Thermochemical Properties of Dissolution of Nicotinic Acid C6H5NO2(s) in Aqueous Solution
Autor: | Dan Xu, Chun-Sheng Zhou, Yu-Xia Kong, You-Ying Di |
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Rok vydání: | 2017 |
Předmět: |
Chromatography
Aqueous solution Chemistry Enthalpy Biophysics 02 engineering and technology Crystal structure 010402 general chemistry 01 natural sciences Biochemistry Standard enthalpy of formation 0104 chemical sciences Enthalpy change of solution Crystal Crystallography 020401 chemical engineering 0204 chemical engineering Physical and Theoretical Chemistry Molecular Biology Dissolution Monoclinic crystal system |
Zdroj: | Journal of Solution Chemistry. 46:886-895 |
ISSN: | 1572-8927 0095-9782 |
Popis: | Nicotinic acid (also known as niacin) was recrystallized from anhydrous ethanol. X-ray crystallography was applied to characterize its crystal structure. The crystal belongs to the monoclinic system, space group P2(1)/c. The crystal cell parameters are a = 0.71401(4) nm, b = 1.16195(7) nm, c = 0.71974(6) nm, α = 90°, β = 113.514(3)°, γ = 90° and Z = 4. Molar enthalpies of dissolution of the compound, at different molalities m/(mol·kg−1) were measured with an isoperibol solution–reaction calorimeter at T = 298.15 K. The molar enthalpy of solution at infinite dilution was calculated, according to Pitzer’s electrolyte solution model and found to be $$ \Delta_{\text{sol}} H_{m}^{\infty } = ( 2 7. 3 \pm 0. 2) $$ kJ·mol−1 and Pitzer’s parameters ( $$ \beta_{{\text{MX}}}^{{\text{(0)}L}} $$ , $$ \beta_{{\text{MX}}}^{{\text{(1)}L}} $$ and $$ C_{{\text{MX}}}^{\phi L} $$ ) were obtained. The values of apparent relative molar enthalpies ( $$ {}^{\phi }L $$ ) and relative partial molar enthalpies ( $$ \overline{{L_{2} }} $$ and $$ \overline{{L_{1} }} $$ ) of the solute and the solvent at different molalities were derived from the experimental enthalpy of dissolution values of the compound. Also, the standard molar enthalpy of formation of the anion $$ {\text{C}}_{ 6} {\text{H}}_{ 4} \text{NO}_{2}^{-} $$ in aqueous solution was calculated to be $$ {\Delta_{\text{f}}^{} H}_{\text{m}}^{\text{o}} ({\text{C}}_{ 6} {\text{H}}_{ 4} {\text{NO}}_{2}^{-} \text{,aq}) = - \left( {603.2 \pm 1.2} \right)\;{\text{kJ}}{\cdot}{\text{mol}}^{-1} $$ . |
Databáze: | OpenAIRE |
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