1.

A quantity of PCl_(5) was heated in a 10 dm^(3) vessel at 250^(@)C PCl_(5(g)) hArr PCl_(3 (g)) + Cl_(2(g)). At equilibrium the vessel contains 0.1 mole of PCl_(5), 0.2 moles of PCl_(3) and 0.2 moles of Cl_2. The equilibrium constant of the reaction is

Answer»

0.025
0.04
0.05
0.02

Solution :Equilibrium conc. Of `PCl_(5) = (0.1)/(10) = 0.01M`
Equilibrium conc. Of `Cl_(2) = PCl_(3) = (0.2)/(10) = 0.02M`
`RARR K_(C) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]) = (0.02 XX 0.02)/(0.01) = (0.0004)/(0.01)`
`rArr K_(C) = 0.04`


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