University College - Chemistry Chem 112A Lecture Notes - Lecture 36: Vapor Pressure, Boiling Point, Stopcock
13 Dec 2016
School
Department
Professor
18 April 2016
Lecture 36: Vapor Pressure
Equation of the Day:
I. Vapor Pressure
A. Measuring Equilibrium Vapor Pressure (Peq)
1. The equilibrium vapor pressure of a liquid is independent of atmospheric pressure from air in
free space
a. Increased surface area in a vessel increases evaporation rate
b. More free space in a vessel increases evaporation rate
c. Peq is a constant for a liquid at a given temperature (volume and other stuff doesn’t
matter)
2. Mercury barometer
a. Can be used to measure the vapor pressure of a liquid
b. There is 1 atm pushing down on a reservoir of mercury and an evacuated tube above it
• This will produce 760 mm of Hg with the stopcock closed (760 mm Hg = 1 atm =
760 torr)
c. If you open the stopcock, the vapor pressure is represented by the change in height of the
mercury column (∆h)
d. Ex: for H2O at 25°C, Peq = 23.7 mm Hg = 23.7 torr *(1 atm/760 torr) = 0.0312 atm
3. A(l) A(g)
a. ∆G = ∆G° + RTlnQ
b. ∆Gvap = ∆Gvap + RTlnP, at equilibrium, ∆Gvap = 0, and P = Peq
c. ∆G°vap = -RTlnPeq → rearrange, lnPeq = -∆G°vap/RT or Peq
• Ex: for H2O, ∆G°vap = ∆G°f (H2O(g)) - ∆G°vap(H2O(l)) = +8.59 kJ/mol
• lnPeq = (-8.59 kJ/mol) / RT → Peq (H2O) = 0.0312 atm
• Note: can only use this for 298K since ∆G°f’s are at 298
4. Peq at other temperatures
a. ∆G°vap = ∆H°vap – T∆S°vap
b. At any temp:
c.
d. Ex: H2O, ∆H°vap = 44.01 kJ/mol, ∆S°vap = 118.81 J/mol•K
• Plugging into equation, we get that Peq (H2O at 298K) = 0.123 atm
5. Remember, ∆H°vap and ∆S°vap are not completely temperature independent
a. Both have a slight temperature dependence and can be adjusted for using Kirchoff’s law:
b.
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