Question

At certain temperature, $K_c$ for the reaction
$PO{Cl}_3\left(g\right)\rightleftharpoons POCl\left(g\right)+{Cl}_2\left(g\right)$
is $0.30$. If $0.6$ mole of $PO{Cl}_3$ is placed in a closed vessel of $3.0$ litre capacity at this temperature, what percentage of it will be dissociated when equilibrium is established?

Answer 1

Out of 0.6 moles of $POC{l}_3$, let us assume that x moles dissociate to form x moles of $POCl$ and x moles of $C{l}_2$
$0.6-x$ moles of $POC{l}_3$ remains.
The equilibrium concentrations are
$\left[POC{l}_3\right]=\frac{\text{(0.6-x ) mol}}{\text{3.0 L}}=\text{( 0.2- 0.333 x ) M}$
$\left[POCl\right]=\frac{\text{( x ) mol}}{\text{3.0 L}}=\text{( 0.333 x ) M}$
$\left[C{l}_2\right]=\frac{\text{( x ) mol}}{\text{3.0 L}}=\text{( 0.333 x ) M}$
$K_c=\frac{\left[POCl\right]\times \left[C{l}_2\right]}{\left[POC{l}_3\right]}$
$0.30=\frac{\text{0.333x}\times \text{0.333x}}{\text{( 0.2- 0.333 x )}}$
$\text{( 0.2- 0.333 x )}=0.3696x^2$
$0.3696x^2+0.333x-0.2=0$
This is quadratic equation with solution
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-0.333\pm \sqrt{(0.333{)}^2-4(0.3696\left)\right(-0.2)}}{2\left(0.3696\right)}$
$x=\frac{-0.333\pm \sqrt{(0.333{)}^2-4(0.3696\left)\right(-0.2)}}{0.739}$
$x=0.412$ or $x=-1.313$
The value $x=-1.313$ is discarded as the number of moles cannot be negative.
Hence, $x=0.412$.
The percentage dissociation $=\frac{0.412}{0.6}\times 100=68.7\%$