Question

Given that $K_c=13.7$ at $546K$ for
$P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$, calculate what pressure will develop in a $10$ litre box at equilibrium at $546K$ when $1.00$ mole of $P{Cl}_5$ is injected into the empty box?

Answer 1

Let $\alpha$ be the degree of dissociation.
Initally, 1.00 mole of $PC{l}_5$ is present.
$\alpha$ moles of $PC{l}_5$ will dissociate to give $\alpha$ moles of $PC{l}_3$ and $\alpha$ moles of $C{l}_2$ at equilibrium.
$1.00-\alpha$ moles of $PC{l}_5$ will remain
at equilibrium.
$K_c=\frac{\left[PC{l}_3\right]\left[PC{l}_2\right]}{\left[PC{l}_5\right]}$
$13.7=\frac{\frac{\alpha \text{moles}}{\text{10 L}}\times \frac{\alpha \text{moles}}{\text{10 L}}}{\frac{\text{1-}\alpha \text{moles}}{\text{10 L}}}$
$137=\frac{\alpha \times \alpha }{1-\alpha }$
${\alpha }^2=137-137\alpha$
${\alpha }^2+137\alpha -137=0$
This is quadratic equation with solution
$\alpha =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\alpha =\frac{-137\pm \sqrt{(137{)}^2-4(1\left)\right(-137)}}{2\left(1\right)}$
$\alpha =\frac{-137\pm \sqrt{(137{)}^2-4(1\left)\right(-137)}}{2}$

$\alpha =-137.99$ or $\alpha =0.9928$

The value $\alpha =-137.99$ is discarded as the degree of dissociation cannot be negative. Hence, $\alpha =0.9928$.

Total number of moles at equilibrium $n=1+\alpha =1+0.9928=1.9928$

Total pressure at equilibrium $P=\frac{n}{V}RT$
$P=\frac{1.9928mol}{10L}\times 0.08206Latm/mol/K\times 546K$
$P=8.93$ atm.