PCl _ 5 dissociates as PCl _ 5 ( g ) PCl _ 3 ( g )+ Cl _ 2 ( g ) … - Vidyadip

MCQ

$PCl _{5}$ 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)

A
$2312$
B
$954$
✓
$1107$
D
$1451$

✓

Answer

Correct option: C.

$1107$

c
Given : $2\, mole$ of $N _{2}$ gas was present as inert gas.

Equilibrium pressure $=2.46\, atm$

$PCl _{5}( g ) \rightleftharpoons PCl _{3}( g )+ C \ell_{2}( g )$

$t =0 \quad\quad\quad 5 \quad\quad\quad\quad 0\quad\quad 0$

$t = Eq ^{ m }\quad\quad 5- x\quad\,\,\,\, x \quad\quad x$

from ideal gas equation

$PV = nRT$

$2.46 \times 200=(5-x+x+x+2) \times 0.082 \times 600$ $x =3$

$K _{ P }=\frac{ n _{ PCl _{3}} \times n _{ Cl _{2}}}{ n _{ PCl _{5}}} \times\left[\frac{ P _{\text {totall }}}{ n _{\text {total }}}\right]$

$\frac{3 \times 3}{2} \times \frac{2.46}{10}=1.107=1107 \times 10^{-3}$

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