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CIE IGCSE Chemistry 0620 Paper 4 Structured Practice Paper (8 Questions, 80 Marks, 75 Minutes)
IGCSE Chemistry 0620 Paper 4 Structured
CIE IGCSE Chemistry 0620 Paper 4 structured practice paper: 8 original questions worth 80 marks in 75 minutes, covering molar volume and combustion calculations.
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Paper details Level: IGCSE (Extended) | Subject: Chemistry 0620 | Time: 75 minutes | Marks: 80 | Questions: 8
Section A Chemical calculations and electrochemistry (40 marks)
Questions 1-4
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) What amount, in moles, is present in 6.0 dm3 of carbon dioxide at room temperature and pressure, and what is its mass? (molar volume 24 dm3/mol, Mr(CO2) = 44) (5 marks) (b) 4.8 g of methane (CH4, Mr = 16) burns completely. What volume of oxygen is needed and what volume of carbon dioxide is formed? (5 marks) |
Answer: (a) amount = 6.0 / 24 = 0.25 mol; mass = 0.25 x 44 = 11 g. (b) n(CH4) = 4.8 / 16 = 0.3 mol; CH4 + 2O2 -> CO2 + 2H2O, so n(O2) = 0.6 mol, volume = 0.6 x 24 = 14.4 dm3; n(CO2) = 0.3 mol, volume = 7.2 dm3. Marking notes: use of molar volume and the 1:2:1 ratio. |
| 2 | (a) A current of 2.0 A is passed through copper(II) sulfate solution for 30.0 minutes using copper electrodes. Calculate the charge passed, the amount of electrons and the mass of copper deposited at the cathode. (1 mol electrons = 96500 C, Ar(Cu) = 63.5) (5 marks) (b) Describe the change in mass of the anode and explain why the concentration of copper(II) ions stays almost constant. (5 marks) |
Answer: (a) Q = 2.0 x 1800 = 3600 C; n(e-) = 3600 / 96500 = 0.0373 mol; Cu2+ + 2e- -> Cu, so n(Cu) = 0.0187 mol and mass = 0.0187 x 63.5 = 1.18 g. (b) The anode dissolves and loses about 1.18 g, equal to the mass deposited; each mole of copper dissolved releases two moles of electrons and each mole deposited uses two, so the ion concentration is unchanged. |
| 3 | (a) Use bond energies to calculate the enthalpy change of combustion of methanol: CH3OH + 1.5O2 -> CO2 + 2H2O. Bond energies in kJ/mol: C-H 413, C-O 358, O-H 464, O=O 498, C=O 805. (5 marks) (b) All the heat from burning 2.4 g of methanol is used to heat 500 g of water. What is the temperature rise? (Mr(CH3OH) = 32, specific heat capacity of water 4.2 J/(g C)) (5 marks) |
Answer: (a) Bonds broken: 3 x 413 + 358 + 464 + 1.5 x 498 = 2808 kJ; bonds formed: 2 x 805 + 4 x 464 = 3466 kJ; enthalpy change = 2808 – 3466 = -658 kJ/mol, so the reaction is exothermic. (b) n = 2.4 / 32 = 0.075 mol; heat = 0.075 x 658 = 49.35 kJ = 49350 J; temperature rise = 49350 / (500 x 4.2) = 23.5 C. |
| 4 | (a) Magnesium reacts with excess dilute hydrochloric acid and the volume of hydrogen collected is 0 cm3 at 0 s, 24 cm3 at 10 s, 40 cm3 at 20 s, 50 cm3 at 30 s, 56 cm3 at 40 s and 60 cm3 at 60 s, after which it stays constant. Calculate the average rate over the first 20 s in cm3/s. (5 marks) (b) Explain why the rate falls with time and calculate the mass of magnesium used from the final volume. (molar volume 24000 cm3/mol, Ar(Mg) = 24) (5 marks) |
Answer: (a) average rate = (40 – 0) / 20 = 2.0 cm3/s. (b) The acid is used up so its concentration falls, collisions become less frequent and the gradient of the curve decreases until the reaction stops at 60 s. n(H2) = 60 / 24000 = 0.0025 mol; Mg + 2HCl -> MgCl2 + H2, so n(Mg) = 0.0025 mol and mass = 0.0025 x 24 = 0.06 g. |
Section B Energy, rates, equilibrium and organic chemistry (40 marks)
Questions 5-8
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (a) The Haber process is N2 + 3H2 ⇌ 2NH3 with an enthalpy change of -92 kJ/mol. State and explain the effect on the yield of ammonia of raising the pressure and of raising the temperature. (5 marks) (b) At equilibrium a 2.0 dm3 vessel contains 1.2 mol N2, 0.6 mol H2 and 1.6 mol NH3. Calculate the equilibrium constant Kc. (5 marks) |
Answer: (a) Raising the pressure shifts the equilibrium towards the side with fewer moles of gas, which is the right side (4 mol to 2 mol), so the yield rises. Raising the temperature shifts the equilibrium in the endothermic direction, to the left, so the yield falls; about 450 C is used in industry to balance rate and yield. (b) [N2] = 0.60, [H2] = 0.30 and [NH3] = 0.80 mol/dm3; Kc = 0.80^2 / (0.60 x 0.30^3) = 0.64 / 0.0162 = 39.5, with no units. |
| 6 | (a) Ethanol reacts with ethanoic acid in an esterification. Write the equilibrium equation and calculate the mass of ethyl ethanoate obtained from 12.0 g of ethanoic acid with excess ethanol at an 80% yield. (Mr(CH3COOH) = 60, Mr(C4H8O2) = 88) (5 marks) (b) Explain why a reflux condenser is used, the role of concentrated sulfuric acid, and why the yield cannot reach 100%. (5 marks) |
Answer: (a) CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O; n(acid) = 12.0 / 60 = 0.20 mol; theoretical mass = 0.20 x 88 = 17.6 g; actual mass = 17.6 x 0.80 = 14.1 g. (b) The condenser returns vapourised ethanol and acid to the flask, reducing loss and allowing heating to speed up the reaction; concentrated sulfuric acid is a catalyst and also absorbs the water formed, shifting the equilibrium to the right; esterification is reversible, so some reactants remain at equilibrium. |
| 7 | (a) 0.0500 mol/dm3 sulfuric acid is used to titrate 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide. Calculate the volume of acid needed to reach the end point in cm3. (5 marks) (b) Calculate the concentration of the acid in g/dm3 and state a suitable indicator with a reason. (Mr(H2SO4) = 98) (5 marks) |
Answer: (a) n(NaOH) = 0.0250 x 0.100 = 0.00250 mol; H2SO4 + 2NaOH -> Na2SO4 + 2H2O, so n(H2SO4) = 0.00125 mol and volume = 0.00125 / 0.0500 = 0.0250 dm3 = 25.0 cm3. (b) concentration = 0.0500 x 98 = 4.90 g/dm3; this is a strong acid with a strong base, so the pH changes sharply through about 7 and either methyl orange (3.1-4.4) or phenolphthalein (8.3-10.0) changes colour within the sharp part of the curve. |
| 8 | (a) Sodium hydroxide solution added to an unknown solution gives a blue precipitate. Name the precipitate and describe how a flame test and excess ammonia solution can confirm copper(II) ions. (5 marks) (b) Calculate the percentage by mass of water of crystallisation in hydrated copper(II) sulfate, CuSO4.5H2O. (Ar: Cu 64, S 32, O 16, H 1) (5 marks) |
Answer: (a) The blue precipitate is copper(II) hydroxide; the flame test gives a blue-green colour; adding excess ammonia dissolves the precipitate to give a deep blue solution of the copper-ammonia complex, so the three observations together confirm copper(II) ions. (b) Mr = 64 + 32 + 4 x 16 + 5 x 18 = 250; mass of water = 5 x 18 = 90; percentage = 90 / 250 x 100 = 36.0%. |
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