中文（繁體）
中文（简体）
English
Form 5 Chemistry Practice Paper: Organic Chemistry, Polymers and Reaction Rate (12 questions, 100 marks, 75 minutes)
Form 5 Chemistry – 12 questions – 100 marks – 75 minutes
Form 5 Chemistry: 12 questions, 100 marks, 75 minutes on alkanes, alkenes, esterification, polymers and reaction rate, with full answers for DSE revision.
Download Question paper PDF | Answer key PDF
Paper details Form 5 Chemistry practice paper: organic chemistry, polymers and reaction rate, 12 questions, 100 marks, 75 minutes, with full answers and marking points.
Section A: Alkanes, alkenes and empirical formulae of organic compounds (Q1-Q4)
Show the formula, substitute the given data and state units; balance all equations and include state symbols.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Pentane (C5H12) is the simplest alkane containing five carbon atoms. (a) State the general formula of the alkane homologous series and draw the structural formula of pentane. (3 marks) (b) Calculate the relative molecular mass of pentane (C=12.0, H=1.0). (2 marks) (c) Write the equation for the complete combustion of pentane and calculate the number of moles of oxygen needed to burn 1.0 mol of pentane completely. (3 marks) | (a) CnH2n+2; structural formula CH3-CH2-CH2-CH2-CH3. (b) 5 x 12.0 + 12 x 1.0 = 72.0. (c) C5H12 + 8O2 -> 5CO2 + 6H2O; oxygen needed = 5 + 12/4 = 8.0 mol. |
| 2 | 28.0 g of ethene (C2H4) is bubbled into bromine water. (a) Calculate the number of moles of ethene (C=12.0, H=1.0). (2 marks) (b) Write the equation for the addition reaction with bromine and calculate the mass of bromine required (Br=80.0). (3 marks) (c) Describe the observation and explain why bromine water can distinguish an alkane from an alkene. (3 marks) | (a) n = 28.0 / 28.0 = 1.0 mol. (b) C2H4 + Br2 -> C2H4Br2; mass of Br2 = 1.0 x 160.0 = 160 g. (c) The orange colour of bromine water decolorises rapidly in the dark. The C=C bond undergoes addition; an alkane only reacts by slow substitution under UV light, so bromine water distinguishes them. |
| 3 | 0.30 g of a hydrocarbon burns completely to give 0.44 g of carbon dioxide and 0.18 g of water. Its relative molecular mass is 42.0 (C=12.0, H=1.0, O=16.0). (a) Calculate the number of moles of carbon and of hydrogen in the products. (3 marks) (b) Deduce the empirical formula. (2 marks) (c) Deduce the molecular formula and draw one possible structure. (3 marks) | (a) n(CO2) = 0.44/44.0 = 0.0100 mol so n(C) = 0.0100 mol; n(H2O) = 0.18/18.0 = 0.0100 mol so n(H) = 0.0200 mol. (b) C:H = 1:2, empirical formula CH2. (c) Empirical mass 14.0; 42.0/14.0 = 3, so molecular formula C3H6, e.g. CH2=CH-CH3 (propene). |
| 4 | 46.0 g of ethanol is heated under reflux with excess ethanoic acid and concentrated sulphuric acid to form an ester. (a) Calculate the number of moles of ethanol (C=12.0, H=1.0, O=16.0). (2 marks) (b) Name the ester formed, write the equation and calculate its relative molecular mass. (3 marks) (c) If the percentage yield is 80%, calculate the mass of ester actually obtained. (3 marks) | (a) n(ethanol) = 46.0/46.0 = 1.0 mol. (b) Ethyl ethanoate: CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O; Mr = 4×12.0 + 8×1.0 + 2×16.0 = 88.0. (c) Theoretical mass = 88.0 g; actual mass = 88.0 x 80% = 70.4 g. |
Section B: Polymerisation, rate of reaction and oxidation of alcohols (Q5-Q8)
Use the experimental data for rate calculations and give units as cm3 s-1; explanations must refer to collision frequency or energy distribution.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A sample of poly(ethene) has a relative molecular mass of 5,600 (C=12.0, H=1.0). (a) Calculate the degree of polymerisation. (4 marks) (b) State two differences between addition polymerisation and condensation polymerisation in terms of reactants and by-products, giving one example of each. (5 marks) | (a) Mass of one repeat unit C2H4 = 28.0; degree of polymerisation = 5,600/28.0 = 200. (b) Addition polymerisation uses monomers with C=C and forms no by-product (e.g. ethene to poly(ethene)); condensation polymerisation uses monomers with two functional groups and eliminates a small molecule such as water, giving ester or amide linkages (e.g. nylon from a diamine and a diacid). |
| 6 | Excess zinc granules are added to 50.0 cm3 of dilute hydrochloric acid and the hydrogen produced is collected in a gas syringe. In the first 30 s, 24.0 cm3 of hydrogen is collected; by 60 s the cumulative volume is 36.0 cm3 (room conditions). (a) Calculate the average rate of reaction in the first 30 s (cm3 s-1). (3 marks) (b) Calculate the average rate between 30 s and 60 s. (3 marks) (c) Explain why the rate falls with time, and how the experiment shows that the acid is the limiting reactant. (3 marks) | (a) Rate = 24.0/30 = 0.80 cm3 s-1. (b) 36.0 – 24.0 = 12.0 cm3 in 30 s, so 0.40 cm3 s-1. (c) The acid concentration falls, so collisions per unit time decrease and the rate drops. The curve flattens while zinc remains, showing the acid is used up, i.e. the limiting reactant. |
| 7 | Equal masses of calcium carbonate are reacted with 1.0 M hydrochloric acid, once as a powder and once as marble chips; the powder experiment is repeated at 40 degrees C. (a) Explain why the powder reacts faster. (3 marks) (b) Explain how a catalyst speeds up a reaction and why it does not change the overall enthalpy change. (3 marks) (c) Explain how a rise in temperature affects collision frequency and the proportion of effective collisions. (2 marks) | (a) A powder has a much larger surface area per unit mass, so more acid particles can collide with the solid per second; the final gas volume is the same. (b) A catalyst provides an alternative pathway of lower activation energy, so a larger fraction of collisions is effective; it is not consumed and does not change the energy difference between reactants and products, so the enthalpy change is unchanged. (c) Higher temperature increases average kinetic energy, raising collision frequency and sharply increasing the fraction of particles with energy above the activation energy, hence more effective collisions. |
| 8 | Ethanol can be oxidised by acidified potassium dichromate solution. (a) Describe the colour change and state the change in oxidation number of chromium. (3 marks) (b) Name the product, draw its structural formula and write one full equation for the oxidation. (3 marks) (c) Explain why the mixture is heated under reflux and why acidified (rather than alkaline) dichromate is used. (2 marks) | (a) Orange changes to green; chromium falls from +6 to +3. (b) Ethanoic acid CH3COOH; 3C2H5OH + 2Cr2O7(2-) + 16H+ -> 3CH3COOH + 4Cr3+ + 11H2O. (c) Reflux prevents loss of ethanol and ethanoic acid by evaporation and keeps the mixture boiling; acidification increases the oxidising power of dichromate and supplies H+ ions. |
Section C: Isomerism, saponification, plastics and the environment (Q9-Q12)
Draw full structural formulae showing all bonds; name and classify compounds using IUPAC rules.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Compounds with molecular formulae C4H10 and C4H9OH each have several structural isomers. (a) Draw all structural isomers of C4H10 and name them. (3 marks) (b) State the number of structural isomers of C4H9OH and draw any two of them. (3 marks) (c) Classify the isomers of C4H9OH as primary, secondary or tertiary alcohols and state the rule used. (3 marks) | (a) Two: CH3CH2CH2CH3 (butane) and CH3CH(CH3)CH3 (2-methylpropane). (b) Four: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol; draw any two. (c) Primary if the carbon bearing -OH is bonded to one other carbon, secondary if two, tertiary if three; classify each isomer accordingly. |
| 10 | Fats and oils are heated with sodium hydroxide solution to make soap. (a) Name the reaction and write a word equation. (3 marks) (b) Use the structure of a soap molecule to explain how it removes grease. (3 marks) (c) Explain why soap works less well in hard water. (2 marks) | (a) Saponification: fat/oil + sodium hydroxide -> soap (sodium salt of a fatty acid) + glycerol. (b) The soap ion has a hydrophilic head (-COO-Na+) and a long hydrophobic hydrocarbon tail; tails dissolve in the grease while heads remain in water, forming micelles that lift the grease into the water and it is rinsed away. (c) Hard water contains Ca2+ and Mg2+ ions which form insoluble calcium/magnesium salts of the fatty acid, wasting soap and leaving scum. |
| 11 | Plastic pollution is a common environmental problem in Hong Kong. (a) Compare the backbone structure of addition polymers and condensation polymers. (3 marks) (b) Explain why poly(ethene) is not readily degraded in the environment. (3 marks) (c) Suggest two practical ways to reduce plastic waste and explain the principle behind each. (3 marks) | (a) Addition polymers have an all-carbon backbone of C-C single bonds; condensation polymers have ester or amide linkages (heteroatoms) in the backbone. (b) The non-polar C-C backbone is strong and has no hydrolysable group, so micro-organisms lack enzymes to break it down. (c) Reduce at source (reusable bags, no disposable cutlery) and recycle (bottle-to-fibre reprocessing); both cut the volume sent to landfill. |
| 12 | A pharmacy sells disinfectant alcohol labelled 75% (v/v) ethanol. (a) Calculate the volume of pure ethanol needed to prepare 500 cm3 of it. (3 marks) (b) Describe the correct dilution procedure and explain why water must not be poured into concentrated ethanol. (3 marks) (c) State two safety precautions for handling and storing it. (2 marks) | (a) Volume of ethanol = 500 x 75% = 375 cm3. (b) Measure 375 cm3 of pure ethanol, then add distilled water slowly up to the 500 cm3 mark and mix. Ethanol and water contract on mixing and release heat, so adding water to concentrated ethanol can cause local boiling and spitting; ethanol must be added to water. (c) Keep away from flames and heat sources (ethanol is flammable) and store in a closed, labelled container in a cool, ventilated place. |
Find a tutor — freeTutor Registration →
A TutorZone practice paper; original questions and passages, not publisher material.