Q6(a). Quantum yields of the photochemical H₂ + Cl₂ and H₂ + Br₂ reactions
The quantum yield for the photo-chemical reaction between H₂ and Cl₂ is 10⁶ whereas that for H₂–Br₂ reaction is almost unity. Account for the observation.
Answer in brief
Both are chain reactions started by one photon. With chlorine both propagation steps are fast (Cl + H₂ is only 5 kJ mol⁻¹ uphill), so each Cl atom makes about 10⁶ HCl before the chain ends. With bromine Br + H₂ is 70 kJ mol⁻¹ uphill and very slow at room temperature, so most Br atoms recombine first and Φ≈1.
Key points
- Cl₂ + hν → 2Cl (bond 242 kJ mol⁻¹, so light of λ≤494 nm works).
- Cl + H₂ → HCl + H Δ H=436-431=+5 kJ mol⁻¹ H + Cl₂ → HCl + Cl Δ H=242-431=-189 kJ mol⁻¹ Both steps have low activation energies; the Cl atom is regenerated and the cycle repeats.
- only when two atoms meet at the wall (2Cl → Cl₂) or an impurity removes them (H + O₂ → HO₂). The chain runs for up to 10⁶ cycles, so Φ reaches 10⁴–10⁶; the mixture can explode in sunlight.
- Initiation: Br₂ + hν → 2Br (bond 193 kJ mol⁻¹).
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
H₂ + Cl₂: a long chain
- Initiation: Cl₂ + hν → 2Cl (bond 242 kJ mol⁻¹, so light of λ≤494 nm works).
- Propagation: Cl + H₂ → HCl + H Δ H=436-431=+5 kJ mol⁻¹ H + Cl₂ → HCl + Cl Δ H=242-431=-189 kJ mol⁻¹ Both steps have low activation energies; the Cl atom is regenerated and the cycle repeats.
- Termination: only when two atoms meet at the wall (2Cl → Cl₂) or an impurity removes them (H + O₂ → HO₂). The chain runs for up to 10⁶ cycles, so Φ reaches 10⁴–10⁶; the mixture can explode in sunlight.
H₂ + Br₂: chains that hardly start
- Initiation: Br₂ + hν → 2Br (bond 193 kJ mol⁻¹).
- First propagation step: Br + H₂ → HBr + H Δ H=436-366=+70 kJ mol⁻¹. Its activation energy is at least this endothermicity, so at room temperature it is extremely slow (the Boltzmann factor e^(-Δ E/RT) is about 10¹¹ times smaller than for the chlorine step).
- Recombination wins: 2Br + M → Br₂ + M needs no activation energy. Most Br atoms recombine before they can attack H₂; each photon gives only about one HBr.
- Inhibition: the H atoms that do form are partly removed by the product, H + HBr → H₂ + Br, which undoes the chain; hence the rate law (d[HBr])/(dt)=k[H₂][Br₂]^(1/2)1+k'[HBr]/[Br₂].
- Summary: Φ depends on the chain length; chain length depends on how fast the slow propagation step is compared with termination. At higher temperatures the Br + H₂ step speeds up and Φ rises above 1.
Application. the same chain logic explains the selectivity of radical halogenation of alkanes: chlorination (exothermic H-abstraction) is fast and unselective, bromination (endothermic H-abstraction) is slow and highly selective for tertiary C–H.
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All parts of Question 6
- 6(a) Quantum yields of the photochemical H₂ + Cl₂ and H₂ + Br₂ reactions15 marks
- 6(b) Rate constants of H⁺ + OH⁻ ⇌ H₂O from the relaxation time20 marks
- 6(c) Tetrasulfur tetranitride: preparation, structure, action of chlorine15 marks