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

(a) Gibbs equation: variation of G, its implications, G–T graph and real-gas effects

(i) Derivation: For a closed system at constant composition with only expansion work, dG = −S dT + V dP. Since G = G(T, P), dG = (∂G/∂T)ₚ dT + (∂G/∂P)_T dP. Comparing the coefficients:

(∂G/∂T)ₚ = −S and (∂G/∂P)_T = V.

Differentiating again: (∂²G/∂T²)ₚ = −Cₚ/T (negative), and (∂²G/∂P²)_T = −Vκ_T (negative). The cross-derivatives give the Maxwell relation (∂S/∂P)_T = −(∂V/∂T)ₚ.

(ii) Implications: G falls with rising T (S is always positive) and rises with rising P (V is positive). The pressure effect is large for gases and small for solids and liquids. Since S(gas) > S(liquid) > S(solid), the slope of G against T is steepest for the gas. The stable phase at given T and P is the one with the lowest G, and two phases are in equilibrium when their G values are equal. In a first-order transition, the first derivatives of G (S and V) change abruptly; the Clapeyron equation dP/dT = ΔH/(TΔV) follows.

(iii) G versus T graph (at constant P): three downward curves (concave), the solid with the smallest slope, the liquid steeper and the gas steepest. The lowest curve is solid at low T, liquid at intermediate T and gas at high T. The intersection of the solid and liquid curves is the melting point T_m; that of the liquid and gas curves is the boiling point T_b. At a pressure below the triple-point pressure the liquid curve never lies lowest, and solid and gas meet directly at the sublimation temperature. The slope changes abruptly at each intersection because S changes.

(iv) Real gas: μ(real) − μ(ideal) = RT ln(f/P) = RT ln φ. (1) Attractive interactions reduce the escaping tendency, so f < P, φ < 1 and the molar Gibbs energy is lower than for the ideal gas. (2) Repulsive interactions raise the escaping tendency, so f > P, φ > 1 and the molar Gibbs energy is higher.

(b) Continuous energy of a free particle versus quantised energy of a particle in a box

For a free particle, V = 0 everywhere and −(ħ²/2m)ψ″ = Eψ, so ψ = Ae^ikx with E = ħ²k²/2m. No boundary condition restricts k, so k and E take any value from 0 upwards.

In a box of length L, the potential is zero inside and infinite outside, so ψ(0) = ψ(L) = 0. The solution ψ = A sin kx must satisfy sin kL = 0, so kL = nπ with n = 1, 2, 3, …, and

Eₙ = ħ²k²/2m = n²h²/(8mL²).

Only standing waves with whole numbers of half-wavelengths fit between the walls, so the energy is quantised; confinement is what causes quantisation. As L increases, the spacing between levels falls as 1/L², and for L → ∞ the continuum returns. The lowest level (n = 1) is not zero, which is the zero-point energy.

(c) Primitive cubic lattice

(i) Eight corner atoms, each shared by 8 cells: 8 × 1/8 = 1 lattice point.

(ii) Each atom touches six neighbours: coordination number 6.

(iii) Edge a = 2r; cell volume = 8r³; atom volume = (4/3)πr³. Packing fraction = π/6 = 0.5236, so the void = (1 − 0.5236) × 100 = 47.6%.

(iv) Given R = 178.1 pm, so a = 2R = 356.2 pm. The body diagonal is √3a, so the distance from the centre to a corner is √3R. For the central sphere to touch the corner atoms, r + R = √3R, and r = (√3 − 1)R = 0.732 × 178.1 = 130.4 pm.

(v) The central sphere touches the eight corner atoms: coordination number 8.

(d) A₂(g) ⇌ 2A(g), 18.5% dissociated at 25 °C and 1 bar

(i) Start with 1 mol A₂ and α = 0.185. At equilibrium, A₂ = 1 − α = 0.815 mol, A = 2α = 0.370 mol, total = 1 + α = 1.185 mol.

Kₚ = p_A²/p_A₂ = 4α²P/(1 − α²) = 4 × (0.185)² × 1 / (1 − 0.0342) = 0.1369/0.9658 = 0.142 (with p° = 1 bar, K° = 0.142).

(ii) Using the van't Hoff equation, ln(K₂/K₁) = (ΔH°/R)(1/T₁ − 1/T₂). The value of ΔH° must be taken from the data given for the reaction; with ΔH° = 57.2 kJ mol⁻¹ (assumed):

1/298.15 − 1/373.15 = 6.741 × 10⁻⁴ K⁻¹, so ln(K₂/K₁) = (57200/8.314)(6.741 × 10⁻⁴) = 4.64, K₂/K₁ = 103, and K₂ = 0.142 × 103 ≈ 14.7 at 100 °C. K rises with T because the reaction is endothermic.

(iii) Δn(gas) = +1, so compression shifts the equilibrium to the side with fewer moles of gas, towards A₂; the degree of dissociation falls. K, which depends only on temperature, is unchanged.