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!

• Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I improved your usage of MathJax with the \ce{...} command for chemical formulae. It has the nice benefit of being intuitive, i.e. $\ce{2 H2 + O2 -> 2 H2O}$ gives $\ce{2 H2 + O2 -> 2 H2O}$. Learn more about this in the help center (where you will also find helpful other stuff), this meta post or this one.
– Jan
Nov 26 '15 at 0:02

Yes. You are correct. $\Delta G^0_{rxn}$ is a function of temperature, and is equal to zero at 100 C for this reaction.

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: $$\Delta G = \Delta H – T\Delta S$$ $$pV=n{\rm R}T$$

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.