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In the reaction PCI(5)(g)hArr PCI(3)(g...

In the reaction
`PCI_(5)(g)hArr PCI_(3)(g)+CI_(2)(g)`
the equilibrium concentrations of `PCI_(3)` and `PCI_(3)` are `0.4` and `0.2 mol^(-1)`, respectively. If the value of `K_(c)` is `0.5`, what id the concentration of `CI_(2)` in moles per litre ?

A

`2.0`

B

`1.5`

C

`1.0`

D

`0.5`

Text Solution

Verified by Experts

The correct Answer is:
C
`PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`
According to the law of chemical equilibrium,
`K_(C)=(C_(PCl_(3))C_(Cl_(2)))/(C_(PCl_(5)))`
or `C_(Cl_(2))=K_(C)(C_(PCl_(5)))/(C_(PCl_(5)))`
`=(0.5)((0.4))/((0.2))`
`=0.1`
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