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One mole of N_(2) and 3.0 moles of PCl_(5) were placedin a 100-liter vessel and heated to 227^(@)C. The equilibrium pressure was 2.05" atm. "Assuming ideal behaviour, calculate X. Where X=1000xxK_(P) of the reaction at 227^(@)C. |
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Answer» Concen. At equilibrium `""(3-3alpha)""3alpha""3alpha` where `alpha` is degree of dissociation of `PCl_(5).` TOTAL moles of GASES in the vessel `{:(=n=N_(2)(1" MOLE")+PCl_(5)(3-3alpha)+PCl_(3)(3alpha)+Cl_(2)(3alpha)),("moles""moles""moles"):}` Or `n=4+3alpha` USING the ideal gas equation `n=(PV)/(RT)=(2.05xx100)/(0.082xx500K)=5.0" moles "` Or `4+3alpha=5" or "3alpha=1" or "alpha=1//3=0.333 ("degree of dissociation of" PCl_(5))` Partial PRESSURE of `PCl_(5)=(2)/(5)xx2.05=0.82" atm."` Partial pressure of `PCl_(3)=(1)/(5)xx2.05=0.41" atm. "` Partial pressure of `Cl_(2)=(1)/(5)xx2.05=0.41` `K_(P)=((0.41" atm")^(2))/((0.82" atm"))=0.205" atm."` `X=0.205xx1000=205` |
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