`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

Options

A. `1.8xx10^(3)"mol L"^(-1)`
B. `1.8xx10^(-3)`
C. `1.8xx10^(-3)"L mol"^(-1)`
D. `0.55xx10^(4)`

Answer

B

Explanation

For the reaction :
`PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`
`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=((1.2xx10^(-3))^(2))/(0.8xx10^(-3))=1.8xx10^(-3)`