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Step-by-Step Solution
Step 1: Write the Balanced Reaction
The given reaction (formation of benzene from acetylene) is:
3\,HC \equiv CH_{(g)} \;\longrightarrow\; C_{6}H_{6(l)}.
Step 2: Identify the Standard Gibbs Free Energies of Formation
From the data provided:
\Delta G^{\circ}_{f}\bigl(HC \equiv CH\bigr) = -\,2.04 \times 10^{5}\,\text{J mol}^{-1}
\Delta G^{\circ}_{f}\bigl(C_{6}H_{6}\bigr) = -\,1.24 \times 10^{5}\,\text{J mol}^{-1}
Step 3: Compute the Standard Gibbs Free Energy Change of the Reaction
The standard Gibbs free energy change for the reaction,
\Delta G^{\circ}_{\text{reaction}},
is given by:
\Delta G^{\circ}_{\text{reaction}}
\;=\;\sum \Delta G^{\circ}_{f}(\text{products})
\;-\;\sum \Delta G^{\circ}_{f}(\text{reactants}).
Here, the product side has 1 mol of C_{6}H_{6} , and the reactant side has 3 moles of HC \equiv CH . Thus,
\Delta G^{\circ}_{\text{reaction}}
= \bigl[\Delta G^{\circ}_{f}(C_{6}H_{6})\bigr]
- \bigl[3 \times \Delta G^{\circ}_{f}(HC \equiv CH)\bigr].
= \bigl(-\,1.24 \times 10^{5}\bigr) \;-\; 3 \times \bigl(-\,2.04 \times 10^{5}\bigr)\,
\text{J mol}^{-1}.
= -\,1.24 \times 10^{5} \;+\; 6.12 \times 10^{5}
= 4.88 \times 10^{5}\,\text{J mol}^{-1}.
This positive value indicates that the reaction (as written) is non-spontaneous under standard conditions, so K will be less than 1.
Step 4: Relate \Delta G^{\circ} to the Equilibrium Constant K
For a reaction at temperature T (in Kelvin), the relationship between the standard Gibbs free energy change and the equilibrium constant K is:
\Delta G^{\circ}_{\text{reaction}} = -\,R\,T \,\ln K.
Rearranging to solve for \ln K gives:
\ln K = -\,\dfrac{\Delta G^{\circ}_{\text{reaction}}}{R\,T}.
Step 5: Substitute the Numerical Values
Use R = 8.314\,\text{J K}^{-1}\text{mol}^{-1} and T = 298\,\text{K} (for 25^{\circ}C ):
\ln K = -\,\dfrac{4.88 \times 10^{5}}{(8.314)\,(298)}
\;\approx\; -\,\dfrac{4.88 \times 10^{5}}{2477.572}
\;\approx\; -\,197.
Then, convert to base-10 logarithm:
\log K = \dfrac{\ln K}{\ln(10)} \;\approx\; \dfrac{-197}{2.302585}
\;\approx\; -85.5.
This shows \log K \approx -85.5, so K is extremely small.
Step 6: Express the Magnitude of \log K in the Form x \times 10^{-1}
The question asks for the magnitude of \log K in the format x \times 10^{-1} . Since
\bigl|\log K\bigr| = 85.5, we can write:
85.5 = 855 \times 10^{-1}.
Hence, the integer value of x (in x \times 10^{-1} ) is:
\boxed{855}.
