`PCl_5,PCl_3` and `Cl_2` are at equilibrium at 500 K in a closed container and their concentrations are `0.8 xx 10^(-3) "mol L"^(-1), 1.2 xx 10^(-3) "mol L"^(-1)` and `1.2 xx 10 "mol L"^(-1)`, respectively. The value of `K_c` for the reaction `PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))` will be

Options

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

Answer

B

Explanation

For the reaction, `PCl_(5) hArr PCl_(3) + Cl_2`
At 500 K in a closed container,
`[PCl_5]=0.8 xx 10^(-3) "mol"^(-1)`
`[PCl_5]=1.2xx10^(-3) "mol"^(-1)`
`[Cl_2]=1.2xx10^(-3) "mol"^(-1)`
`K_c=([PCl_3][Cl_2])/([PCl_5])=((1.2xx10^(-3))xx(1.2xx10^(-3)))/((0.8xx10^(-3)))`
`=1.8xx10^(-3)`