Organic-Water binary Azeotropes

[nberga]

Hello to everybody,
I'm surfing the net (and being lost in library)looking for hints of existance of binary azeotrope between water and any of the following comps:

112Cl3C2H3 - TriChloroEthane
1122Cl4C2H2 - TetraChloroEthane
EthylBenzene
Styrene
Br3ClC - TriBromoMethane
12Br2C2H4 - DyBromoEthane
1-4DyChlorobenzene (para)

Has anybody ever heard of any azeotrope?

Thanks.

 

[moltenmetal]

All of the compounds you've listed are only sparingly soluble in water (<0.1%). Azeotropes are therefore unlikely.

 

[25362]

Perry VI, CRC Handbook and Lange's Handbook give for etylbenzene: 92 deg C, 33% water at 1 atm abs.

 

[moltenmetal]

25362: Thank you very much! You've taught me something. I looked into Perry's reference and indeed it is possible for a two-phase mixture to exhibit an azeotrope (i.e. two liquid phases in equilibrium with a vapour phase of a particular composition). A prof once told us that the behaviour of immiscible liquid phases in terms of vapour/liquid equilibrium is the same as if they were in a container with a divider between them. I guess that was an over-simplification!

 

[25362]

These are called heterogeneous azeotropes, I've seen quite a few in the past.

 

[gilden]

Water-ethylbenzene azeotrope: bp 92°C, 33% water

Further I have (not exactly what you are looking for, but perhaps it can help you):

1,1,1-trichloroethane: Bp 88°C, 4 % water
Dichloromethane: bp 39°C, 2 % water

 

[DickRussell]

Strictly speaking, the above mentioned compositions (for heterogeneous azeotropes) are for the vapor phase, or what would condense from the vapor. For two immiscible liquid phases, the amounts of each liquid are irrelavent, so the gross composition is not useful.

In general, solubility information (for components quite insoluble in each other) is a reasonable way to approximate dilute activity coefficients. For two immiscible components, each exerting its own vapor pressure at the system temperature, the vapor mol fraction of each will be about equal to its vapor pressure (out of the &quot;pure&quot; phase) divided by total pressure. The K value out of the other phase is then K=y/x = (vp/P)/molar solubility. Therefore the activity coefficient in that other phase is about 1/solubility. If the solubility is very low (<<1%), then going lower in concentration (below saturation - the pure phase has disappeared) won't significantly change the activity coefficient - it already is about at the infinite dilution value. This can be useful for approximating solute K values for calculating stripping from a solvent, when no real VLE data are available (being mostly immiscible) but solubility information is available.