Question

${\text{PCl}}_{\text{5}}$ vapour decomposes on heating according to the reaction:

${\mathrm{PCl}}_5(\mathrm{g})\rightleftharpoons {\mathrm{PCl}}_3(\mathrm{g})+{\mathrm{Cl}}_2(\mathrm{g})$

The density of a sample of a partially dissociated ${\text{PCl}}_{\text{5}}$ at $\text{1}\text{.0}$ atm and $\text{500}\text{K}$ was found $\text{4}\text{.8}\text{g/L}\text{.}$ Calculate the $\Delta {\text{G}}^{{}^{\circ }}$ for the reaction at $\text{500}\text{K}\text{.}$

(Given: $\text{R}=0.082{\text{LatmK}}^{-1}{\text{mol}}^{-1}$ $,\text{R}=8.314{\text{JK}}^{-1}{\text{mol}}^{-1},$ $ln\text{x}=2.303lo{g}_{10}\text{x}$ )

Answer 1

Answer: 23.40

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



$M_{avg}=\frac{M_{PC{l}_5}}{1+\alpha }=196.8=\frac{208.5}{1+\alpha }$

$\therefore$ $\alpha =0.06$

$K_P=\frac{{\alpha }^2}{1-{\alpha }^2}P=\frac{(0.06{)}^2}{1-(0.06{)}^2}=3.6\times 10^{-3}$

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

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

$=23.4\text{kJ}$