`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

`K_c=([PCl_3][Cl_2])/([PCl_5])=((1.2xx10^(-3))xx(1.2xx10^(-3)))/(0.8xx10^(-3))`
`=1.8xx10^(-3) "mol L"^(-1)`