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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» Correct Answer - [0205] `{:("The reaction is",,"PCl"_(5)hArr"PCl"_(3)+"Cl"_(2)),(,,(g)" "(g)" "(g)),("Initial concn.",,3"moles"" "0" "0):}` 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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