1.

K_(e ) " for " PCl_(5)(g) hArr PCl_(3) (g)+Cl_(2)(g) is 0.04 at 250^(@)C. How many moles of PCl_(5) must be added to a 3L flask to obtain a Cl_(2) concentration of 0.15 M

Answer»

4.2 MOLES
2.1 moles
5.5 moles
6.3 moles

Solution :At EQUILIBRIUM the moles of `Cl_(2)` MUST be =`0.15xx3=0.45`
`PCl_(5) HARR PCl_(3)+Cl_(2)`
`(x-0.45)/(3) "" (0.45)/(3) (0.45)/(3)` Eqb. CONC.
`K_(e )=([PCl_(3)][Cl_(2)])/([PCl_(5)])`
`therefore 0.04=(0.15xx0.15)/((x-0.45)//3) "" therefore x=2.1` moles


Discussion

No Comment Found

Related InterviewSolutions