S6 Chemistry Paper: Industrial Chemistry, The Haber Process And Electrolysis (100 Marks)

S6 Chemistry Paper: Industrial Chemistry, the Haber Process and Electrolysis (100 marks)

2026-11-12

Chemistry – Industrial Processes and Electrolysis

S6 Chemistry paper: 20 parts, 100 marks, 75 min. Haber process Kc, nitric acid yield, brine electrolysis, electroplating thickness, sulphuric acid yield and energy efficiency.

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Section A: The Haber Process for Ammonia (16 marks)

The Haber process makes ammonia from nitrogen and hydrogen: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ per mol. All parts refer to this reaction.

#	Question	Answer and marking notes
1	(a) Explain why industry uses about 200 atm and about 450 °C, and state the role of the iron catalyst. (5 marks)	Answer: Higher pressure favours the side with fewer gas molecules. There are 4 mol of gas on the left and only 2 mol on the right, so about 200 atm shifts the equilibrium to the right and raises the equilibrium yield of ammonia. The reaction is exothermic, so a low temperature would favour the products, but the rate at low temperature is far too slow; about 450 °C is therefore a compromise between rate and yield. The iron catalyst lowers the activation energy and shortens the time needed to reach equilibrium without shifting the position of equilibrium or changing Kc. Marking: 2 marks for the pressure argument, 2 for the temperature compromise, 1 for the catalyst role.
2	(b) A 1.00 dm³ vessel at equilibrium contains 0.20 mol N2, 0.40 mol H2 and 0.60 mol NH3. Find Kc at this temperature to three significant figures. (6 marks)	Answer: [NH3] = 0.60 mol dm^-3, [N2] = 0.20 mol dm^-3, [H2] = 0.40 mol dm^-3. Kc = [NH3]^2 / ([N2] x [H2]^3) = 0.36 / (0.20 x 0.064) = 0.36 / 0.0128 = 28.1. Marking: 2 marks for the three concentrations, 2 for substituting into the equilibrium expression, 2 for the value and significant figures.
3	(c) An experiment starts with 1.00 mol N2 and 3.00 mol H2 and gives 0.60 mol NH3 at equilibrium. Find the percentage yield of ammonia. (5 marks)	Answer: The equation shows that 1.00 mol N2 would give at most 2.00 mol NH3, so the theoretical yield is 2.00 mol. Percentage yield = 0.60 / 2.00 x 100 = 30.0 per cent. Marking: 2 marks for the theoretical yield, 2 for the ratio, 1 for the percentage.

Section B: The Nitric Acid Industry and Atom Economy (17 marks)

The first stage of nitric acid manufacture is the catalytic oxidation of ammonia: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g). (Relative atomic masses: N = 14.0, H = 1.0, O = 16.0.)

#	Question	Answer and marking notes
4	(a) Explain why the platinum-rhodium catalyst is made into a fine gauze, and why air is supplied slightly in excess of the stoichiometric ratio. (5 marks)	Answer: A catalyst acts only at its surface, so a fine gauze gives a very large surface area for a small mass of metal, increasing contact with the gas mixture and raising the rate while saving expensive metal. A slight excess of air oxidises almost all the ammonia so that little ammonia passes unreacted into later stages, but a large excess would dilute the product and raise heating and compression costs. Marking: 2 marks for the surface area argument, 1 for the rate, 2 for the excess air.
5	(b) By mass, find the atom economy of the first stage with respect to NO, to three significant figures. (6 marks)	Answer: Mass of reactants = 4 x 17.0 + 5 x 32.0 = 68.0 + 160.0 = 228.0 g; mass of the desired product NO = 4 x 30.0 = 120.0 g. Atom economy = 120.0 / 228.0 x 100 = 52.6 per cent, the remaining atoms leaving as steam. Marking: 2 marks for the reactant mass, 1 for the product mass, 2 for the ratio, 1 for the answer.
6	(c) If 1.00 tonne of ammonia is fully oxidised, what mass of pure nitric acid is expected? What mass is obtained at an overall yield of 92.0 per cent? Give three significant figures. (6 marks)	Answer: Nitrogen atoms are conserved, so 1 mol NH3 gives 1 mol HNO3 and the mass ratio is 63.0 / 17.0 = 3.706. Theoretical mass = 1.00 x 3.706 = 3.71 tonnes; actual mass = 3.71 x 0.920 = 3.41 tonnes. Marking: 2 marks for the mass ratio, 2 for the theoretical mass, 1 for applying the yield, 1 for units and significant figures.

Section C: Calculations on the Chlor-alkali Electrolysis (17 marks)

The chlor-alkali industry electrolyses concentrated brine to make chlorine, hydrogen and sodium hydroxide together. A cell runs at 5.00 A for 2.00 hours. (Faraday constant 96500 C per mol; molar volume at room conditions 24.0 dm3 per mol; Na = 23.0, O = 16.0, H = 1.0.)

#	Question	Answer and marking notes
7	(a) Write the equations for the reactions at the anode and at the cathode, and give the overall equation. (5 marks)	Answer: Anode (oxidation): 2Cl^- → Cl2 + 2e^-; cathode (reduction): 2H2O + 2e^- → H2 + 2OH^-. Overall: 2NaCl + 2H2O → 2NaOH + H2 + Cl2. Marking: 2 marks for the anode equation, 2 for the cathode equation, 1 for the overall equation.
8	(b) Find the volume of chlorine, in dm3 to three significant figures, produced at the anode in this time. (6 marks)	Answer: Charge Q = 5.00 x 2.00 x 3600 = 36000.0 C. Moles of electrons = 36000.0 / 96500 = 0.37 mol. From the anode equation 2 mol of electrons give 1 mol of Cl2, so Cl2 = 0.37 / 2 = 0.19 mol. Volume = 0.19 x 24.0 = 4.48 dm3. Marking: 1 mark for Q, 2 for the moles of electrons, 1 for the 1 to 2 ratio, 2 for the volume.
9	(c) Find the mass of sodium hydroxide formed at the same time, in g to three significant figures. (6 marks)	Answer: The cathode equation shows 1 mol of electrons gives 1 mol of OH^-, so moles of OH^- = 0.37 mol. Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g per mol. Mass = 0.37 x 40.0 = 14.92 g. Marking: 2 marks for the 1 to 1 relation, 1 for the molar mass, 2 for the calculation, 1 for the unit.

Section D: Electroplating and Coating Thickness (17 marks)

Copper is electroplated onto a brass sheet at 2.50 A for 30.0 minutes. Molar mass of copper 63.5 g per mol, density 8.96 g per cm3, Faraday constant 96500 C per mol.

#	Question	Answer and marking notes
10	(a) Explain why pure copper is used as the anode and copper(II) sulphate solution as the electrolyte in copper plating. (5 marks)	Answer: The pure copper anode dissolves as Cu^2+ (Cu → Cu^2+ + 2e^-), replacing exactly the Cu^2+ removed at the cathode, so the electrolyte concentration stays almost constant and the mass and thickness of the deposit remain controllable. With an inert anode the Cu^2+ would be used up, the deposit would thin with time, and oxygen would be produced at the anode. Marking: 2 marks for the dissolving anode, 2 for the constant concentration, 1 for the consequence of an inert anode.
11	(b) Find the mass of copper deposited at the cathode, in g to three significant figures. (6 marks)	Answer: Charge Q = 2.50 x 30.0 x 60 = 4500.0 C; moles of electrons = 4500.0 / 96500 = 0.05 mol. As Cu^2+ + 2e^- → Cu, moles of copper = 0.05 / 2 = 0.02 mol, and mass = 0.02 x 63.5 = 1.48 g. Marking: 1 mark for Q, 2 for the moles of electrons, 1 for the 2 to 1 ratio, 2 for the mass.
12	(c) If the plated area is 50.0 cm2, find the average thickness of the coating in micrometres to three significant figures. (6 marks)	Answer: Volume of copper = 1.48 / 8.96 = 0.17 cm3. Thickness = volume / area = 0.17 / 50.0 = 3.30 x 10^-3 cm, which is 3.30 x 10^-3 x 10^4 = 33.0 micrometres. Marking: 2 marks for the volume, 2 for dividing by the area, 1 for converting cm to micrometres, 1 for significant figures.

Section E: Sulphuric Acid Manufacture and Environmental Issues (16 marks)

In the contact process sulphur is burned to sulphur dioxide, oxidised to sulphur trioxide over a vanadium(V) oxide catalyst, and then absorbed in water to give sulphuric acid. A plant burns 1.00 tonne of sulphur ore containing 98.0 per cent sulphur. (Relative atomic masses: S = 32.1, O = 16.0, H = 1.0.)

#	Question	Answer and marking notes
13	(a) Find the mass of sulphur dioxide obtainable, in tonnes to three significant figures. (4 marks)	Answer: Mass of pure sulphur = 1.00 x 0.980 = 0.98 tonnes. From S + O2 → SO2 the mole ratio is 1 to 1 and the mass ratio is 64.1 / 32.1 = 1.997, so the mass of SO2 = 0.98 x 1.997 = 1.96 tonnes. Marking: 1 mark for the pure sulphur, 1 for the mole ratio, 1 for the mass ratio, 1 for the answer.
14	(b) If all the sulphur dioxide is converted and absorbed, what mass of pure sulphuric acid is expected, in tonnes to three significant figures? (4 marks)	Answer: Each sulphur atom ends up in one sulphuric acid molecule, so the mass ratio is 98.1 / 32.1 = 3.056 and the mass of acid = 0.98 x 3.056 = 2.99 tonnes. Marking: 1 mark for conserving sulphur atoms, 2 for the mass ratio, 1 for the answer with unit.
15	(c) The oxidation of sulphur dioxide to sulphur trioxide is run at only a slight overpressure rather than at very high pressure. Explain why. (4 marks)	Answer: In 2SO2 + O2 ⇌ 2SO3 there are 3 mol of gas on the left and 2 mol on the right, so high pressure would raise the equilibrium yield; however conversion already exceeds 97 per cent at only 1 to 2 atm, the extra gain from higher pressure is small, and the cost and energy of compressing gas and of pressure-rated equipment rise sharply. Marking: 1 mark for comparing gas moles, 1 for the already high conversion, 2 for the cost and energy argument.
16	(d) State two industrial measures that reduce sulphur dioxide emissions and explain the principle of each. (4 marks)	Answer: First, flue gas desulphurisation, in which the waste gas is washed with limestone slurry or lime water so that SO2 is absorbed and converted into calcium sulphite or sulphate instead of leaving the stack. Second, raising the conversion and recycling the tail gas, returning unreacted SO2 to the catalyst bed or scrubbing the remaining SO2 with alkali in a tail-gas tower. Marking: 1 mark per measure and 1 per explanation.

Section F: Energy Efficiency and Industrial Decisions (17 marks)

The theoretical enthalpy change for making ammonia is −46.0 kJ per mol of ammonia. A plant actually uses about 28.0 GJ of energy per tonne of ammonia. (N = 14.0, H = 1.0.)

#	Question	Answer and marking notes
17	(a) Find the theoretical enthalpy change per tonne of ammonia, in GJ to three significant figures. (5 marks)	Answer: Moles in one tonne of ammonia = 1.0 x 10^6 / 17.0 = 58823.5 mol. Theoretical enthalpy change = 58823.5 x 46.0 = 2705882 kJ, that is 2.71 GJ. Marking: 2 marks for the moles, 2 for multiplying by the enthalpy, 1 for the unit conversion.
18	(b) Hence find the energy efficiency of the plant, that is theoretical enthalpy change divided by actual consumption, as a percentage to two decimal places. (4 marks)	Answer: Energy efficiency = 2.71 / 28.0 x 100 = 9.66 per cent, so most of the input energy does not end up in the chemical bonds of the product but is used for compression, heating, cooling and separation. Marking: 2 marks for the ratio, 1 for the percentage, 1 for the comment.
19	(c) Explain why the actual consumption is far higher than the theoretical enthalpy change. (4 marks)	Answer: The theoretical enthalpy change covers only the energy difference of the reaction itself, whereas the plant must compress the gas to about 200 atm, heat it to the catalyst temperature, cool and condense the product to separate ammonia, and recycle the unreacted gas, and every step dissipates energy irreversibly. Compressors and heat exchangers also operate below 100 per cent efficiency, so the total consumption greatly exceeds the theoretical value. Marking: 2 marks for naming three process steps, 1 for irreversibility and efficiency losses, 1 for clear organisation.
20	(d) Suggest two specific measures to lower the plant energy use, and state one limitation of each. (4 marks)	Answer: First, install a heat exchanger network so that the reactor outlet gas preheats the inlet gas and sensible heat is recovered; the limitation is the extra exchanger area and capital cost, and the benefit falls when the temperature difference becomes small. Second, use a lower pressure synthesis loop or a more active catalyst to cut compression work; the limitation is that a lower pressure reduces the conversion per pass, so the recycle flow and refrigeration duty rise. Marking: 1 mark per measure and 1 per limitation.

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