Closing the hydrogen cycle with the couple sodium borohydride‐methanol, via the formation of sodium tetramethoxyborate and sodium metaborate

Summary Methanolysis of sodium borohydride (NaBH4) is one of the methods efficient enough to release, on demand, the hydrogen stored in the hydride as well as in 4 equiv of methanol (CH3OH). It is generally reported that, in methanolysis, sodium tetramethoxyborate (NaB(OCH3)4) forms as single component of the spent fuel. It is, however, necessary to clearly investigate some critical aspects related to it. We first focused on the methanolysis reaction where NaBH4 was reacted with 2, 4, 8, 16, or 32 equiv of CH3OH. With 2 equiv of CH3OH, the conversion of NaBH4 is not complete. With 4 to 32 equiv of CH3OH, NaBH4 is totally methanolized (conversion of 100%). The best conditions are those involving 4 equiv of CH3OH as they offer the highest effective gravimetric hydrogen storage capacity with 4.8 wt%, an attractive H2 generation rate with 331 mL(H2) min−1—a performance achieved without any catalyst—and the formation of NaB(OCH3)4 as single product as identified by X‐ray diffraction, Fourier transform infrared spectroscopy, and nuclear magnetic resonance. We then focused on the transformation of this product NaB(OCH3)4 into sodium metaborate (NaBO2), via the formation of sodium tetrahydroxyborate (NaB(OH)4). NaB(OCH3)4 is easily transformed in water, by hydrolysis, at 80°C and for 90 minutes, into NaB(OH)4 and 4 equiv of CH3OH. In doing so, the cycle with CH3OH is closed. Subsequently, NaB(OH)4 is recovered and converted into NaBO2 under heating at 500°C. This reaction liberates 4 equiv of H2O, which allows to close the cycle with water. Based on these achievements, we have finally proposed a triangular recycling scheme aiming at closing the cycle with the protic reactants of the aforementioned reactions. This scheme may be used as base for implementing a closed cycle with the couple NaBH4‐CH3OH. In stoichiometric conditions, methanolysis of NaBH4 led to NaB(OCH3)4 as by‐product. In over‐stoichiometric conditions, methanolated derivatives of NaB(OCH3)4 form. NaB(OCH3)4 transforms in NaB(OH)4 by hydrolysis, which gives NaBO2 at 500°C. NaBO2 being a precursor of NaBH4, the cycle with NaBH4 can be closed. The cycle is neutral with respect to NaBH4, CH3OH, and H2O.