Question

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 answer)
$3HC\equiv C{H}_{(g)}\rightleftharpoons C_6{H}_{6(\ell )}$
$[\text{ Given: }{\Delta }_f{G}^{\circ }(HC\equiv CH)=-2.04\times 10^{5}Jmo{l}^{-1};$
${\Delta }_f{G}^{\circ }\left(C_6{H}_6\right)=-1.24\times 10^{5}Jmo{l}^{-1};R=8.314J{K}^{-1}mo{l}^{-1}]$

Answer 1

Answer: 855

$3HC\equiv C{H}_{(g)}\to C_6{H}_6(\ell ):\Delta G^0=-RT\ln k$

$\Delta G_f^0-2.04\times 10^5\frac{J}{mol}-1.24\times 10^{5}J/mol$

$\Rightarrow \Delta G^0=\sum {\left(\Delta G_f^0\right)}_P-\sum {\left(\Delta G_f^0\right)}_R$

$\Rightarrow -RT\ell nk=1\times \left(-124\times 10^5\right)-\left(-3\times 2.04\times 10^5\right)$

$\Rightarrow -2.303\times R\times T\log k$

$=4.88\times 10^5$

$\Rightarrow \log k=-\frac{4.88\times 10^5}{2.303\times R\times T}$

$=-\frac{488000}{5705.848}=-85.52$

$=855\times 10^{-1}$

$\Rightarrow x=855$