Question
Consider the evaporation of water at 298 K and 1 atm pressure:
H20 (l) <---> H20 (g)
Given that ΔGOrxn= 8.6kJ/mol, what is the equillibrium vapor pressure of water at 298 K?
H20 (l) <---> H20 (g)
Given that ΔGOrxn= 8.6kJ/mol, what is the equillibrium vapor pressure of water at 298 K?
Answers
DrBob222
dGo = -RTlnK
8600 J = -8.314*298*K
Solve for K, then
K = pH2O
This gives pH2O in atmospheres; that x 760 converts to mm Hg. I get about 24 mm or so but that's approximate.
8600 J = -8.314*298*K
Solve for K, then
K = pH2O
This gives pH2O in atmospheres; that x 760 converts to mm Hg. I get about 24 mm or so but that's approximate.