[boltzmann77]

Adding a nonvolatile substance to a solvent always lowers the vapor pressure - from my textbook.

Why does this happen?

[ARGOS++]

Dear Boltzmann77;

Could this already be enough explanation?:   http://en.wikipedia.org/wiki/Boiling-point_elevation

Good Luck!
                    ARGOS⁺⁺

[boltzmann77]

Here is an explanation I thought of:

Consider a volatile solvent in a closed container. The solvent is in equilibrium with the vapor.
Condensation of vapor in equilibrium with evaporation of solvent
Then I disrupt the equilibrium by adding some non volatile solute. This will make available less molecules of the volatile solvent at the surface. This means that there are less molecules going into the vapor phase than earlier. To achieve equilibrium again more vapor molecules will start condensing. Hence the vapor pressure is reduced.
Could someone check to see whether my explanation is correct?

[ARGOS++]

Dear boltzmann77;

I don’t believe that would be a correct explanation why.
Lower number of solvent molecules at the surface means only a “smaller” area for passing in both directions, and that means only that time for equilibrium would increase.

Beside the other explanation of the Chemical Potential from the link above:

A non-volatile solute has a vapor pressure of zero, so the vapor pressure of the solution is the same as the vapor pressure of the solvent. Thus, a higher temperature is needed for the vapor pressure to reach the surrounding pressure, and the boiling point is elevated.

So what’s the only possible consequence for the partial vapour pressure?

I hope to have been of help to you.
Good Luck!
                    ARGOS⁺⁺

[Hunt]

Adding a nonvolatile substance to a solvent always lowers the vapor pressure - from my textbook.

Why does this happen?

There are many ways to see why this happens. A good and generally acceptable method is using the chemical potential , as ARGOS++ pointed out. Adding a non-volatile solute to a solvent decreases the chemical potential of the solvent , which means the stability of the solvent in the liquid phase has increased and therefore its tendency to escape to the gaseous phase is lowered. The result is a lower vapour pressure of the solvent. This can be proven rigorously and is usually covered in physical chemistry textbooks.

A more elementry approach is to consider raoult's law : p = x p*
It is easy to show that a consequence of this expression is that the rate of vaporization is directly proportional to the mole fraction, that is : r = k x while the rate of condensation is directly proportional to the vapour pressure i.e. r = k' p

Adding a solute decreases the rate of vaporization while the rate of condensation is unaffected , so the vapour pressure must decrease.

[boltzmann77]

Adding a solute decreases the rate of vaporization while the rate of condensation is unaffected , so the vapour pressure must decrease. - Hunt

I don't see how the rate of condensation does not change. If there is less vaporisation, then shouldn't the rate of condensation temporarily increase to re-attain equilibrium? (Le Chatelier's principle)

Then at equilibrium it will again be back to condensation=vaporisation, right? Now there is less vaporisation, hence less condensation (compared to before adding the non volatile solute). Hence the vapor pressure is lower.

Please correct me if my reasoning is wrong.

Thanks.

[Hunt]

Adding a solute decreases the rate of vaporization while the rate of condensation is unaffected , so the vapour pressure must decrease. - Hunt

I don't see how the rate of condensation does not change. If there is less vaporisation, then shouldn't the rate of condensation temporarily increase to re-attain equilibrium? (Le Chatelier's principle)

Then at equilibrium it will again be back to condensation=vaporisation, right? Now there is less vaporisation, hence less condensation (compared to before adding the non volatile solute). Hence the vapor pressure is lower.

Please correct me if my reasoning is wrong.

Thanks.

Ok good point. The rate of condensation does not change immediately after the addition of a solute when you consider that the non-equilibrium state attained has still not changed the vapour pressure. As the rate of vaporization decreases the vapour pressure decreases too until finally rate of vap = rate of cond at the new equilibrium stage. This does not contradict le chatelier's principle but agrees with it. Your explanation is correct as long as you assume that ultimately , after some time in the non-equilibrium process, the rate of condensation decreases to finally become equal to the rate of vaporization at eq.

I hope there's less confusion now. You have to remember that this is a very basic method twith many flaws...

"If there is less vaporisation, then shouldn't the rate of condensation temporarily increase to re-attain equilibrium? (Le Chatelier's principle)"

Why increase ? Let r = r' at equilibrium.
if r decreases , r' decreases too to reach a new equilibrium state.