2025 H2 Chem P1 Q19 — Alkane Solution

The answer is B (K initiation, L termination).
Free-radical substitution of ethane with \(^{35}\text{Cl}_2\):
• Initiation: \(\text{Cl}_2 \rightarrow 2\text{Cl}\cdot\) (uv light, homolytic fission)
• Propagation: \(\text{CH}_3\text{CH}_3 + \text{Cl}\cdot \rightarrow \text{CH}_3\text{CH}_2\cdot + \text{HCl}\) and \(\text{CH}_3\text{CH}_2\cdot + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{Cl}\cdot\)
• Termination: \(\text{Cl}\cdot + \text{Cl}\cdot \rightarrow \text{Cl}_2\), \(\text{CH}_3\text{CH}_2\cdot + \text{CH}_3\text{CH}_2\cdot \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\), \(\text{CH}_3\text{CH}_2\cdot + \text{Cl}\cdot \rightarrow \text{CH}_3\text{CH}_2\text{Cl}\)
Reaction K, one species giving two species of exactly half its mass, is the homolytic fission of \(\text{Cl}_2\) into two \(\text{Cl}\cdot\) radicals: initiation.
Reaction L, two species of identical mass combining into one species of twice the mass, is two identical radicals joining, \(\text{Cl}\cdot + \text{Cl}\cdot \rightarrow \text{Cl}_2\) or \(2\text{CH}_3\text{CH}_2\cdot \rightarrow \text{C}_4\text{H}_{10}\): termination.
• A is wrong: L is not propagation. Each propagation step has two reactants and two products, so it never forms one species.
• C is wrong: K is not propagation either, since a propagation step does not split one species into two.
• D is wrong: K is initiation, not propagation.