(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) For a zero-order reaction
A → products,
the rate law is
Rate = -d[A]/dt = k[A]⁰ = k.
Separating variables,
d[A] = -k dt.
Integrate from [A]₀ at t = 0 to [A] at time t:
∫ from [A]₀ to [A] d[A] = -k ∫ from 0 to t dt.
So,
[A] - [A]₀ = -k t
or
[A] = [A]₀ - k t.
Hence the rate constant is
k = ([A]₀ - [A])/t.
Its unit is mol dm⁻³ s⁻¹.

At half-life t₁/₂, [A] = [A]₀/2. Substituting,
[A]₀/2 = [A]₀ - k t₁/₂.
Therefore,
k t₁/₂ = [A]₀/2
and
t₁/₂ = [A]₀/(2k).
Thus t₁/₂ ∝ [A]₀, i.e. half-life is directly proportional to initial concentration. This holds as long as the zero-order rate law remains valid and [A] does not reach zero.

(b) Lambert’s law states that the intensity of monochromatic light absorbed by a homogeneous medium is proportional to the path length. Beer’s law states that absorption is proportional to the concentration of the absorbing species. Combining them gives the Lambert-Beer law:
A = log₁₀(I₀/I) = ε c l,
where I₀ is incident intensity, I is transmitted intensity, c is concentration, l is path length, and ε is molar absorptivity.

Derivation: Consider a thin slab of thickness dx at depth x. Let I be the intensity at x. The fractional decrease in intensity is
-dI/I = α c dx,
where α is a proportionality constant. Integrate from x = 0 to x = l, with I = I₀ at x = 0 and I = I at x = l:
-∫ from I₀ to I dI/I = α c ∫ from 0 to l dx.
So,
ln(I₀/I) = α c l.
Converting to base 10,
log₁₀(I₀/I) = (α/2.303) c l = ε c l.
Thus,
A = ε c l.
Transmittance T = I/I₀ = 10⁻ᴬ. For a mixture,
A_total = Σ εᵢ cᵢ l.
Conditions: monochromatic radiation, dilute solution, no scattering or fluorescence, and non-interacting absorbing species. Units: c in mol dm⁻³, l in cm, ε in dm³ mol⁻¹ cm⁻¹.

(c) Deoxyheme contains Fe(II). Dioxygen binds to Fe(II) to give oxyheme, but the Fe–O₂ unit has resonance forms
Fe(II)–O₂ ⇌ Fe(III)–O₂•⁻.
In free heme or when two heme units approach, the bound dioxygen can bridge two iron centres:
Fe(II)–O₂ + Fe(II) → Fe(III)–O₂²⁻–Fe(III)
forming a μ-peroxo dimer. Protonation and cleavage of the O–O bond then produce Fe(III)–OH, i.e. hematin/methemoglobin, along with H₂O₂ or superoxide radicals. Fe(III) heme cannot bind O₂ reversibly, so oxygen transport is destroyed; hematin may also precipitate and damage membranes.

Living systems avert this in several ways. The globin pocket around heme is hydrophobic and sterically hinders heme–heme contact, preventing μ-peroxo dimer formation. Distal histidine (His E7) hydrogen-bonds to bound O₂, stabilising the Fe(II)–O₂ form and blocking protonation. If Fe(III) is formed, methemoglobin reductase (NADH-cytochrome b5 reductase) reduces it back to Fe(II). Superoxide dismutase, catalase and glutathione remove O₂•⁻ and H₂O₂. Thus reversible dioxygen binding is preserved.

(d) Assume an octahedral complex [M(AB)b₂c₂], where AB is an unsymmetrical bidentate ligand, and b and c are different monodentate ligands. Fix the oriented AB chelate on one edge, with A at one vertex and B at the adjacent vertex. The four remaining vertices are inequivalent under rotations that preserve this oriented edge. Arrange two b and two c among these four positions:
Number = 4!/(2!2!) = 6.
Thus there are 6 stereoisomers under proper rotations.

A mirror plane containing the AB edge and the metal fixes A, B, and the two vertices trans to A and B, while interchanging the other two remaining vertices. This leaves unchanged the two arrangements in which:
b occupies the two trans-to-A/B vertices and c occupies the interchanged pair;
b occupies the interchanged pair and c occupies the two trans-to-A/B vertices.

These are two achiral geometrical isomers. The other four arrangements form two enantiomeric pairs. Hence:
Geometrical isomers = 4; stereoisomers = 6 (2 achiral + 2 enantiomeric pairs).

(e)
 (i) 6XeF₄ + 12H₂O → 2XeO₃ + 4Xe + 3O₂ + 24HF.
 (ii) XeF₆ + POF₃ → XeOF₄ + PF₅.
 (iii) XeOF₄ + XeO₃ → 2XeO₂F₂.
 (iv) 3XeF₂ + 2(SO₃)₃ → 3Xe(SO₃F)₂.
 (v) (C₆H₅)₂S + XeF₂ → (C₆H₅)₂SF₂ + Xe.

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

Model answer

What "Derive" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Chemistry 2021 Paper I