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2023 O Level Chem P1 Q36 — Cracking / Alkenes / Haber Process Solution

Cracking / Alkenes / Haber Process 2023 O Level Pure Chemistry · Paper 1 · Q36

Question

The hydrocarbon C₁₆H₃₄ is cracked to form three products, X, Y and Z. The equation is shown. C₁₆H₃₄ → 3X + C₁₀H₂₀ + Z, where Y is C₁₀H₂₀. Z is used to convert unsaturated vegetable oils into a solid product. Which row is correct? Question diagram

Question diagram

A
B
C
D

Correct answer: B

Video Solution

Explanation

Answer: B. Z hardens unsaturated vegetable oils — this is the hydrogenation reaction, so Z = \(\text{H}_2\). Balance the equation: LHS = \(\text{C}_{16}\text{H}_{34}\); RHS already accounts for \(\text{Y} = \text{C}_{10}\text{H}_{20}\) and \(\text{Z} = \text{H}_2\), leaving \(3\text{X} = \text{C}_6\text{H}_{12}\) → X = \(\text{C}_2\text{H}_4\) (ethene).

Hydrocarbon products: X (\(\text{C}_2\text{H}_4\)) and Y (\(\text{C}_{10}\text{H}_{20}\)) are hydrocarbons; Z (\(\text{H}_2\)) is not.

Decolourises aqueous bromine: alkenes do. X is an alkene; Y fits \(\text{C}_n\text{H}_{2n}\) and is also an alkene from cracking. So both X and Y decolourise bromine. Z is \(\text{H}_2\) and does not.

Used in the manufacture of ammonia: \(\text{H}_2\) (Haber process) — so Z only. All three columns match row B.

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