Assuming ideal behaviour , the magnitude of log K for the following reaction at ` 25^(@)C` is ` x xx 10^(-1)`. The value of x is _______. (Integer answer )
`3HC -= CH_((g)) iff C_6H_(6(l))`
[Given : `Delta_f G^(@) (HC -= CH) = -2.04 xx 10^(5) J "mol"^(-1), Delta_f G^(@)(C_6 H_6) = -1.24 xx 10^(5) J " mol"^(-1) , R = 8.314 JK^(-1) " mol"^(-1)`]

Answer

To solve the problem, we need to calculate the value of \( \log K \) for the given reaction at \( 25^\circ C \) using the provided Gibbs free energy values. Here’s the step-by-step solution:

### Step 1: Write the Reaction
The reaction given is:
\[ 3HC \rightleftharpoons C_6H_6(l) \]

### Step 2: Identify the Gibbs Free Energy Values
From the problem, we have:
- \( \Delta_f G^\circ (HC \rightleftharpoons CH) = -2.04 \times 10^5 \, \text{J/mol} \)
- \( \Delta_f G^\circ (C_6H_6) = -1.24 \times 10^5 \, \text{J/mol} \)

### Step 3: Calculate \( \Delta G^\circ \) for the Reaction
Using the formula:
\[ \Delta G^\circ_{\text{reaction}} = \Delta G^\circ_{\text{products}} - \Delta G^\circ_{\text{reactants}} \]

For our reaction:
\[ \Delta G^\circ_{\text{reaction}} = \Delta G^\circ (C_6H_6) - 3 \times \Delta G^\circ (HC) \]
Substituting the values:
\[ \Delta G^\circ_{\text{reaction}} = -1.24 \times 10^5 - 3 \times (-2.04 \times 10^5) \]

Calculating:
\[ \Delta G^\circ_{\text{reaction}} = -1.24 \times 10^5 + 6.12 \times 10^5 \]
\[ \Delta G^\circ_{\text{reaction}} = 4.88 \times 10^5 \, \text{J/mol} \]

### Step 4: Use the Relationship Between \( \Delta G^\circ \) and \( K \)
The relationship is given by:
\[ \Delta G^\circ = -RT \ln K \]
We can convert this to logarithm base 10:
\[ \Delta G^\circ = -2.303RT \log K \]

### Step 5: Solve for \( \log K \)
Rearranging gives:
\[ \log K = -\frac{\Delta G^\circ}{2.303RT} \]

Substituting the values:
- \( R = 8.314 \, \text{J/(K mol)} \)
- \( T = 298 \, \text{K} \)

Now substituting:
\[ \log K = -\frac{4.88 \times 10^5}{2.303 \times 8.314 \times 298} \]

### Step 6: Calculate the Denominator
Calculating the denominator:
\[ 2.303 \times 8.314 \times 298 \approx 5708.5 \]

### Step 7: Calculate \( \log K \)
Now substituting back:
\[ \log K = -\frac{4.88 \times 10^5}{5708.5} \approx -85.5 \]

### Step 8: Express in Required Format
We need to express this in the form \( x \times 10^{-1} \):
\[ \log K \approx -855 \times 10^{-1} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{855} \]