1.

One moel of N_(2) and 3.0 moles of PCl_(5) were placed in a 100-lotre vessel and heated to 227^(@)C. The equilibrium pressure was 2.05 atm. Assuming ideal behaviour, calculate X. Where X=1000xxK_(p) of reaction at 227^(@)C.

Answer»

Solution :The reaction is `underset((g))(PCl_(5))hArrunderset((g))(PCl_(3))+underset((g))(Cl_(2))`
`{:("INTIAL concn.","3moles",0,0),("Concn. at equilbrium",(3-3alpha),3alpha,3alpha where alpha"degree of dissociation of"PXl_(5)):}`
Total MOLES of gases in the VESSEL
`=n=N_(2)(1"MOLE")+PCl_(5)underset("moles")((3-3alpha))+PCl_(3)underset("moles")((3alpha))+Cl_(2)underset("moles")((3alpha))`
`Or n=4+3alpha`
Using the ideal gas equation `n=(PV)/(RT)=(2.05xx100)/(0.082xx500K)=5.0` moles
Or `4+3alpha=5or 3alpha=1or alpha=1//3=0.333` (degree of dissociation of `PCl_(5))`
Partial pressure of `PCl_(5)=2/5xx2.05=0.41atm.`
Partial pressure of `PCl_(3)=1/5xx2.05=0.41atm.`
Partial pressure of `Cl_(2)=1/5xx2.05=0.41`
`K_(p)=((0.41atm)^(2))/((0.82atm))=0.205atm.`
`X=0.205xx1000=205`


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