1.

1.6 mol of PCl_(5(g)) is placed in 4 dm^(3) closed vessel. When the temperaturee is raised to 500 K, it decomposes and at equilibrium, 1.2 mole of PCl_(5(g)) remains. What is the K_(c) vlaue for the decomposition of PCl_(5(g)) and Cl_(2(g))at 500 K

Answer»

`0.013`
`0.050`
`0.033`
`0.067`

Solution :`1.6` mol of `PCl_(5)` is placed in 4 `dm^(3)` CLOSED vessel. `PCl_(5(G))hArrPCl_(3(g))+Cl_(2(g))`
`{:(1.6mol,0,0,("initially")),((1.6-x)mol,xmol, xmol, ("At equlibrium")):}`
GIVEN that `1.6-x=1.2`
`thereforex=0.4` mol
Therefore, `[PCl_(5)]=(1.2)/(4)=0.3,[PCl_(3)]=(0.4)/(4)=0.1`
`&[Cl_(2)]=(0.4)/(4)=0.1`
`thereforeK_(C)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.1xx0.1)/(0.3)=0.033.`


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