Why is DeltaG Negative for H2O(l) to H2O(g) at Equilibrium?

[Kqwert]

Calculate deltaG for the reaction:

H2O(l) = H2O(g). 100 degrees celsius, water is clean. P(H2O) = 0.1 bar.

Given that it is an equilibrium, I'd think that deltaG would be zero. But the answer is in fact negative. How is that possible?

 

[Ygggdrasil]

What is the equilibrium vapor pressure of water at 100°C?

 

[Kqwert]

1 bar? So it's really not an equilibrium..?

 

[Ygggdrasil]

At equilibrium, the partial pressure of water would be 1 atm. Because P(H2O) is less than that, the forward reaction is favored.

 

[Kqwert]

So it's wrong to call it an equilibrium?

 

[Ygggdrasil]

The reaction is at equilibrium only at a specific P(H2O).

 

[Kqwert]

Thank you!
isn't it wrong to write the equation with a "=" sign between the reactants and products then?

 

[Ygggdrasil]

H₂O_(g) $\rightleftharpoons$ H₂O_(l) is the correct notation.

The reaction is reversible, with vapor molecules continuously condensing into the liquid phase and liquid molecules continuously evaporating into the gas phase. When P(H2O) is less than 1 atm, the rate of evaporation exceeds the rate of condensation. When P(H2O) is greater than 1 atm, the rate of condensation exceeds the rate of evaporation. Only when P(H2O) = 1 atm will evaporation and condensation occur at the same rate, so there will be no net exchange of material between the liquid and gaseous phases.

 

[Chestermiller]

Thank you!
isn't it wrong to write the equation with a "=" sign between the reactants and products then?

I think it would have been better if they had used an arrow. ---->