Evaporation Rates Over Mixtures

When you good folks comment on these blogs I get a much better idea of what you want to hear about in the world of exposure modeling.  I received a number of comments like the one below from anonymous. 

“I would like to see more on vapor generation rates. Calculating this from a chemical's vapor pressure, liquid:vapor equilibrium and evaporation rate can be complicated for chemical (solvent) mixtures. Is there an easier way?”

Indeed, pure substances are relatively rare and we are almost always dealing with mixtures.   Anonymous is correct, the situation is somewhat complicated but I hope to present it in a form that is reasonably easy to follow  over the span of the next few blogs. 

Anonymous is also correct that one of the keys to determining evaporation rate is a chemical’s vapor pressure over and out of the mixture.  So at this point we are going to concentrate on determining the vapor pressure of various compounds over a liquid mixture.  

As an example let’s consider a mixture of toluene, xylene and benzene.   In this particular mixture 50% of the molecules are xylene, 49% of the molecules are toluene and 1% of the molecules are benzene.   This means that the mole fraction of benzene in this mixture is 0.01.  Mole fraction is the number of moles of benzene over the total number of moles in the mixture. Just to review a mole is 6.023 x 10²³ molecules.  The molecular weight (MW) of a compound is the weight of one mole of its molecules.   For benzene its MW is 78.11 g/mole.   So if we know the contributing weight and the MW of all the components we can calculate the mole fraction of any component by adding up all the moles of the compounds and dividing this number into the individual moles of each.  I will present an example later in this blog.

So here we have a mixture in which 1% of the molecules (or moles) are benzene.  Presummably, the benzene is evaporating from the mixture but it is NOT evaporating at nearly the rate that pure benzene would evaporate.  In 1882 some smart guy by the name of François-Marie Raoult came up with the idea that the vapor pressure of benzene over a mixture will be directly proportional to its mole fraction within the mixture.   Raoult’s Law states:

VPM = (VPP)(MF)

VPM = vapor pressure of the compound of interest over the mixture

VPP = vapor pressure of pure compound

MF =  mole fraction of the compound

If Raoult was correct then the vapor pressure of benzene over our mixture will be 1% of the vapor pressure of pure benzene.   The room temperature vapor pressure of pure benzene is 95 mmHg.  Thus, Raoult’s Law would predict 0.95 mmHg vapor of benzene over this mixture.   As it turns out Raoult was essentially correct for this mixture and for any mixture in which the compounds within the mixture were reasonably similar in structure. These are called ideal solutions.  In this case all these compounds are aromatics or benzene derivatives and so it works as pretty close to ideal. The vapor pressures of the xylene and toluene over this mixture are about half of their vapor pressures as pure materials.

Wow, life should be so simple but it’s not.   As soon as some of the major components of the mixture become very different from one another relative to a chemical structure standpoint then Raoult’s Law starts to break down.  In some instances it breaks down dramatically. 

Consider benzene in water.   Benzene has a water solubility of about 1.7g/Liter or about 1.7/78.11 = 0.022 moles in a liter of water which is 55.5 moles of H2O.   Thus the mole fraction of benzene in water at its solubility limit is 0.022/(55.5 + 0.022) = 0.0004.    Raoult’s Law would predict:   VPM  =  (95 mmHg) (0.0004) = 0.038 mmHg.   In reality the VPM of benzene as it approaches the solubility limit in water is essentially as high as the vapor pressure of pure benzene!     Raoult missed by a factor of about 95/0.038 or >2000!

How could Raoult get it so wrong?   The answer lies in the thermodynamic “activity” of the benzene molecule within this aqueous mixture.   In a mixture with its similar “bothers” (e.g., toluene and xylene) the benzene has more or less neutral activity and tends to obey Raoult’s Law.   In water, however, it is so different than the solvent that it is somewhat freaked out and highly active.   It does not want to be in the water and really wants out; thus, it expresses its vapor pressure at a relatively high level. 

High activity relative to the ideal Raoult prediction is not just for organics out of water which is perhaps one of the more extreme examples.   However it can also happen, for example, if you mix dissimilar organics like straight chained or brained hydrocarbons and lower alcohols.   Also intermediate deviations (AC = 2 to 20) from Raoult’s Law will also occur in some mixtures.  As such, we definitely need (and have) a modification and correcting generalization of Raoult’s Law:

VPM = (VPP) (MF) (AC)

VPM = vapor pressure of the compound of interest over the mixture

VPP = vapor pressure of pure compound

MF  = mole fraction of the compound in the mixture

AC = activity coefficient

The activity coefficient (AC) is about 1.0 when the molecule is neutral as the are in ideal mixtures and thus obeys Raoult’s Law.  As we have seen above, it can be very large (e.g. 2000 or more) when the molecule is very different and really wants out of the mixture.  

Getting the AC can be problematic but I will discuss how one might get it in the next blog.  I will also discuss how one can estimate vapor pressure of organics out of water in another way.