Question

Calculate Delta G^0 for H_2O(g) doubleheadarrow H_2O(l) given that the equilibrium vapor pressure of water is 3.168 kPa at 25 degree C.

Answer 1

Consider the vaporization as equilibrium reaction.

The the activities satisfy equilibrium equation
K = a(H₂O(g)) / a(H₂O(l))

The vapor pressure is the partial pressure over gaseous water over pure water.

Since liquid water water is pure , it has activity 1.
a(H₂O(l)) = 1
Assuming ideal gas phase, the activity of gaseous water is equal to ratio of partial pressure to total pressure:
a(H₂O(g)) = p(H₂O(g)) / p°

Hence
K = p(H₂O(g)) / p°
<=>
p(H₂O(g)) = p° · K

Equilibrium constant and ΔG are related as
ΔG = -R·T·ln(K)
<=>
K = e^{-ΔG / (R·T) }

Therefore
p(H₂O(g)) = p° · e^{-ΔG / (R·T) }

3168 pa = 101325Pa · e^{ -ΔG / (8.314472J/molK · 298.15K)}

ΔG = 8.6 kJ/mol