Free Energy Change of Gaseous Equilibrium

For the following reaction at $50^{\circ}\mathrm{C}$ and 2 atm pressure,
$2 \mathrm{N}_2 \mathrm{O}_5(\mathrm{g}) \rightleftharpoons 2 \mathrm{N}_2 \mathrm{O}_4(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})$
$\mathrm{N}_2 \mathrm{O}_5$ is 50
The magnitude of standard free energy change at this temperature is x.
$x = \_\_\_\_ \mathrm{J} \, \mathrm{mol}^{-1}$ [Nearest integer]
Given : $R = 8.314 \mathrm{mol}^{-1} \mathrm{K}^{-1}, \log 2 = 0.30,$
$\log 3 = 0.48, \ln 10 = 2.303, {}^\circ\mathrm{C} + 273 = \mathrm{K}$