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

134 9. Neptunium group 17 (halogen) compounds and complexesshould be noted that although the vapour pressures of UF 6 ,NpF 6 and PuF 6 in the liquidstate follow this expected behaviour, the vapour pressure of NpF 6 (cr) is slightlyhigher than those of UF 6 (cr) and PuF 6 (cr) (which are almost identical) [59WEI/WEA],so the uncertainty in the enthalpy of formation has been increased.The selected value is f H ◦ m (NpF 3, g, 298.15 K) = −(1112 ± 20) kJ·mol −1When combined with the entropy values estimated below, the calculated sublimationpressure for the reactionis closely represented by the expression:NpF 3 (cr) Å NpF 3 (g)from 1000 to 1600 K.9.1.2.3.blog 10 (p/bar) = −20574T −1 + 8.977Standard entropy and heat capacityThe thermal functions of NpF 3 (g) were calculated assuming the molecular parametersgiven by Rand and Fuger [2000RAN/FUG], using the rigid-rotator, harmonic oscillatorapproximation. The molecule is assumed to be pyramidal, with C 3v symmetry.These assumptions giveSm ◦ (NpF 3, g, 298.15 K) =−1(330.5 ± 10.0) J·K−1·molCp,m ◦ (NpF 3, g, 298.15 K) =−1(72.2 ± 5.0) J·K−1·mol9.1.2.4 NpF 4 (cr)9.1.2.4.aEnthalpy of formationThe enthalpy of formation of NpF 4 (cr) is estimated from the correlation betweenthe difference of the enthalpies of formation of the actinide tetrahalides and theaqueous M 4+ ions and the radius of the M 4+ ions [76SHA], as discussed in detailby Fuger et al. [83FUG/PAR, 2000RAN/FUG]. This analysis when combined with f H ◦ m (Np4+ , aq, 298.15 K) =−(556.0 ± 4.2) kJ·mol −1 , selected in this review, gives f H ◦ m (NpF 4, cr, 298.15 K) = −(1874 ± 16) kJ·mol −1which is the selected value.9.1.2.4.bStandard entropy and heat capacityThere are no low-temperature measurements of the heat capacity of NpF 4 (cr), and thestandard entropy is derived from two independent estimates: the spin-only method