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Chemical reactions

IGCSE Chemistry · Topic 6

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6.1

Physical and chemical changes

Syllabus
Core Supplement
1 Identify physical and chemical changes, and describe the differences between them

Source: Cambridge International syllabus

Rust on an iron surface Iron rusting is a chemical change — it forms a new substance.

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
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Physical change vs chemical change

Step through the difference. A physical change makes no new substance and is usually easy to reverse; a chemical change makes a new substance and is hard to undo.

Vocabulary Train
English Chinese Pinyin
physical change 物理变化 wù lǐ biàn huà
chemical change 化学变化 huà xué biàn huà
precipitate 沉淀 chén diàn
6.2

Rate of reaction

Syllabus
Core Supplement
5 Describe collision theory in terms of: (a) number of particles per unit volume (b) frequency of collisions between particles (c) kinetic energy of particles (d) activation energy, $E_a$
1 Describe the effect on the rate of reaction of: (a) changing the concentration of solutions (b) changing the pressure of gases (c) changing the surface area of solids (d) changing the temperature (e) adding or removing a catalyst, including enzymes 6 Describe and explain the effect on the rate of reaction of: (a) changing the concentration of solutions (b) changing the pressure of gases (c) changing the surface area of solids (d) changing the temperature (e) adding or removing a catalyst, including enzymes using collision theory
2 State that a catalyst increases the rate of a reaction and is unchanged at the end of a reaction 7 State that a catalyst decreases the activation energy, $E_a$, of a reaction
3 Describe practical methods for investigating the rate of a reaction including change in mass of a reactant or a product and the formation of a gas 8 Evaluate practical methods for investigating the rate of a reaction including change in mass of a reactant or a product and the formation of a gas
4 Interpret data, including graphs, from rate of reaction experiments

Source: Cambridge International syllabus

Collision theory: energy and orientation

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

Two boxes of particles: the higher-concentration box holds more particles, so collisions happen more often More particles in the same volume collide more often, so the rate is faster

A big lump touched by acid only on its outside, beside the same solid as a powder that acid can touch all over Breaking a solid into a powder exposes much more surface, so the rate is faster

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 are biological catalysts (catalysts that work in living things).

A reaction pathway with a high hill without a catalyst and a lower hill with a catalyst, both ending at the same products A catalyst gives a lower activation energy so more collisions succeed; $\Delta H$ is unchanged

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.

A conical flask of reaction mixture connected by a delivery tube to a gas syringe that fills with gas A gas syringe collects the gas given off; read its volume at regular times to follow the rate

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.

A graph of gas volume against time: steep at first then levelling off, with a faster and a slower curve reaching the same final amount The rate is fastest at the start (steepest) and the line levels off when the reaction finishes

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Rate from a gas-volume graph

The gradient of the volume–time curve is the rate; it is steepest at the start and flattens as reactants run out.

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Rate of reaction

conc = c₀·b

Concentration falls over time — fast at first, then slower.

Explore

Collision theory in action

Watch the particles collide. Heat speeds them up so they collide more often and harder; more particles or a catalyst give more successful collisions — the rate climbs.

Vocabulary Train
English Chinese Pinyin
rate of reaction 反应速率 fǎn yìng sù lǜ
reactants 反应物 fǎn yìng wù
products 生成物 shēng chéng wù
collision theory 碰撞理论 pèng zhuàng lǐ lùn
particles 粒子 lì zi
collide 碰撞 pèng zhuàng
activation energy 活化能 huó huà néng
concentration 浓度 nóng dù
pressure 压强 yā qiáng
surface area 表面积 biǎo miàn jī
kinetic energy 动能 dòng néng
catalyst 催化剂 cuī huà jì
enzymes méi
syringe 注射器 zhù shè qì
6.3

Reversible reactions and equilibrium

Syllabus
Core Supplement
1 State that some chemical reactions are reversible as shown by the symbol $\rightleftharpoons$ 3 State that a reversible reaction in a closed system is at equilibrium when: (a) the rate of the forward reaction is equal to the rate of the reverse reaction (b) the concentrations of reactants and products are no longer changing
2 Describe how changing the conditions can change the direction of a reversible reaction for: (a) the effect of heat on hydrated compounds (b) the addition of water to anhydrous compounds limited to copper(II) sulfate and cobalt(II) chloride 4 Predict and explain, for a reversible reaction, how the position of equilibrium is affected by: (a) changing temperature (b) changing pressure (c) changing concentration (d) using a catalyst using information provided
5 State the symbol equation for the production of ammonia in the Haber process, $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$
6 State the sources of the hydrogen (methane) and nitrogen (air) in the Haber process
7 State the typical conditions in the Haber process as $450\text{ }^{\circ}\text{C}$, $20\,000\text{ kPa}/200\text{ atm}$ and an iron catalyst
8 State the symbol equation for the conversion of sulfur dioxide to sulfur trioxide in the Contact process, $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$
9 State the sources of the sulfur dioxide (burning sulfur or roasting sulfide ores) and oxygen (air) in the Contact process
10 State the typical conditions for the conversion of sulfur dioxide to sulfur trioxide in the Contact process as $450\text{ }^{\circ}\text{C}$, $200\text{ kPa}/2\text{ atm}$ and a vanadium(V) oxide catalyst
11 Explain, in terms of rate of reaction and position of equilibrium, why the typical conditions stated are used in the Haber process and in the Contact process, including safety considerations and economics

Source: Cambridge International syllabus

Le Chatelier's principle
Dynamic equilibrium: the rates converge

Some reactions are reversible 可逆反应: the products can react to form the reactants again. We show this with the symbol $\rightleftharpoons$.

Blue hydrated copper sulfate loses water on heating to white anhydrous; adding water turns it blue again Heating blue copper sulfate drives off water; adding water turns it blue again

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 nothing enters or leaves), 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): more pressure moves the equilibrium to the side with fewer gas molecules.
  • 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:

$$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$
  • 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 catalyst.

The Contact process

The Contact process 接触法 turns sulfur dioxide into sulfur trioxide:

$$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$$
  • The sulfur dioxide comes from burning sulfur (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:

  • A higher pressure would give more product, but very high pressure is dangerous and expensive, so a medium pressure is used.
  • 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.
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Shifting an equilibrium

Change temperature, pressure or concentration and watch a reversible reaction shift to a new balance.

Vocabulary Train
English Chinese Pinyin
reversible reaction 可逆反应 kě nì fǎn yìng
hydrated 水合 shuǐ hé
anhydrous 无水 wú shuǐ
closed system 密闭系统 mì bì xì tǒng
equilibrium 平衡 píng héng
forward reaction 正反应 zhèng fǎn yìng
reverse reaction 逆反应 nì fǎn yìng
position of equilibrium 平衡位置 píng héng wèi zhì
endothermic 吸热反应 xī rè fǎn yìng
exothermic 放热反应 fàng rè fǎn yìng
Haber process 哈伯法 hā bó fǎ
hydrogen 氢气 qīng qì
methane 甲烷 jiǎ wán
nitrogen 氮气 dàn qì
iron tiě
Contact process 接触法 jiē chù fǎ
sulfur liú
oxygen 氧气 yǎng qì
vanadium(V) oxide 五氧化二钒 wǔ yǎng huà èr fán
compromise 折中 zhé zhōng
6.4

Redox

Syllabus
Core Supplement
1 Use a Roman numeral to indicate the oxidation number of an element in a compound
2 Define redox reactions as involving simultaneous oxidation and reduction
3 Define oxidation as gain of oxygen and reduction as loss of oxygen 6 Define oxidation in terms of: (a) loss of electrons (b) an increase in oxidation number
7 Define reduction in terms of: (a) gain of electrons (b) a decrease in oxidation number
4 Identify redox reactions as reactions involving gain and loss of oxygen 8 Identify redox reactions as reactions involving gain and loss of electrons
5 Identify oxidation and reduction in redox reactions 9 Identify redox reactions by changes in oxidation number using: (a) the oxidation number of elements in their uncombined state is zero (b) the oxidation number of a monatomic ion is the same as the charge on the ion (c) the sum of the oxidation numbers in a compound is zero (d) the sum of the oxidation numbers in an ion is equal to the charge on the ion
10 Identify redox reactions by the colour changes involved when using acidified aqueous potassium manganate(VII) or aqueous potassium iodide
11 Define an oxidising agent as a substance that oxidises another substance and is itself reduced
12 Define a reducing agent as a substance that reduces another substance and is itself oxidised
13 Identify oxidising agents and reducing agents in redox reactions

Source: Cambridge International syllabus

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.

One substance passing electrons to another, the first oxidised and the second reduced Electrons pass from the substance that is oxidised (loses them) to the one that is reduced (gains them) — the two always happen together

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 is formed.

A rack of test tubes of potassium manganate(VII), going from deep purple on the right to colourless on the left as it gets more dilute Potassium manganate(VII) is deep purple; as it is reduced (or diluted) the purple fades to colourless

Explore

Oxidation and reduction happen together

Step through a redox reaction. One substance loses electrons while another gains them — you can never have one without the other.

Vocabulary Train
English Chinese Pinyin
oxidation 氧化 yǎng huà
reduction 还原 huán yuán
redox reaction 氧化还原反应 yǎng huà huán yuán fǎn yìng
electrons 电子 diàn zi
oxidation number 氧化数 yǎng huà shù
oxidising agent 氧化剂 yǎng huà jì
reducing agent 还原剂 huán yuán jì
potassium manganate(VII) 高锰酸钾 gāo měng suān jiǎ
potassium iodide 碘化钾 diǎn huà jiǎ
iodine diǎn
6.4

Exam tips

  • 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; raising pressure favours the side with fewer gas molecules; adding a substance shifts the equilibrium away from it.
  • Redox — OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons. They always happen together, and an oxidising agent is itself reduced.

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