Abstract

Spectroscopic method was used to examine the stoichiometry of the reaction of No(VI) with hexamethylenediaminetetraacetic acid (HMDTA, H₄hmdta) in a 0.05 M HClO₄ solution. At an excess of Np(VI), 1 mole of complexon reduces about 4 moles of Np(VI) to Np(V). In 0.1–1.0 M HClO₄ solutions (the ionic strength of 1.0 was maintained with LiClO₄) containing 2–20 mM HMDTA, neptunium(VI) with concentration of 0.3–3.0 M decreases at 35–55°C by the first-order rate law until the instant when less than 20% Np(VI) remains. The initial reaction rate has the first order in [HMDTA] and –2 order in [H⁺]. An activated complex is formed with loss of two H⁺ ions. The activation energy is 102 ± 7 kJ/mol.

中文翻译：

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在HClO 4溶液中用六亚甲基四乙酸还原Np（VI）

摘要

使用光谱法检查在0.05 M HClO ₄溶液中No（VI）与六亚甲基二胺四乙酸（HMDTA，H ₄ hmdta）的反应的化学计量。在过量的Np（VI）下，1摩尔的络合物将约4摩尔的Np（VI）还原为Np（V）。在含有2–20 mM HMDTA的0.1–1.0 M HClO ₄溶液（用LiClO ₄保持离子强度为1.0 ）中，浓度为0.3–3.0 M的n（VI）在35–55°C时一级降低速率定律，直到剩余Np（VI）小于20％的时刻。初始反应速率在[HMDTA]中为一阶，在[H ⁺ ]中为–2阶。失去两个H ⁺形成活化的配合物离子。活化能为102±7kJ / mol。