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S5 Chemistry Practice Paper: Rates of Reaction and Chemical Equilibrium (12 Questions, 100 Marks, 75 Minutes, with Answers)
HKDSE Chemistry: Rates of Reaction and Chemical Equilibrium
A 12-question, 100-mark, 75-minute chemistry paper on rates of reaction, collision theory, catalysts, rate equations and equilibrium constants, with fully worked solutions.
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Paper details Level: S5 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Rates of Reaction (32 marks)
Questions 1 to 4
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Magnesium ribbon reacts with 0.50 mol/dm3 hydrochloric acid and the hydrogen is collected over water. In the first 60 s, 48.0 cm3 of gas is collected; in the next 60 s a further 18.0 cm3 is collected. Find the average rate over each interval in cm3/s and explain why the rate falls. | First interval: 48.0 / 60 = 0.800 cm3/s (2 marks). Second interval: 18.0 / 60 = 0.300 cm3/s (2 marks). The rate falls because the acid is used up, so the concentration of hydrogen ions falls, fewer ions collide with the metal each second and the collision frequency drops (2 marks). |
| 2 | At 25 C, 0.10 mol/dm3 sodium thiosulfate solution is mixed with dilute hydrochloric acid and the mixture turns cloudy after 40 s. The experiment is repeated with 0.20 mol/dm3 sodium thiosulfate, which turns cloudy after 20 s. Taking the rate as 1/t, find the ratio of the rates and explain the difference. | Rate taken as 1/t: first = 1/40 = 0.025 s-1 (2 marks); second = 1/20 = 0.050 s-1 (2 marks); the ratio is 2.0 (2 marks). Doubling the concentration doubles the number of particles per unit volume, so collisions are more frequent and the reaction is faster (2 marks). |
| 3 | A reaction has a rate of 1.0 x 10-3 mol/dm3/s at 20 C. Experiment shows that the rate roughly doubles for every 10 C rise in temperature. Find (a) the rate at 40 C, (b) the rate at 50 C, and explain the effect of temperature using the distribution of particle energies. | (a) From 20 C to 40 C the temperature rises by 20 C, so rate = 1.0 x 10-3 x 2 to the power 2 = 4.0 x 10-3 mol/dm3/s (3 marks). (b) From 20 C to 50 C the rise is 30 C, so rate = 1.0 x 10-3 x 2 to the power 3 = 8.0 x 10-3 mol/dm3/s (3 marks). A higher temperature raises the average kinetic energy, so many more particles have energy above the activation energy, effective collisions increase and the rate rises (3 marks). |
| 4 | For the reaction of A with B to give products, experiment 1 used [A] = 0.10 mol/dm3 and [B] = 0.10 mol/dm3 with an initial rate of 2.0 x 10-3 mol/dm3/s; experiment 2 doubled only [A] and gave 4.0 x 10-3 mol/dm3/s; experiment 3 doubled only [B] and the rate stayed at 2.0 x 10-3 mol/dm3/s. Find (a) the order with respect to A and to B, (b) the rate constant k, (c) the initial rate when [A] = 0.30 mol/dm3 and [B] = 0.10 mol/dm3. | (a) Doubling [A] doubles the rate, so the order with respect to A is one; doubling [B] leaves the rate unchanged, so the order with respect to B is zero (3 marks). (b) The rate equation is rate = k[A], so k = 2.0 x 10-3 / 0.10 = 2.0 x 10-2 s-1 (3 marks). (c) rate = 2.0 x 10-2 x 0.30 = 6.0 x 10-3 mol/dm3/s (3 marks). |
Section B: Collision Theory and Catalysis (36 marks)
Questions 5 to 8
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Marble is mainly calcium carbonate. Equal masses of calcium carbonate, one as a single lump and one as a powder, are reacted with excess 1.0 mol/dm3 hydrochloric acid and the same volume of carbon dioxide is collected. The lump takes 150 s and the powder takes 30 s. Taking the rate as 1/t, find the ratio of the rates and explain the effect of surface area using collision theory. | Rate taken as 1/t: lump = 1/150 = 0.0067 s-1 (2 marks); powder = 1/30 = 0.0333 s-1 (2 marks); the ratio is 5 (2 marks). The powder has a much larger surface area (1 mark), so more calcium carbonate particles are exposed to the acid at the same time (1 mark), collisions per second increase (1 mark) and the reaction is faster (1 mark). |
| 6 | Hydrogen peroxide decomposes slowly at room temperature but much faster after manganese(IV) oxide is added. The activation energy is 75 kJ/mol without the catalyst and 50 kJ/mol with it. Find (a) the fall in activation energy, (b) the fall as a percentage of the original value, (c) how a catalyst increases the rate and whether it changes the enthalpy change of the reaction. | (a) Fall = 75 – 50 = 25 kJ/mol (3 marks). (b) Percentage = 25 / 75 x 100 = 33.3 percent (3 marks). (c) The catalyst provides an alternative route with a lower activation energy, so more particles have enough energy to react (2 marks); it does not change the enthalpy of the reactants and products, so the enthalpy change is unchanged (1 mark). |
| 7 | Nitrogen and hydrogen react to form ammonia in a closed vessel. Explain the effect on the rate of halving the volume of the vessel, and explain why increasing the pressure has little effect on the rate of reactions involving solids and liquids. | Halving the volume doubles the concentration of both gases (2 marks), so the number of molecules per unit volume increases (2 marks); collisions become more frequent and the rate rises (2 marks). Solids and liquids are almost incompressible, so their concentration hardly changes and pressure has little effect on the rate (2 marks). |
| 8 | The concentration of reactant X changes as follows: 0.80 mol/dm3 at 0 s, 0.50 mol/dm3 at 20 s and 0.32 mol/dm3 at 40 s. Find (a) the average rate from 0 to 20 s, (b) the average rate from 20 to 40 s, (c) compare the two rates and explain. | (a) (0.80 – 0.50) / 20 = 0.015 mol/dm3/s (3 marks). (b) (0.50 – 0.32) / 20 = 0.0090 mol/dm3/s (3 marks). (c) The second rate is lower because the concentration of X has fallen, so collisions are less frequent (3 marks). |
Section C: Chemical Equilibrium (32 marks)
Questions 9 to 12
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | At 500 K the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is at equilibrium with [N2] = 0.50 mol/dm3, [H2] = 0.30 mol/dm3 and [NH3] = 0.20 mol/dm3. Find (a) the expression for the equilibrium constant Kc, (b) the value of Kc with its unit. | (a) Kc = [NH3] squared / ([N2] x [H2] cubed) (3 marks). (b) Kc = 0.20 squared / (0.50 x 0.30 cubed) = 0.040 / 0.0135 = 2.96 dm6/mol2 (5 marks). |
| 10 | The synthesis of ammonia is exothermic. State the effect of each change on the mole fraction of ammonia in the equilibrium mixture: (a) raising the temperature, (b) increasing the pressure, (c) continuously removing ammonia. In each case state the direction of the shift and give a reason. | (a) The forward reaction is exothermic, so raising the temperature shifts the equilibrium towards the endothermic reverse reaction and the mole fraction of ammonia falls (3 marks). (b) The forward reaction reduces the number of gas moles from 4 to 2, so increasing the pressure shifts the equilibrium towards the side with fewer gas moles and the mole fraction of ammonia rises (3 marks). (c) Removing ammonia lowers the concentration of the product, so the equilibrium shifts in the forward direction to replace it (2 marks). |
| 11 | At 400 K, Kc = 4.0 for the reaction A(g) + B(g) ⇌ C(g). At a certain instant [A] = 0.10 mol/dm3, [B] = 0.10 mol/dm3 and [C] = 0.40 mol/dm3. Find (a) the reaction quotient Qc, (b) the direction in which the system shifts towards equilibrium. | (a) Qc = [C] / ([A] x [B]) = 0.40 / (0.10 x 0.10) = 40 (4 marks). (b) Qc is greater than Kc (4.0), so the mixture is not at equilibrium and the net reaction proceeds in the reverse direction, with C decomposing into A and B (4 marks). |
| 12 | The Haber process for ammonia uses about 450 C, 200 atm and an iron catalyst. (a) Why is a lower temperature not used, even though it gives a higher yield? (b) What is the role of the catalyst? (c) Why is a high pressure still used? | (a) A low temperature favours the yield because the forward reaction is exothermic, but the rate would then be too slow and equilibrium would take far too long to reach (3 marks). (b) The iron catalyst raises the rate so that equilibrium is reached in a reasonable time at a moderate temperature, but it does not change the position of equilibrium (3 marks). (c) Raising the pressure shifts the equilibrium towards the side with fewer gas moles and so raises the yield of ammonia (2 marks). |
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