(a) Derive an equation for rate constant of a zero-order reaction. Show that half-life period of the reaction is proportional to the initial concentration of reactant. (10 marks)

(b) State and derive Lambert-Beer law for absorption of light by solut

Question

(a) Derive an equation for rate constant of a zero-order reaction. Show that half-life period of the reaction is proportional to the initial concentration of reactant. (10 marks)

(b) State and derive Lambert-Beer law for absorption of light by solutions. (10 marks)

(c) Give the mechanism of fatal formation of hematin in the binding of dioxygen by heme. How can it be averted by living systems? (10 marks)

(d) How many geometrical isomers and stereoisomers are possible in the coordination compounds of the type (AB)Mb₂c₂(AB—bidentate ligand)? (10 marks)

(e) Complete the following reactions:
(i) XeF₄ + 12H₂O —→ ____
(ii) XeF₆ + ____ —→ XeOF₄ + PF₅
(iii) XeOF₄ + ____ —→ 2XeO₂F₂
(iv) 3XeF₂ + 2(SO₃)₃ —→ ____
(v) ____ + XeF₂ —→ (C₆H₅)₂ SF₂ + Xe (10 marks)

UPSC Answer Check · Accepted Answer

(a) Zero-order rate constant and half-life
For A → products, a zero-order rate is independent of concentration: −d[A]/dt = k. Rearranging, d[A] = −k dt. Integrating from [A]₀ at t = 0 to [A] at t: [A]₀ − [A] = kt, so k = ([A]₀ − [A])/t, with units mol L⁻¹ s⁻¹. A plot of [A] against t is a straight line of slope −k.
At the half-life, [A] = [A]₀/2, so [A]₀ − [A]₀/2 = k t₁/₂, giving t₁/₂ = [A]₀/2k. Hence t₁/₂ ∝ [A]₀, whereas for a first-order reaction it is independent of [A]₀. Examples: decomposition of NH₃ on hot tungsten, and enzyme reactions at saturating substrate.

(b) Lambert–Beer law
Statement: when monochromatic light passes through an absorbing solution, A = log(I₀/I) = εcl, where I₀ and I are incident and transmitted intensities, ε the molar absorptivity (L mol⁻¹ cm⁻¹), c the concentration (mol L⁻¹) and l the path length (cm).
Derivation: Lambert’s law states that the fraction of light absorbed by a thin layer dx is independent of the incident intensity: −dI/I = k′dx. Beer’s law states that k′ is proportional to concentration, k′ = k″c. So −dI/I = k″c dx. Integrating from 0 to l, with I = I₀ at x = 0: ln(I₀/I) = k″cl. Converting to base 10, log(I₀/I) = (k″/2.303) cl = εcl, so A = εcl, and I = I₀ 10^(−εcl).
Limitations: deviations occur at high concentrations (above about 0.01 M), with scattering, fluorescence, or polychromatic light and chemical equilibria such as association.

(c) Hematin formation and its prevention
Heme (Fe²⁺ porphyrin) binds O₂ reversibly as Fe(III)–O₂⁻ (superoxide-like). In a free, unprotected heme the bound O₂ is attacked by a second Fe(II) heme, giving a μ-peroxo dimer Fe(III)–O–O–Fe(III), which splits into two ferryl Fe(IV)=O species. These react with more Fe(II) heme to give the stable μ-oxo dimer Fe(III)–O–Fe(III), called hematin. It is irreversibly oxidised and cannot carry oxygen.
Living systems prevent it by holding each heme in a hydrophobic pocket of globin (myoglobin, haemoglobin), so two hemes cannot approach; the proximal histidine binds the iron on one side and the distal histidine hydrogen-bonds to bound O₂ and blocks the other face. Any Fe(III) (methaemoglobin) formed is reduced back by NADH–cytochrome b₅ reductase (methaemoglobin reductase), while superoxide dismutase, catalase and glutathione remove reactive oxygen species.

(d) Isomers of (AB)Mb₂c₂
The complex is octahedral, with the unsymmetrical bidentate AB spanning two cis positions (A and B), and four remaining positions (trans to A, trans to B, and the two mutually trans positions perpendicular to the AB plane) filled by two b and two c.
Arrangements: (1) b trans to A and B, c trans to each other (c,c trans): achiral; (2) c trans to A and B, b,b trans: achiral; (3) b trans to A, c trans to B, b,b cis and c,c cis: chiral, giving a pair of enantiomers; (4) b trans to B, c trans to A: chiral, again a pair.
Hence there are 4 geometrical isomers and 6 stereoisomers in total (two achiral forms plus two enantiomeric pairs).

(e) Xenon reactions
(i) 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂ (hydrolysis, partly disproportionation; Xe(IV) → Xe(0) and Xe(VI)).
(ii) XeF₆ + POF₃ → XeOF₄ + PF₅.
(iii) XeOF₄ + XeO₃ → 2XeO₂F₂.
(iv) 3XeF₂ + 2(SO₃)₃ → 3Xe(OSO₂F)₂ (xenon bis(fluorosulfate)), i.e., XeF₂ + 2SO₃ → Xe(SO₃F)₂.
(v) (C₆H₅)₂S + XeF₂ → (C₆H₅)₂SF₂ + Xe (oxidative fluorination).

(a) Derive an equation for rate constant of a zero-order reaction. Show that half-life period of the reaction is proportional to…

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