(a) A new drug has been synthesized and its phase diagram is explored. It is found that near its triple point, vapour pressure over the liquid (Pl) and over the solid (Ps) are given by : ln Pl = − 3010/T + 13·2 and ln Ps = − 3820/T + 16·1. Calculate

Question

(a) A new drug has been synthesized and its phase diagram is explored. It is found that near its triple point, vapour pressure over the liquid (Pl) and over the solid (Ps) are given by : ln Pl = − 3010/T + 13·2 and ln Ps = − 3820/T + 16·1. Calculate the triple point temperature and pressure. Is the new drug solid, gas or liquid at 1 bar, 298 K ? What is ΔHsublimation ? Explain. (15 marks)

(b) (i) What is polarography ? Explain the concentration polarization at the electrode. Give the labelled diagram of polarographic cell assembly. (15 marks)

(ii) Define concentration cell and mention its types. Justify the statement "Fuel cells are energy conversion devices and not energy storage devices." (5 marks)

(c) (i) For the sequential reaction A → B → C, the rate constants are kA = 5 × 10⁶ s⁻¹ and kB = 3 × 10⁶ s⁻¹. Determine the time when the concentration of B is at a maximum. (10 marks)

(ii) In acidic condition, benzyl penicillin (BP) undergoes the following reaction : P₁ ← BP → P₂ (with k₁, k₂) and BP → P₃ (with k₃). Imagine while swallowing penicillin, pH of the stomach is ~3. At this pH, and temperature 22°C, the rate constants for the processes are : k₁ = 7·0 × 10⁻⁴ s⁻¹, k₂ = 4·1 × 10⁻³ s⁻¹, k₃ = 5·7 × 10⁻³ s⁻¹. What is the yield of P₁ formation ? (5 marks)

UPSC Answer Check · Accepted Answer

(a) At the triple point, the vapour pressures over liquid and solid are equal: Pl = Ps.

−3010/T + 13.2 = −3820/T + 16.1

(3820 − 3010)/T = 16.1 − 13.2 = 2.9

810/T = 2.9

T = 810/2.9 = 8100/29 = 279.31 K

Using the liquid equation:

ln P = −3010/(8100/29) + 13.2 = −8729/810 + 13.2 = 2.42346

P = e^2.42346 = 11.28 (in the pressure unit used in the given log equations; taking it as bar gives 11.28 bar)

At 298 K:

ln Pl = −3010/298 + 13.2 = 3.0993, so Pl ≈ 22.2 bar

ln Ps = −3820/298 + 16.1 = 3.2812, so Ps ≈ 26.6 bar

Both equilibrium vapour pressures exceed 1 bar at 298 K. Also, 298 K is above the triple-point temperature 279.31 K, while the liquid-vapour equilibrium pressure at 298 K is about 22.2 bar, much higher than 1 bar. Hence at 1 bar, 298 K the drug is gas.

For sublimation, ln Ps = −ΔH_sublimation/(RT) + constant. The slope is −3820 K. Therefore:

ΔH_sublimation = 3820 × R = 3820 × 8.314 J mol⁻¹

= 31759 J mol⁻¹ = 31.76 kJ mol⁻¹

This assumes ideal-gas behaviour and constant ΔH_sublimation over the temperature range.

(b)(i) Polarography is a voltammetric electroanalytical method in which current is measured as a function of applied potential using a dropping mercury electrode (DME). The DME gives a fresh mercury surface, and the current is usually diffusion-limited; the diffusion current is proportional to concentration of the electroactive species.

Concentration polarization occurs when mass transport of the electroactive species to or from the electrode is slower than electron transfer. Then the concentration at the electrode surface differs from that in the bulk. The electrode potential no longer corresponds exactly to the bulk Nernst equation, and a concentration overpotential is set up. When the surface concentration becomes effectively zero, the current reaches a limiting diffusion current.

Labelled polarographic cell assembly:

```
Potentiometer/voltage source
 |
 microammeter
 |
 Dropping mercury electrode
 (Hg reservoir → capillary → Hg drop)
 |
 Sample solution + supporting electrolyte
 N₂ inlet; thermostat
 |
 Reference electrode (SCE/Ag-AgCl)
 [Pt auxiliary electrode in 3-electrode mode]
```

(b)(ii) A concentration cell is a galvanic cell in which the two half-cells are chemically identical but differ in concentration of the electroactive species. Its emf arises from the concentration difference.

Types:
Electrode concentration cells, e.g. amalgam or gas electrodes at different concentrations/pressures.
Electrolyte concentration cells, which may be with transference or without transference.

Fuel cells are energy conversion devices, not energy storage devices, because the fuel and oxidant are supplied continuously from outside. They convert chemical energy directly into electrical energy as long as reactants are fed. Unlike a battery, they do not store a fixed amount of chemical energy inside the cell itself.

(c)(i) For A → B → C, with first-order constants kA and kB, and initial [B] = 0:

[B] = [A]₀ kA/(kB − kA) (e^(−kA t) − e^(−kB t))

At maximum [B], d[B]/dt = 0:

kA e^(−kA t) = kB e^(−kB t)

e^((kA − kB)t) = kA/kB

t_max = ln(kA/kB)/(kA − kB) (for kA ≠ kB)

Given kA = 5 × 10⁶ s⁻¹, kB = 3 × 10⁶ s⁻¹:

t_max = ln(5/3)/(2 × 10⁶)

= 0.5108256/(2 × 10⁶)

= 2.554 × 10⁻⁷ s

= 0.2554 μs

(c)(ii) For parallel first-order reactions:

BP → P₁ with k₁, BP → P₂ with k₂, BP → P₃ with k₃

Total rate constant for disappearance of BP is:

Σk = k₁ + k₂ + k₃

= 7.0 × 10⁻⁴ + 4.1 × 10⁻³ + 5.7 × 10⁻³

= 1.05 × 10⁻² s⁻¹

The fraction of BP forming P₁ is:

k₁/Σk = (7.0 × 10⁻⁴)/(1.05 × 10⁻²)

= 0.06667

= 1/15

Therefore, the yield of P₁ formation is:

6.67% or 0.0667 mol P₁ per mol BP at long time, assuming P₁ does not undergo further reaction.

(a) A new drug has been synthesized and its phase diagram is explored. It is found that near its triple point, vapour pressure…

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