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JEE Main 202224 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2022 · 24 June · Evening Shift · Chemistry · Chemical Equilibrium

PCl _ 5 dissociates as PCl _ 5 (g) ⇌ PCl _ 3 (g) + Cl _ 2 (g) 5...

{PCl}_{5}dissociatesasdissociates as{PCl}_{5}\left(g\right) ⇌ {PCl}_{3}\left(g\right) + {Cl}_{2}\left(g\right)55 moles of PCl5{PCl}_{5} are placed in a 200200 litre vessel which contains 22 moles of N2{N}_{2} and is maintained at 600K600K. The equilibrium pressure is 2.46atm2 . 46atm. The equilibrium constant Kp{K}_{p} for the dissociation of PCl5{PCl}_{5} is___×10−3\times {10}^{- 3}. (nearest integer) (Given: R=0.082LR = 0 . 082L atm K−1mol−1{K}^{- 1}{mol}^{- 1}; Assume ideal gas behaviour)

Correct answer

0

Step-by-step solution

PCl5⇌PCl3+Cl25mole−−Intialmoles5−xxxEquilibriumTotalnumberofmoles=5+x+nN2=7+xAsnumberofmolesofNitrogengasaddedis2ntotal=PVRT=2.46X2000.082X600=107+x=10x=3Kp=(310x2.46)(310x2.46)(210x2.46)=1107{PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}5mole - - Intialmoles5 - xxxEquilibriumTotalnumberofmoles = 5 + x + {n}_{{N}_{2}} = 7 + xAsnumberofmolesofNitrogengasaddedis2{n}_{total} = \dfrac{PV}{RT} = \dfrac{2 . 46X200}{0 . 082X600} = 107 + x = 10x = 3{K}_{p} = \dfrac{\left(\dfrac{3}{10}x2 . 46\right)\left(\dfrac{3}{10}x2 . 46\right)}{\left(\dfrac{2}{10}x2 . 46\right)} = 1107
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