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A quantity of PCl_(5) ws heated in a 10 litre vessel at 250^(@)C: PCl_(5_(g))hArrPCl_(3(g))+Cl_(3(g)). At equilibrium the vessel contains 0.1 moles of PCl_(5)0.20 mole of PCl_(3)and 0.2 mole of Cl_(2). The equilibrium constant of the reaction is

Answer»

`0.02`
`0.05`
`0.04`
`0.025`

SOLUTION :`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=((0.2)/(10)XX(0.2)/(10))/([0.1//10])=0.04.`


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