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  4. PCl 5 dissociates as PCl 5( g )leftharpoons PCl 3( g )+ Cl 2( g ) 5 moles of PCl 5 are placed in a 200 litre vessel which contains 2 moles of N 2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant K p for the dissociation of PCl 5 is × 10-3. (nearest integer) (Given: R =0.082 L atm K -1 mol -1: Assume ideal gas behaviour)

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Q. $PCl _{5}$ dissociates as
$PCl _{5}( g )\rightleftharpoons PCl _{3}( g )+ Cl _{2}( g )$
$5$ moles of $PCl _{5}$ are placed in a $200$ litre vessel which contains $2$ moles of $N _{2}$ and is maintained at $600 \,K$. The equilibrium pressure is $2.46 \,atm$. The equilibrium constant $K _{ p }$ for the dissociation of $PCl _{5}$ is ______ $\times 10^{-3}$. (nearest integer)
(Given: $R =0.082\, L\, atm \,K ^{-1} mol ^{-1}$ : Assume ideal gas behaviour)

JEE MainJEE Main 2022Equilibrium

Solution:

Given : $2$ mole of $N_2$ gas was present as inert gas.
Equilibrium pressure $= 2.46\, atm$
image
from ideal gas equation
$ PV = nRT $
$2.46 \times 200=(5- x + x + x +2) \times 0.082 \times 600 $
$ x =3 $
$K _{ P }=\frac{ n _{ PCl _{3}} \times n _{ Cl _{2}}}{ n _{ PCl _{5}}} \times\left[\frac{ P _{\text {total }}}{ n _{\text {total }}}\right] $
$\frac{3 \times 3}{2} \times \frac{2.46}{10}=1.107=1107 \times 10^{-3}$