chemical thermodynamics of neptunium and plutonium - U.S. ...

12.1 Neptunium carbon compounds and complexes 285ε(12.45, q = 1) =−(0.13 ± 0.10) kg·mol −1 and ε(12.45, q = 2) =−(0.56 ±0.19) kg·mol −1 , after increasing the uncertainties by ±0.05 due to the estimation byanalogy. For the 1:3 reaction, we use ε (Np(SCN)+3 ) ≈ ε (AmF + 2 ) = 0.17 kg·mol−1 ,whichresults in ε(12.45, q = 3) =−(0.82 ± 0.13) kg·mol −1 , after increasing the uncertaintyby ±0.05 due to the analogy made. The resulting values at I = 0 are listed inTable 12.8. The values at 25 ◦ C determined with TTA and DNNS agree well. We estimatethat the uncertainties of these values, taking systematic errors into account, arearound ±0.1 logarithmic units for the formation of the 1:1 complex and ±0.2 forthe1:2 and 1:3 complexes. We select the average of these values and assign uncertaintiesthat reflect the fact that no independent study is is available on this system.log 10 β1 ◦ (12.45, q = 1, 298.15 K) = 3.0 ± 0.3log 10 β2 ◦ (12.45, q = 2, 298.15 K) = 4.1 ± 0.5log 10 β3 ◦ (12.45, q = 3, 298.15 K) = 4.8 ± 0.5Rao et al. [78RAO/BAG2] performed their measurements at 10, 25 and 40 ◦ C, cf.Table 12.8. From the temperature dependence of the resulting equilibrium constants,they derived enthalpy values, cf. Appendix A. The values of β 2 at different temperaturesshow no steady variation, which gives an indication of appreciable uncertainty.Due to lack of information, we assume the enthalpies to be independent of the ionicstrength and assign the uncertainties accordingly. r Hm ◦ (12.45, q = 1, 298.15 K) = −(7 ± 3) kJ·mol−1 r Hm ◦ (12.45, q = 2, 298.15 K) = −(9 ± 9) kJ·mol−1 r Hm ◦ (12.45, q = 3, 298.15 K) = −(13 ± 9) kJ·mol−1The thermodynamic formation data are derived from these constants and the selecteddata for Np 4+ and SCN − : f G ◦ m (NpSCN3+ , aq, 298.15 K) = −(416 ± 7) kJ·mol −1 f G ◦ m (Np(SCN)2+ 2, aq, 298.15 K) = −(330 ± 10) kJ·mol−1 f G ◦ m (Np(SCN)+ 3, aq, 298.15 K) = −(241 ± 14) kJ·mol−1 f H ◦ m (NpSCN3+ , aq, 298.15 K) = −(487 ± 7) kJ·mol −1 f Hm ◦ (Np(SCN)2+ 2, aq, 298.15 K) = −(412 ± 13) kJ·mol−1 f Hm ◦ (Np(SCN)+ 3, aq, 298.15 K) = −(340 ± 16) kJ·mol−1Sm ◦ (NpSCN3+ , aq, 298.15 K) =−1−(248 ± 25) J·K−1·molSm ◦ (Np(SCN)2+ 2, aq, 298.15 K) =−1−(90 ± 51) J·K−1·molSm ◦ (Np(SCN)+ 3, aq, 298.15 K) =−1(55 ± 66) J·K−1·mol