A sample of pure `PCl_(5)` was introduced into aln evacuated vessel at `473` K. After equilibrium was attained, concentration of `PCl_(5)` was found to be `0.5 xx 10^(-1) L^(-1)`. If value of `K_(c)` is `8.3 xx 10^(-3)`. What are the concentration of `PCl_(3)` and `Cl_(2)` at equilibrium ?
`PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g)`

Explanation

` PCl_5(g) hArr PCl_3 (g) + Cl_2(g)`
Since, the number of moles of ` PCl_3 (g) and Cl_2(g) ` are equal , their concentrations at equilibrium will also be equal , Suppose , at equilibrium
` [PCl_3 (g) ]=[Cl_2(g) ] = x " mol " L^(-1) `
Given that at equilibrium ` , [PCl_5 ] = 0. 5 xx 10^(-1) " mol " L^(-1) `
` because " "K_c= ([PCl_3(g) ][Cl_2(g)])/([PCl_5(g)])`
` therefore " " 8.3 xx 10^(-3) = (x xx x )/(0.5 xx 10^(-1))`
` or , " " x = (8.3 xx 10 ^(-3) xx 0.5 xx10^(-1) ) ^(1//2) = 0.02`
Hence ` " " [PCl_3(g) ]=[Cl_2(g) ] = 0.02 " mol " L^(-1) `