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This lesson covers cracking of hydrocarbons as required by the Edexcel GCSE Combined Science specification (1SC0). You need to understand why cracking is carried out, the conditions for thermal and catalytic cracking, and the types of products formed.
Fractional distillation of crude oil produces fractions in fixed proportions. However, the demand for different fractions does not match the supply:
Cracking solves this problem by breaking long-chain hydrocarbons into shorter, more useful molecules.
Cracking always produces:
The general idea:
long-chain alkane → shorter alkane + alkene(s)
There are two main types of cracking required by the specification:
| Feature | Detail |
|---|---|
| Temperature | ~500 °C |
| Catalyst | Zeolite (aluminium silicate) or broken pottery (alumina) |
| Pressure | Slight pressure |
| Products | Shorter alkanes + alkenes (often branched) |
Catalytic cracking uses a hot catalyst to break down the long-chain hydrocarbons. The catalyst provides an alternative route with a lower activation energy.
| Feature | Detail |
|---|---|
| Temperature | 800–1000 °C (higher than catalytic) |
| Catalyst | None — heat alone is used |
| Pressure | High pressure, with steam |
| Products | Shorter alkanes + alkenes (including ethene for polymers) |
Thermal cracking uses very high temperatures and high pressure with steam to break the bonds.
C₁₀H₂₂ → C₈H₁₈ + C₂H₄
C₁₀H₂₂ → C₅H₁₂ + C₃H₆ + C₂H₄
This shows that one long molecule can be broken into multiple shorter molecules.
Always check that the number of carbon atoms and hydrogen atoms is the same on both sides of the equation:
| C atoms | H atoms | |
|---|---|---|
| Left: C₁₀H₂₂ | 10 | 22 |
| Right: C₈H₁₈ + C₂H₄ | 8 + 2 = 10 | 18 + 4 = 22 |
graph LR
A["Long-chain alkane<br/>(e.g. C₁₀H₂₂)"] --> B["CRACKING<br/>High temperature<br/>± Catalyst"]
B --> C["Shorter alkane<br/>(e.g. C₈H₁₈)<br/>→ Fuel"]
B --> D["Alkene<br/>(e.g. C₂H₄)<br/>→ Plastics"]
style A fill:#8e44ad,color:#fff
style B fill:#e74c3c,color:#fff
style C fill:#27ae60,color:#fff
style D fill:#2980b9,color:#fff
Exam Tip: In cracking equations, the total number of carbon and hydrogen atoms must be balanced. You can check by counting atoms on each side. At least one product must be an alkene (contains a C=C double bond).
You can tell an alkane from an alkene by their formulae:
| Property | Alkane | Alkene |
|---|---|---|
| General formula | CₙH₂ₙ₊₂ | CₙH₂ₙ |
| Bonding | All single bonds (C–C) | Contains at least one C=C double bond |
| Saturated/unsaturated | Saturated | Unsaturated |
| Test with bromine water | No change (stays orange) | Decolourises (turns colourless) |
If a product has the formula C₃H₆:
Exam Tip: If you are given the formula of a product and asked whether it is an alkane or alkene, use the general formulae. If it fits CₙH₂ₙ₊₂ it is an alkane; if it fits CₙH₂ₙ it is an alkene.
Cracking is economically important because it:
Question: Decane (C₁₀H₂₂) is cracked into octane and one other molecule. Write a balanced equation.
Step 1 — Carbon balance: 10 C on the left. Octane = C₈H₁₈ uses 8 C, leaving 2 C for the other molecule.
Step 2 — Hydrogen balance: 22 H on the left. Octane uses 18 H, leaving 4 H. A molecule with 2 C and 4 H is ethene (C₂H₄) — an alkene.
Equation: C₁₀H₂₂ → C₈H₁₈ + C₂H₄.
Check atoms: left 10 C and 22 H; right (8 + 2) = 10 C and (18 + 4) = 22 H. Balanced.
Question: Cracking hexadecane (C₁₆H₃₄) gives one alkane and propene (C₃H₆). What is the alkane?
Step 1 — Carbons: 16 − 3 = 13 C in the alkane. Step 2 — Hydrogens: 34 − 6 = 28 H in the alkane. Step 3 — Check against alkane general formula (CₙH₂ₙ₊₂): for n = 13, H = 28. Matches.
Answer: The alkane is C₁₃H₂₈ (tridecane).
Question: A cracking reaction produces C₆H₁₄ and another hydrocarbon with formula C₂H₄. Is the second product an alkane or alkene, and how do you know?
Answer: Alkene. The formula C₂H₄ fits the alkene general formula CₙH₂ₙ (for n = 2, H = 4), not the alkane formula CₙH₂ₙ₊₂. C₂H₄ is ethene and contains a C=C double bond (functional group of the alkene homologous series).
Common Mistake: Writing two alkanes as products. Cracking always produces at least one alkene, because breaking a C–C bond in a saturated chain leaves a deficit of hydrogen in one fragment, forcing a C=C double bond.
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