Assuming ideal behaviour, the magnitude of log K for the following reaction at 25$^\circ$C is x $\times$ 10^$-$1. The value of x is ______________. (Integer answer)

$3HC \equiv C{H_{(g)}} \rightleftharpoons {C_6}{H_{6(l)}}$

[Given : ${\Delta _f}{G^o}(HC \equiv CH) = - 2.04 \times {10^5}$ J mol^$-$1 ; ${\Delta _f}{G^o}({C_6}{H_6}) = - 1.24 \times {10^5}$ J mol^$-$1 ; R = 8.314 J K^-1 mol^$-$1]

Reaction,<br/><br/>$$\mathop {3HC \equiv CH(g)}\limits_{Acetylene} \to \mathop {{C_6}{H_6}(l)}\limits_{Benzene} $$<br/><br/>Given, $\Delta G_f^o (CH \equiv CH) = - 2.04 \times {10^5}$ J mol^$-$1<br/><br/>$\Delta G_f^o({C_6}{H_6}) = - 1.24 \times {10^5}$ J mol^$-$1<br/><br/>Gibb's free energy, $\Delta G_f^o = - nRT\ln K$<br/><br/>$\Delta G_f^o = \sum {{{(\Delta G_F^o)}_P} - \sum {{{(\Delta G_F^o)}_R}} }$<br/><br/>$- nRT\ln K = - n'RT\ln {K_p} - ( - n''RT\ln {K_f})$<br/><br/>$$ \Rightarrow - RT\ln K = 1 \times ( - 1.24 \times {10^5}) - ( - 3 \times 2.04 \times {10^5}) $$ <br/><br/>$\Rightarrow$ $- 2.303 \times R \times T\log K = 4.88 \times {10^5}$<br/><br/>$\Rightarrow$ $\log K = - {{4.88 \times {{10}^5}} \over {2.303 \times 8.314 \times 273}}$<br/><br/>$\Rightarrow$ $n\ln K = n'\ln {K_p} - ( - n''ln{K_f})$<br/><br/>$\Rightarrow$ K = 85.52 <br/><br/>$\Rightarrow$ K = 855 $\times$ 10^$-$1<br/><br/>x = 855