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)
$3 HC \equiv CH _{( g )} \rightleftharpoons C _{6} H _{6(\ell)}$
${\left[\right.\text { Given: } \Delta_{f} G ^{\circ}( HC \equiv CH )=-2.04 \times 10^{5} J mol ^{-1} ;}$
$\Delta_{f} G ^{\circ}\left( C _{6} H _{6}\right)=-1.24 \times 10^{5} J mol ^{-1} ; R =8.314 \left. J K ^{-1} mol ^{-1}\right]$

Answer 1

$3 HC \equiv CH _{( g )} \rightarrow C _{6} H _{6}(\ell): \Delta G ^{0}=- RT \ln k$

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

$\Rightarrow \Delta G ^{0}=\sum\left(\Delta G _{ f }^{0}\right)_{ P }-\sum\left(\Delta G _{ f }^{0}\right)_{ R }$

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

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

$=4.88 \times 10^{5}$

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

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

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

$\Rightarrow x =855$