1.

In the reaction PCl_(5(g))hArrPCl_(5(g))+Cl_(2(g)) The equilibrium concentrations of PCl_(5) and PCl_(3) are 0.4 and 0.2 mole/litre respectively. If the value of K_(c) is 0.5 what is the concentration of Cl_(2) in moles/litre

Answer»

`2.0`
`1.5`
`1.0`
`0.5`

Solution :`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.2xx X)/(0.4)=0.5.`


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