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

300 15. Elemental plutoniumfrom 298.15 to 500 K, andC ◦ p,m (Pu, g, T) = (4.592 + 3.31521 × 10−2 T − 5.64466 × 10 −6 T 2from 500 to 1100 K, and+7.0159 × 10 5 T −2 −1) J·K−1·molC ◦ p,m (Pu, g, T) = (−9.588 + 4.60885 × 10−2 T − 8.63248 × 10 −6 T 2from 1100 to 2000 K.15.2.1 Enthalpy of formation+5.01446 × 10 6 T −2 −1) J·K−1·molWork on the vapour pressure prior to 1976 has been summarised by Oetting et al.[76OET/RAN]. There has been one additional study since then (Bradbury and Ohse,[79BRA/OHS]) which was carried out at higher temperatures (giving pressures up to1303 Pa), partially to test the upper limit of the Knudsen effusion regime. The vapourpressure equations from the five studies, and the derived second- and third-law valuesfor f Hm ◦ (298.153 K), are given in Table 15.3; there is a good agreement between thesecond and third law analyses and between the five studies, probably because, unlikethe earlier actinides, the monoxide gas is only a minor species in the evaporation ofoxygen-contaminated plutonium. The selected value for the enthalpy of formation is:Table 15.3: Vaporisation of Pu(g) and the enthalpy of formation at 298.15 KReference T/K log 10 (p/bar) (a) f Hm ◦ /( kJ·mol−1 )A B 2nd-law 3rd-law[56PHI/SEA] 1392-1793 −17587 5.020 355.2 347.6[66MUL] 1133-1792 −17420 4.919 352.1 347.8[69KEN] 1426-1658 −17420 4.871 352.0 349.0[75ACK/RAU] 1210-1620 −17120 4.598 346.2 351.0[79BRA/OHS] 1724-2219 −17066 4.666 343.5 349.4Selected value: (349.0 ± 3.0) kJ·mol −1(a) log 10 (p/bar) = A(T/K) −1 + B f Hm ◦ (Pu, g, 298.15 K) = (349.0 ± 3.0) kJ·mol−1The calculated Gibbs energy of formation thus becomes f G ◦ m (Pu, g, 298.15 K) = (312.415 ± 3.0) kJ·mol−1