Chemical Equilibrium and Sponteneity

Consider the reaction $$\ce{N2O4 <=> 2 NO2}.$$ The forward direction of this reaction is non-spontaneous and under standard conditions $\Delta G^\circ = \pu{4.76 kJ/mol}$. Suppose we begin with $\pu{1.00 bar}$ of $\ce{N2O4}$ and no products. Then, we would expect the reaction to proceed to equilibrium in which some small concentration of products is formed. I am struggling to understand why it is not a contradiction that the reaction is non-spontaneous, but still proceeds in the forward direction (which indicates that the forward direction is spontaneous).




A reaction will proceed to achieve an equilibrium state, this only happens when some amount of both reactants and products are present.




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Consider a general reaction:


$$
\sum\_ir\_i\ce{R}\_i \ce{->} \sum\_jp\_j\ce{P}\_j
$$


The equilibrium constant is given by:


$$
K = \dfrac{\prod\_{j}[\ce{P}\_j]\_\text{eq}^{p\_j}}{\prod\_i[{\ce{R}\_{i}}]\_\text{eq}^{r\_i}}
$$


If no product is formed then:


$$
\prod\_{j}[\ce{P}\_j]\_\text{eq}^{p\_j} = 0 \implies K = 0
$$


Calculating the free energy for such a reaction:


$$
\Delta G^\circ = -\mathrm{R}T\ln{K}
\implies \lim\_{K \rightarrow 0} \Delta G^\circ = \infty
$$


Thus, for a process in which no products are formed, the standard free energy is infinity, which isn't logical, and some products are always formed, regardless of whether the reaction is spontaneous, $\Delta G^\circ <0$, or non-spontaneous, $\Delta G^\circ >0$.


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