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PCl(5), PCl(3) and Cl(2) are at equilibr...

`PCl_(5), PCl_(3)` and `Cl_(2)` are at equilibrium at 500 K in a closed container and their concentrations are `0.8xx10^(-3) mol L^(-1), 1.2xx10^(-3)mol L^(-1)` respectively. The value of `K_(c )` for the reaction
`PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)` will be

A

`1.8xx10^(-3) mol L^(-1)`

B

`1.8xx10^(3)`

C

`1.8xx10^(-3)L mol^(-1)`

D

`0.55xx10^(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(c )=([PCl_(3)][Cl_(2)])/([PCl_(5)])=((1.2xx10^(-3))xx(1.2xx10^(-3)))/((0.8xx10^(-3)))`
`=1.8xx10^(-3) mol L^(-1)`.
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