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For the equilibrium in a closed vessel" ...

For the equilibrium in a closed vessel`" "PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`,
`K_(p)` is found to be double of `K_(e)` . This is attained when `:`

A

`T=2K`

B

`T=12.18K`

C

`T=24.36K`

D

`T=27.3K`

Text Solution

Verified by Experts

The correct Answer is:
C
`PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)`
`K_(p)=K_(c)(RT)^(Deltan)`
Here, `(K_(p))/(K_(c))=2` and `Deltan=2-1=1`
`:. 2=RT`
`T=(2)/(R)=(2)/(0.0821)=24.36K`
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