Sn(alpha)

Formula : Sn

Family: 
Element
Space group: 
Tetragonal - Ditetragonal Dipyramidal H-M Symbol (4/m 2/m 2/m) Space Group: I 41/amd
Dana class: 
1.1.13.1 (1) Native Elements (1.1) with metallic elements other than the platinum group (1.1.13) Dana Group
High temperature: 
Tin in stable at room temperature in acidic medium (HCl).
Geological context: 
In placer sands (Oban, Australia) and in calcite as discrete grains (Beaverlodge, Canada). Reported in vocanic gases.

Dernière mise à jour le 25.01.2021

THERMODYNAMICS

Individual properties (298.15 K)
M
g·mol-1
 V°
cm3·mol-1
 ΔGºf
J·mol-1
 ΔHºf
J·mol-1
 Sº
J·mol-1·k-1
 CP
J·mol-1·K-1
118.710 16.289 (Ref.: 85jac/hel) Jackson, K.J., and Helgeson, H.C., 1985, Chemical and thermodynamic constraints on the hydrothermal transport and deposition of tin: II. Interpretation of phase relations in the Southeast Asian tin belt: Econ. Geol., v. 80, no. 5, pp. 1365-1378. -10 (Ref.: Internal calculation) 0 (Ref.: By convention) 51.212 (Ref.: 85jac/hel) Jackson, K.J., and Helgeson, H.C., 1985, Chemical and thermodynamic constraints on the hydrothermal transport and deposition of tin: II. Interpretation of phase relations in the Southeast Asian tin belt: Econ. Geol., v. 80, no. 5, pp. 1365-1378. 26.35 (Ref.: 85jac/hel) Jackson, K.J., and Helgeson, H.C., 1985, Chemical and thermodynamic constraints on the hydrothermal transport and deposition of tin: II. Interpretation of phase relations in the Southeast Asian tin belt: Econ. Geol., v. 80, no. 5, pp. 1365-1378.
CP(T) = a + b·T + ·c·T-2 (T in Kelvin)
a
J·mol-1·K-1
 b * 103
J·mol-1·K-2
 c * 10-5
J·K·mol-1
 Maier Kelley Ref.
18.49 26.36 0.00 (Ref.: 85jac/hel) Jackson, K.J., and Helgeson, H.C., 1985, Chemical and thermodynamic constraints on the hydrothermal transport and deposition of tin: II. Interpretation of phase relations in the Southeast Asian tin belt: Econ. Geol., v. 80, no. 5, pp. 1365-1378.
Reaction properties

Sn(alpha) + 0.5O2 + 2H+ ⇌ Sn+2 + H2O

298.15 K, 1 bar log10K: 47.81 ΔH°r:-288539 J·mol-1 (Ref.: By convention)
T(°C) 0 25 60 100 150 200 250 300
log10K 52.40 47.81 42.46 37.52 32.59 28.63 25.33 22.49
log10K(T) = A + B·T + C·T-1 + D·log10(T) + E·T−2 (T in Kelvin)
A B C D E
-393.166866 -0.061216 37388.231472 141.323873 -1410679.292660