Chemistry - Chemical Equilibrium Question with Solution | TestHub

${\text{M}}_{\text{avg}}=\frac{\rho \text{RT}}{\text{P}}=\frac{4.8\times 0.082\times 500}{1}=196.8$

$\begin{array}{cccccc}{\text{PCl}}_5\left(\text{g}\right)& \rightleftharpoons & {\text{PCl}}_3\left(\text{g}\right)& +& {\text{Cl}}_2\left(\text{g}\right)& \\ 1-\alpha & & \alpha & & \alpha & \left(\text{Total moles}=1+\alpha \right)\end{array}$

${\text{M}}_{\text{avg}}=\frac{{\text{M}}_{{\text{PCl}}_5}}{1+\alpha }=196.8=\frac{208.5}{1+\alpha }$

$\therefore$$\alpha =0.06$

${\text{K}}_{\text{P}}=\frac{{\alpha }^2}{1-{\alpha }^2}\text{P}=\frac{{\left(0.06\right)}^2}{1-{\left(0.06\right)}^2}=3.6\times {10}^{-3}$

$\Delta {\text{G}}^{\circ }=-\text{RT }\text{ln }\text{K}$

$=-8.314\times 500\times 2.303\text{ log }\left(3.6\times 10^{-3}\right)$

$=23.4\text{kJ}$