Bander - 1-10-2006 at 16:23
2 NaBH4 + 8 H2O2 ==> 2 NaOH + B2O3 + 11 H2O
[-850 kJ (2x NaOH) + -1272.77 kJ (B2O3) + -2660 (11x H2O)] - [ -1528 (8x H2O2) + 214 (2x NaBH4)]
[-4782] - [-1314] = -3468 kJ mol^-1?
I am thinking this is not correct. Can anyone spot where I have gone wrong? Also, it seems likely the boron trioxide will be converted to boric acid
in that wet environment. Is this assumption sane?
12AX7 - 1-10-2006 at 18:48
For that matter, sodium borate solution.
Tim
franklyn - 1-10-2006 at 22:56
I suppose where you've gone wrong is to assume there is a fixed stoichiometry
for what are hypergolic reactants. The products produced in this case are
entirely dependant on the percentage of combined reactants.
Although you don't specify we'll assume 98% strength H2O2. This is immediately
decomposed on contact with a base , whoosh into steam and oxygen , if you're
lucky , if you're not , it will detonate at about 5500 to 6000 meters per second.
NaBH4 is a basic hygroscopic powder , it sucks moisture out of air and continuosly
decomposes when wetted combining with water into sodium borate and liberating
hydrogen gas. Unless it's packed dry , it is only stable in solution in the presense
of another strong water soluble base , NaOH , KOH , Ca(OH)2 , Guanidine hydroxide
or just liquid ammonia or hydrazine. With Potassium or Calcium ions , Sodium will
be replaced as the cation. At the other extreme you may drip , drops of H2O2 onto
the NaBH4 and get small bursts of flame which just may ignite the whole pile ,
it is combustible in air , although not pyrophoric.
The best energy product results from a one to one molar ratio , this yeilds
NaBH4 + H2O2 -> NaBO2 + 3H2
The more H2O2 used the lower the energy produced
per given weight . Complete redox is ,
2 NaBH4 + 8 H2O2 -> Na2O + B2O3 + 12 H2O
The oxides will combine with moisture 4 H2O to form
NaOH and 2 B(OH)3
[Edited on 2-10-2006 by franklyn]
Bander - 2-10-2006 at 07:23
Thanks for your help. I was expecting the hypergolic nature (depending on it, really) but was unsure about how this would effect figuring out the
enthalpy. As such I used the general template of and the associated resources I gathered at the below link(s). It should be noted that in the
references the NaBH4 is probably in a methanol solution. As for peroxide concentration, let's say 40% by volume (http://nervousenergy.net/_/files/lowh2o2.jpg).
H2O2 + 1/4 Metal Hydride -> 11/4 H2O + 1/4 (X)OH + 1/8 Al2O3
http://nervousenergy.net/_/files/metalhydride.jpg
http://nervousenergy.net/_/?q=bipropellant
[Edited on 2-10-2006 by Bander]