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

(a) Triple point, state at 1 bar and 298 K, and ΔH(sublimation)

Triple point. Solid, liquid and vapour coexist, so the two vapour pressures are equal (P_l = P_s):

−3010/T + 13.2 = −3820/T + 16.1

810/T = 2.9, so T_tp = 279.3 K (≈ 6.2 °C)

ln P_tp = −3010/279.3 + 13.2 = −10.777 + 13.2 = 2.424, so P_tp = 11.3 bar

(The equations are taken to give P in bar, the unit in which the question asks about 1 bar.)

Enthalpies from the slopes. The Clausius–Clapeyron equation is ln P = −ΔH/RT + constant, so each slope equals −ΔH/R.

• ΔH_sub = 3820 K × 8.314 J K⁻¹ mol⁻¹ = 3.18 × 10⁴ J mol⁻¹ = 31.8 kJ mol⁻¹
• ΔH_vap = 3010 × 8.314 = 25.0 kJ mol⁻¹
• ΔH_fus = ΔH_sub − ΔH_vap = 6.7 kJ mol⁻¹, which is positive, as it must be. The steeper sublimation line is the solid–vapour curve.

Phase at 1 bar and 298 K.

• At 298 K, above T_tp, the stable condensed phase would be the liquid. Its vapour pressure is ln P_l = −3010/298 + 13.2 = 3.099, so P_l = 22.2 bar.
• The applied pressure (1 bar) is far below this equilibrium vapour pressure, so any liquid would evaporate completely. The drug is a gas at 1 bar and 298 K.
• More generally, the triple-point pressure (11.3 bar) is above 1 bar. Just as for CO₂ (triple point 5.1 bar), the liquid cannot exist at 1 bar. On warming at 1 bar the solid sublimes directly, at the temperature where P_s = 1 bar: T = 3820/16.1 = 237 K (−36 °C).
• A liquid form would need more than 11.3 bar.

Assumptions: ideal vapour, negligible molar volume of the condensed phase, and ΔH constant over the range. If the pressures were in a much smaller unit (Pa or Torr), P_l at 298 K would be below 1 bar and the conclusion would change to "liquid". The pressure unit therefore decides the answer.

(b)(i) Polarography, concentration polarisation and the polarographic cell

Polarography (J. Heyrovsky, 1922; Nobel Prize 1959) is a voltammetric technique. A slowly increasing negative potential is applied to a dropping mercury electrode (DME) relative to a non-polarisable reference (SCE), and the resulting current is recorded. The current–voltage curve (the polarogram) gives:

• the half-wave potential E½, which is characteristic of the species (qualitative analysis);
• the diffusion current i_d, which is proportional to its concentration (quantitative analysis).

Why a DME?

• It presents a fresh, clean, reproducible surface every few seconds.
• Mercury's high hydrogen overvoltage lets ions such as Zn²⁺, Cd²⁺, Pb²⁺ and Ni²⁺ be reduced in aqueous solution before H₂ is evolved.
• The tiny electrode area means very small currents and negligible change in the bulk solution.

Cell components

• A DME fed from a mercury reservoir whose height fixes the drop time.
• A saturated calomel reference electrode (or a large Hg pool), joined through a KCl–agar salt bridge.
• The test solution containing an excess (about 50–100-fold) of supporting electrolyte such as KCl. This carries the current, so the electroactive ion moves only by diffusion and not by electrical migration.
• A trace of gelatin or Triton X-100 to suppress current "maxima".
• An N₂ inlet to remove dissolved O₂, which is itself reduced in two waves.
• A battery with a potential divider, and a galvanometer (or microammeter) in series.

Concentration polarisation

When the DME potential reaches the reduction potential of an ion (for example Cd²⁺ + 2e⁻ + Hg → Cd(Hg)), the ions at the electrode surface are consumed faster than they can be replaced. A thin diffusion layer forms, in which the concentration rises from C(0) at the surface to the bulk value C* over a distance δ.

• The current is set by the rate of diffusion across this layer: i ∝ (C* − C(0))/δ.
• The electrode potential now follows the surface concentration, not the bulk one. It is therefore shifted from its equilibrium value by a concentration overpotential, η_conc = (RT/nF) ln[C(0)/C*] = (RT/nF) ln(1 − i/i_d).
• As the potential becomes more negative, C(0) falls towards zero. The current then reaches a maximum set by diffusion alone, the limiting (diffusion) current, which does not rise further with potential. At this stage the electrode is fully concentration-polarised.
• The half-wave point (i = i_d/2, where C(0) = C*/2) is independent of concentration. The wave follows the Heyrovsky–Ilkovic equation, E = E½ + (0.0592/n) log[(i_d − i)/i] at 25 °C.
• Ilkovic equation: i_d = 708 n D^½ m^⅔ t^⅙ C (maximum current; 607 for the average current). Here i_d is in µA, D in cm² s⁻¹, m (mercury flow) in mg s⁻¹, t (drop time) in s, and C in mmol L⁻¹. Hence i_d ∝ C.
• Concentration polarisation differs from activation polarisation, which comes from a slow electron-transfer step rather than slow mass transport.

(b)(ii) Concentration cells; why fuel cells are converters, not stores

A concentration cell is a galvanic cell whose two half-cells contain the same electrode and electrolyte but differ in the concentration (activity) or pressure of one species. There is no net chemical change: the "reaction" is only the transfer of material from higher to lower concentration. Hence E° = 0 and E = (RT/nF) ln(a₂/a₁).

Types

Electrode concentration cells: two electrodes of the same substance at different concentrations dip into one solution.

• Hydrogen electrodes at different pressures: Pt, H₂(p₁) | HCl | H₂(p₂), Pt, with E = (RT/2F) ln(p₁/p₂).
• Amalgams of different metal content: Zn(Hg)(c₁) | ZnSO₄ | Zn(Hg)(c₂).

Electrolyte concentration cells: identical electrodes dip into solutions of different concentration.

• Without transference (liquid junction removed by a salt bridge): e.g. Zn | Zn²⁺(c₁) || Zn²⁺(c₂) | Zn, with E = (RT/2F) ln(c₂/c₁).
• With transference (direct liquid junction): e.g. Ag | AgNO₃(a₁) | AgNO₃(a₂) | Ag. Here E = 2t₋(RT/F) ln(a₂/a₁) involves the transport number, which is how transport numbers are measured.

Fuel cells as energy-conversion devices

• Open system: fuel (H₂, CH₃OH, natural gas) and oxidant (O₂ or air) are supplied continuously from outside, and the products (H₂O, CO₂) are removed. The electrodes, catalyst and electrolyte are not consumed. They only convert the Gibbs energy of the reaction, for example H₂ + ½O₂ → H₂O with ΔG° = −237 kJ mol⁻¹, into electrical work.
• Energy capacity is decoupled from power: the energy available is set by the size of the external fuel tank, not by the cell. The cell determines only the power, much as an engine does.
• A battery is a closed store: the reactants are sealed inside, the energy is fixed by the mass of active material, and the cell is exhausted when that material is used up (a primary cell), or must be recharged electrically (a secondary cell).
• Direct conversion is efficient: because conversion is direct and not via heat, it escapes the Carnot limit. The theoretical efficiency is ΔG/ΔH = 237/286 ≈ 83% for H₂–O₂, and practical values are about 40–60%.

(c)(i) Time of maximum [B] for A → B → C

With first-order steps and [B]₀ = 0:

• d[A]/dt = −kA[A], so [A] = [A]₀e^(−kA t)
• d[B]/dt = kA[A] − kB[B], so [B] = [A]₀ kA/(kB − kA) · (e^(−kA t) − e^(−kB t))

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

kA e^(−kA t) = kB e^(−kB t), so e^((kA − kB)t) = kA/kB, and t_max = ln(kA/kB)/(kA − kB)

t_max = ln(5 × 10⁶/3 × 10⁶)/(5 × 10⁶ − 3 × 10⁶) = 0.5108/(2 × 10⁶ s⁻¹) = 2.55 × 10⁻⁷ s (0.255 µs)

At this time [B]max = [A]₀(kA/kB)^(kB/(kB − kA)) = (5/3)^(−1.5)[A]₀ = 0.465[A]₀.

(c)(ii) Yield of P₁ from benzylpenicillin at pH ≈ 3

BP decomposes by three parallel first-order paths, so it disappears with k_total = k₁ + k₂ + k₃:

k_total = 7.0 × 10⁻⁴ + 4.1 × 10⁻³ + 5.7 × 10⁻³ = 1.05 × 10⁻² s⁻¹

For parallel first-order reactions, [P₁]/[P₂]/[P₃] = k₁ : k₂ : k₃ at all times, so the fraction of BP that ends up as P₁ is:

Yield of P₁ = k₁/k_total = 7.0 × 10⁻⁴/1.05 × 10⁻² = 0.067 = 6.7%

The other products form in yields of P₂ = 39.0% and P₃ = 54.3%.

Significance: the half-life of BP in the stomach is ln 2/k_total = 0.693/0.0105 ≈ 66 s, so most of an oral dose of penicillin G is destroyed by stomach acid within minutes. This is why benzylpenicillin is given by injection, and why acid-stable analogues (e.g. phenoxymethylpenicillin, penicillin V) are used orally.