Question

At 25°c ∆G° for the process H20(l)=H20(g)is 8.6kj.The vapour pressure of water at this temperature is nearly​

Answer 1

Answer:

23.17 torr

Explanation:

Consider the vaporization as equilibrium reaction. 

The the activities satisfy equilibrium equation 

K=a(H₂O(l)a(H2O(g)

The vapour 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))=pop(H2O(g)) 

Hence 

K=pop(HO(g)) 

 

=p(H2O(g))=po⋅K 

Equilibrium constant and ΔG are related as 

ΔG=−R⋅T⋅ln(K) 

K=eR⋅T−ΔG 

Therefore 

p(H2O(g))=po⋅eR⋅T−ΔG 

=101325Pa⋅e(8.314472J/molK⋅298.15K)−8600J/mol

=3155Pa

converting Pa to torr, we get 23.17 as answer 

Answer 2

Answer:

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