(a) The Maxwell distribution of molecular speed of a gaseous system is dN_c = 4πN ( M / (2πRT) )³/2 c² e^-Mc²/2RT dc (Symbols…

(a) Maxwell speed distribution (15 marks)

The distribution is (1/N)(dN/dc) = 4π(M/2πRT)^(3/2) c² e^(−Mc²/2RT). At small c the factor c² dominates and the curve rises from zero (it is zero at c = 0); at large c the exponential dominates and the curve falls to zero. It therefore has a single maximum at the most probable speed c_mp = (2RT/M)^½, obtained from d/dc = 0. The area under every curve is 1, since the fractions of molecules add up to unity.

(i) Effect of temperature (8 marks). At fixed M, c_mp rises as T^½, so on going from T₁ to T₂ (T₂ > T₁) the maximum moves to higher speed. The peak height is proportional to 1/c_mp ∝ T^(−½), so the curve becomes lower and broader, with a larger fraction of molecules at high speeds and a smaller fraction near c_mp. Mean speed ∝ T^½ and the spread of speeds increases, while the area stays 1.

(ii) Effect of molar mass (7 marks). At fixed T, c_mp ∝ M^(−½), so for M₂ > M₁ the heavier gas has its maximum at lower speed. The peak height ∝ M^½, so the curve is taller and narrower: heavier molecules move more slowly and have a smaller spread of speeds, whereas light molecules have a flatter curve extending to high speeds. Again the area is 1.

(b)(i) Interfacial tension between immiscible liquids (5 marks)

The interfacial tension γ_AB between two mutually saturated immiscible liquids is given approximately by Antonoff's rule: γ_AB = γ_A − γ_B (for γ_A > γ_B), where γ_A and γ_B are the surface tensions of the pure liquids. It is therefore smaller than γ_A, and it is the difference of the two. The reason is that each liquid is partly saturated with the other, and at the interface the attractions across the boundary offset part of the surface tension of each liquid. The rule is approximate; a fuller treatment, γ_AB = γ_A + γ_B − 2√(γ_A γ_B), also gives a value smaller than γ_A + γ_B.

(b)(ii) Pressure difference across an ethanol droplet (10 marks)

For a spherical droplet with a single liquid-vapour surface, the Young-Laplace equation gives ΔP = 2γ/r.

γ = 22.39 mN m⁻¹ = 22.39 × 10⁻³ N m⁻¹ and r = 220 nm = 2.20 × 10⁻⁷ m.

ΔP = 2 × 22.39 × 10⁻³/(2.20 × 10⁻⁷) = 4.478 × 10⁻²/2.20 × 10⁻⁷ = 2.04 × 10⁵ Pa.

The pressure inside the droplet is about 2.04 × 10⁵ Pa (roughly 2 bar or 2 atm) higher than outside; it is the excess pressure on the concave side of the curved surface, and it is large because the radius is very small.

(c)(i) Enthalpy of formation of HI(g) at 225 °C (10 marks)

The reaction is ½H₂(g) + ½I₂(g) → HI(g). The given ΔfH° of HI(g) at 25 °C, 24.7 kJ mol⁻¹, refers to solid iodine, so we first convert I₂(s) to I₂(g) using the sublimation enthalpy 62.3 kJ mol⁻¹:

ΔH(298 K) = ΔfH°(HI) − ½ΔsubH(I₂) = 24.7 − ½ × 62.3 = 24.7 − 31.15 = −6.45 kJ mol⁻¹.

By Kirchhoff's law, ΔH(T₂) = ΔH(T₁) + ΔC_p (T₂ − T₁), with ΔC_p = C_p(HI) − ½C_p(H₂) − ½C_p(I₂) = 29.87 − ½(29.08) − ½(33.56) = 29.87 − 14.54 − 16.78 = −1.45 J K⁻¹ mol⁻¹.

ΔH(498 K) = −6450 + (−1.45)(498 − 298) = −6450 − 290 = −6740 J mol⁻¹ ≈ −6.74 kJ mol⁻¹.

(c)(ii) ΔG° for ½N₂ + 3/2 H₂ → NH₃ at 400 K (10 marks)

ΔS° = S°(NH₃) − ½S°(N₂) − 3/2 S°(H₂) = 192.45 − 95.805 − 196.02 = −99.375 J K⁻¹ mol⁻¹.

At 298 K, ΔH° = ΔG° + TΔS° = −16 496 + 298 × (−99.375) = −16 496 − 29 614 = −46 110 J mol⁻¹ ≈ −46.11 kJ mol⁻¹.

Taking ΔH° and ΔS° as independent of temperature, at 400 K:

ΔG°(400 K) = ΔH° − TΔS° = −46 110 − 400 × (−99.375) = −46 110 + 39 750 = −6 360 J mol⁻¹ ≈ −6.36 kJ mol⁻¹.

The reaction is still spontaneous at 400 K but less favourable than at 298 K, as expected for an exothermic reaction with negative ΔS°.