Q37 · CSIR NET Chemistry, December 2019
Paper: CSIR NET December 2019 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Medium
Substance 'A' is exposed to 600 nm, 100 W light source for 6626 s, with 50% of the incident light being absorbed. 'A' decomposes according to the reaction $\mathrm{A \rightarrow 2B}$. As a result of irradiation, 0.2 mol B is produced. The quantum yield of the reaction is closest to
(a)$6 \times 10^{6}$
(b)$6 \times 10^{-4}$
(c)$6 \times 10^{-2}$
(d)$6$
Answer
Answer: C ✓ checked by 4AB · confidence medium
The source book printed A; on checking, C is correct — see the explanation.
Φ = moles of A decomposed / einsteins absorbed = 0.1/1.66 ≈ 0.06. The nearest printed option is 3.6×10⁻² (c). The stored key A (1.6×10⁶) is physically impossible.
Explanation
Energy absorbed = 0.5 × 100 W × 6626 s = 3.313 × 10⁵ J.
Energy of one photon = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(600 × 10⁻⁹) = 3.313 × 10⁻¹⁹ J. Photons absorbed = 3.313 × 10⁵/3.313 × 10⁻¹⁹ = 10²⁴, i.e. 10²⁴/(6.022 × 10²³) ≈ 1.66 einstein.
A → 2B, so 0.2 mol of B means 0.1 mol of A has decomposed. Φ = 0.1/1.66 ≈ 0.06, which is closest to 3.6 × 10⁻² (c).
Study loop for Chemical Kinetics