Q6(b). Photochemical H₂ + Br₂: rate law and quantum yield
Consider the photochemical reaction H₂(g) + Br₂(g) → 2HBr(g) (hν). Give the mechanism of this reaction. Applying steady-state approximations to [Br] and [H], show that the rate of formation of HBr(g) varies with the square root of the intensity (Ia) of the absorbed radiation. What is the quantum yield for this reaction? Why is the value so low?
Answer in brief
d[HBr]/dt = 2k₂(Ia/k₅)^½[H₂] / (1 + k₄[HBr]/k₃[Br₂]) ∝ Ia^½; quantum yield about 0.01 at room temperature.
Part of VKC Basic
Read the key points and the full model answer with VKC Basic
VKC Basic (₹999 for 6 months) opens every solved answer, the notes, the handbook and the daily answer archive. Every Mains test series includes it.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Covershydrogen-bromine reactionsteady-state approximationquantum yieldphotochemical chain reactionrate lawchain inhibition
All parts of Question 6
- 6(a) Rigid or fluxional: an allyl palladium complex10 marks
- 6(b) Photochemical H₂ + Br₂: rate law and quantum yield20 marks
- 6(c) Langmuir adsorption isotherm20 marks