Question

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

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

Answer 1

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


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

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


Theory:

Writing $K_c$:

Example 1:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
$K_c=\frac{[N{H}_3{]}_{eq}^2}{\left[N_2{]}_{eq}\right[H_2{]}_{eq}^3}$

Example 2:
$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2\rightleftharpoons HI\left(g\right)$

$K_c=\frac{[HI{]}_{eq}}{\left[H_2{]}_{eq}^{\frac{1}{2}}\right[I_2{]}_{eq}^{\frac{1}{2}}}$