(a) (i) Derive the expressions of chemical potential of a component in a mixture system in terms of extensive thermodynamic…

(a)(i) Chemical potential in terms of U, H, A and G (10 marks)

For a mixture, U = U(S, V, n₁, n₂, …), so

dU = T dS − P dV + Σμᵢ dnᵢ,

and hence μᵢ = (∂U/∂nᵢ)_S,V,nⱼ.

Enthalpy H = U + PV gives dH = dU + P dV + V dP = T dS + V dP + Σμᵢ dnᵢ, so μᵢ = (∂H/∂nᵢ)_S,P,nⱼ.

Helmholtz energy A = U − TS gives dA = dU − T dS − S dT = −S dT − P dV + Σμᵢ dnᵢ, so μᵢ = (∂A/∂nᵢ)_T,V,nⱼ.

Gibbs energy G = H − TS gives dG = −S dT + V dP + Σμᵢ dnᵢ, so μᵢ = (∂G/∂nᵢ)_T,P,nⱼ.

Thus μᵢ = (∂U/∂nᵢ)_S,V,nⱼ = (∂H/∂nᵢ)_S,P,nⱼ = (∂A/∂nᵢ)_T,V,nⱼ = (∂G/∂nᵢ)_T,P,nⱼ. Only the last is a partial molar quantity (at constant T and P, the usual laboratory conditions), and it is the one usually called the chemical potential: μᵢ = Ḡᵢ.

(a)(ii) Cadmium-bismuth phase diagram (10 marks)

Cd and Bi are completely miscible as liquids and completely immiscible as solids, so the system is a simple eutectic. On the diagram, the liquidus runs from the melting point of Cd (321 °C) and from that of Bi (271 °C), each falling as the second component is added, and the two lines meet at the eutectic point E (61% Bi, 146 °C). A horizontal line through E at 146 °C is the eutectic (solidus) line. The regions are: above the liquidus, a single liquid melt; to the left of E, below the liquidus, solid Cd + melt; to the right of E, solid Bi + melt; below 146 °C, solid Cd + solid Bi.

Cooling curves. Using the condensed phase rule F = C − P + 1 (pressure fixed), with C = 2:

• Melt of composition 30% Bi (Cd-rich): the temperature falls smoothly until the liquidus is reached, where Cd begins to solidify. The slope changes (a kink) because latent heat is released, and F = 1. As Cd separates, the melt becomes richer in Bi until it reaches the eutectic composition at 146 °C. Then P = 3 (Cd, Bi, melt) and F = 0, so the temperature stays constant (a horizontal halt) while the eutectic solidifies; after this only solids remain and cooling resumes.
• Eutectic melt (61% Bi): the temperature falls without a kink until 146 °C, then shows a single long halt until solidification is complete (F = 0), and then falls again.
• Melt richer in Bi (for example 85% Bi): the behaviour is the same as for the Cd-rich melt, with Bi separating at the kink, followed by the eutectic halt at 146 °C.

The length of the eutectic halt is greatest for the eutectic composition and decreases towards the pure components, which show only a single halt at their melting points.

(a)(iii) Acetone-dry ice freezing bath (5 marks)

Dry ice is solid CO₂, which sublimes at about −78 °C (195 K) at 1 atm and absorbs a large latent heat of sublimation (about 25 kJ mol⁻¹) as it does so. Acetone freezes only at about −95 °C, so it remains liquid at this temperature and acts as a fluid medium that makes good thermal contact with the flask and with all the pieces of dry ice. Heat flowing in from the vessel is used to sublime CO₂, which escapes as gas, so the bath stays at a nearly constant temperature of about −78 °C for as long as solid CO₂ remains, and cannot go lower because the CO₂ is at its sublimation point at 1 atm. The temperature is fixed by the phase change of CO₂, not by a eutectic, and acetone serves as the heat-transfer liquid. It is used to condense volatile substances and to cool reactions.

(b) Hydrated Zn²⁺ ion in an electric field (10 marks)

Data: E = 5 V cm⁻¹ = 500 V m⁻¹, v = 2.74 × 10⁻⁵ m s⁻¹, η = 0.890 mPa s = 8.90 × 10⁻⁴ Pa s, z = 2, T = 298 K.

(i) At steady speed the electric force equals Stokes' viscous drag: zeE = 6πηrv.

r = zeE/(6πηv) = (2 × 1.602 × 10⁻¹⁹ × 500)/(6π × 8.90 × 10⁻⁴ × 2.74 × 10⁻⁵) = 1.602 × 10⁻¹⁶/4.60 × 10⁻⁷ = 3.5 × 10⁻¹⁰ m ≈ 0.35 nm (350 pm).

This is larger than the ionic radius of Zn²⁺ (about 0.074 nm), as expected for a hydrated ion.

(ii) By the Stokes-Einstein relation, D = kT/(6πηr) = (1.381 × 10⁻²³ × 298)/(6π × 8.90 × 10⁻⁴ × 3.49 × 10⁻¹⁰) = 4.12 × 10⁻²¹/5.85 × 10⁻¹² ≈ 7.0 × 10⁻¹⁰ m² s⁻¹.

Check with the Nernst-Einstein relation: the mobility is u = v/E = 5.48 × 10⁻⁸ m² V⁻¹ s⁻¹, and D = uRT/(zF) = 5.48 × 10⁻⁸ × 8.314 × 298/(2 × 96 485) = 7.0 × 10⁻¹⁰ m² s⁻¹, in agreement.

(c)(i) Energy of one einstein of 800 nm radiation (5 marks)

One einstein is one mole of photons. The energy of one photon is E = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸)/(800 × 10⁻⁹) = 2.48 × 10⁻¹⁹ J.

Energy per einstein = N_A hc/λ = 6.022 × 10²³ × 2.48 × 10⁻¹⁹ = 1.495 × 10⁵ J mol⁻¹ ≈ 149.5 kJ per einstein (about 150 kJ).

(c)(ii) Photodecomposition of HI (10 marks)

Steps: (1) HI + hν → H + I, rate I_a (the rate of absorption of photons, in moles per unit volume per second); (2) H + HI → H₂ + I, rate constant k₂; (3) I + I → I₂, rate constant k₃.

Steady state for H: d[H]/dt = I_a − k₂[H][HI] = 0, so k₂[H][HI] = I_a.

Steady state for I: d[I]/dt = I_a + k₂[H][HI] − 2k₃[I]² = 0, so 2k₃[I]² = 2I_a and [I] = (I_a/k₃)^½.

HI is consumed in steps (1) and (2): −d[HI]/dt = I_a + k₂[H][HI] = I_a + I_a = 2I_a.

Quantum yield φ = (molecules of HI reacting)/(photons absorbed) = 2I_a/I_a = 2. One photon decomposes two HI molecules.

Order: the rate = 2I_a does not depend on [HI] itself, so the reaction is zero order in HI when I_a is constant (all the incident light is absorbed); when light is only weakly absorbed, I_a ∝ [HI] and the reaction shows apparent first order. The rate is first order in absorbed light intensity.