Question

$4.0moles$ of $PC{l}_5$ dissociate at $760K$ in a $2$ litre flask. $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ at equilibrium. $0.8mole$ of $C{l}_2$ was present in the flask. The equilibrium constant would be

• A. $1.0\times {10}^{-1}$
• B. $1.0\times {10}^{-4}$
• C. $1.0\times {10}^{-2}$
• D. $1.0\times {10}^{-3}$

Answer 1

The correct option is A $1.0\times {10}^{-1}$

Given $\to$

$4$ moles of ${PCl}_5$ is dissociated in $2$ litre flask

At equilibrium, moles of ${Cl}_2$ present $=0.8$

${PCl}_5\left(g\right)\rightleftharpoons {PCl}_3\left(g\right)+{Cl}_2\left(g\right)$

initial $4$ $0$ $0$

at $t=0$

at equilibrium $4-x$ $x$ $x$

$x=0.8$

So, $4-0.8$ $0.8$ $0.8$

$K=\frac{\frac{\left[{PCl}_3\right]}{V}\frac{\left[{Cl}_2\right]}{V}}{\frac{\left[{PCl}_5\right]}{V}}$

$K=\frac{\frac{0.8}{2}\times \frac{0.8}{2}}{\frac{3.2}{2}}=\frac{0.64}{6.4}=0.1$

Equilibrium constant, $K=1\times {10}^{-1}$