜色变化来识别氧化还原反应 | |
| 11 将氧化剂定义为能氧化其他物质且自身被还原的物质 | |
| 12 将还原剂定义为能还原其他物质且自身被氧化的物质 | |
| 13 在氧化还原反应中识别氧化剂和还原剂 |
来源:Cambridge International 教学大纲
Oxidation 氧化 and reduction 还原 always happen at the same time, in what is called a redox reaction 氧化还原反应. ('Redox' is short for reduction–oxidation.)
There are three ways to describe oxidation and reduction:
- Oxygen: oxidation is the gain of oxygen; reduction is the loss of oxygen.
- Electrons 电子: oxidation is the loss of electrons; reduction is the gain of electrons.
- Oxidation number 氧化数: in oxidation the oxidation number goes up; in reduction it goes down.
A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain — of electrons.

Oxidation number rules
The oxidation number is shown by a Roman numeral, as in iron(II) and iron(III). The rules are:
- An element that is not combined has an oxidation number of $0$.
- A single-atom ion has an oxidation number equal to its charge (so $\text{Na}^{+}$ is $+1$).
- The oxidation numbers in a compound add up to $0$.
- The oxidation numbers in an ion add up to the charge on the ion.
Worked example. Find the oxidation number of manganese in the manganate(VII) ion, $\text{MnO}_4^{-}$. Each oxygen is $-2$, and there are four of them, giving $4 \times (-2) = -8$. The oxidation numbers in an ion must add up to the charge on the ion, which here is $-1$. So if manganese is $x$, then $x + (-8) = -1$, giving $x = +7$ - exactly what the (VII) in the name tells you. Set the total to the ion's charge, not to zero: zero is only for a neutral compound.
Oxidising and reducing agents
- An oxidising agent 氧化剂 oxidises another substance, and is itself reduced.
- A reducing agent 还原剂 reduces another substance, and is itself oxidised.
Some redox reactions show clear colour changes:
- Acidified potassium manganate(VII) 高锰酸钾 is purple. When it acts as an oxidising agent it is reduced, and the purple colour fades to colourless.
- Potassium iodide 碘化钾 is colourless. When it is oxidised, red-brown iodine 碘 is formed.

氧化和还原一起发生
逐步看一个氧化还原反应。一种物质失去电子而另一种获得它们——你永远不能有一个而没有另一个。
| English | 中文 | 拼音 |
|---|---|---|
| oxidation/ˌɒksɪˈdeɪʃn/ | 氧化 | yǎng huà |
| reduction/rɪˈdʌkʃn/ | 还原 | huán yuán |
| redox reaction/rɪˈdɒks rɪˈækʃn/ | 氧化还原反应 | yǎng huà huán yuán fǎn yìng |
| electrons/ɪˈlektrɒnz/ | 电子 | diàn zǐ |
| oxidation number/ˌɒksɪˈdeɪʃn ˈnʌmbə/ | 氧化数 | yǎng huà shù |
| oxidising agent/ˈɒksɪdaɪzɪŋ ˈeɪdʒənt/ | 氧化剂 | yǎng huà jì |
| reducing agent/rɪˈdjuːsɪŋ ˈeɪdʒənt/ | 还原剂 | huán yuán jì |
| potassium manganate(VII)/pəˈtæsɪəm ˈmæŋɡəneɪt/ | 高锰酸钾 | gāo měng suān jiǎ |
| potassium iodide/pəˈtæsɪəm ˈaɪədaɪd/ | 碘化钾 | diǎn huà jiǎ |
| iodine/ˈaɪədiːn/ | 碘 | diǎn |
6.4
考试技巧
- Explain a faster rate with collision theory: successful collisions happen more often. Concentration, pressure and surface area all raise how often particles collide; a higher temperature does that and gives more particles enough energy.
- A catalyst lowers the activation energy and speeds the reaction up, but is not used up and does not change $\Delta H$ or the position of equilibrium.
- At equilibrium the forward and reverse rates are equal and the concentrations stop changing — it does not mean the amounts of reactant and product are equal.
- Use Le Chatelier's principle: raising temperature favours the endothermic direction; compressing gases favours the side with fewer gas molecules, if the numbers differ; adding a reacting substance shifts the equilibrium to consume some of it.
- Redox — OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons. They always happen together, and an oxidising agent is itself reduced.
该知识点的互动课程
逐步完成,配合即时检查练习。