Question

Under what pressure must an equimolar mixture of $PC{l}_3$ and $C{l}_2$ be placed at ${250}^{\circ }C$ in order to obtain $PC{l}_5$ at $1atm$?
($K_p$ for dissociation of $PC{l}_5=1.78)$.

• A. $3.5atm$
• B. $4.6atm$
• C. $2.5atm$
• D. None of these

Answer 1

Answer: (B)

$4.6atm$

The given reaction is :-

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

Let, the equilibrium constant for this reaction is $Kp1$.

Given that, for dissociation of ${PCl}_5,Kp=1.78$

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

$Kp=1.78$

So, $Kp1=\frac{1}{Kp}=\frac{1}{1.78}=0.56$

Now, $Kp1=\frac{P{pcl}_5}{P{pcl}_3.{Pcl}_2}$

where, ${Ppcl}_5,{Ppcl}_3,{Pcl}_2$ are partial pressure at equilibrium of ${PCl}_5,{PCl}_3\&{Cl}_2$.

If $P$ is initial pressure of ${PCl}_3\&{Cl}_2$ that at eqn, its partial pressure.

So, ${Ppcl}_3={Pcl}_2=(P-1)$ atm ($\because$ partial pressure of ${PCl}_5$ formed is $1$ atm)

$\Rightarrow 0.56=\frac{1}{(P-1)(P-1)}$

$\Rightarrow {(P-1)}^2=\frac{1}{0.56}=1.78$

$\Rightarrow P-1=\sqrt{1.78}=1.33$

$\Rightarrow P=2.33$ atm

$2P=2\times 2.33=4.66atm$

Option B is correct.