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?

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.
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