The questions are on the chemical equilibrium test. Take it under real conditions before reading the mark scheme.
Quick answer: A reversible reaction can go in both directions, and in a closed system it reaches dynamic equilibrium when the forward and reverse rates are equal and concentrations stay constant. This mark scheme gives the model answers and the marks for the 8 questions of the Chemistry 9-10 test on chemical equilibrium, from key terms to use of ammonia, with the equivalent answers to accept.
Award marks as indicated. For shift questions, a mark is given for the direction and another for the reason, so a correct direction without a reason does not earn full marks. Accept clear alternative wording.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS1-6
Question 1 Key terms (3 marks)
- A reaction in which the products can react to re-form the reactants. (1 mark)
- The forward and reverse reactions both continue at the same rate in a closed system, so the concentrations stay constant. (2 marks)
Marking: 1 mark for both reactions continuing, 1 mark for equal rates or constant concentrations.
Question 2 Concentration graph (4 marks)
- About 30 s. (1 mark)
- [A] = 0.40 mol/dm³ and [B] = 0.60 mol/dm³. (1 mark)
- The forward and reverse reactions continue; their rates are equal, so A is made as fast as it is used up. (2 marks)
Marking: (c) 1 mark for both reactions continuing, 1 mark for equal rates.
Question 3 Phosphorus pentachloride (3 marks)
- Shifts left, because there is 1 gas molecule on the left and 2 on the right, and the equilibrium moves to the side with fewer molecules. (2 marks)
- Shifts right, because the forward reaction is endothermic. (1 mark)
Marking: 1 mark for direction, 1 mark for the counting of molecules; 1 mark for (b).
Question 4 Conditions in the Haber process (3 marks)
A lower temperature favours the exothermic forward reaction and would give a higher yield. But the rate would be far too slow because few particles have the activation energy. 450 °C is a compromise that gives a reasonable rate and an acceptable yield, helped by the catalyst. Compromise between rate and yield.Marking: 1 mark for yield at low temperature, 1 for too slow, 1 for compromise.
Question 5 Catalyst (2 marks)
The position does not change, because the catalyst speeds up the forward and reverse reactions equally. Equilibrium is reached sooner. No shift; faster.Marking: 1 mark for no change in position, 1 mark for reached sooner.
Question 6 Ammonia calculation (2 marks)
N₂ reacted = 0.20 × 60 = 12 mol, so N₂ left = 48 mol. NH₃ = 2 × 12 = 24 mol.Marking: 1 mark for 48 mol N₂, 1 mark for 24 mol NH₃.
Question 7 Recycling (2 marks)
Only a fraction of the gases reacts on each pass. Returning them gives them another chance to react, which saves raw materials and increases the overall yield. Recycling raises the overall yield.Marking: 1 mark for only part reacts, 1 mark for the saving or higher yield.
Question 8 Use of ammonia (1 mark)
Making fertilisers (for example ammonium nitrate). Fertilisers.Marking: 1 mark.
