Assuming ideal behaviors, the magnitude of log K for the following reaction at 25^oC is
$-$x x 10^-1. The Value of x is _____. (Integer answer)
$3\mathrm{HC}\equiv {\mathrm{CH}}_{\left(\mathrm{g}\right)}\stackrel{ }{\rightleftharpoons }{\mathrm{C}}_6{\mathrm{H}}_{6\left(\mathrm{l}\right)}$
[Given: $∆$_fG^o (HC$\equiv$CH) = -2.04 x 10⁵ J mol^-1 ;
$∆$_fG^o (C₆H₆) = -1.24 x 10⁵ J mol^-1 ; R=8.314 K^-1 mol^-1]

$$\begin{gathered} \begin{array}{l}3\mathrm{HC}\equiv {\mathrm{H}}_{\left(\mathrm{g}\right)}\to {\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)\\ {\mathrm{\Delta G}}^{\mathrm{o}}=-\mathrm{RTIn}\mathrm{K}=-2.303\mathrm{RTlog}\mathrm{K}\end{array} \\ \mathrm{Given}, {\mathrm{\Delta G}}_{\mathrm{f}}^{\mathrm{O}}\left({\mathrm{C}}_6{\mathrm{H}}_6\right)=-1.24\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1} \\ \begin{array}{l}{\mathrm{\Delta G}}_{\mathrm{f}}^{\mathrm{O}}(\mathrm{CH}\equiv \mathrm{CH})=-2.04\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1}\\ {\mathrm{\Delta G}}_{\mathrm{r}}^{\circ }={\mathrm{\Delta G}}_{\mathrm{f}}^{\circ }\left({\mathrm{C}}_6{\mathrm{H}}_6\right)-3{\mathrm{\Delta G}}_{\mathrm{f}}^{\circ }\left({\mathrm{C}}_2{\mathrm{H}}_2\right)\\ {\mathrm{\Delta G}}_{\mathrm{r}}^{\circ }=-1.24\times {10}^5-3\times -\left(2.04\times {10}^5\right)\\ {\mathrm{\Delta G}}_{\mathrm{r}}^{\circ }=4.88\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1}\end{array} \end{gathered}$$
So, 4.88 x 10⁵ = -2.303 x 8.314 x 298 logK
$\log \mathrm{K}=\frac{-4.88\times {10}^5}{5705.848}=-85.52=-855\times {10}^{-1}$
X=855