S5 Chemistry Paper: Chemical Equilibrium, Reaction Rates And Industrial Processes (100 Marks)

S5 Chemistry Paper: Chemical Equilibrium, Reaction Rates and Industrial Processes (100 marks)

2026-11-11

S5 Chemistry – equilibrium and rates

S5 Chemistry paper: six questions on reaction rates, collision theory, Kc calculations, Le Chatelier shifts, Contact and Haber conditions, and yield sums. Full answers.

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Paper details　S5 Chemistry practice paper with six questions on measuring reaction rates, collision theory, equilibrium constant Kc calculations, Le Chatelier principle, the Contact and Haber processes, and mole and percentage yield calculations. Full answers with marking notes.

Section A: Measuring the Rate of Reaction (16 marks)

A piece of magnesium ribbon is added to excess 1.0 mol dm^-3 hydrochloric acid and the volume of hydrogen collected is recorded every 20 s in cm^3: 0 at 0 s, 28 at 20 s, 48 at 40 s, 62 at 60 s, 72 at 80 s and 78 at 100 s. Give units in every answer.

#	Question	Answer and marking notes
1	Find the average rate of gas production over the first 20 s.	Answer: the average rate is 28 divided by 20, that is 1.4 cm^3 s^-1. (Marking: 1 mark for the formula, 1 mark for substitution, 2 marks for the value with unit.)
2	Find the average rate from 60 s to 100 s and explain why the rate falls.	Answer: the rate is 16 divided by 40, that is 0.4 cm^3 s^-1; the acid concentration falls as it is used up, so the collision frequency drops and the rate slows down. (Marking: 2 marks for the calculation, 2 marks for the explanation.)
3	How many times faster is the first 20 s than the 60 s to 100 s period? Give the number.	Answer: 1.4 divided by 0.4 gives 3.5 times. (Marking: 2 marks for the ratio, 2 marks for noting the rate decreases with time.)
4	If the experiment were run at 40 degrees Celsius instead of 25, would the final gas volume change? Explain.	Answer: the final volume stays at 78 cm^3 because the amount of magnesium is unchanged; the higher temperature raises the average kinetic energy, so a larger fraction of collisions exceeds the activation energy and the reaction finishes sooner. (Marking: 2 marks for unchanged volume, 2 marks for the kinetic explanation.)

Section B: Concentration, Surface Area and Collision Theory (16 marks)

Three experiments each produce 0.020 mol dm^-3 of hydrogen. Experiment A uses 1.0 mol dm^-3 acid and magnesium ribbon and takes 25.0 s; experiment B uses 2.0 mol dm^-3 acid and ribbon and takes 12.5 s; experiment C uses 1.0 mol dm^-3 acid and magnesium powder and takes 6.3 s.

#	Question	Answer and marking notes
5	Find the average rate in experiments A and B.	Answer: A gives 0.020 divided by 25.0, that is 8.0 times ten to the minus four mol dm^-3 s^-1; B gives 0.020 divided by 12.5, that is 1.6 times ten to the minus three mol dm^-3 s^-1. (Marking: 2 marks each, units required.)
6	Compare A and B: what does doubling the concentration do to the rate? Give the rate ratio.	Answer: the ratio is 1.6 times ten to the minus three divided by 8.0 times ten to the minus four, that is 2.0, so doubling the concentration roughly doubles the rate. (Marking: 2 marks for the ratio, 2 marks for the conclusion.)
7	Use collision theory to explain the effect of concentration on rate.	Answer: at higher concentration there are more reacting particles per unit volume, so collisions are more frequent and more effective collisions occur per unit time, giving a faster rate. (Marking: 2 marks for particle number, 2 marks for collision frequency.)
8	Find the rate in experiment C and explain the difference from A.	Answer: C gives 0.020 divided by 6.3, that is 3.2 times ten to the minus three mol dm^-3 s^-1, about 4.0 times that of A; the powder exposes far more surface area to the acid, so collisions are more frequent and the rate is higher. (Marking: 2 marks for the rate, 2 marks for the surface area explanation.)

Section C: The Equilibrium Constant Kc (18 marks)

At 700 K, 0.100 mol of hydrogen and 0.100 mol of iodine vapour are sealed in a 1.0 dm^3 vessel. When H2(g) + I2(g) gives 2HI(g) reaches equilibrium the concentration of hydrogen iodide is 0.160 mol dm^-3.

#	Question	Answer and marking notes
9	Find the equilibrium concentrations of hydrogen and iodine vapour.	Answer: 0.160 mol dm^-3 of HI means 0.080 mol dm^-3 of H2 has reacted, so each equilibrium concentration is 0.100 minus 0.080, that is 0.020 mol dm^-3. (Marking: 2 marks for the change, 1 mark for each equilibrium concentration.)
10	Calculate Kc at 700 K.	Answer: Kc equals the square of 0.160 divided by the product of 0.020 and 0.020, that is 0.0256 divided by 0.00040, which gives 64. (Marking: 2 marks for the expression, 2 marks for substitution, 1 mark for the value; the mole numbers of gas are equal on both sides so Kc has no unit.)
11	If 0.050 mol of hydrogen is added to the equilibrium mixture, how does the position of equilibrium shift and how does the HI concentration change?	Answer: the extra hydrogen makes the system shift to the right to use it up, so the hydrogen iodide concentration rises while that of iodine vapour falls. (Marking: 3 marks for the direction, 2 marks for the concentration change.)
12	The forward reaction is exothermic with a ΔH of minus 53 kJ mol^-1. How does Kc change when the temperature rises?	Answer: Kc falls, because a higher temperature favours the endothermic reverse reaction, so the equilibrium shifts left, HI decreases and the reactants increase. (Marking: 2 marks for Kc falling, 2 marks for the explanation.)

Section D: Le Chatelier Principle and the Contact Process (18 marks)

The key step of the Contact process for making sulfuric acid is 2SO2(g) + O2(g) giving 2SO3(g) with a ΔH of minus 197 kJ mol^-1. Industrial conditions are about 1 to 2 atm, 450 degrees Celsius and a vanadium pentoxide catalyst.

#	Question	Answer and marking notes
13	What is the effect of raising the pressure on the conversion of sulfur dioxide? Explain in terms of mole numbers of gas.	Answer: a higher pressure shifts the equilibrium right and raises the conversion, because there are 3 mol of gas on the reactant side and only 2 mol on the product side, so pressure favours the side with fewer moles of gas. (Marking: 2 marks for the direction, 2 marks for the mole explanation.)
14	The forward reaction is exothermic, so why is a temperature far below 450 degrees Celsius not used?	Answer: cooling shifts the equilibrium right, but it slows the reaction sharply and the catalyst is not active enough at low temperature, so equilibrium would take too long; 450 degrees is a compromise between conversion and rate. (Marking: 2 marks for rate, 2 marks for catalyst activity, 1 mark for the compromise.)
15	What effect does the vanadium pentoxide catalyst have on the position of equilibrium and on Kc?	Answer: neither changes; the catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner, but the equilibrium concentrations and hence Kc stay the same. (Marking: 2 marks for unchanged position, 3 marks for unchanged Kc with explanation.)
16	Why does industry use only about 1 to 2 atm instead of a very high pressure?	Answer: very high pressure needs thicker, costlier plant, much more compression energy and carries greater safety risk, while 1 to 2 atm with a catalyst already gives a high enough conversion. (Marking: 2 marks for cost, 2 marks for safety or benefit.)

Section E: The Haber Process and Equilibrium Calculations (16 marks)

In a 1.0 dm^3 vessel, 1.0 mol of nitrogen and 3.0 mol of hydrogen are mixed. When N2(g) + 3H2(g) gives 2NH3(g) reaches equilibrium, the amount of ammonia is 1.2 mol.

#	Question	Answer and marking notes
17	Find the amounts of nitrogen and hydrogen at equilibrium.	Answer: 1.2 mol of ammonia uses 0.60 mol of nitrogen and 1.80 mol of hydrogen, so 0.40 mol of nitrogen and 1.20 mol of hydrogen remain. (Marking: 2 marks for the amounts used, 1 mark for each equilibrium amount.)
18	Calculate Kc at this temperature.	Answer: Kc equals 1.44 divided by the product of 0.40 and 1.20 cubed, that is 1.44 divided by 0.6912, which gives 2.08. (Marking: 2 marks for the expression, 2 marks for substitution, 1 mark for the value; the unit is mol^-2 dm^6.)
19	Find the percentage conversion of nitrogen.	Answer: the conversion is 0.60 divided by 1.0 times 100 per cent, that is 60 per cent. (Marking: 2 marks for the method, 2 marks for the value.)
20	Why does industry use about 450 degrees Celsius rather than room temperature?	Answer: at room temperature the reaction is far too slow and the catalyst is not active enough; 450 degrees lets equilibrium be reached in a reasonable time, so output is higher even though the conversion is slightly lower. (Marking: 2 marks for rate, 1 mark for the output conclusion.)

Section F: Mole Calculations and Percentage Yield (16 marks)

A plant makes ammonia from 2.8 kg of nitrogen and excess hydrogen (relative atomic masses: N 14.0, H 1.0) and obtains 2.72 kg of ammonia.

#	Question	Answer and marking notes
21	Find the number of moles in 2.8 kg of nitrogen.	Answer: the molar mass is 28.0 g mol^-1, so the amount is 2800 divided by 28.0, that is 100 mol. (Marking: 2 marks for the molar mass, 2 marks for the value with unit.)
22	Find the theoretical mass of ammonia in kg.	Answer: one mole of nitrogen gives two moles of ammonia, so 200 mol is expected; the mass is 200 times 17.0, that is 3400 g or 3.40 kg. (Marking: 2 marks for the mole ratio, 1 mark for the amount, 1 mark for the mass.)
23	Calculate the percentage yield.	Answer: the yield is 2.72 divided by 3.40 times 100 per cent, that is 80 per cent. (Marking: 2 marks for the method, 2 marks for the value.)
24	Suggest two industrial measures that raise the ammonia yield and explain each with equilibrium ideas.	Answer: first, raise the pressure, since 4 mol of gas becomes 2 mol, which favours the right; second, liquefy and remove the ammonia to keep the product concentration low so the equilibrium shifts right; recycling unreacted nitrogen and hydrogen also helps. (Marking: 1 mark per measure, 1 mark per explanation.)

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