Worked examples: Manufactured Substances in Industry

Work through each example by covering the answer, attempting it yourself, then checking your method against the solution. Every example flags the slip that most often costs marks. These are original questions in SPM style, not past-year papers.

Example 1, Naming the catalyst and conditions

Question. In the Contact process, state the catalyst and the operating temperature for the conversion of sulfur dioxide to sulfur trioxide, and write the equation.

Solution. The catalyst is vanadium(V) oxide, V₂O₅, and the temperature is about 450–550 °C. The equation is 2SO₂ + O₂ ⇌ 2SO₃. The reaction is reversible and exothermic, so the ⇌ sign must be shown.

Common slip: writing a single arrow. This is an equilibrium, so use ⇌.

Example 2, Explaining the temperature choice

Question. Explain why a temperature of about 450–550 °C is used in the Contact process rather than a much lower or much higher temperature.

Solution. The forward reaction is exothermic, so a lower temperature would move the equilibrium towards more sulfur trioxide (higher yield), but the rate of reaction would be too slow to be economical. A higher temperature would speed up the reaction but shift the equilibrium back, lowering the yield. The chosen temperature is a compromise that gives a high yield within a short enough time.

Common slip: saying only “to speed up the reaction”. A full answer names the trade-off: rate versus yield.

Example 3, Explaining the pressure choice

Question. The forward reaction 2SO₂ + O₂ ⇌ 2SO₃ produces fewer gas molecules than the reactants. Explain why the Contact process is nevertheless run at about 1 atmosphere.

Solution. Because the forward reaction reduces the number of gas moles (3 → 2), a high pressure would favour the forward reaction and raise the yield. However, at about 1 atmosphere the yield of sulfur trioxide is already very high, so building and running high-pressure plant would add large costs for very little extra yield. Atmospheric pressure is therefore the economical choice.

Common slip: concluding that high pressure “must” be used because moles decrease, the yield is already high, so it is not worth the cost.

Example 4, Comparing Contact and Haber pressures

Question. Both the Contact and Haber processes have an exothermic forward reaction with fewer gas moles on the product side. Why does the Haber process use a very high pressure while the Contact process uses about 1 atmosphere?

Solution. In the Contact process the yield is already very high at atmospheric pressure, so extra pressure is not worth the cost. In the Haber process the yield of ammonia at low pressure is poor, so a high pressure of about 200–300 atmospheres is needed to shift the equilibrium and obtain an acceptable yield. The pressure decision depends on how much the yield actually improves.

Common slip: assuming the two processes must use the same conditions because their equations look similar.

Example 5, Alloy hardness in terms of structure

Question. Explain, in terms of arrangement of atoms, why steel is harder than pure iron.

Solution. In pure iron the atoms are the same size and arranged in orderly layers that can slide over one another when a force is applied, so pure iron is soft. In steel, carbon atoms of a different size are present among the iron atoms. They disrupt the orderly arrangement and prevent the layers from sliding, so steel is harder and stronger.

Common slip: saying steel is harder “because carbon is hard”. The reason is the disruption of layers, not the hardness of carbon itself.

Example 6, Choosing an alloy for a use

Question. Suggest a suitable alloy for making cutlery and knives, and give one reason.

Solution. Stainless steel (iron with carbon, chromium and nickel) is suitable because it is hard and, importantly, it resists rusting, so cutlery stays clean and safe for contact with food.

Common slip: naming plain steel, it rusts, so it is unsuitable for cutlery.

Example 7, Reasoning about a composite

Question. Explain why reinforced concrete is used for the beams of a building instead of plain concrete.

Solution. Plain concrete is strong when compressed but weak and likely to crack when stretched (in tension). Steel bars are strong in tension. Placing steel bars inside the concrete produces a composite that is strong under both compression and tension, so the beams can carry heavy loads without cracking.

Common slip: saying reinforced concrete is “just stronger” without naming compression and tension, the marks are in the comparison.

Example 8, Matching glass to a use

Question. A cooking dish must not crack when moved from a hot oven to a cool surface. State a suitable type of glass and explain your choice.

Solution. Borosilicate glass is suitable because it can withstand sudden changes in temperature without cracking. This property comes from its low expansion when heated, which is exactly what a dish moved between hot and cool conditions requires.

Common slip: choosing soda-lime glass, it is cheaper but cracks under sudden temperature change.

Example 9, Ceramic properties and uses

Question. Ceramics are used both as electrical insulators on power lines and as linings for the inside of furnaces. Explain how two properties of ceramics make them suitable for these two uses.

Solution. First, ceramics do not conduct electricity (they are good insulators), so they stop current from flowing to the metal pylon, which makes them suitable as electrical insulators. Second, ceramics have a very high melting point and resist heat, so they do not melt or break down at the high temperatures inside a furnace. Each use is matched to a specific property, not just to “ceramics are hard”.

Common slip: giving one general property for both uses. Each use must be matched to the property that actually explains it.

Using these examples

Notice that every strong answer does one of two things: it states a condition and then explains its effect on both rate and yield, or it names a property and links it to a use. Those are the two habits this chapter rewards. Once the reasoning feels automatic, move on to the practice questions and mark yourself the same way. A one-to-one teacher can check that your compromise answers name both sides of the trade-off, which is exactly where marks are gained or lost in the SPM Chemistry written papers.

From examples to marks

These eight examples span the whole chapter. Together they train the same core moves: write ⇌ for a reversible reaction, explain temperature and pressure as rate-versus-yield compromises, explain hardness through disrupted sliding layers, and justify every material choice by a named property. Rehearse them until the reasoning is automatic, then test yourself with the practice questions for this chapter and check each answer against the method shown here.