Q8(d). Steady-state approximation; photochemical H₂ + Cl₂; quantum yield
What is meant by steady-state approximation? How does this approximation help in deriving the kinetics of the following photochemical reaction? H₂(g) + Cl₂(g) → 2HCl(g). The quantum yield of this reaction is extremely large. Justify or criticize this statement.
Answer in brief
For a reactive intermediate d[I]/dt ≈ 0; applied to Cl and H it gives d[HCl]/dt = (4k₂/k₄)Iₐ[H₂] and Φ = 4k₂[H₂]/k₄ ≫ 1; the statement is correct, Φ being 10⁴ to 10⁶ in pure gases.
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Coverssteady-state approximationquantum yieldhydrogen–chlorine photochemical reactionchain reactionphotochemical equivalence
All parts of Question 8
- 8(a) The more stable linkage isomer of a cyanide-bridged complex10 marks
- 8(b) Haemocyanin: active site, colours and functions10 marks
- 8(c) Stationary and branching chain reactions; conditions for explosion10 marks
- 8(d) Steady-state approximation; photochemical H₂ + Cl₂; quantum yield10 marks
- 8(e) Langmuir isotherm and the kinetics of surface reactions10 marks