Initially, 0.8 mole of `PCl_(5)` and 0.2 mol of `PCl_(30` are mixed in one litre vessel. At equilibrium, 0.4 mol of `PCl_(3)` is present. The value of `K_(c )` for the reaction `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` would beA. `0.1 mol L^(-1)`B. `0.05 mol L^(-1)`C. `0.013 molL^(-1)`D. ` 0.66 mol L^(-1)`

Initially, 0.8 mole of `PCl_(5)` and 0.2 mol of `PCl_(30` are mixed in

1.	Initially, 0.8 mole of `PCl_(5)` and 0.2 mol of `PCl_(30` are mixed in one litre vessel. At equilibrium, 0.4 mol of `PCl_(3)` is present. The value of `K_(c )` for the reaction `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` would beA. `0.1 mol L^(-1)`B. `0.05 mol L^(-1)`C. `0.013 molL^(-1)`D. ` 0.66 mol L^(-1)`
Answer» Correct Answer - A `K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.4 xx 0.2 )/(0.6)=0.13 mol L^(-1)`

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