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


$$\ce{H2O (l) -> H2O (g)}$$


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.


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The change in state equation of water is a “half reaction”, but your equation is a phase change reaction.


SI units are confusing with gases:


Temperature is actually 298K in this case (it’s a trick because standard conditions mean temperature is set at 298K and pressure at 1 atm)


Reaction quotant = [products ÷ reactants] = as a ratio 
 $$\rm R = 8.314~J K^{-1} mol^{-1}$$
$$\rm \Delta G = \Delta G^\circ~of~8.59~\frac{kJ}{mol}~ – RTln1 = 8.59~\frac{kJ}{mol}$$


[the little degree symbol means heat of formation / formation energy [Δ= products - reactants] (in this instance I'm guessing its the evaporation energy for the change in state ? anyway) because evaporation of $\ce{H2O}$ is an exothermic reaction and 
Other useful Eqs:
$$\Delta G = \Delta H – T\Delta S$$
$$pV=n{\rm R}T$$


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You are right $\Delta G^0\_{rxn}$ is not temperature independent. 
$$\Delta G (T,P)=\Delta G^0 (T) + RTln(Q)$$
When we say standard most of the time we mean a static point, but in this case it's a dynamic line along P=1bar.


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