Assuming ideal behavior, the magnitude of log K for the following reaction at ${25}^{0}C$, $x\times {10}^{–1}$. The value of x is ___________ (Integer answer)

$3HC\equiv C{H}_{\left(g\right)}\rightleftharpoons C_6{H}_{6\left(l\right)}$

[Given: ($\mathrm{equivCH}$)  = $–2.04\times {10}^{5}Jmo{l}^{–1}$;

${\Delta }_r{G}^0{C}_6{H}_6$ = $–1.24\times {10}^{5}Jmo{l}^{–1}$;

R = 8.314 $J{K}^{–1}mo{l}^{–1}$]

 ${\Delta }_r{G}^0$ = $-RTln{K}_{eq}$

${\Delta }_{reaction}{G}^0$ = $\sum \Delta {G_F^0}_{\left(Product\right)}-\sum \Delta {G_F^0}_{\left(Reactant\right)}$

$-RTln{K}_{eq}$ = [$1\times$ ($–1.24\times {10}^5)$]  - [$3\times$($–2.04\times {10}^5)$]

$-RTln{K}_{eq}$ = [$–1.24\times {10}^5)$] – [$-6.12\times {10}^5)$]

$-RTln{K}_{eq}$ = $4.88\times {10}^5$

-log K = $\frac{4.88\times {10}^5}{2.303\times 8.314\times 298}$

log K = -85.526 = $-8.5\times {10}^{-1}$.

Hence, the answer is $-8.5\times {10}^{-1}$.