Distillation Theory Part IV: Vapor Pressure of Mixtures

Distillation Theory Part IV –  

In part I, we defined distillation.  In Part II, we explained vapor pressure and defined boiling as the point where the vapor pressure is equal to the atmospheric pressure.  In part III, we added the concept of vacuum boiling, where boiling happens when the vapor pressure of a liquid is equal to pressure of the environment the liquid is in.

 Up to this point, this discussion has been about boiling a pure liquid.  This part teaches what happens when the liquid is a mixture.  How do we determine the boiling point of a mixture?  Each component will have a different boiling point, so what happens?

 What chemists have discovered is actually very simple.  The vapors above the liquid mixture is composed of each of the liquids in the mixture.  But how much of each component is in the vapor phase depends on two factors.

 Factor 1: the vapor pressure of the pure component at the temperature of the liquid.  Each component, if pure, has a specific vapor pressure at each temperature.  Review Part II to understand this.  So the first factor is the vapor pressure of the pure component at that temperature.

 Factor 2: the mole ratio of the component in the mixture.  Mole ratio is a chemistry term that means the percentage of molecules of the component in the mixture expressed as a ratio.  For example, if I have a binary mixture where 60% of the molecules are water, the mole ratio of the water would be 0.6.

 The partial vapor pressure above a liquid from a single component of a mixture is calculated by multiplying the two factors.

 In our binary example, at 20°C, the partial vapor pressure from the water is equal to the pure vapor pressure of water at 20°C (17.54 mmHg) times the mole ratio (0.6).  This comes to 10.54 mmHg (11 mmHg). 

 This makes sense, 60% of the mixture is water molecules, so the vapor phase will have water.  It just has less of it than if it was a pure water sample.

 The overall vapor pressure above a liquid mixture is equal to the sum of all the partial vapor pressures.  Let’s say that the other liquid was methanol.  The vapor pressure of pure methanol at 20°C is 96 mmHg.  Therefore the partial vapor pressure of the methanol in this mixture is 96 mmHg x 0.4 = 38 mmHg.

 The overall vapor pressure of this mixture is therefore 11 mmHg + 38 mmHg = 49 mmHg.  This equation works for any number of components.  You just need to calculate the partial vapor pressure of each of the components, no matter how many, and add all of those up.

 If the goal was to boil this mixture at 20°C, our chemist would need to apply a pressure (vacuum) of 49 mmHg.

 In the next post, I will discuss how all of this leads to purifying a material.  I will explain how distillation actually works.

 Part 4 in theory of distillation series.