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Question #44807

Difficulty: 8/10 | 🟢 Free

The Question

The photochemical dechlorination of chlorobenzene (\ceC6H5Cl\ce{C6H5Cl}) in hexane at 298 K298 \text{ K} is driven by continuous UV irradiation (λ=254 nm\lambda = 254 \text{ nm}). The strong C-Cl bond, stabilized by resonance (+M+M effect of Cl), necessitates this high-energy initiation. The reaction mixture absorbs UV photons at a constant rate of 1.0×10−3 Einstein L−1 s−11.0 \times 10^{-3} \text{ Einstein L}^{-1} \text{ s}^{-1}. Given that the quantum yield (Φ\Phi) for the formation of phenyl radicals is 0.100.10 and the reaction follows zero-order kinetics with respect to the substrate, determine the rate constant kk in mol L−1 min−1\text{mol L}^{-1} \text{ min}^{-1} and the half-life (t1/2t_{1/2}) of the reaction if the initial concentration of \ceC6H5Cl\ce{C6H5Cl} is 0.12 M0.12 \text{ M}.

Options

  • A.

    6.0×10−3 mol L−1 min−16.0 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1} and 10 min10 \text{ min}

  • B.

    1.0×10−4 mol L−1 min−11.0 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1} and 600 min600 \text{ min}

  • C.

    6.0×10−3 mol L−1 min−16.0 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1} and 115.5 min115.5 \text{ min}

  • D.

    6.0×10−2 mol L−1 min−16.0 \times 10^{-2} \text{ mol L}^{-1} \text{ min}^{-1} and 1 min1 \text{ min}

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Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Zero Order Rate Integration|Aryl Halide Resonance|Units of Rate Constant (k)

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Photochemical Kinetics and Zero-Order Integration.

The 'Chemical Exception' is the resonance stabilization of the aryl halide (C-Cl bond), which prevents standard thermal degradation and forces reliance on UV photon flux, leading to zero-order kinetics.

The primary trap is the reflexive use of the first-order half-life formula (t1/2=0.693/kt_{1/2} = 0.693/k) for a degradation reaction (Option C).

A secondary trap involves failing to convert the rate constant units from seconds to minutes (Option B) or ignoring the quantum yield entirely (Option D).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterChemical Kinetics - Class 12
ConceptZero Order Rate Integration
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree