Given
$$ \begin{align} \Delta_\mathrm f H^\circ(\ce{N2O4}) &= \pu{9.16 kJ mol^-1} &\quad \Delta_\mathrm f H^\circ(\ce{NO2}) &= \pu{33.18 kJ mol^-1} \\ S^\circ(\ce{N2O4}) &= \pu{304.3 J K^{-1} mol^{-1}} &\quad S^\circ(\ce{NO2}) &= \pu{204.1 J K^-1 mol^-1} \end{align} $$
for the reaction $\ce{N2O4 <=> 2 NO2},$ find the percentage of dissociation of $\ce{N2O4}$ at $\pu{1 bar}$ and $T = \pu{333 K}.$
I formed the equilibrium:
$$ \begin{array}{ccc} \ce{&N2O4 &<=> &2 NO2,} \tag{1} \\ &n_0\left(1 - \frac{\alpha}{2}\right) & & n_0\alpha \end{array} $$
where $n_0 $ are the initial amount of $\ce{N2O4}$ and $\alpha$ the dissociation value:
$$\alpha = \frac{2x}{n_0}. \tag{2}$$
After doing
$$\Delta G^\circ = -RT\ln K_p \quad\implies\quad K_p = \exp{\frac{-\Delta G^\circ}{RT}} = \pu{7.32E21}, \tag{3}$$
and having to find $\alpha,$ I get stuck with two unknown variables $n_0$ and $\alpha$:
$$K_p = p_\mathrm{tot}\frac{x(\ce{NO2})^2}{x(\ce{N2O4})} = \ldots = p_\mathrm{tot}\frac{n_0\alpha^2}{1 - \alpha/2}. \tag{4}$$
How to proceed?