GCSE (9–1) · GCSE Chemistry (Foundation and Higher)
Chapter 13 Practice Paper
Reversible reactions, equilibrium and industry
READ THESE INSTRUCTIONS FIRST
Original practice material written for the MOE GCSE Chemistry syllabus. It is not an official MOE or school examination paper. Chapter 13: reversible reactions and dynamic equilibrium, Le Chatelier’s principle (Higher), the Haber process and its compromise conditions, percentage yield, and (Chemistry only) atom economy and NPK fertilisers.
| For Examiner’s Use | |
|---|---|
| Section A | / 10 |
| Section B | / 20 |
| Total | / 30 |
Section A: Multiple-choice questions[10 marks]
Choose one answer for each question. Each question is worth 1 mark.
At equilibrium the rate of the forward reaction is
Increasing the pressure on N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Adding a catalyst to an equilibrium mixture changes
The forward reaction is exothermic. Raising the temperature shifts the equilibrium
A run collects 12 g of product where 60 g was theoretically possible. The percentage yield is
A reaction is guaranteed an atom economy of 100% when
The catalyst used in the Haber process is
The Haber process is not run at a much lower temperature because
Ammonia is removed from the Haber recycle loop by
The main plant nutrient supplied by ammonium nitrate is
Section B: Structured questions[20 marks]
Answer all questions in the spaces provided.
This question is about hydrated copper(II) sulfate: hydrated CuSO₄ ⇌ anhydrous CuSO₄ + water.
State the colours of the hydrated and anhydrous salts.
State why the mixture in a sealed tube can reach dynamic equilibrium.
Describe what is meant by dynamic equilibrium in this tube.
This question is about the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), forward reaction exothermic.
State the temperature, pressure and catalyst used.
Explain why the temperature is a compromise.
Explain why increasing the pressure increases the equilibrium yield, and why 200 atm is chosen rather than 1000 atm.
A cement works decomposes calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g). (Mr: CaCO₃ = 100, CaO = 56, CO₂ = 44)
Work out what percentage of the product mass is the desired calcium oxide.
100 g of calcium carbonate theoretically makes 56 g of CaO; a kiln run makes 42 g. Calculate the percentage yield.
Explain the difference between the atom economy and the percentage yield of this process.
— End of paper —