Add time:08/20/2019         Source:sciencedirect.com

The partial pressure of thallium(I) hydroxide TlOH in equilibrium with gaseous thallium(I) oxide Tl2O and H2O(g) was studied by measuring the rate of loss of mass of crystalline thallium(III) oxide Tl2O3 into flowing (water + oxygen)(g) and (water + oxygen + argon)(g) streams at atmospheric pressure and temperatures of 940 K to 1020 K. Crystalline Tl2O3 volatilized through the reaction: Tl2O3(c) = Tl2O(g)+O2(g); Ko1 . Thallium(I) hydroxide is formed by the reaction: Tl2O(g)+H2O(g) = 2TlOH(g); Ko2 . Ko2 was calculated from the rate of mass loss of Tl2O3 in the presence of H2O(g) using values of Ko1 measured previously in the absence of H2O(g). At T = 980 K, Ko2 = (0.83±0.12) and ΔrHom = (54.86±7.29) kJ·mol-1. The standard molar enthalpy of formation of TlOH at T = 980 K: ΔfHom(TlOH, g, 980 K), is estimated to be - 104.2 kJ·mol-1, and its value at T = 298.15 K: ΔfHom(TlOH, g, 298.15 K), is estimated to be -90 kJ·mol-1. The molar enthalpy of sublimation of TlOH: ΔgcrHom is estimated to be 149 kJ·mol-1.

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