The most common Manufactured Substances errors and their fixes: one-sided rate-and-yield explanations, claiming a catalyst raises yield, forcing high pressure onto the Contact process, calling an alloy a compound, and property-to-use answers that never name the property.
This chapter loses marks in predictable places. For each mistake below we show what students write, why it loses marks, and the correct version. Group your revision around these and you will keep the marks the marking scheme is looking for.
8.1 & 8.2, Contact and Haber processes
1. Writing a single arrow for a reversible reaction. Students write 2SO₂ + O₂ → 2SO₃ or N₂ + 3H₂ → 2NH₃. This loses marks because both reactions are equilibria. Correct version: use the reversible sign, 2SO₂ + O₂ ⇌ 2SO₃ and N₂ + 3H₂ ⇌ 2NH₃.
2. Explaining temperature as “just to speed up the reaction”. A one-sided answer ignores the effect on yield. Because the forward reaction is exothermic, a high temperature lowers the yield even though it raises the rate. Correct version: the moderate temperature (about 450–550 °C) is a compromise, high enough for a fast rate, low enough to keep the yield acceptable.
3. Forcing high pressure onto the Contact process. Students see fewer gas moles on the product side (3 → 2) and conclude that high pressure “must” be used. Correct version: at about 1 atmosphere the yield of sulfur trioxide is already very high, so high-pressure plant is not worth the cost. Atmospheric pressure is the economical choice.
4. Claiming the catalyst increases the yield. Writing that V₂O₅ or iron “produces more product” confuses rate with yield. Correct version: a catalyst speeds up the rate and helps the reaction reach equilibrium faster, but it does not change the position of equilibrium or the final yield.
5. Swapping the two catalysts. Naming iron for the Contact process, or vanadium(V) oxide for the Haber process, loses the mark. Correct version: V₂O₅ for the Contact process (SO₂ → SO₃); iron for the Haber process (N₂ + H₂ → NH₃).
6. Dissolving sulfur trioxide directly in water. Writing SO₃ + H₂O → H₂SO₄ as the industrial step misses why oleum is made. Dissolving SO₃ directly in water produces a dense acid mist that is hard to condense. Correct version: SO₃ is dissolved in concentrated sulfuric acid to form oleum (SO₃ + H₂SO₄ → H₂S₂O₇), which is then diluted with water.
7. Getting the Haber ratio or formula wrong. Answers such as N₂ + H₂ → NH₃ (unbalanced) or writing NH₄ instead of NH₃ lose marks. Correct version: N₂ + 3H₂ ⇌ 2NH₃, with ammonia as NH₃.
8. Forgetting that unreacted gases are recycled. Students describe the Haber process as if all the gas reacts in one pass. Correct version: only part of the nitrogen and hydrogen reacts each pass; the unreacted gases are cooled, separated from liquid ammonia, and recycled to raise the overall conversion.
9. Swapping the raw-material sources. Students write that hydrogen comes from air and nitrogen from natural gas. Correct version: nitrogen is obtained from the fractional distillation of liquefied air; hydrogen is obtained from natural gas.
10. Thinking a higher temperature raises the yield. Because the forward reaction is exothermic, students sometimes assume more heat gives more product. Correct version: raising the temperature shifts an exothermic equilibrium backwards, so the yield falls; the higher temperature is used only for a faster rate.
11. Not stating a use when the question asks for one. A question that asks for a use of sulfuric acid or ammonia needs a named use, not a description of the process. Correct version: sulfuric acid, making fertilisers, detergents or accumulator acid; ammonia, making nitrogenous fertilisers and nitric acid.
8.3, Alloys
12. Explaining hardness as “carbon is hard”. Saying steel is harder because carbon is a hard element misses the structural reason. Correct version: the foreign atoms are a different size, so they disrupt the orderly layers of metal atoms and stop the layers from sliding, which makes the alloy harder.
13. Calling an alloy a compound. An alloy is not a compound and has no fixed formula. Correct version: an alloy is a mixture of two or more elements, the main one a metal; the components are not chemically bonded in fixed proportions.
14. Choosing an alloy that does not suit the use. Suggesting plain steel for cutlery ignores that it rusts. Correct version: match the property to the use, stainless steel for cutlery because it resists rusting; duralumin for aircraft because it is light and strong.
8.4, Composite materials
15. Defining a composite as “just a mixture”. Leaving out the purpose loses the mark. Correct version: a composite combines two or more materials so the product has better properties than the individual components.
16. Not naming both compression and tension for reinforced concrete. Saying reinforced concrete is “stronger” is too vague. Correct version: concrete is strong in compression but weak in tension; steel is strong in tension; the composite is strong in both, so it does not crack under load.
17. Confusing fibreglass with fibre optic. These are different composites. Correct version: fibreglass (glass fibre in resin) is light, strong and corrosion-resistant, used for boat and car bodies; fibre optic is very pure glass that carries data as light.
8.5, Glass and ceramics
18. Choosing the wrong glass for heat. Picking soda-lime glass for cookware or laboratory heating loses the mark because it cracks under sudden temperature change. Correct version: borosilicate glass withstands sudden temperature change and is the correct choice for heating.
19. Giving a property that does not explain the use. Writing “ceramics are hard” to explain why they are used as electrical insulators does not answer the question. Correct version: ceramics are used as insulators because they do not conduct electricity, and as furnace linings because they have a very high melting point, match each use to the property that explains it.
How to use this list
Before an exam, read each mistake and write the correct version from memory. The single most valuable habit for this chapter is to give both sides of every rate-and-yield answer: what the condition does to the rate, and what it does to the yield, then the decision. A one-to-one teacher can mark your compromise answers and point out any half-answers, which is exactly where students lose easy marks in the SPM Chemistry papers. Fix these fifteen and you will have removed most of the avoidable losses in this chapter.
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