Sciencemadness Discussion Board - Thermodynamics of nitric acid distillation

Thermodynamics of nitric acid distillation

vulture - 15-2-2003 at 09:46

I noticed that when people are discussing the distillation of nitric acid they are concerned with some factors of thermodynamics but they forget others.

Example:

Distilling under reduced pressure is favourable, because it lowers the boiling point of nitric acid and thus decreases decomposition.

But there is also a disadvantage when distilling under reduced pressure.
If pressure drops inside a vessel where a chemical reaction is moving towards an equilibrium, the equilibrium will move to the side of the reaction where the gas volume will be the highest.

Now let's take a look towards the decomposition reaction of nitric acid:

4HNO₃ --> 2H₂O + 4NO₂ + O₂

At the boiling point of nitric acid, there are 4 gas molecules on the left side and 5 (assuming that the 2H₂O condenses) on the right side.

So, if I am not mistaken, lower pressure will also push the equilibrium towards decomposition of the nitric acid.

Now, like in all chemical processes, the question is to find the best of both worlds.

Any thoughts?

rikkitikkitavi - 15-2-2003 at 10:24

yes that is true, according to leChateliers law, that a reduced pressure shifts the temperature to the left. However , this is only true for gaseous HNO3.

Usually , reactions taking place at ambient (for most common activation energies) , reaction speed doubles at an 10 C increase in temperature. Thus, lowering the boiling point from 84 C to about 40 C, means that decomposition speed decreases atleast 16 times.

I think that without burying myself in incomplete calculations, the lowered reaction speed means more gain than the shift in equlibrium , since otherwise, why can you make white HNO3 by vaccum distillation, but doint the same at 1 atm gives a red HNO3.

There is a third way however. By heating HNO3 / water absorbant to roughly 40C, and slowly blowing an inert gas through, if the gas/HNO3 have good contact (i e low mass transfer resistance, i e large contact surface) the gas will be saturated with HNO3 when leaving. (RH = 100 % )

The saturation pressure of HNO3 can be found in many tabels, it equates to the vapour pressure at said temperature.

F e x , @ 40 C, 1 atm ambient , Pv = 0,3 atm, meaning that for every mole of gas leaving the contact area, 0,3 moles will be HNO3. (the gas law, you know, P/Ptot = n/ntot)

By doing the same thing as in a A/C , cooling the gas, vapour pressure of HNO3 decreases and HNO3 condenses on the surfaces. The major disadvantage is that heat transfer in the gas is low, due to present inert, demanding a large cooling surface. You also need strong cooling to get a good yeild.

With distillation, heat transfer is much higher due to no inert being present.

At the E&W forum a topic describing how HNO3 was made by a method used for puryfying urine for drinking, f e x in the desert (you probably know which topic, placing HNO3 in a glassjar, put a lid of plastic, heating under it e t c)

Or you could build a more sofisticated equipment of course.

/rickard

rikkitikkitavi - 15-2-2003 at 10:26

edit/
'I now saw that the first sentence didnt quite make sence, of course I meant
that the eq. shifts to the right by reduced pressure.

/rickard

vulture - 16-2-2003 at 07:02

I have "professional" distilling equipment, but when distilling under reduced pressure I still get lot's of NO₂ so I was wondering if the cure wasn't worse than the disease, because most information on the forums does not really mention al factors of thermodynamics involved.

Microtek - 16-2-2003 at 07:16

I use a simple retort at atmospheric pressure and a sandbath on a thermostatically controlled hotplate. With careful experimentation I have found a setting which allows distillation of practically colourless acid. It does take quite a long time to distil even moderate amounts of HNO3 and there is also the heating cycle to get the sandbath warmed up but this slowness of operation also means that there is no problem with condensing the vapours.

rikkitikkitavi - 16-2-2003 at 07:18

yes, I have had a similar problem. I think that it is very difficult to get a no-NOx HNO3 from XNO3/H2SO4 , even under vacuum, because the mixture probably needs to be heated to far.
My own experiments seems to indicate this, that the "sludge" has low heat transfer and easily get hot spots.
Patience, Patience and a warm water bath, termostatic controlled is probably the way to go.

But with HNO3/H2SO4/H2O distillation as far as I know.

I have also seen references here and there about distilling @ 1atm, but very slowly and under a tight temperature control ( 85 C) so that little HNO3 decomposes.

But really, I think that doing any serious calculation we need reaction constants and equlibrium data, which makes it a difficult system of differential equations to solve (numerically). This way you could optimize with respect to pressure and temperature. However, I have no idea where to get this data

/rickard

Microtek - 16-2-2003 at 23:34

I forgot to note that I always distil a mix of equal volumes of 62 % HNO3 and 96 % H2SO4. Where I am, a chain of stores sell 500 mL bottles of 62 % HNO3, so I never have to use fertiliser.

vulture - 17-2-2003 at 12:42

I'll try distilling with 40% HNO3 and 98% H2SO4 under reduced pressure and see if that works.

Maybe stirring the mixture with a magnetic stirrer? I have an idea about this because I've got a stirring "stick" and a fairly powerful magnet. I'll let you know if I find the time to test it.

lucifer - 17-2-2003 at 15:00

Vulture, why don’t you distill the 40% first up to ~65%,
With fractional distilling that shouldn’t’ be a problem.
It will make distilling 100% HNO3 easier.
Or else you will need a lot of 99% H2SO4 to get a reasonable output of the HNO3

Polverone - 17-2-2003 at 15:44

Quote:

But really, I think that doing any serious calculation we need reaction constants and equlibrium data, which makes it a difficult system of differential equations to solve (numerically). This way you could optimize with respect to pressure and temperature. However, I have no idea where to get this data

Ahh, but can't you compute the equilibrium data just from thermodynamic information? And by Hess's law you don't really need all that many experimentally determined thermodynamic quantities to start with to determine heats of formation for many compounds. Also, solving differential equations numerically isn't so bad. Writing code to do so (inefficiently) is easy, if you don't already have a powerful math package lying around. It's solving differential equations by hand that would frighten me