(a) Using the Gibbs equation for a closed system in the absence of non-expansion work at constant composition, answer the…

(a)(i) For a closed system of constant composition and no non-expansion work, the Gibbs equation is dG = −S dT + V dP. But G = G(T, P), so its total differential is dG = (∂G/∂T)_P dT + (∂G/∂P)_T dP. Comparing coefficients gives (∂G/∂T)_P = −S and (∂G/∂P)_T = V. Also, (∂²G/∂T²)_P = −(∂S/∂T)_P = −Cₚ/T, (∂²G/∂P²)_T = (∂V/∂P)_T = −V κ_T, and the cross-derivative gives the Maxwell relation (∂S/∂P)_T = −(∂V/∂T)_P. For a phase change, the Gibbs–Helmholtz relation is ∂(ΔG/T)/∂T = −ΔH/T². These relations are valid for a closed system, constant composition, and only P–V work.

(a)(ii) At constant P, G decreases as T increases because S is positive. At constant T, G increases as P increases because V is positive. The pressure effect is large for gases, small for liquids and solids. Since entropy follows S_gas > S_liquid > S_solid, the G–T slope −S is most negative for the gas and least negative for the solid. At a given T and P, the stable phase is the one with the lowest G. A phase transition occurs when the G curves of two phases intersect, i.e. when ΔG = 0. For a first-order transition, S and V change discontinuously, so the slope of G changes abruptly. The Clapeyron equation follows: dP/dT = ΔS/ΔV = ΔH/(T ΔV).

(a)(iii) At constant pressure, the required G versus T graph has three curves: solid (least steep negative slope), liquid (intermediate slope), and gas (steepest negative slope). At low T the solid curve is lowest; at intermediate T the liquid curve is lowest; at high T the gas curve is lowest. The intersections are the phase transition temperatures:

• Tₘ where G_solid = G_liquid, the melting point.
• T_b where G_liquid = G_gas, the boiling point. If the pressure is below the triple-point pressure, the solid and gas curves intersect directly at the sublimation temperature T_sub. At the triple point, all three curves meet. A sketch is: G ↑ | S | \ | L | V | \ |_______________→ T 0 Tₘ T_b At Tₘ and T_b, ΔG = 0 and the slope changes discontinuously because entropy changes.

(a)(iv) For a real gas at temperature T and pressure P, μ_real = μ° + RT ln(f/p°), while for the ideal gas, μ_ideal = μ° + RT ln(P/p°). Hence ΔGₘ = G_real − G_ideal = RT ln(f/P) = RT ln φ.

1. Attractive interactions lower the escaping tendency, so f < P and φ < 1. Hence ln φ < 0, and the molar Gibbs free energy is lower than the normal ideal-gas value.
2. Repulsive interactions increase the escaping tendency, so f > P and φ > 1. Hence ln φ > 0, and the molar Gibbs free energy is higher than the normal ideal-gas value. Equivalently, at the same T and P, attractive forces reduce V relative to ideal, giving a negative contribution to ∫V dP; repulsive forces increase V, giving a positive contribution.

(b) For a free particle, V = 0 everywhere. The time-independent Schrödinger equation in one dimension is −(ħ²/2m)(d²ψ/dx²) = Eψ. Its solutions are plane waves ψₖ(x) = A e^ikx, with k any real number. The energy is E = ħ²k²/(2m). Since k is not restricted by any boundary condition, E can take any value from 0 to ∞ continuously. Momentum p = ħk is also continuous. A single plane wave is not normalizable, but normalizable wave packets can be formed by superposing many k values.

For a particle in a one-dimensional box of length L, the potential is zero inside 0 < x < L and infinite outside. The boundary conditions are ψ(0) = 0 and ψ(L) = 0. Therefore ψ(x) = A sin(kx), and sin(kL) = 0, so kL = nπ, n = 1, 2, 3, ... Thus kₙ = nπ/L, and Eₙ = ħ²kₙ²/(2m) = n²h²/(8mL²). Only discrete energies are allowed because the confined wave must form standing waves with an integer number of half-wavelengths between the walls. As L → ∞, the spacing ΔE = (2n + 1)h²/(8mL²) → 0, so the continuum of the free particle is recovered.

(c)(i) A primitive cubic unit cell has atoms only at the eight corners. Each corner is shared by eight unit cells, so the number of lattice points per unit cell is 8 × 1/8 = 1.

(c)(ii) In primitive cubic packing, each atom touches six nearest neighbours along the three axes. Hence the coordination number is 6.

(c)(iii) For touching atoms, edge length a = 2r. The unit-cell volume is a³ = 8r³. One atom occupies (4/3)πr³. Packing fraction = [(4/3)πr³]/8r³ = π/6 = 0.5236. Therefore percentage void volume = (1 − π/6) × 100 = (6 − π)/6 × 100 = 47.64%.

(c)(iv) Given R = 178.1 pm. The cube edge is a = 2R = 356.2 pm. The distance from the body centre to a corner atom is (√3/2)a = √3 R. If the central sphere just touches the corner atoms, r + R = √3 R. Hence r = R(√3 − 1) = 178.1 × 0.7320508 pm = 130.38 pm.

(c)(v) The central sphere touches the eight corner atoms of the primitive cubic unit cell. Therefore its coordination number is 8.

(d)(i) Let initial moles of A₂ be 1. Degree of dissociation α = 0.185. At equilibrium: A₂ = 1 − α = 0.815, A = 2α = 0.370, total moles = 1 + α = 1.185. At total pressure P = 1 bar, p_A₂ = [(1 − α)/(1 + α)]P, p_A = [2α/(1 + α)]P. Thus Kₚ = (p_A)²/p_A₂ = 4α²P/(1 − α²). Putting α = 0.185, α² = 0.034225, Kₚ = 4 × 0.034225 × 1 / (1 − 0.034225) = 0.1369 / 0.965775 = 0.14175 bar. Taking p° = 1 bar, K° ≈ 0.1418.

(d)(ii) Use the van’t Hoff equation: ln(K₂/K₁) = −ΔH°/R (1/T₂ − 1/T₁) = ΔH°/R (1/T₁ − 1/T₂). T₁ = 298.15 K, T₂ = 373.15 K, ΔH° = 57.2 kJ mol⁻¹ = 57200 J mol⁻¹, R = 8.314 J K⁻¹ mol⁻¹. 1/298.15 − 1/373.15 = 6.7413 × 10⁻⁴ K⁻¹. Therefore ln(K₂/K₁) = (57200/8.314)(6.7413 × 10⁻⁴) = 4.638. So K₂/K₁ = e⁴.638 ≈ 103.3. K₂ = 0.14175 × 103.3 ≈ 14.65 bar. Thus K_eq at 100 °C is about 14.65 (or dimensionless relative to p° = 1 bar). The increase with temperature is expected because the reaction is endothermic.

(d)(iii) The reaction A₂(g) → 2A(g) increases the number of gaseous moles: Δn_gas = 2 − 1 = +1. By Le Chatelier’s principle, compression increases pressure and shifts the equilibrium towards the side with fewer gas molecules, i.e. towards A₂. Thus the degree of dissociation of A₂ decreases. At constant temperature, Kₚ itself remains unchanged; only the equilibrium position changes.