1.

`PCl_(3), PCl_(3)` and `Cl_(2)` are at eqilibrium at 500 K and above have concentration 1.59 M for `PCl_(3), 1.59 M` for `Cl_(2)` and 1.41 M for `PCl_(5).` Calculate `K_(c)` for the reaction : `PCl_(5) hArr PCl_(3) + Cl_(2)`

Answer» `K_(c) = [[PCl_(3)][Cl_(2)]]/[[PCl_(5)]] = ((1.59 M)xx(1.59M))/((1.41M)) = 1.79`


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