Assuming ideal behaviour, the magnitude of log K for the following reaction at 25°C is $$x \times 10^{-1}$$. The value of x is ______. (Integer answer)
$$3HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(l)}$$
[Given: $$\Delta_f G^{\circ}(HC \equiv CH) = -2.04 \times 10^5$$ J mol$$^{-1}$$; $$\Delta_f G^{\circ}(C_6H_6) = -1.24 \times 10^5$$ J mol$$^{-1}$$; R = 8.314 J K$$^{-1}$$ mol$$^{-1}$$]

For the reaction $$3HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(l)}$$, we first calculate the standard Gibbs free energy change.

$$\Delta_r G^{\circ} = \Delta_f G^{\circ}(C_6H_6) - 3 \times \Delta_f G^{\circ}(HC \equiv CH)$$

$$\Delta_r G^{\circ} = (-1.24 \times 10^5) - 3 \times (-2.04 \times 10^5)$$

$$\Delta_r G^{\circ} = -1.24 \times 10^5 + 6.12 \times 10^5 = 4.88 \times 10^5 \text{ J/mol}$$

Using the relationship $$\Delta_r G^{\circ} = -2.303 RT \log K$$:

$$\log K = \frac{-\Delta_r G^{\circ}}{2.303 RT}$$

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

$$\log K = \frac{-4.88 \times 10^5}{5705.8} = -85.52$$

The magnitude of $$\log K$$ is $$85.5 = 855 \times 10^{-1}$$.

Therefore, the value of $$x$$ is $$\textbf{855}$$.