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Prove that the pressure necessary to obtain 50% dissociation of `PCl_(5)` at 500 K is numerically three times the value of `K_(p)`. |
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Answer» `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Iitial molar conc".,1,,0,,0),("Eqm. molar conc".,(1-0.5),,0.5,,0.5):}` Total no. of moles at eqm. Point =1-0.5 +0.5 = 1.5 mol. `K_(p)= (pPCl_(3)xxpCl_(3))/(pPCl_(5))=(((0.5p)/(1.5))xx((0.5P)/(1.5)))/(((0.5P)/(1.5)) )= (0.5)/(1.5) P` `K_(p) =1/3 P or P = 3 K_(p)` |
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