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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. |
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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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