Question

A quantity of $P{Cl}_5$ was heated in a $10$ litre vessel at ${250}^{o}C$
$P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$
At equilibrium the vessel contains $0.25$ moles of $P{Cl}_5$, $0.3$ mole of $P{Cl}_3$, $0.30$ moles of ${Cl}_2$
(a) Compute the equilibrium constant for the formation of $P{Cl}_5$ at ${250}^{o}C$
(b) What is $\Delta G^o$ for the dissociation of $P{Cl}_5$? [$\log 0.036=-1.443$]

Answer 1

(a) A quantity of ${PCl}_5$ was heated in a 10 litre vessel.

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

at equilibrium 0.25 0.3 0.3

$\left[{PCl}_5\right]=\frac{0.25}{10}mol{L}^{-1}\left[{PCl}_3\right]=\frac{0.3}{10}mol{L}^{-1},\left[{CL}_2\right]=\frac{0.30}{10}mol{L}^{-1}$

$K_C=\frac{\left[{PCl}_3\right]\left[{CL}_2\right]}{\left[{PCl}_5\right]}$

$=\frac{\frac{0.3}{10}\times \frac{0.3}{10}}{\frac{0.25}{10}}=0.036$

(b) we know that, Since, $T=523K$

${△G}^0=-RTln{K}_{eq}$

$=-RT\times 2.303lng{K}_{eq}$

$=-8.314\times 523\times 2.303\times -1.443$

$=14450.13g$