Gibbs Free Energy and Equlibrium Constant I have some problems understanding the relationship between $\Delta$$G°$ and $K$. For example, in the reaction of $$\ce{N2 + H2 &lt;=&gt; 2NH3}$$ where $\Delta$$G°= - 33.5$ kJ/mol. 

$\Delta$$G°=- 33.5$ kJ/mol is obtained from the $\sum G^°_{\text{f, products}} – \sum G^°_{\text{f, reactants}}$
. This means that the complete conversion of 1 molar of $N_2$ gas and 1 molar of $H_2$ gas gives 2 molars of $NH_3$ gives -33.5 kJ/mol of free energy. This is assuming that the reaction goes fully to completion, yet in reality it doesn't. 

From this formula below

$\Delta{G^°} = –RT \ln K_p \tag{5-6}$

$K_p$ can be obtained as $7*   10^5$, assuming at 298K and 1 atm.  My question is, since the complete reaction corresponds to  $\Delta$$G°= - 33.5$ kJ/mol of free energy, why should there be $K$ in the first place? $- 33.5$ kJ/mol tells me this value is for the complete reaction, so why should we insert this value into the equation just to get $K$ which tells me otherwise that the reaction doesn't go to completion? It doesn't make sense to me? In fact, shouldn't have K be infinity since the value that we had inserted is one obtained for a complete reaction?

Also, from this equation $\Delta{G} = \Delta{G^°} + RT \ln Q \tag{5-5}$

does having 1 mol of $N_2$ , 1 mol of $H_2$ and 1 mols $NH_3$ of in the reaction correspond to $- 33.5$ kJ/mol, since that makes $Q = 1$? And since $Q$ slowly reaches the value of $K$,  $\Delta{G} = 0$, then  $- 33.5$ kJ of free energy must have corresponded from the initial state to the equlibrium state. Yet again to me this doesn't make sense, because my understanding is that $- 33.5$ kJ is a value obtained only for the ideal scenario that the reaction goes to completion, and not from its initial standard states to equilibrium.

How does this all work out?
Q is supposed to be calculated in terms of partial pressures, not moles.  $\Delta G^0$ is based on a reversible process you have devised to take the pure reactants in stoichiometric proportions and 1 atm (actually, 1 bar) and 298 K and convert them to pure products in stoichiometric proportions, also at 1 atm. and 298 K.