The ductility and malleability typical of metals has made metals essential structural elements in mo — Physical Chemistry — Thermodynamics Chemistry Question
The Tin Pest: Solid State Structure and Phase Equilibrium
The ductility and malleability typical of metals has made metals essential structural elements in modern construction. The thermodynamically stable form of elemental tin at 298 K and ambient pressure is white tin, which has mechanical properties typical of metals and therefore can be used as a building material. At lower temperatures, however, a second allotrope of tin, gray tin, becomes thermodynamically stable. Because gray tin is much more brittle than white tin, structural elements made of tin that are kept at low temperatures for prolonged periods may crumble and fail. Because this failure resembles a disease, it has been termed the "tin pest".
Given the thermodynamic data below, calculate the temperature at which gray Sn is in equilibrium with white Sn (at 1 bar = 105 Pa pressure).
Model Answer
The two phases are in equilibrium if ∆Gº = 0 for Sn (white) → Sn (gray).
∆H° = (–2.016 kJ mol –1) – (0.00 kJ mol–1) = –2.016 kJ mol–1
∆S° = 44.14 J mol –1 K–1 – 51.18 J mol–1 K–1 = –7.04 J mol–1 K–1
At equilibrium, ∆G° = 0 = ∆H° – T∆S° = –2016 J mol –1 – T(–7.04 J mol–1 K–1)
T = (2016 J mol–1)/(7.04 J mol–1·K–1) = 286.4 K = 13.2 °C
Crystalline white tin has a somewhat complex unit cell. It is tetragonal, with a = b = 583.2 pm and c = 318.1 pm, with 4 atoms of Sn per unit cell. Calculate the density of white tin in g cm–3.
Model Answer
The volume of the tetragonal unit cell is 583.2 pm × 583.2 pm × 318.1 pm = 1.082 · 108 pm3 = 1.082 · 10–22 cm3. Since there are 4 atoms of tin per unit cell, the mass of the Sn contained in the unit cell is (4 × 118.71 g mol–1) / (6.022 · 1023 mol–1) = 7.885 · 10–22 g. Thus the density of white tin is 7.885 · 10–22 g / 1.082 · 10–22 cm3 = 7.287 g cm–3.
Gray tin adopts a face-centered cubic structure called the diamond lattice, illustrated below. When a crystalline sample of gray tin is examined by X-ray diffraction (using CuKα radiation, λ = 154.18 pm), the lowest-angle reflection, due to diffraction from the (111) family of planes, is observed at 2θ = 23.74°. Calculate the density of gray tin in g/cm3.
Model Answer
From Bragg's Law, nλ = 2 d sinθ.
For the lowest-angle reflection, n = 1, so d = λ / (2 sinθ) = 374.8 pm. The spacing between the (1 1 1) planes in a cubic lattice is a / √3 , where a is the length of the unit cell edge. So a = d √3 = 649.1 pm, V = a3 = 2.735 pm3 = 2.735 · 10–22 cm3. From the picture, there are 8 Sn atoms in a unit cell of gray tin, so the density is 5.766 g cm–3.
The pressure at the bottom of the Mariana Trench in the Pacific Ocean is 1090 bar. Will the temperature at which the two allotropes of tin are in equilibrium increase or decrease at that pressure, and by how much? In your quantitative calculations, you may assume that the energy (E), entropy (S), and molar volume of the two phases of tin are independent of temperature and pressure.
Model Answer
Qualitatively, increasing the pressure will increase the stability of the denser phase. Since white tin is substantially denser than gray tin, white tin would be more stable at high pressure, and the temperature at which gray tin spontaneously converts to white tin would therefore decrease.
Quantitatively, the pressure-dependence of the entropy is (by assumption) negligible. While the pressure dependence of the energy change is assumed to be negligible, the enthalpy change depends on both ∆E° and pV explicitly:
∆H° = ∆E° + ∆(pV) = ∆E°+ p ∆Vrxn (for reactions at constant pressure)
So as the pressure changes, ∆H° for the phase change will change accordingly:
∆H°1090 bar = ∆E° + (1090 bar)∆Vrxn
∆H°1090 bar = ∆H°1 bar + (1089 bar)∆Vrxn
since neither ∆E° nor ∆Vrxn is assumed to change with pressure (or temperature).
From the densities of the two allotropes determined in the previous two parts, the molar volumes of white tin and gray tin are 16.17 cm3 mol–1 and 20.43 cm3 mol–1, respectively. So for Sn (white) → Sn (gray), ∆Vrxn = + 4.26 cm3 mol–1 = 4.26 · 10–6 m3 mol–1.
∆H°1090 bar = ∆H°1 bar + (1089 bar)∆Vrxn
∆H°1090 bar = ∆H°1 bar + (1.09 · 108 Pa)(4.26 · 10–6 m3 · mol–1)
∆H°1090 bar = ∆H°1 bar + 464 J·mol–1
Since Teq = ∆H° / ∆S° , then
Teq, 1090 bar = ∆H°1090 bar / ∆S°1 bar = ∆H°1 bar / ∆S°1 bar + (464 J mol–1 / –7.04 J mol–1 K–1).
Teq, 1090 bar = Teq, 1 bar – 66.0 K = –52.8 °C
Despite the relatively low temperatures at the bottom of the ocean, the stable allotrope of tin will be white tin because of the high pressure.