`PCI_(5), PCI_(3)` and `CI_(2)` are in equilibrium at 500 K in a closed container and their concentration are `0.8 gt 10^(-3) " mol" L^(-1) " and " 1.2 xx 10^(-3) "mol" L^(-1) " and " 1.2 xx 10^(-3) "mol" L^(-1)` respectively. The value of `K_(c)` for the reaction `PCI_(5) (g) hArr PCI_(3) (g) + CI_(2) (g)` will be

Answer

B

Explanation

For the reaction, `PCl_(5) Leftrightarrow PCI_(3)+Cl_(2)`
At 500K in a closed conatiner, `[PCl_(5)]=0.8 X 10^(-3)" mol L"^(-1)`
`[PCl_(3)]=1.2 X 10^(-3)" mol L"^(-1)`
`[Cl_(2)]=1.2 X 10^(-3)" mol L"^(-1)`
`K_(C)=([PCl_(3)] [Cl_(2)])/([PCl_(5)]) =((1.2 xx 10^(-3)) xx (1.2 xx 10^(-3)))/((0.8 xx 10^(-3)))=1.8 xx 10^(3)`