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

152 9. Neptunium group 17 (halogen) compounds and complexesfrom 298.15 to 975 K, and henceCp,m ◦ (NpBr 3, cr, hex, 298.15 K) =−1(103.8 ± 6.0) J·K−1·mol9.1.4.1.dFusion dataThe fusion data are taken to be close to the mean values for UBr 3 (cr) [63RAN/KUB]and PuBr 3 (cr) (this review, Section 18.1.4.2.a), giving:T fus = (975 ± 30) K fus Hm ◦ (NpBr 3, cr, ortho, 975 K) = (48 ± 8) kJ·mol −1The heat capacity of the liquid has been estimated to be similar to those of the liquidlanthanide tribromides, which lie between 145 and 155 J·K −1·mol−1 .C ◦ p,m (NpBr −13, l) = (150 ± 8) J·K−1·mol9.1.4.2 NpBr 3 (g)There are no experimental thermodynamic data involving NpBr 3 (g) and since the vapourpressures of UBr 3 (cr) and PuBr 3 (cr) at 1000 K calculated from the assessed data([92GRE/FUG] forUBr 3 , this review, Section 18.1.4.2 for PuBr 3 ) differ by a factorof more than 100, and also both have large uncertainties, we have not attempted toestimate any data for this species.9.1.4.3 NpBr 4 (cr)9.1.4.3.aEnthalpy of formationFuger and Brown [73FUG/BRO] measured the enthalpy of solution of NpBr 4 (cr) tobe −(258.78 ± 0.50) kJ·mol −1 in1MHCland−(199.20 ± 0.42) kJ·mol −1 in 6 MHCl. With the partial molar enthalpy of formation of Br − in these solutions, assumedto be the same as those in HBr solutions of the same molality, −(119.40 ± 0.15) and−(109.55 ± 0.20) ,and f H ◦ m (Np4+ ) in these solutions −(551.9 ± 1.7) kJ·mol −1and −(532.9 ± 2.1) kJ·mol −1 [2000RAN/FUG], the calculated values for the enthalpyof formation of NpBr 4 (cr) become −(770.7 ± 1.9) kJ·mol −1 (from 1 M HCl) and−(771.9 ± 2.3) kJ·mol −1 (from 6 M HCl). The selected value is the weighted mean: f H ◦ m (NpBr 4, cr, 298.15 K) = −(771.2 ± 1.8) kJ·mol −19.1.4.3.bStandard entropy and heat capacityThere are no low-temperature measurements of the heat capacity of NpBr 4 (cr),and the standard entropy is derived from two reasonably consistent estimates by[2000RAN/FUG]. The selected value is thus the rounded mean of the two estimates:Sm ◦ (NpBr −14, cr, 298.15 K) = (243 ± 10) J·K−1·mol