Water-water vapour equilibrium above the boiling point

Question

Consider $\ce{H2O_{(g)} -> H2O_{(l)}}$ at $\pu{105^{\circ} C}$. Under these conditions it is false that:

(1) $T\Delta S_r$ is negative.
(2) Relative humidity is 100%.
(3) $|\Delta H_r|>|T\Delta S_r|$
(4) $P_{\ce{H2O_{(g)}}}>\pu{1 atm}$
(5) $|{\Delta G_f}_{\ce{H2O_{(g)}}}|=|{\Delta G_f}_{\ce{H2O_{(l)}}}|$

Been trying to understand this question. Here is my reasoning:

1. True, because $T$ is positive $\pu{(105 + 273)K}$ while $ΔS$ should obviously be negative because the final entropy is lower than the initial entropy. We are after all going from water's gaseous state to its liquid state. There are fewer degrees of freedom in water's liquid state.

2. If there is an equilibrium between water vapor and water liquid, then by definition relative humidity is 100%. The air is saturated with water vapor to the point where it can't even hold the water vapor and water vapor exists in a dynamic equilibrium with liquid water. From Wikipedia:

Saturated air is really just air which contains water vapor in equilibrium with a liquid water source.

3. False, because $ΔG$, not $ΔG^{\circ}$. $ΔG$ for a system at equilibrium is always $0$. $ΔG^{\circ}$, not necessarily $0$. So we have $0 = ΔG = ΔH - TΔS$. This implies $ΔH = TΔS$.

4. True. For liquid water to boil, vapor pressure must equal atmospheric pressure. Similarly for water to condense, atmospheric pressure must equal vapor pressure. Also, vapor pressure is solely dependent on temperature of the pure liquid. The below statement is from UC Davis' ChemWiki. I would think that adding solute would change the vapor pressure (and yes it does, and it's discussed further down the page of the site).

Vapor pressures are dependent only on temperature and nothing else.

Taking into consideration that water boils at $\pu{100^{\circ}C}$, we may conclude that at $\pu{105^{\circ}C}$, the vapor pressure of liquid water is greater than $\pu{1 atm}$. And therefore, for this equilibrium to exist, both the vapor pressure of liquid water and the atmospheric pressure of gaseous water must be equal and thus greater than $\pu{1 atm}$.

5. This is true. This would make $ΔG = 0$ since $ΔG = G_{\text{(products)}} - G_{\text{(reactants)}}$.

I would like to know if my reasoning is correct.

Answer 1

All the arguments are perfect except the fourth.

The statement is true, but the premise that the vapour pressure of water must equal the atmospheric pressure is incorrect. A phase equilibrium between the gaseous and the liquid phase can be established at a particular temperature without the liquid boiling. Boiling is a special case of this equilibrium when the atmospheric pressure is entirely constituted by water vapour in the vicinity of the liquid surface and hence the vapour pressure can no longer rise at that pressure. In normal cases, as is possible in this case also, vapour pressure of water is just one component of the atmospheric pressure and the liquid need not be boiling.

A more accurate sentence would be $P_{atm}\geq P_{vap}$. This would still mean that both vapour pressure and the atmospheric are higher than 1 atm but not necessarily equal.