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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2018 · Paper I

Q6(b)(ii). Quantum yield of the photobromination of cinnamic acid

  • 5 marks
  • Answered
Question

In the photobromination of cinnamic acid to dibromocinnamic acid using light of wavelength 4358 Å a light intensity of 14,000 ergs sec⁻¹ produced, and a decrease of 0.075 millimole of bromine during an exposure for 1105 sec. The solution absorbed 80.1% of the light passing through it. Calculate the quantum yield.

Answer in brief

Einsteins absorbed =4.51×10⁻⁶; Br₂ used =7.5×10⁻⁵ mol; quantum yield Φ=16.6.

Key points

  • 14 000 erg s⁻¹×1105 s=1.547×10⁷ erg =1.547 J; absorbed =0.801×1.547=1.239 J
  • E=(N_Ahc)/(λ)=(6.022×10²³)(6.626×10⁻³⁴)(2.998×10⁸)435.8×10⁻⁹=2.745×10⁵ J mol⁻¹
  • Einsteins absorbed: 1.2392.745×10⁵=4.514×10⁻⁶ mol of photons
  • Bromine reacted: 0.075 mmol =7.5×10⁻⁵ mol

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. Energy absorbed: 14 000 erg s⁻¹×1105 s=1.547×10⁷ erg =1.547 J; absorbed =0.801×1.547=1.239 J
  2. One einstein at λ=4358 Å =435.8 nm: E=(N_Ahc)/(λ)=(6.022×10²³)(6.626×10⁻³⁴)(2.998×10⁸)435.8×10⁻⁹=2.745×10⁵ J mol⁻¹
  3. Einsteins absorbed: 1.2392.745×10⁵=4.514×10⁻⁶ mol of photons
  4. Bromine reacted: 0.075 mmol =7.5×10⁻⁵ mol
  5. Φ=7.5×10⁻⁵4.514×10⁻⁶ Φ=16.6
  6. Φ≫1: one absorbed photon removes about 17 molecules of Br₂. The reaction is a chain: Br₂ →2Br•, then Br atoms add to the C=C bond and the radical abstracts Br from Br₂, regenerating Br.

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Model answer to Q6(b)(ii), 2018 Paper I: solution page 38
Solution page 38 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 6