Reversing Bottoms/Distillate in simple distillation? 2

[ec2003]

I'm having a little trouble understanding the following, perhaps someone here can help me out.

Nitric acid/water is an azeotropic system that has the azeotrope at ~70% HNO3 and boils at 122C, 1 atm.

Now, if you wanted to distill a 30% by wt mixture in a column to a 50% by weight mixture, you are safely underneath the azeotrope and can use simple distillation methods. However, by doing this, HNO3 is the light key (bp 84 C, 1 atm) and H2O is the heavy key (bp 100 C, 1 atm). In other words, HNO3 wants to come out in the distillate stream.

Now I have read in many places (albeit generally) that by using simple distillation as above, they are able to obtain the desired wt % HNO3 in the BOTTOMS and pure water in the DISTILLATE.

I'm confused and wonder: Just how do they accomplish this? Is there any good reference material to read online/hardcopy?

Thanks!

E

 

[25362]

When two liquids form an azeotrope (aka constant boiling mixtures-CBM) they do not act as normal liquids would.
When starting with 30% HNO[sub]3[/sub] the light key component is water and the heavy key is the azeotrope.
When starting with, say, 80% acid, the light key is acid and again the heavy key is the azeotrope. This is common with high boiling azeotropes.
Many inorganic acids have high CBM with water. A good example would be HCl (BP:-85[sup]o[/sup]C) having an atmospheric azeotrope with water at 108.6[sup]o[/sup]C and 20.2% acid.

When looking at an isobaric (say, 1 atm abs) temperature/composition diagram one can discern two pair of vapour/liquid equilibrium curves, to left and to the right of the azeotrope located at 120.7[sup]o[/sup]C and 67.4% nitric acid.
As a pair of unequal eyeglasses.
The equilibrium vapour and liquid compositions are connected by horizontal (isothermal) lines.

Try to construct such a diagram on cartesian coordinates (temperatures on the ordinate) and you'll make the above highly explicit. Azeotropes change with pressure.

In your sample case an atmospheric column operating with a bottom temperature of 121[sup]o[/sup]C would have a bottom product consisting of about 60% acid. The top could be pure water, or a very water-rich vapour, depending on the number of contact stages and the top temperature.

As a last note, water cannot be totally separated from nitric acid by simple distillation. Industrial methods for achieving that are either extractive distillations or a reaction with additional nitrogen oxides, or both.

 

[ec2003]

Thanks. That clarifies things a bit.

 

[25362]

Most common azeotropes are of the low boiling type. But the principle is the same.

I'll give you an example of a low boiling azeotrope with which I had some experience in the past: furfural + water. Furfural boils at 161[sup]o[/sup]C and is taken as vapour in a column where water is the bottom liquid. The overhead is the low boiling azeotrope containing furfural. Just the opposite of what we'd think based on the boiling points of the pure components.

Happily furfural-water make an heterogeneous azeotrope meaning that on condensation they separate in two phases (one rich in furfural, the other in water) a fact that helps to get their full separation in two towers.
To our regret, nitric acid and many others, form homogeneous (totally soluble) CBM in the liquid phase, and are more difficult to separate. Extractive distillations are used for the purpose.

I know that it is difficult to grasp, but the general point is that when two liquids form an azeotrope (high boiling or low boiling) the azeotrope becomes one key component in distillation. The other depends on the starting concentration of the mix, if richer in water (than the CBM), then the other key is water. If richer (than the CBM) in the other component, this one will be the other key.

I hope I've been of help.

 

[ec2003]

Definitely.

To clarify to myself then (tough as it is to understand the wonderful world of homogeneous azeotropes), HNO3 going in at a lower-than-azeotropic concentration will make water be the LK. The result of this makes the azeotrope (~70 wt% acid) the HK. Therefore, using a simple column (packed most likely for small separations), you may distill pure water out of the top and 70 wt % acid (about 0.4 mol %) out the bottom. So, in essence, it acts like a glorified evaporator.

I'm still confused about the separation equations, etc. for these azeotropic systems. They're non-ideal, which makes the simple (perhaps overly so) McCabe-Thiele diagram unapplicable (especially considering how the equilibrium line falls under the operating line from 0-40 mole %) for tray estimation or further design (diameter, height, etc.). What kind of considerations need to be taken into account when designing a column for this very separation of high boiling azeotrope under its azeotropic point? Concentrating the acid further (to 99% for example) would require extractive distillation using an entrainer like sulfuric acid. But I just cannot get my head around the simple side of it (!). Give me "ideal" binary separations over this stuff anyday... LOL.

You've been a great help. I appreciate the clarification on this issue.

E