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Initially 0.8 mole of PCl(5) and 0.2 mol...

Initially 0.8 mole of `PCl_(5)` and 0.2 mole of `PCl_(3)` are mixed in one litre vessel . At equilibrium 0.4 mole of `PCl_(3)` is present . The value of `K_(C)` for the reaction `PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) ` is

A

0.13 `molL^(-1)`

B

0.05 `molL^(-1)`

C

0.065 `molL^(-1)`

D

0.1 `molL^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A

`K_(c)=([PCl_(3)](Cl_(2)])/([PCl_(3)])=0.13`
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