Assuming ideal behaviour,the magnitude of log | vedclass

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(A) For the reaction: $3 HC \equiv CH_{(g)} ightleftharpoons C_6H_{6(\ell)}$$\Delta G^{\circ} = \sum \Delta_f G^{\circ}(	ext{products}) - \sum \Del

Vedclass · Accepted Answer

$855$

Vedclass · Answer

(A) For the reaction: $3 HC \equiv CH_{(g)} ightleftharpoons C_6H_{6(\ell)}$$\Delta G^{\circ} = \sum \Delta_f G^{\circ}(	ext{products}) - \sum \Delta_f G^{\circ}(	ext{reactants})$$\Delta G^{\circ} = [1 	imes (-1.24 	imes 10^5)] - [3 	imes (-2.04 	imes 10^5)]$$\Delta G^{\circ} = -1.24 	imes 10^5 + 6.12 	imes 10^5 = 4.88 	imes 10^5 \, J \, mol^{-1}$We know that $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$$4.88 	imes 10^5 = -2.303 	imes 8.314 	imes 298 	imes \log K$$\log K = -\frac{4.88 	imes 10^5}{2.303 	imes 8.314 	imes 298} \approx -85.52$The magnitude of $\log K$ is $85.52$.Given $\log K = x 	imes 10^{-1}$,so $85.52 = x 	imes 10^{-1} \Rightarrow x = 855.2 \approx 855$.

Similar Questions

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