Asked by anna
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
Answered by
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.