Question

Derive the expressions for $K_c$ and $K_p$ for the decomposition of $P{Cl}_5$.

Answer 1

The dissociation equilibrium of $P{Cl}_5$ is represented as
$P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_5\left(g\right)+{Cl}_2\left(g\right)$
Let '$a$' moles of $P{Cl}_5$ be present in '$V$' litres initially. If $x$ moles of $P{Cl}_5$ dissociate to $P{Cl}_3$ and ${Cl}_2$ then molar concentrations of $P{Cl}_5$, $P{Cl}_3$ and ${Cl}_2$ at equilibrium will be
$\frac{a-x}{V},\frac{x}{V}$ and $\frac{x}{V}$ respectively.
$K_c=\frac{\left[P{Cl}_3\right]{\left[{Cl}_2\right]}_e}{{\left[P{Cl}_5\right]}_2}$
$K_c=\frac{\frac{x}{V}\times \frac{x}{V}}{\frac{a-x}{V}}=\frac{x^2}{V^2}\times \frac{V}{(a-x)}$
$\frac{x^2}{(a-x)V}$
$x$ is known as degree of dissociation
$x=\frac{No.ofmolesdossociated}{Totalno.ofmolespresentintitially}$
if initially $1$ mole of $P{Cl}_5$ is present then
$K_c=\frac{x^2}{(1-x)V}=\frac{x^{2}P}{(1-x)RT}(\therefore V=\frac{RT}{P})$
In terms of partial pressures of $P{Cl}_5$, $P{Cl}_3$ and ${Cl}_2$
$K_p=\frac{P_{P{Cl}_5}\times P_{{Cl}_2}}{P_{P{Cl}_5}}atm{K}_p=\frac{x^{3}p}{{(1-x)}^2}atm$
If $x$ is small compare to unity, then $(1-x)$ is approximately equal to $1.0$
$\therefore K_c=\frac{x^2}{V}$ or $x^2=K_c\times V$
$=x\alpha \sqrt{V}$
$=x\alpha \sqrt{1/P}(\therefore V\alpha \frac{1}{P})$
Where $x$ is small, degree of dissociation varies inversely as the square root of $P$ (or) varies directly as the square root of volume of the system.