What is the source of discrepancy in the calculation of Standard State Gibbs Free Energy change of the reaction of HCl (l) dissolving in water? I have calculated $\Delta_\text{r} G^o$ two different ways and got completely different results. Why? What assumptions have I made incorrectly, particularly in the first calculation shown below? 

Okay, so everyone knows that liquid Hydrochloric acid is a strong acid so it completely dissociates in water according to the following chemical equation:

$\ce{HCl (l) + H2O (l) -&gt; H3O+ (aq) + Cl- (aq)}$

If I used tabulated values of Standard State Gibbs Free Energy of formation $\Delta_\text{f} G^o$ to calculate $\Delta_\text{r} G^o$, standard State Gibbs Free Energy formation change of the reaction.

Standard State Gibbs Free Energies of formation:
$\begin{array}{lr}
\ce{HCl (l)} &amp; \pu{-131.17 kJ mol^{-1}}\\
\ce{H2O (l)} &amp; \pu{-237.19 kJ mol^{-1}}\\
\ce{H3O+(aq)} &amp; \pu{-237.19 kJ mol^{-1}}\\
\ce{Cl-} &amp; \pu{-131.17 kJ mol^{-1}}
\end{array}$

Source (I used this one because I couldn't find any that listed $\ce{HCl}$ as a liquid and not a gas):
http://www.vias.org/genchem/standard_ent...

Then, if you have one mole of each constituent according to the chemical equation, the standard state Gibbs Free Energy of the reaction = 0 kJ

by means of calculating ΔGoreaction = ΣΔGoformation of products - ΣΔGoformation of reactants

So ΔGoreaction should be less than zero if the reaction is favorable! But we know that HCl completely dissociates in water into Chloride ion and hydronium. What is going on here!?

If I calculate the standard state Gibbs Free Energy of the chemical reaction a different way.
That is that

$\Delta G^\circ_\text{reaction} = -R \cdot T \cdot \ln(\ce{Keq})$ at equilibrium, where 

for the acid $\ce{HCl}$, $\ce{Keq = Ka} = \pu{1.3e6}$,

Then the Gibbs Free Energy change of the chemical reaction at standard state

$\Delta G^\circ_\text{reaction} = (\pu{-8.314 J K-1 mole-1}) \cdot (\pu{1 mole}) \cdot (\pu{298 K}) \cdot \ln(\pu{1.3e6})$

$= \pu{-34.88 kJ}$,

which is less than zero; therefore, the reaction is favorable. What is going on!? See the difference in the calculations!?

It seems that the Ka of strong acids like HCl is usually assumed to be too large for most websites to even bother displaying. But, I found a table of Ka's of strong acids here: http://depts.washington.edu/eooptic/links/acidstrength.html
Using the $\ce{\Delta{G^{\circ}}}$ value of liquid $\ce{HCl}$ ($-131.17$ kJ/mol) is wrong. Simply put, we are indeed interested in gaseous $\ce{HCl}$. The normal boiling point of $\ce{HCl}$ is $-85^\circ$ C. On the other hand, the standard state of Gibbs Free Energy is $25^\circ$ C for what you are calculating. (Having liquid $\ce{HCl}$ entails water being below that temperature as well, in which case it would freeze. Additionally, the equilibrium constant for the dissociation of $\ce{HCl}$ is also measured at $25^\circ$ C). Finally, though this is unrelated to your question, if $\ce{HCl}$ was liquid, it would not be included in the $K_a$ expression.

Obtaining the $\ce{\Delta{G^{\circ}}}$ value of $-95.30$ kJ/mol for $\ce{HCl}$ $(g)$, we find that $\ce{\Delta{G^{\circ}}}$ for the reaction is $-35.87$ kJ/mol. Applying the equation $K = e^{-\ce{\frac{\Delta{G^{\circ}}}{RT}}}$, we find that $K_a$ for $\ce{HCl}$ is $1.9 * 10^6$ (within magnitude of your given value). 

What this looks like physically is that $\ce{HCl}$ gas is bubbled up into water, and it dissolves (dissociates) into the water as it mixes.