The standard free energy change $\left(\mathrm{\Delta G}°\right)$ for 50 % dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ into ${\mathrm{NO}}_2$ at $27°\mathrm{C}$ and 1 atm pressure is $-\mathrm{x} \mathrm{J}{ \mathrm{mol}}^{-1}$. The value of x is _________.(Nearest Integer)
[Given : $\mathrm{R}=8.31 \mathrm{J}{ \mathrm{K}}^{-1}{ \mathrm{mol}}^{-1}$,$\log 1.33=0.1239$ $\ln 10=2.3]$

$$\begin{gathered} {\mathrm{N}}_2{\mathrm{O}}_4 \rightleftharpoons 2{\mathrm{NO}}_2 \\ 1 \mathrm{mol} 0 \\ (1-0.5)\mathrm{mol} 0.5\times 2\mathrm{mol} \\ 0.5 \mathrm{mol} 1 \mathrm{mol} \\ {\mathrm{k}}_{\mathrm{P}}=\frac{{\left(\frac{1}{1.5}\times 1\right)}^2}{\left(\frac{0.5}{1.5}\times 1\right)}=\frac{1}{0.75}=\frac{100}{75} \\ =1.33 \\ {\mathrm{\Delta G}}^0=-\mathrm{RT\ell }{\mathrm{nk}}_{\mathrm{P}} \\ {\mathrm{\Delta G}}^0=-\mathrm{RT}(2.3) \log k_{\mathrm{P}} \\ =-8.31\times 300\times \ln (1.33)=-710.45\mathrm{J}/\mathrm{mol} \\ =-710\mathrm{J}/\mathrm{mol} \end{gathered}$$