Pearson Edexcel · International GCSE · Chemistry

The language of chemistry

Everything in this course is written in one language: particles, formulae and equations. Section 1 teaches it — how particles arrange themselves in solids, liquids and gases; how atoms are built and lined up in the Periodic Table; how to turn a sentence about a reaction into a balanced equation and a mass calculation; and how the three kinds of bonding give every substance its properties.

Edexcel International GCSE Chemistry 4CH1, Section 1 (statements 1.1–1.60C) is the largest section and the foundation for the other three. C-suffixed statements are the separate-science (Chemistry-only) extension.

States of matter and solutions

Solid: particles vibrate about fixed positions in a regular pattern. Liquid: particles touch but move past each other at random. Gas: particles far apart, fast, random. Interconversions: melting 熔化 / freezing 凝固, boiling 沸腾 / condensation 凝结, evaporation 蒸发 (surface only, below boiling), sublimation 升华 (solid → gas directly). Energy is absorbed to separate particles and released as they bond again.

A solvent 溶剂 dissolves the solute 溶质 to make a solution 溶液; a saturated solution 饱和溶液 holds all the solute it can at that temperature. (C) Solubility is the mass dissolved per 100 g of solvent at a stated temperature, read from a solubility curve 溶解度曲线 — points on the line are saturated; below it, unsaturated.

Dilution of coloured solutions and gas diffusion (bromine spreading through air) both show particles moving from where they are many to where they are few.

Elements, compounds, mixtures and separation

An element 元素 contains one type of atom; a compound 化合物 holds different atoms chemically joined; a mixture 混合物 is not chemically joined. A pure substance has a sharp melting/boiling point; mixtures melt and boil over a range — the test for purity.

Separation techniques, matched to the job: filtration 过滤 (insoluble solid from liquid); crystallisation 结晶 (evaporate a solution to leave crystals); simple distillation 蒸馏 (liquid from dissolved solid — evaporate then condense); fractional distillation 分馏 (liquids with close boiling points, using a fractionating column); paper chromatography 纸层析 (solutes that dissolve differently in a solvent).

Rf = distance moved by the spot ÷ distance moved by the solvent. The same substance always has the same Rf in the same solvent — comparing with known values identifies the components of the mixture.

Atoms and the Periodic Table

Atomic number 质子数 = protons; mass number 质量数 = protons + neutrons. Isotopes 同位素 have the same protons but different neutrons. Relative atomic mass ($A_r$) is the weighted mean over the isotopes.

Worked check. Chlorine: 75 % Cl-35, 25 % Cl-37 → $A_r = (75 \times 35 + 25 \times 37)/100 = 35.5$.

The Periodic Table lists elements by atomic number into groups 族 (columns, same outer electrons) and periods 周期 (rows). The first 20 electronic configurations follow 2,8,8,2 shells (e.g. sodium 2.8.1). Metals sit left/centre: their oxides are basic and they conduct; non-metal oxides are acidic. Group members behave alike because they have the same number of outer electrons; Group 0 noble gases have full outer shells and do not readily react.

Formulae, equations and calculations

Balance equations with state symbols 状态符号: (s), (l), (g), (aq). $M_r$ sums the $A_r$ values. The mole 摩尔 is the amount of substance: $n = m/M_r$ (mass in grams).

Worked check (reacting masses). What mass of magnesium oxide forms from 6.0 g of magnesium? $2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}$: $n(\mathrm{Mg}) = 6.0/24 = 0.25$ mol → $n(\mathrm{MgO}) = 0.25$ mol → mass $= 0.25 \times 40 = 10$ g.

Percentage yield 产率 $=$ actual ÷ theoretical × 100 %. Empirical formula 实验式 is the simplest whole-number ratio (divide each mass by its $A_r$, scale to integers); the molecular formula is the true count ($M_r$ ÷ empirical mass gives the multiplier).

Worked check. A hydrocarbon is 85.7 % C, 14.3 % H: C 85.7/12 = 7.14, H 14.3/1 = 14.3 → ratio 1:2 → empirical CH₂. With $M_r = 56$: 56/14 = 4 → molecular C₄H₈.

(C) Solutions: $n = c \times V$ with $c$ in mol/dm³ and $V$ in dm³ (cm³ ÷ 1000). Gases: molar volume 24 dm³ at rtp, so $n = V/24$. (C-practical) A metal oxide's formula comes from combustion (weigh magnesium, burn, weigh the oxide) or reduction (weigh copper(II) oxide, reduce with hydrogen, weigh the copper).

Bonding

Ionic bonding 离子键: metals lose electrons to non-metals; the charged ions attract electrostatically in a giant lattice — dot-and-cross diagrams show the transfer (Groups 1–3 with 5–7). High melting points (strong attractions); conduct only when molten or dissolved (ions free to move). Learn the charges: Groups 1/2/3 → 1+/2+/3+; Groups 5/6/7 → 3−/2−/1−; Ag⁺, Cu²⁺, Fe²⁺, Fe³⁺, Pb²⁺, Zn²⁺; OH⁻, NH₄⁺, CO₃²⁻, NO₃⁻, SO₄²⁻. Formulae balance the charges: calcium nitrate Ca(NO₃)₂, ammonium sulfate (NH₄)₂SO₄.

Covalent bonding 共价键: shared pairs of electrons (dot-and-cross for H₂, O₂, N₂, halogens, HCl, H₂O, NH₃, CO₂ and small organics). Simple molecular substances have weak intermolecular forces — low melting points, rising with $M_r$; they do not conduct (no ions, no free electrons). Giant covalent: diamond (each C bonded to 4 — very hard, non-conductor), graphite (3 bonds + one free electron per C, layers with weak forces — soft, conducts, slippery), C₆₀ fullerene (simple molecular cage).

(C) Metallic: a lattice of positive ions in a sea of delocalised electrons — electrostatic attraction; the free electrons conduct electricity, and layers slide (malleable).

Electrolysis (C)

Electrolysis 电解 uses electricity to split molten or dissolved ionic compounds: the cathode 阴极 (negative) attracts cations 阳离子; the anode 阳极 (positive) attracts anions 阴离子. Molten PbBr₂ gives Pb at the cathode (Pb²⁺ + 2e⁻ → Pb) and Br₂ at the anode (2Br⁻ → Br₂ + 2e⁻). In aqueous solution, hydrogen forms at the cathode if the metal is more reactive than hydrogen; at the anode, halide ions give the halogen, otherwise oxygen (4OH⁻ → O₂ + 2H₂O + 4e⁻). Loss of electrons is oxidation 氧化, gain is reduction 还原.

Check yourself

1. Describe the particle arrangement and movement in a liquid.
2. Define saturated solution, and state the units of solubility.
3. Which technique separates two miscible liquids with close boiling points?
4. An isotope has 17 protons and 20 neutrons. Give its mass number and the numbers of each particle.
5. Calculate $A_r$ of an element with 80 % of mass 20 and 20 % of mass 22.
6. Balance with state symbols: magnesium + oxygen → magnesium oxide.
7. 8.4 g of the oxide of calcium forms from 6.0 g of calcium. Show the empirical formula of the oxide.
8. Write the formula of ammonium carbonate.
9. Explain why graphite conducts but diamond does not.
10. (C) What forms at each electrode in the electrolysis of molten potassium chloride?

Answers: 1 particles touch but are disordered, moving randomly past each other; 2 a solution containing the maximum mass of dissolved solute at that temperature; g per 100 g of solvent; 3 fractional distillation; 4 mass number 37; 17 protons, 20 neutrons, 17 electrons; 5 $(80 \times 20 + 20 \times 22)/100 = 20.4$; 6 2Mg(s) + O₂(g) → 2MgO(s); 7 Ca 6.0/40 = 0.15 mol; O mass 8.4 − 6.0 = 2.4 g → 2.4/16 = 0.15 mol; ratio 1:1 → empirical formula CaO; 8 (NH₄)₂CO₃; 9 graphite uses three bonds per carbon, leaving one delocalised electron per carbon free to move between layers — diamond uses all four in fixed bonds; 10 (C) potassium at the cathode (K⁺ + e⁻ → K) and chlorine at the anode (2Cl⁻ → Cl₂ + 2e⁻).

El mundo reactivo de la química inorgánica

Si introduces litio en agua, effervesce suavemente; si introduces potasio, se enciende. Esta sección sigue los patrones detrás de esas observaciones — las familias de la Tabla Periódica, el aire que respiramos, la escalera de reactividad que determina cómo se extrae cada metal y por qué el hierro se oxida — y luego el enorme centro práctico del curso: ácidos, bases, preparación de sales e identificación de pruebas.

Edexcel International GCSE Chemistry 4CH1, Sección 2 (declaraciones 2.1–2.50) proporciona la mayoría de las preguntas de recuerdo y prácticas en ambos exámenes. Las declaraciones con sufijo C son solo de Química.

Grupo 1 y Grupo 7

Grupo 1 (Li, Na, K) reaccionan con el agua para formar el hidróxido y el hidrógeno — efervescencia, flotación, fusión (Na), llama lila (K). Las reacciones se vuelven más rápidas y vigorosas hacia abajo del grupo: una familia con una tendencia. (C) La reactividad aumenta porque el electrón externo está más lejos del núcleo y se pierde con mayor facilidad.

Grupo 7 (Cl₂ gas verde pálido, Br₂ líquido naranja-marrón, I₂ sólido gris-negro) oscurecen hacia abajo del grupo. Un halógeno más reactivo desplaza la sal de uno menos reactivo: el cloro desplaza al bromo del bromuro de potasio (la solución se vuelve naranja). (C) La reactividad disminuye hacia abajo del grupo — la capa externa está más lejos, por lo que ganar un electrón es más difícil.

Aire y gases

Aire seco: aproximadamente 78 % nitrógeno, 21 % oxígeno, 0.96 % argón, 0.04 % dióxido de carbono (los cuatro más abundantes). El porcentaje de oxígeno puede medirse consumiéndolo — oxidación del hierro en aire húmedo sobre agua, o combustión de fósforo, por lo que el agua sube aproximadamente un quinto. Los elementos arden en oxígeno: el magnesio produce una llama blanca brillante y MgO blanco; el hidrógeno arde con un pop hasta vapor; el azufre da una llama azul y SO₂ sofocante. El carbonato de cobre(II) se descompone al calentar: CuCO₃ → CuO + CO₂ (el agua de cal se vuelve lechosa). CO₂ es un gas de efecto invernadero cuyo aumento contribuye al cambio climático.

Serie de reactividad y metales

Orden para aprender: potasio, sodio, litio, calcio, magnesio, aluminio, (carbono), zinc, hierro, (hidrógeno), cobre, plata, oro. Los metales por encima del hidrógeno — todo hasta el hierro — reaccionan con HCl/H₂SO₄ diluidos dando hidrógeno; cuanto más reactivo, más rápido la efervescencia. Un metal desplaza cualquier metal menos reactivo de su solución salina u óxido.

Óxido necesita oxígeno y agua; se previene mediante métodos de barrera (pintura, aceite, plástico), galvanizado (capa de zinc) y protección sacrificial (un metal más reactivo como zinc o magnesio se corroe en su lugar). Oxidación es pérdida de electrones o ganancia de oxígeno; reducción lo contrario; una reacción redox tiene ambas; el agente oxidante se reduce, el agente reductor se oxida.

(C) Extracción coincide con la reactividad: metales poco reactivos (oro, plata) ocurren nativos; aquellos por debajo del carbono (hierro) son reducidos por carbono en un alto horno; aquellos por encima (aluminio) necesitan electrólisis de mineral fundido. El aluminio (baja densidad, resiste la corrosión), el cobre (conduce), el hierro y los aceros (fuertes; el inoxidable resiste el óxido) se eligen por sus propiedades. Aleaciones mezclan un metal con otros elementos y son más duras porque los átomos de diferente tamaño alteran el deslizamiento de las capas.

Ácidos, bases y sales

Los ácidos proporcionan H⁺(aq); las bases solubles proporcionan OH⁻(aq). Indicadores: litmo (rojo en ácido, azul en base soluble), fenolftaleína (incoloro → rosa), naranja de metilo (rojo → amarillo). La escala de pH pH 0–14 clasifica soluciones (0–3 fuertemente ácidas … 11–14 fuertemente alcalinas); el indicador universal mide el pH aproximado. Neutralización: ácido + base soluble → sal + agua. (C) Los ácidos son donantes de protones, las bases son aceptores de protones; las bases solubles son bases solubles. (C) Titulación: volumen fijo de base soluble con indicador en un matraz, ácido desde una bureta gota a gota hasta el cambio de color, repetir para resultados concordantes.

Reglas de solubilidad: sales de sodio, potasio y amonio — solubles; todos los nitratos — solubles; cloruros — solubles excepto AgCl y PbCl₂; sulfatos — solubles excepto BaSO₄, CaSO₄ y PbSO₄; carbonatos — insolubles excepto Na⁺, K⁺, NH₄⁺; hidróxidos — insolubles excepto Na⁺, K⁺ (Ca(OH)₂ ligeramente soluble).

Ácido + metal / base / carbonato produce una sal: p. ej. Zn + H₂SO₄ → ZnSO₄ + H₂; CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O. Sal soluble a partir de un reactivo insoluble (CuO + H₂SO₄ diluido tibia): añadir exceso de base, filtrar, evaporar, cristalizar. (C) A partir de ácido + base soluble: titular, luego evaporar. (C) Sal insoluble: mezclar dos soluciones para que se forme un precipitado (Pb(NO₃)₂ + Na₂SO₄ → PbSO₄↓ + 2NaNO₃), filtrar, lavar, secar.

Pruebas químicas

Gases: hidrógeno — una cerilla encendida hace pop; oxígeno — aviva una cerilla incandescente; dióxido de carbono — enturbia el agua de cal; amoníaco — torna el litmo húmedo rojo en azul; cloro — decolora el papel de litmo húmedo a blanco. Cationes: pruebas de llama (Li⁺ rojo, Na⁺ amarillo, K⁺ lila, Ca²⁺ naranja-rojo, Cu²⁺ azul-verde); NH₄⁺ calienta con NaOH(aq) dando amoníaco; Cu²⁺/Fe²⁺/Fe³⁺ dan precipitados azul/verde/marrón oxidoso con NaOH(aq). Aniones: Cl⁻/Br⁻/I⁻ dan precipitados blanco/crema/amarillo pálido con nitrato de plata acidificado; SO₄²⁻ da un precipitado blanco con cloruro de bario acidificado; CO₃²⁻ burbujea con HCl diluido dando CO₂. Agua: torna el sulfato de cobre(II) anhidro blanco → azul; una muestra de agua pura hierve bruscamente a 100 °C.

Compruébalo tú mismo

1. Describe la tendencia en la reactividad hacia abajo del Grupo 1 y da la razón (C) electrónica.
2. ¿Qué se observa al añadir agua de cloro a bromuro de potasio y por qué?
3. Nombra los cuatro gases más abundantes en el aire seco con porcentajes aproximados.
4. Establece las dos condiciones necesarias para que el hierro se oxide y dos métodos de prevención.
5. Define oxidación y reducción en términos de electrones.
6. (C) ¿Por qué se extrae el aluminio por electrólisis pero el hierro por reducción con carbono?
7. ¿Cuáles cloruros son insolubles y cuáles sulfatos?
8. Escribe la ecuación para el óxido de cobre(II) reaccionando con ácido sulfúrico.
9. Describe los colores de la prueba de llama para sodio, potasio y cobre.
10. Describe la prueba para iones sulfato y el resultado positivo.

Respuestas: 1 la reactividad aumenta hacia abajo del grupo (B1); (C) el electrón externo está en una capa más lejos del núcleo, por lo que se pierde más fácilmente; 2 la solución se vuelve naranja — el cloro desplaza al bromo porque es el halógeno más reactivo; 3 nitrógeno ≈78 %, oxígeno ≈21 %, argón ≈0.96 %, dióxido de carbono ≈0.04 %; 4 oxígeno y agua; métodos de barrera (pintura/grasa/plástico) y galvanizado o zinc sacrificial; 5 la oxidación es pérdida de electrones, la reducción es ganancia; 6 (C) el aluminio se sitúa por encima del carbono en la serie de reactividad, por lo que el carbono no puede reducir su óxido — se necesita electrólisis de alúmina fundida; el hierro se sitúa por debajo del carbono, por lo que el carbono lo desplaza de su óxido; 7 AgCl y PbCl₂; BaSO₄, CaSO₄ y PbSO₄; 8 CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l); 9 sodio amarillo, potasio lila, cobre azul-verde; 10 acidificar con ácido clorhídrico diluido, añadir solución de cloruro de bario — un precipitado blanco de BaSO₄ confirma el sulfato.

Energy, speed and balance

Three questions decide whether a reaction is useful: does it heat or cool, how fast is it, and does it finish? Physical chemistry answers them — calorimetry and Q = mcΔT, collision theory with its four levers, and the dynamic equilibrium a reversible reaction reaches in a closed vessel.

Edexcel International GCSE Chemistry 4CH1, Section 3 (statements 3.1–3.22C) is short but calculation-heavy. C-suffixed statements are Chemistry-only.

Energetics

Exothermic 放热 reactions give out heat (combustion, neutralisation, most displacements) — the mixture warms, ΔH is negative. Endothermic 吸热 reactions take heat in (thermal decomposition, some dissolving) — the mixture cools. Simple calorimetry 量热: measure the temperature change of a known mass (polystyrene cup for solutions; spirit burner under a water beaker for combustion).

Worked check. 50 g of water rises 24 °C in a spirit-burner trial: $Q = 50 \times 4.2 \times 24 = 5040$ J. If 0.010 mol of fuel burned: $\Delta H = -5040/0.010 = -504\,000\ \mathrm{J\,mol^{-1}} = -504\ \mathrm{kJ\,mol^{-1}}$.

(C) Bond breaking is endothermic; bond making is exothermic. ΔH = bonds broken − bonds formed. Worked check. H₂ + Cl₂ → 2HCl: broken 436 + 242 = 678; formed 2 × 431 = 862; ΔH = 678 − 862 = −184 kJ/mol (exothermic).

Rates of reaction

Experiments follow a reaction that is visible: gas volume (marble + acid), mass loss on a balance, or disappearing cross (sodium thiosulfate + acid). Four levers: surface area of a solid, concentration (or gas pressure), temperature, and a catalyst 催化剂.

Collision theory 碰撞理论 explains them all: reactions need collisions with at least the activation energy 活化能. Powder gives more exposed surface → more collisions per second; higher concentration or pressure packs particles closer → more frequent collisions; higher temperature makes particles faster → collisions both more frequent and more energetic — the only lever that raises the fraction of successful collisions. A catalyst provides an alternative pathway with lower activation energy — more collisions succeed — and is chemically unchanged at the end; it does not change the amount of product, only how fast it appears.

Reversible reactions and equilibria

A reversible reaction 可逆反应 (⇌) runs both ways: hydrated copper(II) sulfate ⇌ anhydrous salt + water (heat drives it forward, water reverses it); ammonium chloride ⇌ ammonia + hydrogen chloride on heating. (C) In a sealed container it reaches dynamic equilibrium 动态平衡: forward and reverse rates are equal, and all concentrations stay constant — both reactions still run.

(C) Temperature up → equilibrium shifts in the endothermic direction; pressure up → shifts towards the side with fewer moles of gas; a catalyst speeds both directions equally and leaves the position unchanged.

Check yourself

1. Classify as exo- or endothermic: combustion, thermal decomposition, neutralisation.
2. 100 g of water cools by 8.0 °C in a dissolving experiment. Calculate Q.
3. (C) State the sign of ΔH for bond breaking and bond making.
4. Use bond energies to find ΔH for 2H₂ + O₂ → 2H₂O (H–H 436, O=O 498, O–H 463 kJ/mol).
5. Why does powdered marble react faster with acid than one large chip?
6. Which rate lever increases the fraction of successful collisions, and why?
7. Define a catalyst.
8. (C) State the two characteristics of dynamic equilibrium.
9. (C) N₂ + 3H₂ ⇌ 2NH₃: what does raising the pressure do, and why?
10. (C) Why does a catalyst not change the position of equilibrium?

Answers: 1 exothermic (combustion), endothermic (thermal decomposition), exothermic (neutralisation); 2 $Q = 100 \times 4.2 \times 8.0 = 3360$ J taken in — the water cools, so the dissolving is endothermic; 3 (C) breaking positive (endothermic), making negative (exothermic); 4 broken: 2(436) + 498 = 1370; formed: 4(463) = 1852; ΔH = 1370 − 1852 = −482 kJ/mol (per 2 mol H₂O); 5 powder exposes far more surface, so more acid collisions per second; 6 temperature — faster particles collide more often AND more collisions exceed the activation energy; 7 a substance that increases the rate but is chemically unchanged at the end; 8 (C) forward and reverse rates equal; concentrations of all substances constant; 9 (C) shifts right — 4 moles of gas on the left, 2 on the right, so higher pressure favours the fewer-mole side.

Answers continued: 10 (C) it lowers the activation energy of forward and reverse equally, speeding both without changing their balance.

Carbon's family tree

Crude oil is a mixture of hydrocarbons — compounds of carbon and hydrogen only — built from the smallest molecules in the gas fraction to the tars of bitumen. This section organises them into families, follows crude oil from well to fraction, and then meets each family in turn: the unreactive alkanes, the double-bonded alkenes, and (C) the alcohols, acids, esters and polymers that come after.

Edexcel International GCSE Chemistry 4CH1, Section 4 (statements 4.1–4.50C) closes the course. C-suffixed statements are Chemistry-only.

Introduction and crude oil

A hydrocarbon 烃 holds hydrogen and carbon only. Molecules are drawn as empirical, molecular, general, structural and displayed formulae 结构式. A homologous series 同系列 shares a general formula and functional group 官能团, members differing by CH₂; isomerism 同分异构现象 — isomers 同分异构体 share a molecular formula with different structures. IUPAC naming runs to six carbons.

Crude oil separates by fractional distillation 分馏: the furnace vaporises the oil, and vapours rise up the column, cooling as they go — small molecules with low boiling points collect at the top, large viscous molecules at the bottom.

Fractions darken and thicken down the column (rising boiling point and viscosity). Complete combustion gives CO₂ + H₂O; incomplete combustion 不完全燃烧 also gives carbon monoxide — poisonous because it reduces the blood's oxygen-carrying capacity — and soot. In engines, hot air makes nitrogen and oxygen combine into oxides of nitrogen; sulfur impurities burn to SO₂; both gases dissolve in rain making it acidic (acid rain damages leaves, lakes and buildings). Cracking 裂解 converts long-chain alkanes to alkenes + shorter alkanes over silica/alumina at 600–700 °C — because the refinery's supply of long fractions outstrips demand while short fractions are wanted.

Alkanes and alkenes

Alkanes 烷烃 ($C_nH_{2n+2}$) are saturated 饱和 — only single C–C bonds. Methane, ethane, propane, butane, pentane. With halogens in ultraviolet light they substitute one hydrogen: CH₄ + Cl₂ → CH₃Cl + HCl (substitution 取代).

Alkenes 烯烃 ($C_nH_{2n}$) carry C=C — unsaturated 不饱和. With bromine they add across the double bond to a dibromoalkane (addition 加成): C₂H₄ + Br₂ → C₂H₄Br₂. Bromine water distinguishes them: decolourised (orange → colourless) by an alkene, unchanged by an alkane.

Alcohols, acids and esters (C)

Alcohols 醇 carry –OH: methanol, ethanol, propan-1-ol, butan-1-ol. Ethanol oxidises to ethanoic acid by complete combustion, by microbial oxidation in air, or by warming with potassium dichromate(VI) in dilute sulfuric acid. Ethanol is made by ethene + steam over a phosphoric acid catalyst (≈300 °C, 60–70 atm) or by fermentation 发酵 of glucose with yeast enzymes in the absence of air at ≈30 °C — the optimum for the enzymes, and no oxygen so the yeast produces ethanol not water.

Carboxylic acids 羧酸 carry –COOH: methanoic, ethanoic, propanoic, butanoic. Their solutions react with metals (giving H₂) and carbonates (giving CO₂) like other acids; vinegar is dilute ethanoic acid.

Esters 酯 carry –COO–: ethanol + ethanoic acid, with an acid catalyst, give ethyl ethanoate 乙酸乙酯 and water (esterification). Given an alcohol and an acid, join the acid's –COOH to the alcohol's –O–. Esters are volatile, sweet-smelling, used as flavourings and perfumes.

Synthetic polymers

Addition polymers 加成聚合物 form when many alkene monomers 单体 join, opening their C=C bonds.

Draw the repeat unit of poly(ethene), poly(propene), poly(chloroethene), PTFE; from a repeat unit, close the two bonds back into a C=C to recover the monomer. Disposal problems: polymers are inert and do not biodegrade, and burning can release toxic gases. (C) Condensation polymerisation: a dicarboxylic acid + a diol link ester bonds, releasing water — a polyester 聚酯; some (biopolyesters) biodegrade.

Check yourself

1. Define hydrocarbon, homologous series and isomerism.
2. Name the six main fractions of crude oil in boiling-point order.
3. Write the products of complete and incomplete combustion of a hydrocarbon.
4. Give two pollutant gases from burning fuels and one source of each.
5. Why is cracking necessary?
6. Write the equation for methane reacting with chlorine in UV light, naming the reaction type.
7. How does bromine water distinguish an alkene from an alkane?
8. (C) Name the two methods of manufacturing ethanol with one condition for each.
9. (C) Which alcohol and acid make ethyl ethanoate, and what else forms?
10. (C) What monomers make a polyester, and what small molecule is released?

Answers: 1 a compound of hydrogen and carbon only; a family with the same general formula and functional group, differing by CH₂; molecules with the same molecular formula but different structures; 2 refinery gases, gasoline, kerosene, diesel, fuel oil, bitumen; 3 complete: carbon dioxide + water; incomplete adds carbon monoxide (and soot); 4 carbon monoxide — incomplete combustion; sulfur dioxide — burning sulfur impurities; oxides of nitrogen — hot engine air; 5 supply of long-chain fractions exceeds demand while short fractions and alkenes are in short supply; 6 CH₄ + Cl₂ → CH₃Cl + HCl — substitution (photo, UV); 7 the alkene decolourises orange bromine water; the alkane leaves it orange; 8 (C) ethene + steam, phosphoric acid catalyst, ~300 °C, 60–70 atm; fermentation of glucose with yeast, no air, ~30 °C; 9 (C) ethanol + ethanoic acid — water also forms (acid catalyst); 10 (C) a dicarboxylic acid and a diol; water is released at each ester link.