Question

$PC{l}_3\rightleftharpoons PC{l}_3+C{l}_2$ in the reversible reaction the moles of $PC{l}_3$ and $C{l}_2$ are $a,b$ and $c$ respectively and total pressure is $P$, then value of $K_p$ is:

• A. $\frac{bc}{a}$$.RT$
• B. $\frac{a}{a+b+c}$ . P
• C. $\frac{bc.P}{a(a+b+c)}$
• D. $\frac{c}{(a+b+c)}$.P

Answer 1

Answer: (C)

Given

Moles of $PC{l}_5$=a

Moles of $PC{l}_3$=b

Moles of $C{l}_2$=c

Total pressure =P

Solution

$PC{l}_5=PCl3+C{l}_2$

$Kp=\frac{p\left(PC{l}_3\right)\ast p\left(C{l}_2\right)}{p\left(PC{l}_5\right)}$

Total no. of moles = a+ b+ c

$p\left(PC{l}_5\right)=\frac{a}{a+b+c}\ast P$

$p\left(PC{l}_3\right)=\frac{b}{a+b+c}\ast P$

$p\left(C{l}_2\right)=\frac{c}{a+b+c}\ast P$

Putting values

$K_{sp}=\frac{\left[\frac{b}{a+b+c}\right]\ast \left[\frac{c}{a+b+c}\right]}{\frac{a}{a+b+c}}\ast P$

$K_{sp}=\frac{bc}{a(a+b+c)}\ast P$

The correct option is C