`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) and 1.2xx10^(-3) ` mol `L^(-1)` respectively. The value of `K_(c)` for the reaction `PCl_(5)(g) hArrPCl_(3)(g)+Cl_(2)(g)` will be

Options

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

Answer

B

Explanation

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