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Analysis and Thermochemistry of the Vapor of the NaBr ‐ DyBr3 System

“…The I(DyBr~)//(Dy2Br~) ratio increases if there are essential additional contributions to the DyBr~ intensity from NaDyBr4(g) on vaporizing the NaBr/DyBr3 sample. By comparison the mass spectra on vaporizing pure DyBr3(s) and the NaBr/DyBr3 sample the fraction originating from NaDyBr~(g) is obtained from the following equation I*(DyBr~ from NaDyBr4(g)) I~ = I*(DyBr~, total) iO(Dy2Br~ ) I*(Dy2Br~) [1] where I*(DyBr~, total) = I*(DyBr~ from DyBr3(g)) + I*(DyBr~ from NaDyBr~(g)) [2] I* denotes the intensity over the NaBr/DyBr~ sample and I ~ over pure DyBr3(s). The fraction of the DyBr~ ion intensity originating from NaDyBr~(g) on studying the NaBr-DyBr~ system is obtained as described above for DyBr~.…”

“…The gaseous species NaBr, Nal, (NaBr)2, Na2Brl, (NaI)2, SnBr2, SnBrI, Snip, NaSnBr3, NaSnBr21, NaSnBrI2, and NaSnI3 were detected in the vapor. The enthalpy and entropy changes of the reactions NaSnBr2I(g) = NaI(g) + SnBr2(g) [i], NaSnBrI2(g) = NaBr(g) + SnI2(g) [2], and SnI2(g) + 2NaBr(g) = SnBr2(g) + 2NaI(g) [3] resulted as AH~ = 172.9 + 6.4 kJ tool -1, AS~ = 140.5 • 9.9 J mol -~ K ~ for reaction Eq. 1, hH~6 = 181.6 + 5.8 kJ mo1-1, hS~gs = 143.1 +_ 9.3 J tool 1 K -1 for reaction Eq.…”

“…The gaseous species NaBr, (NaBr)2, DyBr3, (DyBr3)2, NaDyBr4, and Na2DyBr5 were detected in the vapor. The enthalpy and entropy changes of the reactions NaDyBr4(g) = NaBr(g) + DyBr3(g) [1], NaDyBr4(g) = NaBr(g) + DyBr3(s) [2], and 2NaDyBr4(g) = 2NaBr(g) + (DyBr3)2(g) [3], as well as the enthalpy change of the reaction Na2DyBrs(g) = 2NaBr(g) + DyBr3(g) [4], resulted as AH$98 = 242.7 -+ 5.6 kJ reel ~,…”

Study of the Vapor of the NaI ‐ ScI3 System: Vapor Composition, Thermochemistry of the Gaseous Heterocomplexes and , and Their Potential for Chemical Vapor Transport

“…The resulting value b = 0.1I -+ 0.01 agrees within the given error limits with that obtained according to Eq. [2]. Figure 2 summarizes the quantitative results on fragmentation obtained from the two quotients and the intensities from the various runs.…”

“…The following values of the equilibrium constant of the dissociation reactions Eq. [1] and [2] (Table II) were obtained from the selected values of h~/~ and A~S%m log Kp = -11,652 1/(T/K) + 8.0533 (613-848 K) [9] for Eq. [1], Table II and log Kp = -19,747 1/(T/K) + 15.088 (623-848 K) [10] for Eq.…”