Question

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

Dissociation constant $\left(K_c\right)$ of $PC{l}_5$ is $4$. How many moles of $PC{l}_5$ should be taken in $5$L vessel to obtain $0.15$M of $PC{l}_3$ at equilibrium?

• A. 0.778
• B. 0.6
• C. 0.899
• D. 0.305

Answer 1

The correct option is B 0.6

The given reaction is:-

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$

Initial moles : $M$ $0$ $0$ (let)

At eqm : $(M-m)$ $m$ $m$ (let)

Given, that at equilibrium, the concentration of ${CO}_2=0.15M$

$\Rightarrow 0.15$ moles of ${CO}_2$ per litre of solution

Now, the volume of the vessel is $5L$.

So, at equilibrium, $\left[{PCl}_5\right]=\frac{M-0.15}{5}$

$\left[{PCl}_3\right]=\frac{0.15}{5}=0.03;\left[{Cl}_2\right]=0.03$

Now, $K_C=\frac{\left[{PCl}_3\right]\left[{Cl}_2\right]}{\left[{PCl}_5\right]}$

$\Rightarrow 4=\frac{0.3\times 0.3\times {10}^{-2}}{\left(\frac{M-0.15}{5}\right)}$

$\Rightarrow 4(M-0.15)=0.45\times {10}^{-2}$

$\Rightarrow 4M-0.6=0.0045$

$\Rightarrow M=\frac{0.6045}{4}=0.151125$

Now, $\left[{PCl}_5\right]=M-m=0.151125-0.03$

$=0.121125$ mole/litre

So, no. of moles to be taken in $5L=0.121125\times 5$

$=0.605625$

Hence, the correct option is $B$