Question

$2\text{mol}$ of $PC{l}_5$ were heated to $327{}^{o}C$ in a closed $2L$ vessel and when equilibrium was achieved, $PC{l}_5$ was found to be $40\%$ dissociated into $PC{l}_3$ and $C{l}_2.$ What will be the equilibrium constant $\left(K_C\right)$ for the reaction:

• A. $1.267{\text{mol L}}^{-1}$
• B. $0.267{\text{mol L}}^{-1}$
• C. $1.967{\text{mol L}}^{-1}$
• D. $2.967{\text{mol L}}^{-1}$

Answer 1

The correct option is B $0.267{\text{mol L}}^{-1}$

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2\text{Initially}:200$
Degree of dissociation: $\left(\alpha \right)=40$ %$=0.4$
After dissociation,
Moles at equilibrium:
$PC{l}_5=2-(2\times 0.4)=1.2\text{mol}PC{l}_3=2\times \alpha =0.8\text{mol}C{l}_2=2\times \alpha =0.8\text{mol}$
so,
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
At
equilibrium: $1.20.80.8$
as $\text{Concentration}=\frac{\text{Moles}}{\text{Volume}}\left[PC{l}_5\right]=\frac{1.2}{2}=0.6{\text{mol L}}^{-1}\left[PC{l}_3\right]=\frac{0.8}{2}=0.4{\text{mol L}}^{-1}\left[C{l}_2\right]=\frac{0.8}{2}=0.4{\text{mol L}}^{-1}$

$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$
$=\frac{0.4\times 0.4}{0.6}$
$=\frac{1.6}{6}=\frac{0.8}{3}$
$=0.267{\text{mol L}}^{-1}$