Question

$0.1$ mole of $P{Cl}_5$ is heated in a litre vessel at $533K$. Determine the concentration of various species present at equilibrium, if the equilibrium constant for the dissociation of $P{Cl}_5$ at $533K$ is $0.414$.

Answer 1

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
Initially 0.1 mole of $PC{l}_5$, 0 moles of $PC{l}_3$ and 0 moles of $C{l}_2$ were present.
To reach equilibrium, x moles of $PC{l}_5$ dissociate to form
x moles of $PC{l}_3$ and x moles of $C{l}_2$.
$0.1-x$ moles of $PC{l}_5$ remains.
The equiibrium concentrations are
$\left[PC{l}_5\right]=\frac{\text{0.1-x mol}}{\text{1 L}}=\text{0.1-x M}$
$\left[PC{l}_3\right]=\frac{\text{x mol}}{\text{1 L}}=\text{x M}$
$\left[C{l}_2\right]=\frac{\text{x mol}}{\text{1 L}}=\text{x M}$
The equilibrium constant
$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$
$0.414=\frac{\text{x M}\times \text{x M}}{\text{0.1-x M}}$

$x^2=0.414(0.1-x)$
$x^2=0.0414-0.414x$

$x^2+0.414x-0.0414=0$

This is quadratic equation with solution
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-0.414\pm \sqrt{(0.414{)}^2-4(1\left)\right(-0.0414)}}{2\left(1\right)}$
$x=\frac{-0.414\pm \sqrt{(0.414{)}^2-4(1\left)\right(-0.0414)}}{2\left(1\right)}$
$x=0.08326$ or $x=-0.4973$
The value $x=-0.4973$ is discarded as it will lead to negative value of concentration. Hence, $x=0.08326$
The equiibrium concentrations are
$\left[PC{l}_5\right]=\text{0.1-x M}=\text{0.1-0.08326 M}=\text{0.0167 M}$
$\left[PC{l}_3\right]=\text{x M}=\text{0.08326 M}$
$\left[C{l}_2\right]=\text{x M}=\text{0.08326 M}$