For the following reaction at 50^° C and at 2 atm pressure, 2 N_2 O_5( g ) ⇌ 2 N_2 O_4( g )+ O_2( g ) N_2 O_5 is 50 % dissociated. The magnitude of…

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