# Gibbs free energy problem

Please post this type of questions in HW section using the template.
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?

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Ygggdrasil
Gold Member
2019 Award
What is the equilibrium vapor pressure of water at 100°C?

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

Ygggdrasil
Gold Member
2019 Award
At equilibrium, the partial pressure of water would be 1 atm. Because P(H2O) is less than that, the forward reaction is favored.

So it's wrong to call it an equilibrium?

Ygggdrasil
Gold Member
2019 Award
The reaction is at equilibrium only at a specific P(H2O).

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

Ygggdrasil
Gold Member
2019 Award
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
Mentor
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. ---->