Question

a moles of $PCl _{5}$ are heated in a closed container to equilibriate $PCl _{5}( g ) \rightleftharpoons PCl _{3}( g )+Cl _{2}( g )$ at a pressure of $P$ atm. If $x$ moles of $PCl _{5}$ dissociate at equilibrium, then:

• A. $\frac{ x }{ a }=\frac{ K _{ p }}{ K _{ p }+ P }$
• B. $\frac{ x }{ a }=\left(\frac{ K _{ p }+ P }{ K _{ p }}\right)^{\frac{1}{2}}$
• C. $\frac{ x }{ a }=\left(\frac{ K _{ p }}{ P }\right)^{\frac{1}{2}}$
• D. $\frac{ x }{ a }=\left(\frac{ K _{ p }}{ K _{ p }+ P }\right)^{\frac{1}{2}}$

Answer 1

Answer: (D)

$PCl _{5}( g ) \leftrightharpoons PCl _{3}( g )+ Cl _{2}( g )$
Moles of $PCl _{5}$ dissociated $= x$
Moles of $PCl _{5}$ left $=( a - x )$
Moles of $PCl _{3}$ formed $= x$
Moles of $Cl _{2}$ formed $= x$
Total moles $=(a+x)$
Degree of dissociastion $\alpha=\frac{x}{a}$
Pressure due to $PCl _{5}=\frac{(1-\alpha)}{(1+\alpha)} P$
Pressure due to $PCl _{3}=\frac{\alpha}{(1+\alpha)} P$
Pressure due to $Cl _{2}=\frac{\alpha}{(1+\alpha)} P$
Now,
$K _{ p }=\frac{ X _{ PCl _{3}} \cdot X _{ Cl _{2}}}{ X _{ P cl _{5}}} P =\frac{\alpha^{2}}{1-\alpha^{2}} P$
$K _{ p }\left(1-\alpha^{2}\right)= K _{ p }- K _{ p } \alpha^{2}=\alpha^{2} P$
$K _{ p }=\left( K _{ p }+ P \right) \alpha^{2} $
$\Rightarrow \alpha=\sqrt{\frac{ K _{ p }}{ K _{ p }+ P }}=\frac{ x }{ a }$
$\Rightarrow \frac{ x }{ a }=\sqrt{\frac{ K _{ P }}{ K _{ P }+ P }}$