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1.6 moles of PCl(5)(g) is placed in 4d...

1.6 moles of `PCl_(5)(g)` is placed in `4dm^(3)` closed vessel . When the temperature is raised to 500 K, it decomposes and at equilibrium 1.2 moles of `PCl_(5)(g)` remains . What is the `K_(c)` value for the decomposition of `PCl_(5)(g)` to `PCl_(3)(g)andCl_(2)(g)` at 500 K?

A

a) 0.013

B

b) 0.05

C

c) 0.033

D

d) 0.067

Text Solution

Verified by Experts

The correct Answer is:
C
`PClhArrPCl_(3)+Cl_(2)`
`{:(1.6,0,0,"Initially"),("(1.6)"-x,x,x,"At equilibrium"),(((1.6-x)/(4)),(x)/(4),(x)/(4),):}`
Given , `1.6-x=1.2`
`x=0.4`
`[PCl_(5)]=(1.2)/(4)`
`[PCl_(3)]=(0.4)/(4)`
`[Cl_(2)]=(0.4)/(4)`
`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])`
`=(((0.4)/(4))((0.4)/(4)))/(((1.2)/(4)))`
`=(0.1xx0.1)/(0.3)`
`=0.033`
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