1.

In the reaction PCl_(5(g))iffPCl_(3(g))+Cl_(2(g)), the equilibrium concentrations of PCl_(5)andPCl_(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)])i.e.,0.5=(0.2xx[Cl_(2)])/(0.4)`
or, `[Cl_(2)]=1.0molL^(-1)`


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