Change in Gibbs free energy of vaporization reaction In my chemistry book, I see the equation

$$\Delta G_{rxn} = \Delta G_{rxn}^0 + RT \ln(Q)$$

where $\Delta G_{rxn}^0$ is the change in Gibbs free energy under standard conditions (1 atm pressure in this case), R is the gas constant, T is the absolute temperature, and Q is the reaction quotient. Consider the reaction

$$\ce{H2O (l) -&gt; H2O (g)}$$

Clearly in this example, $Q = P_{\ce{H_2 O}}$, the pressure of water vapor.

Here is my confusion. The earlier expression for $\Delta G_{rxn}$ implies that anytime $P_{\ce{H_2 O}} = 1\ \mathrm{atm}$, we have $\Delta G_{rxn} = \Delta G_{rxn}^0$, which is $8.59 \mathrm{kJ/mol}$ for this reaction. We should see this regardless of temperature, according to this expression.

However, when $T = 100\ \mathrm{^\circ C}$, liquid water and water vapor are in equilibrium at $P_{\ce{H_2 O}} = 1\ \mathrm{atm}$! And if they are in equilibrium, it stands to reason that we should see $\Delta G_{rxn}=0$, which clearly we do not.

Could anyone help me understand this apparent contradiction? My only guess at the moment is that it arises because $\Delta G_{rxn}^0$ is not temperature-independent and at the higher temperatures, perhaps the value of $8.59\ \mathrm{kJ/mol}$ becomes inapplicable. If anyone could explain this to me I would be very grateful!
Yes. You are correct. $\Delta G^0_{rxn}$ is a function of temperature, and is equal to zero at 100 C for this reaction.