# Gibbs free energy problem

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1. Mar 16, 2019 at 7:15 AM

### Kqwert

• 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?

2. Mar 16, 2019 at 8:15 AM

### Ygggdrasil

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

3. Mar 16, 2019 at 9:14 AM

### Kqwert

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

4. Mar 16, 2019 at 10:08 AM

### Ygggdrasil

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

5. Mar 16, 2019 at 10:23 AM

### Kqwert

So it's wrong to call it an equilibrium?

6. Mar 16, 2019 at 10:34 AM

### Ygggdrasil

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

7. Mar 16, 2019 at 10:37 AM

### Kqwert

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

8. Mar 16, 2019 at 3:14 PM

### 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.

9. Mar 17, 2019 at 7:00 AM

### Staff: Mentor

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