I'm trying to work out the numbers for thermodynamic likelihood of KBF4 + H2SO4 ---> BF3 + HF + KHSO4. At standard temps here is what I have for
deltaGo:
KBF4: -1784.9 kJ/mol
H2SO4: -690 kJ/mol
BF3: -1119.4 kJ/mol
HF: -275.4 kJ/mol
KHSO4: -1031.3 kJ/mol
( -1031.3 + -275.4 + -1119.4 )-(-1784.9 + -690) = 48.8kJ/mol , unfavorable.
I got these from the CRC handbook. Where am I going wrong with this? Is it because the energies for BF3 and HF are gas phase when they would likely
be in solution? I checked what effect increased temp would have, but it didn't show an improvement at reasonable temperatures (<373K). blogfast25 - 16-2-2014 at 07:56
No, assuming the values for BF3 and HF are for gases at STP, then the value for:
KBF4(s) + H2SO4(l) ---> BF3(g) + HF(g) + KHSO4(s)
of + 48.8 kJ/mol is correct.
But would you carry this out as solid (KBF4)/liquid (pure H2SO4) or in a solution? Solvated the calculation becomes quite different because the
starting values used would not apply.
Edit: thinking about it, these reactions would be the displacement of one strong acid with another, not expected to be highly favourable. But since as
you're removing volatile reaction products, you'd be pushing the equilibrium to the right anyway. It's similar to the displacement of HCl in NaCl +
H2SO4 reactions: the HCl leaves thus making the reaction possible.
[Edited on 16-2-2014 by blogfast25]roXefeller - 16-2-2014 at 11:49
If I were to carry it out ... I'm not sure. It seems like a difficult one to execute. Separating the HF from the BF3 in the gas products would prove
difficult. The KBF4 isn't terribly soluble in much at ambient temperatures. In solution, perhaps diethyl ether could be used to capture the gases
and leave a BF3 etherate. Both are soluble in ether though, leaving purification difficult. Perhap using boron trioxide to convert the HF over, but
would the resulting H2O (B2O3 + 6HF-->2BF3 + 3H2O) decompose the BF3 or scuttle away to the H2SO4 layer?
I looked at the full neutralization to K2SO4, but it wasn't any more favorable. It was +148.8kJ/mol at ambient, and elevated temperatures seems
dangerous for such chemicals. Strangely enough though, HF is weaker than HCl, and is the only halide acid that doesn't fully dissociate. So how
would that change things?
How is the solvation effect quantified in this analysis? Is it just a matter of appropriating the solvated standard energies? I haven't been able to
find this, though it may be a matter of search terms.plante1999 - 16-2-2014 at 12:11
Simply heating the tetrafluoroborate will realease the boron trifluoride.blogfast25 - 16-2-2014 at 12:41
How is the solvation effect quantified in this analysis? Is it just a matter of appropriating the solvated standard energies? I haven't been able to
find this, though it may be a matter of search terms.
Yes, you should find values for standard G for solvated species like BF<sub>4</sub><sup>-</sup>. In the CRC I think.
But Plante is right, KBF4 should undergo pyrolysis:
KBF4(s) === > KF(s) + BF3(g), thermodynamically favourable or not because the BF3 leaves the reaction mix. Not sure about temperatures though. At
least 700 C I would think...chornedsnorkack - 24-2-2014 at 01:29
If I were to carry it out ... I'm not sure. It seems like a difficult one to execute. Separating the HF from the BF3 in the gas products would prove
difficult.
Boiling point as pure liquids:
HF - +19 Celsius
BF3 - -100 Celsius.
What is the full phase diagram? Presumably some higher boiling azeotrope?
