Vapor Pressure: Solvent versus Solution ```Name:Michael Status: student Grade: 9-12 Location: N/A Country: USA Date: November 2007 ``` Question: Why is the vapor pressure of a solution is lower than vapor pressure of a pure solvent? What is happening on the molecular level? Replies: Michael, Imagine the entropy of the vapor from the pure solvent and that of the solution. Since the solute does not have an appreciable vapor pressure, then we can say that only the solvent is in vapor form above both the pure solvent and the solution. In this case we can say that the entropy of the two systems are practically the same. Now imagine the entropy of the pure solvent and that of the solution. Since there are more microstates (possible arrangements) in the solution than in the pure solvent, we say that the solution is more disorganized, has a higher entropy. So now think of the change in entropy in going from a pure solvent to the vapor phase, and that of the solution going to the vapor phase - remembering that the vapor entropy in both systems are the same, and that the pure solvent has a lower entropy than that of the solution. This means that the change in entropy from the pure solvent to its vapor is higher than that of the solution to its vapor. If the change in entropy is higher and the final entropy state is the same, then the process with a higher entropy change is more favored, tends to be more spontaneous. Thus, we can expect that the pure solvent will spontaneously form more vapor (at a particular temperature) when compared to the solution. Greg (Roberto Gregorius) Your inquiry can be addressed at several levels of complexity. The simplest, and probably adequate, is that the vapor pressure of a solvent depends upon how many solvent molecules are present at the surface of the liquid. If the number of molecules at the surface is reduced by the presence of a solute, then the partial pressure of the solvent will be reduced in proportion to the number of solute molecules that replace the solvent molecules. This simple approach "explains" Raoult's Law that: P = Po*X where P is the partial pressure of solvent at mol fraction, X. and Po is the vapor pressure of the pure solvent. Building off of that oversimplified picture you can see how various interactions at the molecular level can increase, or decrease, the relative number of mols of solute and/or solvent that could be present in any particular set of conditions. Vince Calder Click here to return to the Chemistry Archives

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