Determination of the standard molar enthalpy of formation of the ternary complex of neodymium with vitamin B3 and 8-hydroxylquinoline by microcalorimetry

• Published in: Journal of thermal analysis and calorimetry 2013-06, Vol.112 (3), p.1533-1538
• Main Authors: Xiao, Sheng-Xiong, Li, Ai-Tao, Li, Chuan-Hua, Xiao, Hang-Ying, Xu, Xiao-Yan, Gu, Hui-Wen, Li, Qiang-Guo
• Format: Article
• Language: English
• Subjects: Analytical Chemistry; Calorimetry; Chemistry; Chemistry and Materials Science; Chlorides; Coordination compounds; Enthalpy; Exact sciences and technology; Inorganic Chemistry; Inorganic chemistry and origins of life; Magnetic alloys; Measurement Science and Instrumentation; Neodymium; Niacin; Nitrogen; Nitrogen atoms; Physical Chemistry; Polymer Sciences; Preparations and properties; Rare earth metals; Spectra; Vitamins
• ISSN: 1388-6150; 1588-2926; 1572-8943
• DOI: 10.1007/s10973-012-2644-4

A novel solid ternary complex, [Nd(C 6 H 4 NO 2 ) 2 ·C 9 H 6 NO·2H 2 O], was synthesized in a water bath (333.15 K) by three kinds of reagents: neodymium chloride, vitamin B3 (C 6 H 5 NO 2 ), and 8-hydroxylquinoline (C 9 H 7 NO). Its composition and structure were characterized by elemental analysis, IR spectra, UV spectra, molar conductance, and thermogravimetric analysis. During the process of coordination, C 6 H 5 NO 2 was bidentate-coordinated with Nd 3+ in the form of an acidic group by removing the proton; hydroxyl oxygen atom and heterocyclic nitrogen atom of C 9 H 7 NO formed a chelate ring. Particularly, in this article, a thermochemical cycle in the calorimetric solvent ( V HCl : V DMF : V EtOH  = 3:1:1) was designed on the basis of Hess’s law. At 298.15 K, the dissolution enthalpies of the reactants and products were determined by a advanced solution–reaction isoperibol microcalorimeter, respectively. According to the above results and relevant literature data, the standard molar enthalpy of formation of [Nd(C 6 H 4 NO 2 ) 2 ·C 9 H 6 NO·2H 2 O], was estimated to be [[Nd(C 6 H 4 NO 2 ) 2 ·C 9 H 6 NO·2H 2 O(s)], 298.15 K] = −(2,129.1 ± 2.5) kJ mol −1 .