HKDSE Chemistry Practice Paper: Chemical Equilibrium And Industrial Processes

HKDSE Chemistry Practice Paper: Chemical Equilibrium and Industrial Processes

2026-11-04

HKDSE Chemistry: Chemical Equilibrium and Industrial Processes

HKDSE Chemistry practice paper: 12 questions, 100 marks in 75 minutes on Kc, reaction quotients, Le Chatelier and the Haber and Contact processes, with worked answers.

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Paper details　Level: S4-S6 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12

Section A (Questions 1 to 4, 32 marks)

Read each question, quote the formula, substitute the values and give the unit.

#	Question	Answer and marking notes
1	The synthesis of ammonia, N2(g) + 3H2(g) ⇌ 2NH3(g), reaches equilibrium at a certain temperature in a closed vessel of volume 2.0 dm3 containing 0.50 mol N2, 1.00 mol H2 and 1.00 mol NH3. (a) Write the Kc expression in terms of concentrations. (b) Find the equilibrium concentration of each gas. (c) Calculate Kc at this temperature and state its unit. (d) The vessel is compressed to 1.0 dm3 at constant temperature; state the direction of the shift and explain.	(a) Kc = [NH3]^2 / ([N2] x [H2]^3), with all concentrations in mol/dm3. [2] (b) Concentration = moles / volume: N2 = 0.50 / 2.0 = 0.25 mol/dm3, H2 = 1.00 / 2.0 = 0.50 mol/dm3, NH3 = 1.00 / 2.0 = 0.50 mol/dm3. [2] (c) Kc = 0.50^2 / (0.25 x 0.50^3) = 0.25 / 0.03125 = 8.0, unit mol^-2 dm^6. [2] (d) Halving the volume doubles every concentration, so Qc = 1.00^2 / (0.50 x 1.00^3) = 2.0, which is less than Kc = 8.0, so the net reaction goes forward and the equilibrium shifts right, increasing the amount of NH3. [2] Marking: Kc depends on temperature only, not on volume; the total number of gas moles falls from 4 to 2, so pressure favours the forward reaction.
2	The first step of the Contact process is 2SO2(g) + O2(g) ⇌ 2SO3(g), delta H = -197 kJ/mol. At equilibrium at a certain temperature the concentrations are [SO2] = 0.40 mol/dm3, [O2] = 0.20 mol/dm3 and [SO3] = 0.80 mol/dm3. (a) Write the Kc expression. (b) Calculate Kc at this temperature and state its unit. (c) If the temperature is raised, does Kc increase or decrease? Explain. (d) Explain why industry still uses about 450 °C instead of room temperature.	(a) Kc = [SO3]^2 / ([SO2]^2 x [O2]). [2] (b) Kc = 0.80^2 / (0.40^2 x 0.20) = 0.64 / 0.032 = 20.0, unit mol^-1 dm^3. [2] (c) The forward reaction is exothermic, so raising the temperature shifts the equilibrium to the left, lowering the product and raising the reactant concentrations, and Kc therefore decreases. [2] (d) At room temperature the reaction is far too slow to give useful output, however favourable the position; at about 450 °C the vanadium catalyst is active enough for equilibrium to be reached in an acceptable time, which is the compromise between rate and yield. [2] Marking: the unit follows from the change in gas moles (2 – 3 = -1); the temperature effect on an exothermic reaction opposes the rate effect.
3	For H2(g) + I2(g) ⇌ 2HI(g) at 855 K, Kc = 54.0. At one moment a 1.0 dm3 vessel contains [H2] = 0.100 mol/dm3, [I2] = 0.100 mol/dm3 and [HI] = 0.500 mol/dm3. (a) Write the Kc expression and explain why Kc for this reaction has no unit. (b) Calculate the reaction quotient Qc at that moment. (c) State the direction of the net reaction. (d) When equilibrium is reached, will [HI] be higher or lower than 0.500 mol/dm3? Explain.	(a) Kc = [HI]^2 / ([H2] x [I2]); there are two gas moles on each side, so the concentration units cancel and Kc has no unit. [2] (b) Qc = 0.500^2 / (0.100 x 0.100) = 0.25 / 0.0100 = 25.0. [2] (c) Qc = 25.0 is less than Kc = 54.0, so the net reaction proceeds from left to right. [2] (d) The forward reaction continues, so [HI] keeps rising above 0.500 mol/dm3 while [H2] and [I2] fall, until Qc equals Kc. [2] Marking: direction comes from comparing Qc with Kc, not from the size of one concentration alone.
4	1.00 mol N2 and 3.00 mol H2 are mixed in a closed vessel of volume 2.0 dm3 at constant temperature. At equilibrium for N2(g) + 3H2(g) ⇌ 2NH3(g) the vessel contains 0.80 mol NH3. (a) Find the equilibrium amounts of N2 and H2 in moles. (b) Find the equilibrium concentration of each gas. (c) Calculate Kc and state its unit. (d) Find the percentage conversion of N2.	(a) Forming 0.80 mol NH3 uses 0.40 mol N2 and 1.20 mol H2, so at equilibrium N2 = 0.60 mol and H2 = 1.80 mol. [2] (b) The concentrations are N2 0.30 mol/dm3, H2 0.90 mol/dm3 and NH3 0.40 mol/dm3. [2] (c) Kc = 0.40^2 / (0.30 x 0.90^3) = 0.16 / 0.2187 = 0.732, unit mol^-2 dm^6. [2] (d) Conversion of N2 = N2 reacted / N2 initially x 100% = 0.40 / 1.00 x 100% = 40.0%. [2] Marking: use the stoichiometric ratio to find amounts used; conversion is quoted for one reactant and uses the initial amount as the denominator.

Section B (Questions 5 to 8, 36 marks)

Answer the following, showing full working and the formula used.

#	Question	Answer and marking notes
5	Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g), delta H = -92 kJ/mol. (a) Explain how raising the temperature affects the position of equilibrium and Kc, and why industry still uses about 450 °C. (b) Explain the effect of a higher total pressure on the position of equilibrium. (c) Explain the purpose of continuously liquefying and removing the ammonia formed. (d) Explain the role of the iron catalyst and state whether it changes Kc.	(a) The forward reaction is exothermic, so a higher temperature shifts the equilibrium left, lowers the ammonia yield and lowers Kc; industry still uses about 450 °C because at low temperature the rate and the catalyst activity are both too low, so 450 °C is the compromise between yield and rate. [3] (b) The forward reaction takes the gas moles from 4 to 2, so a higher total pressure shifts the equilibrium towards the side with fewer gas moles, that is to the right, raising the yield of ammonia. [2] (c) Continuously liquefying and removing NH3 lowers the product concentration so that Qc falls below Kc and the equilibrium keeps shifting right, converting more of the reactants. [2] (d) The iron catalyst provides a route of lower activation energy so equilibrium is reached faster, but it speeds up the forward and reverse reactions equally, so it does not change the position of equilibrium or Kc. [2] Marking: a catalyst changes rate only; a pressure effect follows from the change in gas moles.
6	2.0 mol SO2 and 1.0 mol O2 are mixed in a 1.0 dm3 vessel. At equilibrium for 2SO2(g) + O2(g) ⇌ 2SO3(g) the vessel contains 1.60 mol SO3. (a) Find the equilibrium amounts of SO2 and O2 in moles. (b) Taking the volume as 1.0 dm3, calculate Kc and state its unit. (c) Find the percentage conversion of SO2. (d) Explain why a real plant cannot convert all the SO2 and how the overall use of raw materials is improved.	(a) Forming 1.60 mol SO3 uses 1.60 mol SO2 and 0.80 mol O2, so at equilibrium SO2 = 0.40 mol and O2 = 0.20 mol. [2] (b) Kc = 1.60^2 / (0.40^2 x 0.20) = 2.56 / 0.032 = 80.0, unit mol^-1 dm^3. [2] (c) Conversion of SO2 = 1.60 / 2.0 x 100% = 80.0%. [2] (d) The reaction is reversible and at equilibrium the forward and reverse rates are equal, so the reactants are never used up completely; the plant recycles unreacted SO2 and O2 back to the converter and removes SO3, which keeps shifting the equilibrium right and raises the overall use of raw materials. [3] Marking: conversion has a ceiling for a reversible reaction; overall efficiency comes from recycling and removing product, not from adding more catalyst.
7	For PCl5(g) ⇌ PCl3(g) + Cl2(g), 1.0 mol PCl5 is placed in a 5.0 dm3 vessel. At equilibrium at a certain temperature 0.60 mol PCl5 remains. (a) Find the equilibrium amounts of PCl3 and Cl2 in moles. (b) Find the equilibrium concentration of all three species. (c) Calculate Kc and state its unit. (d) At the same temperature the volume is changed to 2.5 dm3; predict how the extent of decomposition of PCl5 changes and explain.	(a) 0.40 mol PCl5 decomposes to give 0.40 mol PCl3 and 0.40 mol Cl2. [2] (b) The concentrations are PCl5 0.12 mol/dm3, PCl3 0.08 mol/dm3 and Cl2 0.08 mol/dm3. [2] (c) Kc = (0.08 x 0.08) / 0.12 = 0.0064 / 0.12 = 0.0533, unit mol dm^-3. [2] (d) Halving the volume doubles every concentration, so Qc becomes twice Kc and the equilibrium shifts towards the side with fewer gas moles, that is to the left; the extent of decomposition falls and more PCl5 remains. [3] Marking: decomposition produces two molecules, so pressure works against it; Kc does not depend on volume.
8	A study compares the Haber process at 200 atm: the equilibrium yield of ammonia is 63 per cent at 400 °C, 45 per cent at 450 °C and 26 per cent at 500 °C, while the time needed to reach equilibrium falls clearly as the temperature rises. (a) Explain why the equilibrium yield falls as the temperature rises. (b) Calculate the fall in yield from 400 °C to 500 °C in percentage points and as a percentage. (c) Explain why industry still chooses about 450 °C rather than 400 °C. (d) Explain why plants use 200 atm rather than a much higher pressure.	(a) The synthesis is exothermic, so a higher temperature shifts the equilibrium left and the equilibrium yield of ammonia falls. [2] (b) The fall in percentage points is 63 – 26 = 37 percentage points; the percentage fall is 37 / 63 x 100% = 58.7%. [3] (c) The yield at 400 °C is higher, but the rate and the catalyst activity are much lower and equilibrium takes far too long; 450 °C gives a useful yield in an acceptable time, a compromise between rate and yield. [2] (d) 200 atm already raises the yield a great deal and further pressure gives diminishing gains, while the cost of compression, the thicker vessels needed and the safety risk all rise sharply, so 200 atm is the practical balance point. [2] Marking: percentage points and percentage change are different; the benefit of high pressure is limited by cost.

Section C (Questions 9 to 12, 32 marks)

Show fuller working and explanation, with units in every answer.

#	Question	Answer and marking notes
9	Methanol can be made from carbon monoxide and hydrogen: CO(g) + 2H2(g) ⇌ CH3OH(g). At equilibrium at a certain temperature in a 2.0 dm3 vessel, [CO] = 0.50 mol/dm3, [H2] = 0.25 mol/dm3 and [CH3OH] = 0.75 mol/dm3. (a) Write the Kc expression. (b) Calculate Kc and state its unit. (c) Assuming only reactants were present at the start, find the initial amounts of CO and H2 in moles. (d) Explain why a higher pressure raises the equilibrium yield of methanol.	(a) Kc = [CH3OH] / ([CO] x [H2]^2). [2] (b) Kc = 0.75 / (0.50 x 0.25^2) = 0.75 / 0.03125 = 24.0, unit mol^-2 dm^6. [2] (c) At equilibrium there are 1.50 mol CH3OH and 1.00 mol CO; making 1.50 mol CH3OH uses 1.50 mol CO and 3.00 mol H2, so initially there were 2.50 mol CO and 3.50 mol H2. [2] (d) The forward reaction takes the gas moles from 3 to 1, so a higher pressure shifts the equilibrium towards the side with fewer gas moles, that is to the right, raising the equilibrium yield of methanol. [2] Marking: when working back to the start, add the amounts used, in the stoichiometric ratio, to the equilibrium amounts.
10	For each change below, state the direction of the shift in equilibrium and explain: (a) some SO3 is removed from 2SO2(g) + O2(g) ⇌ 2SO3(g); (b) more catalyst is added to N2(g) + 3H2(g) ⇌ 2NH3(g); (c) the vessel volume is increased for PCl5(g) ⇌ PCl3(g) + Cl2(g); (d) the temperature is raised for H2(g) + I2(g) ⇌ 2HI(g), which is exothermic.	(a) Removing the product SO3 lowers its concentration, so Qc falls below Kc and the equilibrium shifts right to replace it. [2] (b) A catalyst speeds up the forward and reverse reactions equally, so it only shortens the time to reach equilibrium; the position of equilibrium does not move and Kc is unchanged. [2] (c) Increasing the volume lowers every concentration, so the equilibrium shifts towards the side with more gas moles, that is to the right, and the extent of decomposition of PCl5 rises. [2] (d) The reaction is exothermic, so raising the temperature shifts the equilibrium left, lowering [HI] and lowering Kc. [2] Marking: decide the direction from the change in gas moles, and the temperature effect from whether the reaction is exothermic or endothermic.
11	Compare, in at least four respects, the time taken to reach equilibrium with the position of equilibrium, and explain separately how a catalyst and temperature affect each of them.	(a) Rate: a catalyst lowers the activation energy and speeds up both directions, so equilibrium is reached sooner, but the position of equilibrium and Kc are unchanged. [2] (b) Temperature: a higher temperature speeds up both directions so equilibrium is reached sooner, but it shifts the equilibrium in the endothermic direction and changes Kc. [2] (c) Position of equilibrium: it is fixed by temperature and Kc only, and is unaffected by a catalyst or the size of the vessel; changes in concentration or pressure move the equilibrium but do not change Kc. [2] (d) The two can pull in opposite directions: a higher temperature reaches equilibrium faster but may lower the yield of an exothermic reaction, so industry must trade rate against yield. [2] Marking: the answer must treat rate and position of equilibrium separately and state that a catalyst leaves Kc unchanged while temperature changes it.
12	(a) Explain why an ammonia plant recycles unreacted N2 and H2 back to the reactor. (b) Explain why the Contact process absorbs SO3 in concentrated sulphuric acid rather than in water. (c) Give one measure used in each of the Haber and Contact processes to improve energy efficiency. (d) Explain why lowering the temperature alone cannot be used to raise the equilibrium yield of an industrial reaction.	(a) Recycling returns unreacted N2 and H2 to the reactor so they can react again, avoiding waste and giving an overall conversion far above the single-pass value. [2] (b) SO3 reacts violently with water to form a dense acid mist that is hard to absorb and corrodes the plant; concentrated sulphuric acid absorbs SO3 smoothly to give oleum, which is then diluted to the required acid strength. [2] (c) Haber: an iron catalyst with about 450 °C and 200 atm, using the heat of reaction to warm the incoming gases; Contact: heat exchangers use the heat of reaction to preheat the incoming gases and to raise steam. [2] (d) Lowering the temperature raises the yield of an exothermic reaction, but the rate and the catalyst activity fall at the same time, so equilibrium takes far too long and output drops; temperature, pressure and catalyst must be balanced, and cooling alone is not enough. [2] Marking: recycling, removing product and heat exchange are the usual efficiency measures; temperature effects must include rate.

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