Chapter C6: Making useful chemicals
Chapter C6 focuses on the production of useful chemicals, specifically covering the synthesis of salts from acid reactions and the management of reaction rates. It explores how chemists control conditions in both laboratory and industrial settings to optimize yields and efficiency, including the use of catalysts and reversible reactions.
Quick Revision Summary (Key Takeaway)
Chapter C6: Making useful chemicals covers the industrial production of key chemicals, including the Haber process for ammonia, the Contact process for sulfuric acid, and the production of fertilisers. It focuses on reversible reactions, equilibrium, and the optimisation of conditions (temperature, pressure, catalyst) to maximise yield and rate, linking to sustainability and economic considerations.
Topic Overview
This chapter explores how chemists make useful chemicals on an industrial scale, focusing on two major processes: the Haber process for making ammonia and the Contact process for making sulfuric acid. These chemicals are essential for producing fertilisers, which help increase crop yields and feed a growing global population. Understanding these processes involves applying principles of reversible reactions and dynamic equilibrium, as well as considering the economic and environmental factors that influence industrial choices.
The Haber process combines nitrogen from the air with hydrogen from natural gas to produce ammonia. This is a reversible reaction that reaches equilibrium, and the conditions (temperature, pressure, catalyst) are carefully chosen to maximise profit. Similarly, the Contact process oxidises sulfur dioxide to sulfur trioxide, which is then dissolved in water to make sulfuric acid. Both processes illustrate the trade-offs between rate, yield, and cost, and highlight the importance of catalysts in making reactions faster without being used up.
This topic builds on earlier work about rates of reaction and energy changes, and it connects to broader themes of sustainability and the environmental impact of chemical production. You will need to interpret graphs and data, explain how changing conditions affect equilibrium, and evaluate the choices made by industry. Mastering this chapter will help you understand how the chemicals we rely on every day are manufactured and why the conditions are optimised.
Key Concepts
Core ideas you must understand for this topic
- →Reversible reactions and dynamic equilibrium: In a closed system, the forward and backward reactions occur at the same rate, and the concentrations of reactants and products remain constant.
- →Le Chatelier's principle: If a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium shifts to counteract the change.
- →The Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = -92 kJ/mol, using iron catalyst, 450°C, 200 atm, with nitrogen from air and hydrogen from natural gas.
- →The Contact process: 2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = -197 kJ/mol, using vanadium(V) oxide catalyst, 450°C, 2 atm, with sulfur dioxide from burning sulfur and oxygen from air.
- →Compromise conditions: Industrial processes use conditions that balance rate, yield, and cost, often not the conditions that give the highest yield alone.
What You Need to Demonstrate
Key skills and knowledge for this topic
- Correct identification of products from acid reactions with metals, hydroxides, and carbonates.
- Accurate description of laboratory procedures for salt preparation (filtration, evaporation, crystallisation, drying).
- Correct use of pH scale and understanding of H+ ion concentration.
- Explanation of factors affecting reaction rates (temperature, concentration, pressure, surface area) using collision theory.
- Description of catalytic action in terms of activation energy.
- Understanding of dynamic equilibrium in reversible reactions.
- Prediction of equilibrium shifts based on changes in conditions.
Marking Points
Key points examiners look for in your answers
- Correct identification of products from acid reactions with metals, hydroxides, and carbonates.
- Accurate description of laboratory procedures for salt preparation (filtration, evaporation, crystallisation, drying).
- Correct use of pH scale and understanding of H+ ion concentration.
- Explanation of factors affecting reaction rates (temperature, concentration, pressure, surface area) using collision theory.
- Description of catalytic action in terms of activation energy.
- Understanding of dynamic equilibrium in reversible reactions.
- Prediction of equilibrium shifts based on changes in conditions.
Examiner Tips
Expert advice for maximising your marks
- 💡When asked about rate of reaction, always refer to collision frequency and energy.
- 💡Ensure balanced symbol equations include state symbols where required.
- 💡Use the term 'dynamic equilibrium' when discussing reversible reactions in closed systems.
- 💡Practice calculating pH changes based on H+ concentration shifts.
- 💡Clearly distinguish between the effect of conditions on rate versus the effect on equilibrium position.
- 💡Always refer to 'equilibrium position' when discussing changes in conditions, and use the terms 'shifts to the left' or 'shifts to the right' to show direction.
- 💡When explaining industrial conditions, mention both rate and yield, and justify why a compromise is used – this shows higher-level understanding.
- 💡For 6-mark questions, structure your answer with clear paragraphs: state the process, the conditions, and then explain each condition using Le Chatelier's principle and rate considerations.
Common Mistakes
Pitfalls to avoid in your exam answers
- Confusing 'strong' and 'weak' acids (degree of ionisation) with 'concentrated' and 'dilute' (amount of substance).
- Incorrectly describing the effect of catalysts as changing the yield rather than just the rate.
- Failing to mention that dynamic equilibrium only occurs in closed systems.
- Misinterpreting the relationship between H+ ion concentration and pH (factor of 10 per pH unit).
- Incomplete descriptions of collision theory (e.g., omitting 'frequency' or 'energy' of collisions).
- Misconception: Increasing temperature always increases the yield of a reaction. Correction: For an exothermic forward reaction, increasing temperature decreases the yield because the equilibrium shifts to the left (endothermic direction).
- Misconception: A catalyst increases the yield of a reaction. Correction: A catalyst speeds up the rate of reaching equilibrium but does not change the position of equilibrium, so the yield remains the same.
- Misconception: In the Haber process, high pressure is used because it gives the highest yield. Correction: While high pressure does increase yield, very high pressures are expensive and dangerous, so a compromise pressure of 200 atm is used.
Revision Plan
How to revise this topic in 1–2 weeks
- 1Week 1: Review the basics of reversible reactions and equilibrium. Use flashcards to memorise the key equations and conditions for the Haber and Contact processes.
- 2Week 1: Practice explaining how changes in temperature, pressure, and concentration affect equilibrium using Le Chatelier's principle. Do at least 5 practice questions.
- 3Week 2: Focus on the industrial contexts: why compromise conditions are used, the role of catalysts, and the economic and environmental considerations. Create a comparison table of the two processes.
- 4Week 2: Attempt past exam questions on this topic, especially 6-mark extended response questions. Mark your answers using the mark scheme to identify gaps.
- 5Week 2: Use active recall to test yourself on key facts, and teach the topic to a friend or family member to reinforce understanding.
Exam Question Types
How this topic typically appears in the exam
- 📋Multiple choice questions asking about the effect of changing conditions on equilibrium (e.g., 'What happens to the yield of ammonia if pressure is increased?').
- 📋Short answer questions requiring you to state the conditions used in the Haber or Contact process and explain why they are used.
- 📋Data analysis questions where you interpret graphs of yield vs temperature or pressure and suggest optimal conditions.
- 📋6-mark extended response questions asking you to evaluate the choice of conditions for an industrial process, considering rate, yield, and cost.
Command Word Expectations (OCR)
What examiners look for when using specific command words in this specification
Give a reason or set of reasons for a phenomenon, using scientific principles. In OCR GCSE, you must link cause and effect, e.g., 'Explain why high pressure increases the yield of ammonia' – you must state that pressure increases, equilibrium shifts to side with fewer moles, so yield increases.
Weigh up the pros and cons of a decision or process, and come to a judgement. For example, 'Evaluate the use of high pressure in the Haber process' – you must discuss advantages (higher yield) and disadvantages (cost, safety), then conclude whether it is worth it.
Give a brief, factual answer without explanation. For example, 'State the catalyst used in the Contact process' – answer: vanadium(V) oxide. No extra detail needed.
How Students Lose Marks (Examiner Pitfalls)
Common mark loss traps and how to write 100% full-mark answers
Step-by-Step Worked Solutions
Detailed solution breakdown for typical exam problems
Question: In the Haber process, nitrogen and hydrogen react to form ammonia: N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = -92 kJ/mol. If the pressure is increased from 200 atm to 400 atm, what happens to the equilibrium yield of ammonia? Explain your answer.
- 1.Step 1: Identify the number of gas molecules on each side: reactants have 4 moles of gas (1 N2 + 3 H2), products have 2 moles of gas (2 NH3).
- 2.Step 2: Apply Le Chatelier's principle: increasing pressure shifts equilibrium to the side with fewer gas molecules to reduce pressure.
- 3.Step 3: Since products have fewer moles, the equilibrium shifts right, increasing the yield of ammonia.
Question: In the Contact process, sulfur dioxide is converted to sulfur trioxide: 2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = -197 kJ/mol. The process uses a vanadium(V) oxide catalyst. Explain why a temperature of 450°C is used despite the forward reaction being exothermic.
- 1.Step 1: State that the forward reaction is exothermic, so lower temperatures would favour the forward reaction and increase yield.
- 2.Step 2: However, lower temperatures slow down the rate of reaction, making it uneconomical.
- 3.Step 3: A compromise temperature of 450°C is chosen to give a reasonable rate while still achieving a good yield, and the catalyst helps speed up the reaction without affecting the equilibrium position.
Active Recall Memory Test
Test your memory before revealing the key facts
Frequently Asked Questions
Common questions students ask about this topic
Before You Start
Prior knowledge that will help with this topic
- •Rates of reaction (factors affecting rate: temperature, pressure, surface area, catalysts).
- •Energy changes in reactions (exothermic and endothermic reactions).
- •Reversible reactions and the concept of equilibrium (from earlier in the course).
Likely Command Words
How questions on this topic are typically asked
Ready to test yourself?
Practice questions tailored to this topic