IGCSE Chemistry 0620 Paper 4 Practice Paper: Rates, Equilibrium and Energy Changes (100 marks, 60 min)

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 60 minutes | Total: 100 marks | Parts: 26

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (18 marks) – Measuring the rate of reaction

(a) Experiment 1: magnesium reacts with dilute hydrochloric acid and 40 cm3 of hydrogen is collected in 20 s. Calculate the average rate of reaction in cm3/s. (3 marks)

(b) Experiment 2: the same apparatus collects 60 cm3 of hydrogen in 24 s. Calculate the average rate in experiment 2 and state which experiment is faster. (3 marks)

(c) On a graph of volume against time the gradient gradually decreases. Explain this using collision theory. (4 marks)

(d) A further experiment uses 2.4 g of magnesium (Ar = 24) with excess dilute acid and finishes in 80 s. Calculate (i) the total volume of hydrogen (molar volume 24000 cm3/mol at r.t.p.); (ii) the average rate over the whole reaction. (4 marks)

(e) Compare the effects of adding a catalyst and raising the temperature on the rate, and state how they differ in their effect on the final amount of product. (4 marks)

Question 2 (18 marks) – Concentration, surface area and temperature

(a) 25.0 cm3 of 0.50 mol/dm3 hydrochloric acid is mixed with 25.0 cm3 of distilled water. Calculate the concentration of the diluted acid. (3 marks)

(b) Explain why crushing the same piece of marble into powder greatly speeds up its reaction with dilute acid. (3 marks)

(c) A reaction takes 160 s at 20 C. The rate roughly doubles for every 10 C rise. Estimate the time needed at 50 C. (4 marks)

(d) Explain why a catalyst increases the rate but does not change the enthalpy change of the reaction. (4 marks)

(e) Two experiments use identical temperature, volume and acid concentration, but one uses magnesium ribbon and the other magnesium powder. Sketch both volume-time curves and state the difference. (4 marks)

Question 3 (16 marks) – Energy changes

(a) Burning 0.64 g of methanol (CH3OH, Mr = 32) releases 14.5 kJ. Calculate the enthalpy change of combustion of methanol in kJ/mol. (4 marks)

(b) In the same experiment 100 g of water is heated from 20.0 C to 45.0 C. Using a specific heat capacity of 4.2 J/(g C), calculate the heat absorbed by the water in kJ. (4 marks)

(c) Use bond enthalpies to calculate the enthalpy change of H2(g) + Cl2(g) giving 2HCl(g). H-H = 436, Cl-Cl = 243, H-Cl = 431 kJ/mol. (4 marks)

(d) An energy profile shows the products at a lower energy than the reactants. State whether the reaction is exothermic or endothermic, and explain which part of the profile changes when a catalyst is added. (4 marks)

Question 4 (16 marks) – Chemical equilibrium

(a) Explain what is meant by dynamic equilibrium and state the relationship between the rates of the forward and reverse reactions at equilibrium. (4 marks)

(b) Ammonia synthesis: N2(g) + 3H2(g) giving 2NH3(g), dH = -92 kJ/mol. Explain why industry does not use a very low temperature even though the yield would be higher. (4 marks)

(c) For the same reaction, state the effect of increasing the total pressure on the position of equilibrium and explain it in terms of numbers of molecules. (4 marks)

(d) Iron catalyst is added to the ammonia synthesis. State its effect on (i) the time needed to reach equilibrium and (ii) the yield of ammonia at equilibrium. (4 marks)

Question 5 (16 marks) – Fuels and combustion

(a) Propane C3H8 (Mr = 44) burns completely: C3H8 + 5O2 giving 3CO2 + 4H2O. Calculate the volume of oxygen needed to burn 2.2 g of propane completely (molar volume 24 dm3/mol). (4 marks)

(b) What volume of carbon dioxide is produced when the same amount of propane burns completely under the same conditions? (4 marks)

(c) Burning a fuel in a limited supply of air produces carbon monoxide. Explain why carbon monoxide is harmful to humans and state how it can be avoided. (4 marks)

(d) Compare hydrogen and propane as fuels, giving two advantages or disadvantages based on combustion products and storage. (4 marks)

Question 6 (16 marks) – Practical analysis

(a) Describe a method for measuring the rate of reaction between a metal and an acid, and state what you would measure. (4 marks)

(b) A student collects 90 cm3 of gas in 30 s. Calculate the average rate and state one condition that must be controlled for a fair comparison. (4 marks)

(c) Give two reasons why the measurements might be inaccurate. (4 marks)

(d) If the temperature is raised by 10 C, predict how the shape of the curve changes and explain why. (4 marks)

Answers and marking notes

Question 1 – Measuring the rate of reaction

(a) Rate = volume / time = 40 / 20 = 2.0 cm3/s (3; subtract 1 for an inverted division or a missing unit). [3 marks]

(b) Rate = 60 / 24 = 2.5 cm3/s (2); experiment 2 is faster (1). [3 marks]

(c) As the reaction proceeds the concentration of acid falls (1); fewer collisions per unit time (1); the frequency of successful collisions drops (1); so the rate (gradient) decreases (1). [4 marks]

(d) (i) n(Mg) = 2.4 / 24 = 0.10 mol, so n(H2) = 0.10 mol and V = 0.10 x 24000 = 2400 cm3 (2); (ii) rate = 2400 / 80 = 30 cm3/s (2). [4 marks]

(e) A catalyst lowers the activation energy so more collisions are successful (1); heating raises the average kinetic energy so a larger fraction of collisions succeed (1); both increase the rate (1); neither changes the final amount of product because the amount of reactant is the same (1). [4 marks]

Question 2 – Concentration, surface area and temperature

(a) n(HCl) = 0.50 x 0.0250 = 0.0125 mol (1); total volume = 50.0 cm3 (1); concentration = 0.0125 / 0.050 = 0.25 mol/dm3 (1). [3 marks]

(b) The surface area increases (1); more particles are exposed to the acid (1); so collisions per unit time increase and the rate rises (1). [3 marks]

(c) A 30 C rise is 3 intervals of 10 C (1); the rate becomes 2^3 = 8 times greater (2); time = 160 / 8 = 20 s (1). [4 marks]

(d) A catalyst provides an alternative route of lower activation energy (2); the energies of reactants and products are unchanged, so the enthalpy change is unchanged (2). [4 marks]

(e) The powder curve is steeper with a larger initial gradient (2); both curves finish at the same plateau because the total volume of hydrogen is the same (2). [4 marks]

Question 3 – Energy changes

(a) n = 0.64 / 32 = 0.02 mol (2); dH = -14.5 / 0.02 = -725 kJ/mol (2; subtract 1 for a missing negative sign). [4 marks]

(b) q = mcDT = 100 x 4.2 x 25.0 = 10500 J (2) = 10.5 kJ (2). [4 marks]

(c) Bonds broken = 436 + 243 = 679 kJ (1); bonds formed = 2 x 431 = 862 kJ (1); dH = 679 – 862 = -183 kJ/mol (2). [4 marks]

(d) Exothermic (1); the peak (activation energy) is lowered (2); the energy difference between reactants and products, and hence dH, is unchanged (1). [4 marks]

Question 4 – Chemical equilibrium

(a) Both forward and reverse reactions continue (1); their rates are equal (2); the concentrations of reactants and products stay constant (1). [4 marks]

(b) A lower temperature shifts the equilibrium right and raises the yield (1); but the rate falls greatly (1); equilibrium would take too long to reach (1); so a moderate temperature near 450 C is used with a catalyst (1). [4 marks]

(c) It shifts to the right, towards the products (2); there are 4 mol of gas on the left and 2 mol on the right, so higher pressure favours the side with fewer moles of gas (2). [4 marks]

(d) (i) The time is shortened (2); (ii) the yield is unchanged because a catalyst does not shift the position of equilibrium (2). [4 marks]

Question 5 – Fuels and combustion

(a) n(C3H8) = 2.2 / 44 = 0.05 mol (1); n(O2) = 5 x 0.05 = 0.25 mol (2); V = 0.25 x 24 = 6.0 dm3 (1). [4 marks]

(b) n(CO2) = 3 x 0.05 = 0.15 mol (2); V = 0.15 x 24 = 3.6 dm3 (2). [4 marks]

(c) CO binds to haemoglobin and reduces oxygen transport (2); ensure a sufficient supply of air and good ventilation (2). [4 marks]

(d) Hydrogen burns to give only water, with no carbon dioxide (2); but hydrogen is harder and more costly to store and transport (2). [4 marks]

Question 6 – Practical analysis

(a) Collect the hydrogen in a gas syringe (1); record the volume at fixed intervals such as every 10 s (2); find the rate from the gradient of the volume-time graph (1). [4 marks]

(b) Rate = 90 / 30 = 3.0 cm3/s (2); the temperature, acid concentration or surface area of the metal must be kept the same (2). [4 marks]

(c) Gas leaking from the syringe or tubing (2); reading the volume at an angle or timing inaccurately (2). [4 marks]

(d) The curve is steeper with a larger initial gradient (2); the final plateau is unchanged because the amount of reactant is the same (1); greater kinetic energy gives more successful collisions (1). [4 marks]

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