# Assuming ideal behaviour, the magnitude

Question:

Assuming ideal behaviour, the magnitude of $\log \mathrm{K}$ for the following reaction at $25^{\circ} \mathrm{C}$ is $\mathrm{x} \times 10^{-1}$. The value of $\mathrm{x}$ is - (Integer answer)

$3 \mathrm{HC} \equiv \mathrm{CH}_{(\mathrm{g})} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6(\ell)}$

$\left[\right.$ Given: $\Delta_{f} \mathrm{G}^{\mathrm{o}}(\mathrm{HC} \equiv \mathrm{CH})=-2.04 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}$

$\Delta_{f} \mathrm{G}^{\mathrm{o}}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=-1.24 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1} ; \mathrm{R}=8.314$ $\left.\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right]$

Solution:

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

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

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

$\Rightarrow-R T \ell n k=1 \times\left(-124 \times 10^{5}\right)-\left(-3 \times 2.04 \times 10^{5}\right)$

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

$\Rightarrow \log \mathrm{k}=-\frac{4.88 \times 10^{5}}{2.303 \times \mathrm{R} \times \mathrm{T}}=-\frac{488000}{5705.848}=-85.52$

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

$\Rightarrow x=855$

Leave a comment

Click here to get exam-ready with eSaral