Q8(b). Moles of HI decomposed from the quantum yield
The quantum yield for photo decomposition of HI(g) at wavelength 254 nm is 1.99. If a total of 3070 J of energy is absorbed then how many moles of HI(g) decomposed?
Answer in brief
n(HI)=Φ× einsteins absorbed =1.99×6.514×10⁻³=1.296×10⁻² mol, about 0.013 mol.
Key points
- Φ=(moles of HI decomposed)/(einsteins (moles of photons) absorbed)
- E=(hc)/(λ)=(6.626×10⁻³⁴)(3×10⁸)2.54×10⁻⁷=7.826×10⁻¹⁹ J
- N_AE=(6.022×10²³)(7.826×10⁻¹⁹)=4.713×10⁵ J =471.3 kJ
- 30704.713×10⁵=6.514×10⁻³ einstein (3.923×10²¹ photons)
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Given: λ=254 nm =2.54×10⁻⁷ m; Φ=1.99; Eₐbs=3070 J; h=6.626×10⁻³⁴ J s, c=3×10⁸ m s⁻¹, N_A=6.022×10²³ mol⁻¹.
- Quantum yield: Φ=(moles of HI decomposed)/(einsteins (moles of photons) absorbed)
- Energy of one photon: E=(hc)/(λ)=(6.626×10⁻³⁴)(3×10⁸)2.54×10⁻⁷=7.826×10⁻¹⁹ J
- Energy of one einstein: N_AE=(6.022×10²³)(7.826×10⁻¹⁹)=4.713×10⁵ J =471.3 kJ
- Einsteins absorbed: 30704.713×10⁵=6.514×10⁻³ einstein (3.923×10²¹ photons)
- HI decomposed: n(HI)=1.99×6.514×10⁻³=1.296×10⁻² mol≈0.013 mol
- Why Φ≈2: primary step HI + hν → H + I; secondary step H + HI → H₂ + I; then I + I → I₂. One photon removes two HI molecules, and there is no chain, so Φ=2 (1.99 measured).
Application. the HI photolysis is used as a chemical actinometer in the gas phase: since Φ is known and almost independent of conditions, the amount of H₂ or I₂ formed measures the light absorbed.
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All parts of Question 8
- 8(a) Electrode reactions and Nernst equation of a Zn | calomel cell10 marks
- 8(b) Moles of HI decomposed from the quantum yield10 marks
- 8(c) IUPAC names of five compounds10 marks
- 8(d) Four short and two long Cu–F bonds in CuF₂: Jahn–Teller distortion10 marks
- 8(e) Magnetic moment of Gd³⁺10 marks