Abstract
Neptunium (IV) is oxidized to Np(V) with nitric acid in the presence of U(VI) under conditions of low acidity (<∼0.1 M). The reaction rate is described by the equation d[Np(V)]/dt = k 1[Np(IV)]/[H+]2 + k 2[Np(IV)]2[U(VI)]/[H+]3, in which k 1 = (2.0±0.3) × 10−5 mol2 l−2 min−1 and k 2 = (5.50±0.47) × 10−2 mol l−1 min−1 at 50°C and solution ionic strength μ = 0.5. The activation energies of the two pathways are 148±31 and 122±12 kJ mol−1. The reaction along the main pathway (with the rate constant k 2) is limited by disproportionation of Np(IV) involving NpOH3+ and Np(OH)2UO 4+2 complex ions.
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Translated from Radiokhimiya, Vol. 47, No. 3, 2005, pp. 228–233.
Original Russian Text Copyright © 2005 by Koltunov, Taylor, Marchenko, Savilova, Dvoeglazov, Zhuravleva.
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Koltunov, V.S., Taylor, R., Marchenko, V.I. et al. Kinetics and Mechanism of Np(IV) Oxidation with Nitric Acid. Radiochemistry 47, 252–257 (2005). https://doi.org/10.1007/s11137-005-0082-x
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DOI: https://doi.org/10.1007/s11137-005-0082-x