Abstract
The exothermic process that occurs around 700 K during calcination of ZrO2−x(OH)2x, associated with the crystallization of the low-temperature tetragonal metastable phase of ZrO2, was analyzed using x-ray diffraction, high-resolution thermogravimetric analysis (TGA), nitrogen adsorption, and modulated differential scanning calorimetry (MDSC). High-resolution TGA allowed us to determine the water loss, resulting from condensation of OH− groups. The amount was 0.137 wt% in our case, equivalent to 1.7 × 10−2 mol of H2O/mol of ZrO2. That corresponds to about one −OH group per nm2 being lost in that process. By using MDSC we determined that the change in enthalpy (∆Hglobal = −15.49 kJ/mol of ZrO2) was the result of two parallel contributions. One of them was reversible and endothermic (∆Hrev = 0.11 kJ/mol of ZrO2), whereas the other was irreversible and exothermic (∆Hirrev = −15.60 kJ/mol of ZrO2). The variability and magnitude of the exotherm, as well as the fact that the accompanying weight loss is so small, are consistent with a mechanism involving the formation of tetragonal nuclei, rather than global crystallization, and hence depend on the number of nuclei so formed.
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