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: $\text{R} = 0.082 \, \text{LatmK}^{- 1} \text{mol}^{- 1}$ $, \, \text{R} = 8.314 \, \text{JK}^{- 1} \text{mol}^{- 1},$ $ln \text{x} = 2.303 log_{10} \text{x}$ )

Answer 1

$\text{M}_{\text{avg}} = \frac{\rho \text{RT}}{\text{P}} = \frac{4 . 8 \times 0 . 0 8 2 \times 5 0 0}{1} = 1 9 6 . 8$



$M _{ avg }=\frac{ M _{ PCl _{5}}}{1+\alpha}=196.8=\frac{208.5}{1+\alpha}$

$\therefore $ $\alpha = 0 . 0 6$

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

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

$= - 8 . 3 1 4 \times 5 0 0 \times 2 . 303 \text{ log } \left(3 . 6 \times 1 0^{- 3}\right)$

$= 23.4 \, \text{kJ}$