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

96 7. Neptunium aqua ions(∂ E ◦′ /∂T ) = −(0.00027 ± 0.00005) V ·K −1 , r H m (7.4, 1 M HClO 4 ) =−(117.47 ± 1.44) kJ·mol −1 is calculated. The enthalpy of transfer to infinitedilution is assumed to be zero, and the value of r Hm ◦ is the same as the value of r H m (7.4, 1MHClO 4 ). This value, derived from electrochemical data, agrees with r Hm ◦ =−(117.4 ± 0.6) kJ·mol−1 obtained by Brand and Cobble [70BRA/COB]from calorimetric measurements of the NpO 2+2reduction to NpO + 2by hydrogenperoxide. These values were adjusted to infinite dilution. The weighted average ofthese two independent values is r Hm ◦ (7.4, 298.15 K) =−(117.4 ± 0.6) kJ·mol−1 .Taking this, and using r G ◦ m (7.4, 298.15 K) =−(111.8 ± 0.4) kJ·mol−1 as acceptedabove, r Sm ◦ =−(18.9 ± 2.3) J·K−1·mol−1 . This value is required in the followingsection, and corresponds to (∂ E ◦ /∂T ) = −(0.00019 ± 0.00002) V·K −1 for theNp(VI)/Np(V) couple. From r Hm ◦ (7.4) and from fHm ◦ (NpO+ 2, aq, 298.15 K) =−(978.2 ± 4.6) kJ·mol −1 (cf. Section 7.3), we determine the selected value for theenthalpy of Reaction 7.5to beNp(cr,α)+ O 2 (g) + 2H + → NpO 2+2+ H 2 (g) (7.5) f Hm ◦ (NpO2+ 2, aq, 298.15 K) = −(860.7 ± 4.7) kJ·mol−1Similarly, using the selected value, f G ◦ m (NpO+ 2, aq, 298.15 K) = −(907.8 ±5.6) kJ·mol −1 (cf. Section 7.3), we determine the selected value f G ◦ m (NpO2+ 2, aq, 298.15 K) = −(795.9 ± 5.6) kJ·mol−17.3 NpO + 2The thermodynamic quantities for NpO + 2are derived from the value of S◦ m (NpO2+ 2 ),and from the values of the standard potential based on electrochemical potentialmeasurements. The temperature coefficients for the VI/V and V/IV potentials are usedto calculate sequentially Sm ◦ (NpO+ 2 ) and S◦ m (Np4+ ), and from the latter f G ◦ m (Np4+ ).The value of the V/IV potential is then used to determine f G ◦ m (NpO+ 2 ) and f Hm ◦ (NpO+ 2 ).Cohen and Hindman [52COH/HIN2] determined the formal potential in 1 molalHClO 4 for the reactionNpO + 2 + 3H+ + 1 2 H 2(g) Å Np 4+ + 2H 2 O(l) (7.6)with junction-free emf measurements. They derived E ◦′ = (0.7391 ±0.0010) V, and(∂ E ◦′ /∂T ) =−(0.00269 ± 0.00006) V·K −1 . This review corrects the formal potentialfrom 1 m to 1 M and 1 bar pressure and derives the NEA recommended values E ◦′ (7.6,298.15 K) = (0.7431 ±0.0010) V, r G m (7.6, 1MHClO 4 , 298.15 K) =−(71.7 ±0.1) kJ·mol −1 , r S m (7.6, 1MHClO 4 , 298.15 K) =−(259.5 ± 5.8) J·K −1·mol −1 ,